1

Ch6

6.1

a) Consider the FBD at right.

Use � � 0 when the block is motionless (largest force BEFORE it starts to slide).

From this we know the right & left forces must balance as do the up & down forces. ��:� � �:� � �

Since the block is just about to slide, it is on the verge of slipping.

It is ok to use � ���.

In this case we know � � � from the vertical force equation.

Plugging it all together we find � � � � ��� � � ���

� � 0.5 �1.00kg��9.8 ms��

Block is on the verge of slipping when �� ! � "#�$ � %. &'(.

b) For � ) �*+, , the FBD is unchanged!

Since the block is no longer on the verge of slipping, DO NOT use � ���.

One finds � � � 3.00N and � � 0.

c) For � / �*+, , the FBD must have non-zero acceleration! The force equations become ��:� 0 � �� �:� � �

Also, because the block now slides relative to the floor, use � �1�!

Similar to part a one finds

� �1� � �1� � 0.3�1.00kg� �9.8 ms�� � 2.94N

� � �� 0 �

� � �� 0 �1

� � 1.96 ms�

For reference, I made plots of 56. � and �56. �. In question 6.26 I show you how to make these plots yourself.

0.00

1.00

2.00

3.00

4.00

5.00

0.00 2.00 4.00 6.00 8.00 10.00

f(N

)

F (N)

f vs F

0.00

2.00

4.00

6.00

8.00

0.00 2.00 4.00 6.00 8.00 10.00

a(m

/s2)

F (N)

a vs F

�

�

� � 0

�

2

6.2

a) I think it will take less force to cause the block to slide. Why? Now you pulling up on the block a little bit

with the vertical component of �. Less normal force is required to support the block. This means the max

possible value of friction given by � ��� should decrease. Furthermore, because the angle is small,

most of � is pretty much still pulling the block horizontally. I think we still get most of the force applied

horizontally plus a small reduction in max possible friction. Combined I think it should be easier to make

the block slide. Therefore �*+, should be smaller than in the previous problem. Let’s find out.

b) Consider the FBD at right. Because the block is motionless (largest force BEFORE it

starts to slide) we know � � 0. Form this we know the right & left forces must

balance as do the up & down forces. ��:� cos 9 � �:� : � sin 9 � �

Since the block is just about to slide it is on the verge of slipping. Ok to use � ���.

In this case we know � � � 0 � sin 9 from the vertical force equation. � cos 9 � � cos 9 � ��� � cos 9 � ���� 0 � sin 9� � cos 9 � ��� 0 ��� sin 9 � cos 9 : ��� sin 9 � ��� ��cos 9 : �� sin 9� � ���

� � ��� cos 9 : �� sin 9

� � 0.5 �1.00kg��9.8 ms��cos 9 : 0.5 sin 30.0°

� � 0.5 �1.00kg��9.8 ms��cos 30.0° : 0.5 sin 30.0° Block is on the verge of slipping when �� ! � "#�$ � %. >&(. When lifting up slightly on the block it

is easier to get it moving!!! Recall, in 6.1 the horizontal applied force required 4.91 N to move block.

c) For � ) �*+, , the FBD is unchanged!

Since the block is no longer on the verge of slipping, DO NOT use � ���.

One finds ? � � @AB C � D. E'NNNN and F � '. Notice frictional force is lower than the previous case. The

friction need not balance the entire applied force…just the horizontal component of applied force.

d) For � / �*+, , the FBD must have non-zero acceleration! The force equations become ��:� cos 9 0 � �� �:� : � sin 9 � �

Also, because the block now slides relative to the floor, use � �1�!

Similar to part a one finds

� �1� � �1�� 0 � sin 9� � 0.3 G�1.00kg� �9.8 ms�� 0 �4.91N� sin 30.0°H � 2.20N

� � � cos 9� 0 �

� � � cos 9� 0 �1�� 0 � sin 9��

� � �� cos 9 : �� �1 sin 9 0 �1

� � �� �cos 9 : �1 sin 9� 0 �1

� � 2.05 ms�

Notice acceleration is larger than the previous problem even though the applied force has equal magnitude.

�

�

� � 0

�

� cos9

� sin 9

3

6.3

a) For part a, we expect the system is on the verge of slipping.

We may use � ���&� � 0.

Block �J Block �D

��:K 0 � �L�� K 0 � 0

��:�� 0 K � ���� �� 0 K � ��0

Σ�:� 0 �L � �L� � 0 �L � 0

From �� for �L: K �

Now use � ���: K � ���

Now use � from block �L and �� from block ��: �� � ���L

�D� ! � "#�J

Solution continues on the next page…

�

�L

� � 0

K

� � 0

��

K

4

b) In part b we are told �� � 2��*+, � 2���L

We expect the system will accelerate (� M 0) and kinetic friction applies ( � �1�).

Block �J Block �D

��:K 0 � �L� ��:�� 0 K � ���

Σ�:� � �L

Stacking the horizontal forces equations and adding gives

�� 0 K � ���

:K 0 � �L�

�� 0 � ��L : ����

�� 0 �1� � ��L : ����

�� 0 �1�L � ��L : ����

Now I will plug in �� � 2��*+, � 2���L.

�2���L� 0 �1�L � ��L : 2���L��

Notice every single term has �L in it…all the �L’s cancel out!!!

2�� 0 �1 � �1 : 2����

F � $ND"# 0 "OJ : D"# P

�

�L

�

K

�

��

K

5

6.3½ part a) Perhaps you have noticed it is convenient to include acceleration in the FBD even if � � 0…

Block �J Block �D

��:� cos 9 0 K 0 � �L�

Note: in this scenario � � 0.

� cos 9 0 K 0 � 0

��:K 0 �� � ���

Note: in this scenario � � 0.

K 0 �� � 0

Because � � 0 I often use

“the ups” = “the downs”

Σ�:� : � sin 9 � �L

Procedure from here:

1. Get an equation for the normal force from Σ� for block �L.

2. Plug that equation for � into � ��� to get an algebraic expression for .

3. Plug that equation for into �� for block �L.

4. Use �� for block �� to eliminate K in �� for block �L.

5. Solve algebraically for �. Because we used � ���, this is the special applied force �*+,.

� � �L 0 � sin 9

� ��� � ����L 0 � sin 9�

� cos 9 0 K 0 ����L 0 � sin 9� � 0

� cos 9 0 �� 0 ����L 0 � sin 9� � 0

� cos 9 0 �� 0 ���L : ��� sin 9 � 0

� cos 9 : ��� sin 9 � �� : ���L

��cos 9 : �� sin 9� � �� : ���L

�� ! � �D$ : "#�J$@AB C : "# BQR C

Solution continues on the next page…

�

�L

�

K

�

��

K �

� cos 9 9 � sin 9

6

6.3½ part b) If � � 2�*+,, the system will accelerate.

Use the same FBD as in part a.

This time, however, we must use � M 0& � �1�.

Stacking the equations and adding them is often the easiest way to determine acceleration.

� cos 9 0 K 0 � �L�

:K 0 �� � ���

� cos 9 0 0 �� � ��L : ����

� � � cos 9 0 0 �� �L : ��

� � � cos 9 0 �1� 0 �� �L : ��

Now plug in � � �L 0 � sin 9:

� � � cos 9 0 �1��L 0 � sin 9� 0 �� �L : ��

Now group the � terms together:

� � ��cos 9 : �1 sin 9� 0 �1�L 0 �� �L : ��

Now plug in � � 2�*+, � 2� *STUVW*XTYZ[\UVW []^\�

F � D � �D$ : "#�J$@AB C : "# BQR C� �@AB C : "O BQR C� 0 "O�J$ 0 �D$�J : �D

Technically, this is a correct final answer (all terms on the right side were given in the problem statement).

I think the least you could do is factor out the .

F � $D��D : "#�J� � @AB C : "O BQR C@AB C : "# BQR C� 0 "O�J 0 �D�J : �D

Also, it perhaps looks prettier if you divide all terms by �L.

� � 2 ����L : ��� � cos 9 : �1 sin 9cos 9 : �� sin 9� 0 �1 0 ���L1 : ���L

After doing this, let _ � *X*S (let “gamma” equal the mass ratio).

� � 2�_ : ��� � cos 9 : �1 sin 9cos 9 : �� sin 9� 0 �1 0 _1 : _

Well…I guess this one just looks insanely ugly.

7

6.4 ZERO frictional force is required to keep the box at rest! Clearly ) ���.

6.5 Box in relative motion with the floor. Friction points to the left with � �1�.

6.6 On the verge of slipping up the plane. Friction points down the plane with � ���.

6.7 I deliberately left this wording a bit vague. As worded, to me, it sounds like the box remains stationary on the

back of the truck as it accelerates to the right. If this is the case, friction is keeping the box from sliding off the back

of the truck. No indication is made that the box is on the verge of slipping. Assume friction points to the right on

the box with ) ���. Notice, by Newton’s third law, friction must point to the left on the truck with magnitude ) ���. To be clear, for this problem it is very unlikely we are at the maximum possible frictional force given by � ��� . If you want to include the possibility of the block being on the verge of slipping you could say ≤ ���.

6.8 There is a normal force upwards from the floor on the lower block �. The friction between the floor and the

lower block is . The lower block moves relative to the floor. Assume friction from the floor points to the left on

the lower block with � �1�.

There is a normal force between the two blocks �L�. The friction between the two blocks is L�. In this case the

problem says “largest acceleration for which…move together”. This wording indicates the upper block is on the

verge of slipping. Use L� � ���L�. The frictional force between the blocks points to the right on the upper block

AND to the left on the lower block! Notice �L� & L� are internal forces to the two block system.

6.9 From the floor on the lower block we still have friction pointing left on the lower block with � �1�. In this

problem statement, no mention is made of the blocks being on the verge of slipping but we are told they move

together. Even though both blocks are moving, there is no relative velocity between the blocks. Static friction

applies but we must use L� ) ���L�. To be clear, for this problem L� M ���L . Friction points to the left on the

upper block AND to the right on the lower block. Compare this to the previous problem…

6.10 In this case, the frictional force points up the wall and prevents the zombie from sliding down! The normal

force points towards the center perpendicular to the wall. No mention is made of being on the verge of slipping. In

this case ) ���. In fact, we know friction must balance the zombie’s weight so we could actually use � � !

6.11

a) As the angle increases the force �a becomes less vertical. The normal force decreases.

b) As the angle increases �a becomes more horizontal. The frictional force must increase if the block is always

stationary!

c) The maximum possible frictional force is given by the equation � ���. Notice this goes down since it is

proportional to �!

6.12

a) Relative to the truck, the box moves to the left. Friction points opposite the relative motion between the

two surfaces. Here you would use � �1� to the left on the truck and to the right on the box.

b) Relative to the earth, the box is still moving forwards (albeit not at the same rate as the truck). Both the

velocity of the block relative to the earth and friction on the block point to the right.

6.13 I deliberately left this wording a bit vague. As worded, to me, it sounds like the block is held up with the

assistance of friction. If the minimum possible � is applied, friction points up the plane with � ���. If the person

exerts a slightly larger force, friction still points the same way but we no longer require the maximum possible

friction force to keep the block in place ( ) ���). Depending on the applied force, it is possible that no friction is

required! Going even further, it is possible that, if you push really hard on the block, friction points down the plane

to prevent you from sliding the block up the plane. Once again, depending on the magnitude of �, it is possible

friction is either less than or equal the maximum possible friction force. Surprisingly complicated…

8

6.14

a) On the verge of slipping; use � ���.

b) FBD to the right. Force equations give ��:� sin 9b 0 � 0 �:� 0 � cos 9b � 0

For fun I also showed the graphical vector addition

using the tail to tip method. The vectors sum to zero

when there is no acceleration. Notice the coordinate

system helps us see that and n need not be split into

components. Notice the components of mg are split

parallel to the coordinate system…not straight up and

down! The relative sizes of the arrows in this figure

are approximately correct.

c) �� � tan 9b.

d) The critical angle equation is independent of mass! Change the mass and you should get the same 9b.

e) Above 9b the block will slide relative to the plane…use � �1�.

f) Below 9b the block is at rest relative to the plane AND not at the onset of slipping…use ) ���.

6.15

a) Figure at right. The relative sizes of the arrows in

this figure are approximately correct. The force

equations give ��:� sin 9 0 � �� �:� 0 � cos 9 � 0

Notice in the graphical vector addition the resultant

vector is not zero but in fact ��a.

b) From the force equations one finds

�1 � tan 9 0 � cos 9

Since all the forces in the problem are constant, we

expect acceleration is constant. In this case

∆f � 5+�g : 12 ��g�

h � 12 �g�

� � 2hg�

Putting it all together gives

�1 � tan 9 0 2h g� cos 9

The units check. Also, consider the case of � � 0 to check our result. When � � 0 static friction applies

and we get back the critical angle equation. Looks pretty good.

Notice we now have a method for determining both coefficients of friction in the lab. Use problem 6.14 to

determine the static coefficient and use problem 6.15 to determine the kinetic coefficient.

�

�

� � 0

�

�

Vector Addition

Tail-to-tip Method

� sin9b

� cos9b 9b

FBD

�

�

�

�

�

Vector Addition

Tail-to-tip Method

� sin9

� cos9 9

FBD

��

9

6.16

a) If you assume pulley is massless & axle friction is negligible, tension is same on both sides of pulley.

b) If we assume the string is inextensible the acceleration magnitudes of the block are equal.

c) If the string is massless the acceleration is constant.

d) Expect �� / K if �� is to accelerate downwards.

e) See below. The relative sizes of the arrows in this figure are approximately correct.

��:K 0 �L sin 9 0 � �L� �:� 0 �L cos 9 � 0

Since �L slides relative to plane use � �1� K 0 �L sin 9 0 �1�L cos 9 � �L�

��:�� 0 K � ���

f) Use a coordinate system that rotates with the acceleration (keeps acceleration aligned with positive x-axis).

Stack the equations and add them. K 0 �L sin 9 0 �1�L cos 9 � �L�

:�� 0 K � ���

____________________________________________________

�� 0 �L sin 9 0 �1�L cos 9 � ��� : �L��

Why? I get an additional way to check my work. You should always end up with �ijikl� on the right side

and no internal forces on the left side. If this doesn’t occur, you made a mistake in the FBD. This

technique spots mistakes before tons of algebra. This technique works with more than two blocks as well.

Now we find the acceleration is

� � �� 0 �L sin 9 0 �1�L cos 9��� : �L�

Since acceleration is constant 5m� � 5+� : 2�∆f 5m� � 2�n

5m � √2�n � p2 n �� 0 �L sin 9 0 �1�L cos 9��� : �L�

Notice ∆f � :n and 5m � :√2�n because the coordinate system shows down as the positive direction!!!

g) If the problem asks for the minimum mass to slide upwards we are on the verge of slipping. We may still

use � �� if we are at the onset of slipping as long as we remember to switch from �1to ��. Also, for this

case, verge of slipping also implies � � 0 (not always true). This gives�� � �L�sin 9 : �� cos 9�.

�L �L sin 9

�L cos 9 9

��

K

�

�

� K

10

6.17

a) FBDs are shown below.

Force Eqt’ns �D � D� Force Eqt’ns �J � � Force Eqt’ns �qrq � >� f:� cos 9 0 � 0 �L� 0 �� sin9 � ���� s:�� 0 � sin9 0 �� cos9 � ���

f:�L� 0 L 0 �L sin9 � �L�� s:�L 0 �L cos9 � �L�

f:� cos 9 0 0 �iji sin 9 � �iji�� s:� 0 � sin 9 0 �iji cos9 � �iji�

I hope you see how important it is to do a good job on the pictures before getting the equations. Note: perhaps you

already set the accelerations to zero since �� � 0 & � � 0 for all terms. That is a good idea. I left them as �� and � in case you wanted to see what to write for some other problem with non-zero acceleration.

Notice how convenient the system FBD is (if it is appropriate to use). The problem statement wording implied the

blocks are on the verge of slipping. We can solve for � in the s-equation from the system FBD. Then plug in this

crazy looking algebraic result into � ��� from the f-equation FBD. � � � sin 9 : �iji cos 9 � cos 9 0 ���� sin 9 : �iji cos 9� 0 �iji sin 9 � 0 � cos 9 0 ��� sin 9 0 ���iji cos 9 0 �iji sin 9 � 0 ��cos 9 0 �� sin 9� � �iji ��� cos 9 : sin 9�

� � �qrq$"# @AB C : BQR C@AB C 0 "# BQR C

b) Once we have use the system FBD to learn about the unknown �, we can plug that into either of the other two

FBDs to learn about the internal force �L�. Notice the system FBD is easier to start with but it will never help you

learn about internal forces. On the contrary, the two separate FBDs will always give you the same info as the

system FBD…but sometimes the algebra gets a lot messier.

From the �L FBD: s:�L � �L cos 9

Since on the verge of slipping we know L � ���L � ���L cos 9. f:�L� � L : �L sin 9 �L� � ���L cos 9 : �L sin 9 !JD � �J$�"# @AB C : BQR C�

FBD �D � D�

FBD �J � �

FBD �qrq � �J : �D

��

�L

�gtg

9 9 9

9 �

�� �

9 �

�L

� L

�L�

�L�

It is ok to use a system

FBD when the blocks

move as a single unit.

Notice the internal force �L� is not present on the

system FBD.

� � 0 � � 0 � � 0

11

6.18

a) The direction of friction would change in each FBD. This change would in turn affect signs in the force

equations. Personally, I would start the problem over from scratch to ensure I don’t screw things up. Some

people might be clever and simply flip the signs of each friction term and each arrow…that said, I

recommend starting over from scratch to clearly indicate you are analyzing a different scenario.

b) Once the incline is angled, a horizontally applied force includes a small component of force up the plane.

For large incline angles, increasing the horizontally applied force tends to increase the normal force more

than the component of applied force up the plane.

As a result, you are also increasing the maximum possible friction force down the incline.

c) We want to determine and angle for the incline such that we are just barely able to cause the blocks to

accelerate up. This means our previous problem force equations are still valid but we want to solve for 9

instead of �. From a step in the previous problem we found � cos 9 − ��� sin 9 − ���iji cos 9 − �iji sin 9 = 0

cos 9 �� − ���iji � − sin 9 ���� + �iji � = 0

sin 9 ���� + �iji � = cos 9 �� − ���iji �

tan 9 = � − ���iji ��� + �iji

tan 9 = ��iji − ���� ��iji + 1

9 = tanuL v ��iji − ���� ��iji + 1w

Looks like the size of the applied force � (relative to the total weight 3� ) and the frictional coefficient

matter in determining the max possible incline angle.

12

6.18½

a) A system FBD may be used when two objects move together as a single unit.

b) FBD and force equations are shown below.

The relative sizes of the arrows in this figure are approximately correct.

Block 1 Block 2 1 & 2 System

Σ��:� − L − �L� = �L� Σ��:�L� − � = ��� Σ��:� − iji = ��L + ����

�:�L = �L �:�� = �� �:�iji = ��L + ���

When the blocks are moving relative to the floor we may use = �1�.

In this particular case, it allows us to say iji = �1�iji,L = �1�L , and � = �1��.

Part b asks us to determine acceleration magnitude.

The given variables are the masses, the frictional coefficients, and .

I look at each set of equations and realize the system force equations (and iji = �1�iji) will get the job done.

The horizontal system force equation gives: � − iji = ��L + ����

� = � − iji�L + ��

Now I use the friction equation iji = �1�iji: � = � − �1�iji�L + ��

Now I use vertical system force equation:

� = � − �1��L + ��� �L + ��

� = ��L + �� − �1��L + ��� �L + ��

F = ��J + �D − "O$

�L ��j,L

�L

�Lj,� �

��

��

� L

��L + ���

�iji

� iji

� � �

13

Solution continues on the next page…

Part c:

In part c we were asked to determine the normal force (magnitude) acting between the blocks.

You can choose either FBD (block 1 or block 2) and solve for �L� now that we know � from part b!

I will choose to use the block 2 equations (because they have one less term…one less possible thing to screw up). �L� − � = ���

Using the friction relationship � = �1��: �L� − �1�� = ���

Using the vertical force equation (for block 2): �L� − �1��� � = ���

First solve this simpler result for �L�, then plug in the big messy result for�: �L� = �1��� � + ��� �L� = �1��� � + �� N ��L + �� − �1 P

Now try to clean it up a bit: �L� = �1�� + ����L + �� − �1��

WHOA! Notice the terms with a frictional coefficient on it cancel out! !JD = �D�J + �D �

Part d:

We were told �L = 1.00kg,�� = 2.00kg, � = 40.0N, �� = 0.777&�1 = 0.666.

This gives F ≈ E. {'| }BD

!JD = DE. E|NNNN

Part e:

We are asked to determine the minimum force required to cause the blocks to slide.

Our previous work must be modified in two ways:

1. The block is on the verge of slipping. Use = ��� instead of = �1�.

2. When on the verge of slipping, acceleration is zero!

If allowable, the system FBD is a good thing to try first.

Because these particular blocks move in unison we may use the system FBD.

The horizontal force equation give: �*+, − iji = ��L + ���0 �*+, = iji �*+, = ��� �*+, = ����L + ��� �� ! = DD. {%(

Notice the value used in part d is clearly larger than �� !.

14

6.19

a) 45See below. The relative sizes of the arrows in this figure are approximately correct.

Block 1 Block 2 1 & 2 System

Σ��:L� = �L� Σ��:� − − L� = ��� Σ��:� − = ��L + ����

�:�L� = �L �:� = �� + �L� �:� = ��L + ���

b) no mention is made of the block being on the verge of slipping!

We expect L� ) ���L�.

However, we were told � = 5.5 ~[S.

Use the horizontal force equation from block 1 to find L� = �L�.

Furthermore, because block 2 is moving (relative to the floor) friction from the floor on block 2 is = �1� = �1��L + ��� Plug in these expressions into the horizontal force equation for block 2 and solve for �. � − − L� = ��� � − �1��L + ��� − �L� = ��� � = �1��L + ��� + �L� + ��� � = �1��L + ��� + ��L + ���� � = ��L + ����� + �1 � � ≈ 31.5N

Recall for this problem we were told to use = 10 ~[S.

Also note: the system FBD should give identical results.

Note: we could’ve stacked the horizontal force equations for blocks 1 & 2 and added as well.

Check: the maximum possible value is L�*k� = ���L� = ���L = 6.0N.

We have verified L� ` L�*k� �j��+�l� .

Solution continues on the next page…

�L�

L�

�L �L�

L�

��

�

�

��L + ���

�

�

� � �

15

Continuation of 6.19…force equation shown again to reduce scrolling

Block 1 Block 2 1 & 2 System

Σ��:L� = �L� Σ��:� − − L� = ��� Σ��:� − = ��L + ����

�:�L� = �L �:� = �� + �L� �:� = ��L + ���

c) The problem statement asks about the maximum possible � for which the block move as a single unit.

THINK: if too much force is applied, block 2 will slide out from underneath block1!

When the maximum possible � is applied to block 2, block 1 is on the verge of slipping.

In this special case we use L� = ���L� = ���L

WATCH OUT!

We are applying more force than in the previous scenario (part b).

We no longer know the block’s acceleration!

HOWEVER, we can use the horizontal force equation from block 1 to determine �. L� = �L� ���L = �L� � = ��

Now plug L� = ���L & � = �� into the horizontal equation for the 1 & 2 system (or block 2 eqt’n). � − = ��L + �����

Block 2 is still moving relative to the floor; we can still use = �1��L + ��� � − �1��L + ��� = ��L + ����� � = ��L + ����� + �1��L + ��� � = ��L + ��� ��� + �1� �*k� ≈ ��L + ��� ��� + �1� = 33.0N

Again, for this problem we were told to use = 10 ~[S.

Notice the result for part b had � ) �*k� which is another way to check our previous result.

d) The table below gives action reaction pairs.

Action (Forces exerted on �D) Reaction (Forces exerted by �D)

Earth pulls down on �� using gravitational force. �� pulls up on earth using gravitational force.

�L pushes down on �� using normal force. �� pushes up on �L using normal force.

�L pulls left on �� using frictional force. �� pulls right on �L using frictional force.

Floor pushes up on �� using normal force. �� pushes down on floor using normal force.

Floor pulls left on �� using frictional force. �� pulls right on floor using frictional force.

Zombie pushes right on �� (using normal force). �� pushes left on zombie (using normal force).

16

6.20

a) This part is nearly identical to part a of 6.14. 9b = tanuL �� ≈ 42°. b) Since we are less than the critical angle, the component of weight down the plane is not

sufficient to cause the block to slide. The FBD would look like the one at right. While

not perfect, the sizes of the arrows are roughly to scale. Note: at onset of slipping we

assume � = 0 and = �6�. You should find � = � ��� cos 9 − sin 9�

c) In this case and � would switch directions in the FBD. You would still assume � = 0

and = �6� since we are once again at onset of slipping. Find � = � ��� cos 9 + sin 9�

Not quite the same. When pulling down gravity helps the invisible woman slide the block. The force

magnitude she must apply is friction minus a little bit. When pulling up the plane the invisible woman

fights against gravity and friction. Her force magnitude is friction plus a little bit.

6.21 FBDs are shown for 1, 2, and 1-2 system. The relative sizes of the arrows are approximately correct.

Box (mass D�) Sled (mass �) Box-Sled System (mass >�)

Σ��:� cos 9 − L� = 2�� Σ��:L� = �� Σ��:� cos 9 = 3��

Σ�:�L� + � sin 9 = 2� Σ�:� = � + �L� Σ�:� + � sin 9 = 3�

a) � = 0.693�

b) L� = 0.200�

c) �L� = 1.654�

d) The maximum possible frictional force is given by L�*k� = ���L� = 1.49� . Notice L� ) ���L�.

e) If the zombie is included in the system the total mass 7�. We know the system accelerates to the right

with magnitude � = T�. Since the sled experiences negligible friction, the ground must somehow exert a net

horizontal force of ��,�i = �ijikl� = 7� �T�� = 1.40� . Perhaps this happens because the zombie is

wearing spiked shoes which prevents his feet from sliding? If you are still confused, draw a new system

FBD. Notice the tension force becomes internal…

f) Internal forces are aL� and ��aL�. In case you are wondering why the arrows suddenly appeared, remember aL� is the vector while L� is just the magnitude.

g) Earth pulls down on box with gravitational force; box pulls up on earth with gravitational force.

�L�

L�

2� �L�

L�

�

�

3�

�

� � �

�

� cos9 � sin 9 � sin 9 � cos9

� � sin 9

� cos9 9

� ≈ 0 �

�

17

6.22 Consider FBDs below. I tried to draw them roughly to scale…the angle isn’t that close but the rest of the stuff

is in the right ballpark.

Note: this problem indicates the blocks are sliding relative to the surface; use = �1�.

Even though the blocks move together as a single unit the � & 3� system FBD is problematic.

The frictional force comes only from the block of mass 3�.

One must use the normal force exerted only on 3� when plugging into = �1�. � = Nsin 9 − 34 �1 cos 9P = 0.297

K = 34 �1� cos 9 = 0.276�

The bonus question says show the blocks will slide for all angles greater than 20.5°. This implies you are to assume the system on the verge of slipping.

Assume � = 0 and = ~�� �Z[[]��� = ���.

From there solve for the critical angle just like a single block on an incline (I think that was problem 6.14 or so).

�

��

3� cos 9

3� sin 9

9

FBD 3m

K

�L

� cos 9

� sin 9

FBD �

K

f

s �

f

s �

18

6.23

a) See the upper FBD at right.

If the block is lowered that means velocity is down the plane.

The lowering rate is increasing; this implies the speed is increasing.

Increasing speed implies �a points same direction 5a.

Friction will point opposite the direction of motion (opposite 5a).

Because block and incline in relative motion, use = �1�.

Force equations are � = � cos 9 + � sin 9 � sin 9 − − � cos 9 = �� = �1�� cos 9 + � sin 9�

b) See the lower FBD at right.

If the lowering occurs with decreasing rate, the velocity is still down the

hill but it is slowing down. This implies �a points up the plane.

Friction still points opposite the direction of motion (opposite 5a).

Because block and incline in relative motion, use = �1�

Force equations are slightly modified. See below. � = � cos 9 + � sin 9 −� sin 9 + + � cos 9 = �� = �1�� cos 9 + � sin 9�

� sin 9

�

� cos9

� cos 9

� sin 9

f

s �

� sin 9

�

� cos9

� cos 9

� sin 9

f s

�

19

6.24 Note: in this problem masses �&4� move in unison…it is appropriate to use a system FBD if desired.

a) We don’t know the acceleration but we do know the blocks accelerate to the right…

Force equations for %� Force equations for %� Force equations for System Σ��:K − �L� = �4��� Σ��:�L� = ���� Σ��:K = �5��� Σ�:� − L� − �4�� = 0 Σ�:L� − ��� = 0 Σ�:� − �5�� = 0

From the horizontal system force equation � = K5�

Plug this into the horizontal force equation for the simpler of the two individual block FBDs.

In this case, the simpler FBD between �&4� is the FBD for �. �L� = � N K5�P = K5

Since we are told to find the minimum thrust, we know block � must be on the verge of slipping! L� = ���L� = ��K5

Finally from the vertical force equation on the simpler FBD (block � FBD) L� = � ��K5 = �

�� ! = ��$"

Note: People tend to forget downwards L� on block 4� for some reason.

b) Friction force remains constant! If the thrust force is doubled then �L� (the normal force between the blocks) will

double, the maximum possible frictional force will double, but the actual friction force remains the same. Think

about it: if the friction force also doubled the friction force would be twice as big as the weight of the smaller block.

This would, in turn, imply the friction on the smaller block accelerates the small block upwards. This makes no

sense. Watch out for the subtle difference between frictional force and max possible friction force.

�

�L�

L�

4�

K �L�

L�

�

� �

FBD for %� FBD for � System FBD

for �&%�

�

5�

K

�

20

6.25 a) Nothing has indicated the block is on the verge of slipping. Assume ) ���.

b) Action: The earth exerts a gravitational force directed downwards on the block.

Reaction: The block exerts a gravitational force directed upwards on the earth.

c) Weight remains constant. Weight is the force of gravity exerted by the earth on the block and this has

nothing to do with the angle of the board!

Component of weight parallel to the incline decreases as angle decreases.

Component of weight perpendicular to the incline increases as angle decreases.

Normal force increases (to offset the increasing component of weight perpendicular to the incline).

Frictional force decreases (less static friction required because component of weight parallel to incline is

decreasing).

Max possible frictional force increases. The max possible frictional force is given by *k� = ���. The

normal force has increased; so does the max possible frictional force. Note: in this case ) *k�.

21

6.26 Friction Graph #1 (FBDs and force equations shown in 6.1)

e) Block is on the verge of slipping when �*+, = ��� = 4.90N.

f) For ) �*+,, one finds = � and � = 0.

g) For > �*+,, one finds = �1� and � = �* − �1 .

h) The data and plots are shown below. The step size I chose, 0.50 N, was overkill but who cares. Notice the

highlighted values of near Fmin. I added in extra values of F just before and after Fmin to ensure the

transition looks smooth in the graph.

i) This is why it is nice to do things algebraically. Looking at parts a) through c) we see Fmin increases which

shifts the transition point in both graphs to the right. For � > �*+, the slope of f vs F will increase while

the slope of a vs F will decrease (hint: consider the �* term which determines the slope).

j) Decreasing �� shifts the transition point to the left. None of the slopes are affected before or after the

transition point.

k) Since Fmin changes I would need to change the equations for both f and a for values of F near Fmin.

µs µk m (kg) g (m/s2)

0.5 0.3 1.00 9.8

F (N) f (N) a (m/s2)

0.00 0.00 0

0.50 0.50 0

1.00 1.00 0

1.50 1.50 0

2.00 2.00 0

2.50 2.50 0

3.00 3.00 0

3.50 3.50 0

4.00 4.00 0

4.50 4.50 0

4.89 4.89 0

4.90 4.90 0

4.91 2.94 1.97

5.00 2.94 2.06

5.50 2.94 2.56

6.00 2.94 3.06

6.50 2.94 3.56

7.00 2.94 4.06

7.50 2.94 4.56

8.00 2.94 5.06

8.50 2.94 5.56

9.00 2.94 6.06

9.50 2.94 6.56

10.0 2.94 7.06

0.00

1.00

2.00

3.00

4.00

5.00

0.00 2.00 4.00 6.00 8.00 10.00

f(N

)

F (N)

f vs F

0.00

2.00

4.00

6.00

8.00

0.00 2.00 4.00 6.00 8.00 10.00

a(m

/s2)

F (N)

a vs F

22

6.27 Friction Graph #2 (FBDs and force equations similar to 6.14)

a) The block is on the verge of slipping when 9 = 9b = tanuL���� ≈ 25.6°. b) For 9 ) 9b one finds = � sin 9 and � = 0.

c) For 9 > 9b one finds = �1� cos 9 and � = �sin 9 − �1 cos 9�.

d) Data table and graphs are shown below. Notice the extra values of 9 near 9b = 25.6°. These extra values

are used to make the plots look pretty.

e) The critical point is not affected by mass! A 5 kg block will slip at the same angle as a 10 kg block. The

magnitude of the frictional force is proportional to the mass. The magnitude of the acceleration is also

unaffected by the change!

f) If �� = 0.4 then 9b = tanuL���� ≈ 21.8°. This shifts the transition point in the graphs to the left creating

new data between 21.8° and 25.6°. Since �� does not appear in parts b) or c), the rest of the graph remains

exactly the same!

µs µk m (kg) g (m/s2)

0.5 0.3 1.00 9.8

θ (°) f (N) a (m/s2)

0.0 0.00 0

5.0 0.85 0

10.0 1.70 0

15.0 2.54 0

20.0 3.35 0

25.0 4.14 0

26.0 4.30 0

26.5 4.37 0

26.6 2.63 1.8

27.0 2.62 1.8

30.0 2.55 2.4

35.0 2.41 3.2

40.0 2.25 4.0

45.0 2.08 4.9

50.0 1.89 5.6

55.0 1.69 6.3

60.0 1.47 7.0

65.0 1.24 7.6

70.0 1.01 8.2

75.0 0.76 8.7

80.0 0.51 9.1

85.0 0.26 9.5

90.0 0.00 9.8

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

0 10 20 30 40 50 60 70 80 90

f (N

)

θ (°)

f vs θ

0

2

4

6

8

10

0 10 20 30 40 50 60 70 80 90

a(m

/s2)

θ (°)

a vs θ

23

6.28

a) Fall 2018 students…start at part b… When the block lifts off the surface, the normal force

must be zero. Use forces in the vertical direction but set � = 0 and keep � = 0. Why? This

gives the smallest possible component � sin 9 which completely supports the object’s

weight. If � sin 9 = � one then finds 9*k� = sinuL � � ≈ 78.5° b) Determining the onset of slipping angle is surprisingly tricky! First assume �� = 0. The

horizontal force equation becomes � cos 9 − ��� + ��� sin 9 = 0. To do the problem

analytically, isolate the cosine term then square both sides of the equation. Next use cos� 9 = 1 − sin� 9 to get a quadratic formula for sin 9. Finally, don’t forget to take

inverse sine of your quadratic formula result. For my quadratic formula I found

sin 9 = −N−2� ���� P � pN−2� ���� P� − 4�1 + ���� N�� ��� �� − 1P2�1 + ����

Don’t be surprised if yours looks different; we may have grouped terms differently. While it is possible to

clean this up a little bit, the result will always be ugly. I decided to plug in numbers to find sin 9 = 0.98 � 1.0762.5

If you are like me, you are wary of such a complicated looking equation and would want to check your

work. Fortunately, this problem can be solved without the quadratic formula mess. First create a function s�9� = � cos 9 − ��� + ��� sin 9. The solution of the problem occurs when s�9� = 0. Make a table

of 9 and s�9�. Look in the data table for a sign change in s�9�. Alternatively, plot s�9� vs. 9 and look for

the point when the graph crosses the horizontal axis. There is a lesson to be learned here; sometimes

problems can and should be solved using numerical methods.

c) For 9 ) 9b the block slides! One finds = �1�� − � sin 9� and � = �* �cos 9 + �1 sin 9� − �1� .

d) For 9 > 9b the block is at rest. One finds = � cos 9 and � = 0.

e) A data table and graphs are shown on the next page.

f) The maximum acceleration occurs when �k�\ = 0. One finds sin 9 −�1 cos 9 = 0 or 9 = tanuL �1. Notice

this uses �1 not �� as in Friction Graph #2! This gives an angle of 9 ≈ 16.7° with �*k� = 2.28 ~[S. This

matches with the graph of � vs 9 shown below. I added in some sig figs for a few rows to clearly show this

in the data set.

�

�

� = 0

�

� cos9

� sin 9

�� =?

24

µs µk m (kg)

0.5 0.3 1.00

g (m/s2) F (N)

9.8 5.00

θ (°) f (N) a (m/s2)

0.0 2.94 2.06

5.0 2.81 2.17

10.0 2.68 2.24

15.0 2.55 2.2779

16.7 2.51 2.2802

16.9 2.50 2.2801

20.0 2.43 2.27

25.0 2.31 2.23

30.0 2.19 2.14

40.0 1.98 1.85

50.0 1.79 1.42

55.0 1.71 1.16

55.3 1.71 1.14

55.4 2.84 0

55.5 2.83 0

60.0 2.50 0

65.0 2.11 0

70.0 1.71 0

78.5 1.00 0

0.00

0.50

1.00

1.50

2.00

2.50

3.00

0 10 20 30 40 50 60 70f

(N)

θ (°)

f vs θ

0.00

0.50

1.00

1.50

2.00

2.50

0 10 20 30 40 50 60 70

a(m

/s2)

θ (°)

a vs θ

25

6.29 Friction Graph #4

Σ��:K − �L sin 9 − = �L� Σ�:� − �L cos 9 = 0 �L slides relative to plane: = �1� K − �L sin 9 − �1�L cos 9 = �L�

Σ��:�� − K = ���

a) The system experiences no friction when �� = �L sin 9.

b) For small ��, friction points up the plane giving ��*+, = �L�sin 9 − �� cos 9� ≈ 0.616kg.

c) For large ��, friction points down the plane giving ��*k� = �L�sin 9 + �� cos 9� ≈ 1.116kg.

d) The magnitude of friction is given by = �1�L cos 9 and points up the plane. Assume acceleration is

down the plane for �Land upwards for �� with magnitude a. � = �L�sin 9 − �1 cos 9� − ���L + �� ≈ 0.716kg − ��1.00kg + ��

e) The magnitude of friction is given by = �1�L cos 9 and points down the plane. Assume acceleration is

up the plane for �Land downwards for �� with magnitude a. � = �� − �L�sin 9 + �1 cos 9��L + �� ≈ �� − 1.02kg1.00kg + ��

f) For the entire range of masses � = 0. The friction force up the plane is = �L sin 9 − �� . Notice that

for �� > �L sin 9 the result is negative. This implies the frictional force points down the plane as

expected.

g) In part e) � ≈ when �� → ∞. Both parts d) and e) check out when �1 = 0 and when 9 = 90°. h) A partial data table and the plots are shown on the next page. Notice the above equations for a and f are

magnitudes. For parts c) and e) friction points down the plane and negative values were used in the data

set. For parts b) and d) acceleration points down the plane and negative values were used. On the friction

plot I ignored masses greater than 2 to show the interesting portion of the graph in greater detail. Notice

the acceleration plot begins to asymptotically approach + for large �� as we expect.

�L �L sin 9

�L cos 9 9

��

K

�

�

� K

26

µs µk m1 (kg) g (m/s2) θ (°)

0.5 0.3 1.00 9.8 60.0

m2 (kg) f (N) a (m/s2)

0.000 1.47 -7.02

0.200 1.47 -4.21

0.400 1.47 -2.21

0.500 1.47 -1.41

0.600 1.47 -0.71

0.616 1.47 -0.61

0.617 2.38 0.00

0.700 1.56 0.00

0.800 0.57 0.00

0.866 -0.09 0.00

0.900 -0.42 0.00

1.000 -1.41 0.00

1.115 -2.55 0.00

1.116 -2.56 0.00

1.117 -1.47 0.45

1.118 -1.47 0.45

1.200 -1.47 0.80

1.300 -1.47 1.19

1.400 -1.47 1.55

1.500 -1.47 1.88

5.000 -1.47 6.50

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

0.0 0.5 1.0 1.5 2.0

f (N)

m2 (kg)

f vs m2

-8.0

-4.0

0.0

4.0

8.0

0 1 2 3 4 5

a (m/s2)

m2 (kg)

a vs m2

27

6.30 If you were thinking this problem seems implausible,

you are correct. I drew an FBD below assuming this thing

was accelerating to the right. I tried to make the forces

roughly to scale based on a small angle. To make the figure

possible, I had to ensure that the vertical component of �

was at least as large as � . My magnitude for � is about

4.5 times larger than the mass. Even with this huge applied

force the normal force is still less than half of � !!! Notice

the net force parallel to the plane (and thus the acceleration)

is massive. Clearly this scenario can only work in the most

special of circumstances. That said, it is physically

possible (probably only for a very short time interval in the real world).

a) �*+, = *T��^\. To figure this out I assumed the thing would start to lose contact if � = 0. I did the forces

perpendiuclar to the plane, set � = 0, then solved for �. If � were any smaller, the normal force would be

negative (which doesn’t make any sense).

b) From forces parallel to the plane one finds � = �* �cos 9 − � sin 9� + �� cos 9 + sin 9�.

c) �*+, = 3.657N compared to a weight of � = 0.980N.

Corresponding acceleration is � = 10.3 ~[S which implies the block must accelerate faster than freefall.

This might work while you extend your arm but it would be impossible for a human to keep this up for any

significant amount of time without some mechanical advantage of some kind…

�

�

�

�

9

9

28

6.31

a) FBDs are shown at right. Forces equations become Σ��:� cos 9 − = �� Σ�:� cos 9 + � − � = 0

When � = 0 block lifts off the surface.

Therefore, set � = 0 and solve for �. ��F��%� ?qr?? = �$BQR C

b) Answer in problem statement.

c) For 9 = 0° we get �*+, = ��� which matches a previous result.

For 9 = 90° we get �*+, = � . Notice this would lift the block off the table.

d) � = �* �cos 9 + �1 sin 9� − �1

e) Take �k�\ = 0 and solve for 9. Cr�q ��� = ��RuJ "O

6.32

a) FBDs are shown at right. Forces equations become Σ��:� cos 9 − = �� Σ�:� − � cos 9 − � = 0

�� = � cos 9� sin 9 + �

b) For 9 = 0° we get �� = �*T which matches a previous result. For 9 ≈ 90° we get �� ≈ 0.

This makes because 9 ≈ 90° implies you are pushing almost straight down.

If the block is going to move the surface had better be frictionless.

c) As � → 0 the coefficient drops to 0 as well.

This makes sense because the block is expected to move with almost no force.

The surface has better be frictionless if we wish the block to move with almost zero applied force.

d) � = �* �cos 9 − �1 sin 9� − �1

e) Take �k�\ = 0 & solve for 9. 9j�i+*�* = tanuL�−�1� Cr�q ��� = − ��RuJ "O

Notice this gives a negative angle.

f) Pushing down at a negative angle is the same thing as pulling up with a positive angle. The result from

both questions match.

�

�

� = 0

�

� cos9

� sin 9

�� = � =?

�

�

� = 0

�

� cos9 � sin 9

�� = � =?

29

6.33 I expect the heavier block (2) should go downhill. I set up my coordiantes with the upper block (1) and rotate

them around the pulley to match the other block. While it looks a bit strange to have the positive y-direction

pointing down for block 2, it is no big thing. Since there is no acceleartaion in the s-direction we can still use the

ups equal the downs (thus ignoring the confusion associated with the upside down coordinates on 2). To be clear,

when I say ups equal the downs for this one I am assuming you know I mean the forces perpendicular to the plane.

WATCH OUT! Even though the blocks slide with same magnitude of acceleration, they are not moving together as

a single unit. I would advise against doing a system FBD. Friction IS NOT an internal force in this case.

Notice, friction between the blocks points opposite acceleration in both FBDs. It does not make sense to treat it as

an internal force. Separate FBDs for each object is always valid…the system FBD is sometimes valid.

a) In this case expect heavier object (2) to go downhill.

b) Setting up the equations to stack them and add them K − �L sin 9 − ��L cos 9 = �L�

+�� sin 9 − K − ��L cos 9 = ���

____________________________________________________

�� sin 9 − �L sin 9 − 2��L cos 9 = ��� + �L��

On the verge of slipping (min angle to slide) use � = 0 and � = �� giving 9 = tanuL � 2���L�� − �L  ≈ 5.7°

Solution continues on the next page…

K

�L�

12

�L

K

��

12

�2

� �

�12

30

c) Same equations as before except � = �1 and we are solving for � (instead of setting it equal to zero). One

finds � = 0.4996 and K = 1.416� . As a at least a small check, notice the tension is greater than

component of weight pulling 1 downhill and less than the component of weight pulling 2 downhill as to be

expected. � = ��� − �L� sin 9 − 2��L cos 9�� + �L

� = �3�� sin 9 − 2�� cos 95�

� = 3 sin 9 − 2� cos 95

� = 3 sin�60°� − 2�0.10� cos�60°�5 � = 0.4996 K = N 2���� + �LP�L sin 9 + N�� − �L�� + �LP ��L cos 9 = 1.416�

Note: if we set friction to zero and the angle to 90°, how does this compare to an Atwood’s machine? Do

the formulas make sense in this limit?

d) g = 0.452sec. Hint: you need the acceleration of �L relative to ��. This acceleration has twice the

magnitude of either object alone. Therefore � = 0.9992 . The distance is still 1.0 m. If released from

rest the initial speed is zero. From there do kinematics of constant acceleration.

Note: when the length of �� is much greater than �L we can ignore the small timing error introduced as �L tips over and falls off of ��.

31

6.34 This is another one where the system FBD might lead you astray. The friction comes from block 1; therefore

you need and FBD with the normal force from the floor on block 1. To save time force vectors NOT to scale…

Look at the FBD of 1 only. Notice blocks 2 & 3 exert normal forces parallel to the plane. This implies the normal

force between 1 and the floor has magnitude �L = �L cos 9. On the verge of slipping this system has � = 0 and L = ��L cos 9. Then consider the system FBD. Since the acceleration is zero, we expect L = �ijikl sin 9 ��L cos 9 = �ijikl sin 9 9 = tanuL N� �L�ijiklP

• As a way to check your answer, consider what happens if blocks 2 and 3 have no mass. The problem

should reduce to 6.18. Does this more complicated result recue to the result of 6.18 as we expect?

• As another check on your answer, consider what should happen if block one is very small compared to the

other two blocks. Do you think the critical angle should be large or small? Does the above equation reflect

what your intuition suggests?

The normal forces should have the following magnitudes:

Between floor and block 1 �L = �L cos 9

Between floor and block 2 �� = ��� + ��� cos 9

Between block 2 and block 3 ��� = �� cos 9

Between block 1 and block 2 �L� = �� sin 9

Between block 1 and block 3 �L� = �� sin 9

For angles greater than the critical angle, the blocks should accelerate. Normal forces parallel to the plane are

affected. For example �� sin 9 − �L� = ��� giving �L� = �� sin 9 − ���…a little smaller than the previous

result! This makes sense. Similarly �L� = �� sin 9 − ���.

FBD 1

�L

�13 �12

1

�L

9

� = 0 FBD 1-2-3

�gtg�¢1

�gtg�¢

9

� = 0

32

6.35

a) K = � . Hint: consider forces on block � in the vertical direction.

b) � = T�. Hint: consider forces on block 2� in the horizontal direction. Plug in � determined from last part.

c) Kn£¤6g = ¥� � . Hint: forces on system FBD in horizontal and vertical. Plug in � = T� again.

d) Consider an FBD of the pulley. Tension appears twice since the string acts on the

pulley at each point it loses contact. You should also include the weight of the pulley

and a support force on the pulley in both the x and y direction! In chapter 12 we will

learn it is standard practice to call the forces at the pivot ¦� and ¦. That is why I

chose to label them that way. You could’ve just as easily called it a normal force or

some other letter. Net support force on pulley is 1.63� directed 47.5° above the

positive f-axis (I figured out each component then did the pythagorean thm).

e) It is worth it to consider if the blocks will slide if � = 0. They do. I then found �*+, = 0.222 and �*k� = 0.875 .

�L�

K

�

FBD for �

�23

K

2�

FBD for D� �

�

6�

FBD for �/D�/>�

System

FBD

for >�

�

�

K K �23

�13

3�

�

� � �

K

K

� = 2

FBD for pulley

15�

¦s

¦f

33

6.36

a) Assume the big block is object 1 with mass ¨ and the table cloth is object 2 with mass �. In case you are

wondering why I used a system FBD I would think about it like this. On the verge of slipping, the big

block ¨ is still moving essentially in unison with the tablecloth beneath it. For all intents and purposes, the

two blocks have identical accelerations. For any slightly larger acceleration, the block and the tablecloth

would have significantly different accelerations and the � + ¨ system FBD would no longer make sense.

Σ��:L� = ���L� = ¨� Σ��:� − = �¨ + ��� Σ�:�L� = ¨ Σ�:� = �¨ + �� L� = ���L� Block M not sliding relative to m, on the verge of slipping

= �1� ¨ + � system slides relative to floor

Note the internal forces L� & �L� do not appear in the system FBD.

From the big block FBD you can find � = �� .

From there you can plug that acceleration into the system FBD to find � − = �¨ + ���� .

Then you can also use = �1� = �1�¨ + �� .

I found � = ��� + �1��¨ + �� ≈ 8.6N.

b) The free body diagram for M is the same as in part a) with one exception. With a large force applied, the

block will slip relative to the tablecloth. The frictional force is now L� = �1�L� (uses �1 not ��). The

acceleration is � = �1 ≈ 2.94 ~[S.

Continued next page…

�

L�

�L�

¨

�

�

�¨ + ��

FBD for ¨ FBD for ¨ + �

�

34

c) We can no longer use the system FBD since � and © no longer move together as a single unit! The

two blocks have different accelerations! The friction force is tricky: use �L� for L� but use � for . Note: I

used �L� for acceleration of 1 relative to the earth. Similarly ��� is acceleration of 1 relative to the earth.

I found ��� = �uVª��«U*�T* ≈ 388.3 ~[S. This corresponds to about 39 ’s!

I found �L� = �1 ≈ 2.94 ~[S. Solution continues on the next page.

BEFORE GOING ON LET’S DISCUSS WHAT’S HAPPENING.

You apply a medium sized force (10 lbs.) to a very light tablecloth (100 grams is about ¼ lbs). There is friction

between the massive block (1 kg is about 2.2 lbs.) and the tablecloth but, because the force is so big relative to the

table cloth, the friction is insufficient to keep the block at rest relative to the tablecloth. As a result, the table cloth

flies out with a massive acceleration of about 39 ’s while the massive block barely accelerates at a fraction of a .

d) Use ∆f = L� �¬�lg�. Notice 5+ = 0 since the system starts from rest.

If we think about the situation from the block’s perspective (instead of the earth), the table cloth accelerates

forwards with magnitude �a�L = �a�� + �a�L = �a�� − �aL� = 385.4 ~[S ­.̂ The tablecloth moves distance ∆f = 6 relative to block at the instant the block falls off. One finds

g = p 26�¬�l ≈ p2�0.20�385.4 = 32.22msec

e) The speed of the block is 9.47 cm/s at the instant the tablecloth is fully removed. It has moved 1.53 mm

RELATIVE TO THE TABLE at this point. Remember: relative to the table means essentially the same

thing as relative the earth. Why? Because the table is not moving relative to the earth! We know the

acceleration of the block relative to the earth is �L� = �1 ≈ 2.94 ~[S. We can do kinematics using this

acceleration but we do know the time is the same as the previous part!

f) Once the tablecloth is gone, the block lands on the table. A frictional force of �1¨ directed to the left

causes the big block to slow down. The block accelerates to the left with magnitude � = �1 ≈ 2.94 ~[S.

The initial speed is 0.0947 m/s to the right. The block travels an additional 1.53 mm relative to the earth

for a total of 3.06 mm relative to the earth.

OBJECT 1 (the big block of mass ©) OBJECT 2 (the tablecloth of mass �) Σ��:L� = ¨�L� Σ��:� − − L� = ���� Σ�:�L� = ¨ Σ�:� = �L� + � = �¨ + �� = �1�&L� = �1�L� = �1¨ = �1� = �1�¨ + ��

���

�

�

�L� + �

FBD for �

+ L�= �

�L�

12

�12

¨

FBD for ¨

35

6.37

a) From forces on all three blocks we find 2� sin 9 = 3� giving � = �*T� []^\.

From forces on just the Crap, we see the normal force between the blocks must be � = � cos 9 = �*T� ��^\.

From forces on More Crap, we know the 2 = � . The factor of two comes from the fact that the forces

from Crap and Chemistry are equal. On the verge of slipping we may use = ���.

Combining gives 9 = tanuL�3��� = 56.3°. The force is thus � = 3� 2 sin�56.3°� = 1.80�

Check: the total force upwards on all three books is 2� sin 9 = 3.60� sin�56.3°� = 3.00� .

Check: the normal force is � = 1.00� . We find ��� = 0.5� . Looks consistent with 2 = � .

b) The vertical forces must always balance or the books would be accelerating upwards. As you press harder

and harder you must angle less and less (make the applied force more horizontal). The vertical force

equation on all three masses gives 9 = sinuL ��*T�� �. The vertical forces on More Crap must also balance.

This implies 2 = � (or = *T� ). If more horizontal force is applied, nothing changes with the frictional

forces upwards on More Crap. We simply changed from the onset of slipping to ) ���.

6.38

a) First consider the case when friction is zero. By doing this (and a system FBD) you should find the block

with the larger � sin�angle� should win the tug of war. We now know the blocks should tend to slide

that way and friction points the opposite way. Unfortunately, we still do not know if the blocks will stay at

rest or move yet. We do not yet know if we can use = ���, = �1�, or neither….

b) First determine the maximum possible friction force from both blocks combined. You should find ~�� �Z[[]��� = ��L�L cos 9 + ����� cos 9

From the system FBD we expect the net force in the direction of acceleration is �L sin 9 − �� sin °

Note: I’m assuming block 1 one the frictionless tug-of-war from the previous part. Flip the signs if 2 won

the tug of war in part a. We know the blocks will accelerate if �L sin 9 − �� sin ° > ~�� �Z[[]��� �L sin 9 − �� sin° > ��L�L cos 9 + ����� cos 9

36

6.39

a) The FBD and force equations are shown at right.

Think: if there was insufficient frictional force, the box would slide off the back of the truck.

This implies friction must point forwards (to the right) on the box.

Furthermore, the wording in part a of the problem statement never indicates the box is on the

verge of slipping.

This implies frictional force is LESS than the maximum possible value of ���.

Notice: in problems where ) ��� we typically know acceleration.

This allow one to compute the frictional force as shown below.

Using the horizontal force equation we quickly find = ��

? = �$�

b) Part b now asks for the largest possible acceleration which allows the block to stay on the truck.

The wording “largest possible” is your clue to know the problem in part b is on the verge of slipping.

In this scenario we CAN use = ���!

I will set up the equations and take a ratio to eliminate both �&�: � = ��� ���� = ���

�� = �

F = "#$

F ≈ %. & �#D

Block 1

��: = ��

Σ�:� = �

�

�

�

37

6.40

a) The weight is constant.

b) Normal force is constant.

c) The horizontal applied force ��a has increasing magnitude �.

While the block is at rest, frictional force must balance this increasing force. As a result, the frictional

force is increase until the moment the block is on the verge of slipping.

Once moving the frictional force assumes a constant value.

When the block is on the verge of slipping, ≈ *k��j��+�l� = ���.

Just after the transition, when the block is sliding, = 1+,�i+b = �1�.

Unless otherwise specified, one typically assumes �1 ) ��.

Therefore, during the transition from static to kinetic friction, the magnitude of friction decreases slightly.

In summary: we expect the frictional force to gradually increase until the block starts to slide.

Once sliding, frictional force drops to a constant value slightly lower than max possible frictional force.

d) The maximum possible frictional force (*k��j��+�l� = ���) does not change in this scenario because the

normal force is not changing in this scenario. Note: in general the actual friction force exerted on the block

depends on whether or not the block is

a. sliding ( = 1+,�i+b = �1�)

b. at rest but on the verge of sliding ( = *k��j��+�l� = ���)

c. at rest & not on the verge of sliding ( ` *k��j��+�l�).

e) Just before the transition, when the block is on the verge of sliding, ≈ *k��j��+�l� = ���.

Just after the transition, when the block is sliding, = 1+,�i+b = �1�.

Unless otherwise specified, one typically assumes �1 ) ��.

Therefore, during the transition from static to kinetic friction, the magnitude of friction decreases slightly.

f) The coefficient of friction is a property of the interface between the two objects and the state of relative

motion between the two objects. Until the box begins to slide, the coefficient of friction is constant at 0.5.

Once it slides the coefficient becomes the new constant value of 0.3.

Notice the differences between frictional force, maximum possible frictional force, and frictional coefficient.

6.41 Run the simulation and see.

6.42

a) Weight remains constant.

b) The component of weight down the incline increases.

c) The normal force decreases.

d) The frictional force increases.

e) The maximum possible frictional force decreases.

f) During the transition the magnitude of the frictional force decreases.

g) The coefficient of friction is a property of the interface between the two objects and the state of relative

motion between the two objects. Until the box begins to slide, the coefficient of friction is constant at 0.5.

Once it slides the coefficient becomes the new constant value of 0.3.

Notice the differences between frictional force, maximum possible frictional force, and frictional coefficient.

6.43 Run the simulation and see.

6.44 It is possible both in the simulation and in real life. No incline is needed. Give the block a hard push to get it

moving quickly. Stop pushing on it. The block slides to the right and has friction pointing to the left. The block is

slowing down while moving to the right; acceleration also points to the left.

38

6.45 While possible in real life, I don’t believe you can make it happen with this simulation. Consider a person

walking. Friction between the floor and her feet pushes her forwards. In this case both velocity and friction point in

the same direction.

6.46

a) One must use an angle less than the critical angle of 26.6°. Get the block moving on the flat portion and

stop pushing on it. Once you release the block it should hopefully slide up and stop. It can be done.

b) Before coming to rest, friction points down the plane. After coming to rest, friction points up the plane and

the magnitude increases slightly!

39

6.47 See the figure at right. Notice I added the vectors 5am and −5a+ because ∆5a = 5am − 5a+ = 5am + �−5a+�. Since �a ≈ ∆±�a∆i we see that the acceleration at the top

of the circle must point in the same direction as ∆5a…towards the center of the

circle.

6.48

a) Now the final speed is greater than the initial.

Notice there is now a horizontal component of ∆5a (tangent to the circle).

b) See rightmost figure. Notice tangential acceleration

and velocity point the same direction when the

object is speeding up.

6.49

a) Now the final speed is less than the initial. Notice

there is still a horizontal component of ∆5a (tangent

to the circle).

b) See rightmost figure. Notice tangential acceleration

and velocity point in opposite directions when the

object is slowing down.

6.50 Change your speed with the brake or gas pedal.

Notice, however, if you maintain constant speed you can still accelerate!

Any time your velocity vector changes direction you accelerate.

Turning the wheel at constant speed also causes acceleration.

−5a+ 5am ∆5a

−5a+ 5am ∆5a �ik, �b �a

−5a+ 5am ∆5a �ik,

�b �a

40

6.51 a) Red arrows indicate �ik,.

b) Blue arrows indicate �b.

c) The magnitude of the total

acceleration is given by

Fqrq = ²�b� + �ik,� .

Notice Fqrq is ever-changing but never

equal to zero.

At instants A & E, Fqrq = �ik,.

At instant C, Fqrq = �b .

d) I interpreted the wording of this

problem as speeding up at A & B,

slowing down at D & E, and constant

speed for a split second at point C.

The split second after it is released at

A we know it will speed up.

The split second before it reaches E it

is still, ever so slightly swinging upwards and slowing to rest.

This was my interpretation. Some might argue at point E the ball is about to start swinging back down and

speeding up…that is a fine interpretation as well.

I suppose the important thing is to notice �ik, M 0 at points A & E.

e) A recommended simulation is found on robjorstad.com.

A

B

C D

E

�ik,

�ik, �ik, = 0 �ik,

�ik, �b �b �b �b = 0 �b = 0

Fqrq Fqrq

41

6.52

a) The top figure shown at right uses the standard directions for the coordinate

system (£̂£�³´�¢¢st¤gµ�£³6, 9¶g�� ·�ggt¸´£¸¢· t´� ¸t¤�g·£¸¢t¸¹µ´6·�.

The force equations are Σ�\º :− � cos 9 = ��−�ik,� Σ�¬̂ :− K − � sin 9b = ��−�b�

Remember, according to the convention, n�ik, and �b are the magnitudes of the

tangential and centripetal acceleration. The minus signs on the right side of each

equation arise because the acceleration components point opposite the directions

of the coordinate system.

The lower figure shows how you could flip the coordinate system.

In this coordinate system the force equations are Σ�\º :� cos 9 = ��ik, Σ�b̂:K + � sin 9b = ��b

Upon solving for the acceleration magnitudes, the results will be the same.

Physical results are independent of coordinate systems.

b) �ik, = cos 9 = 8. 49 ~[S. I used = 9.8 ~[S.

c) �b = ±S¬ = ±S» = 25. 0 ~[S

d) �ijikl = ²�b� + �ik,� = 26. 4 ~[S.

6.53

a) The distance around is 2¼£. Speed is distance over time. We find ½ = �¾¬± .

b) ¿ = �¾½ = ±¬

c) 10RPMs = 10 Ã�Ä~]^ Å �ÆÃ�ÇLÃ�Ä Å L~]^ÈÉ[�Y = 1.047 Ã�Ç[ = 1.05 Ã�Ç[ . Notice a good quick estimate is to divide by

10. d) 5.98 s

e) 3.14 m/s. Notice it is wise to avoid intermediate answer whenever possible to avoid round-off error.

Another good reason to do work algebraically as far as possible before plugging in numbers.

� cos9 9

�b

K � sin 9 �

�ik,

�aiji 9¶ £̂

STYLE 1

� cos9 9

�b

K � sin 9 �

�ik, �aiji 9¶ ¸̂ STYLE 2

�b �ik,

�aiji = 26. 4 ms�

71.2° 18.8°

42

6.54 HEADS UP! For this problem I used $ = J' }BD and only worried about 2 sig figs on final answers.

a) Up and to the left. For the rider’s path to curve to the left, there must be a force to the left (towards the

center of the circle).

b) In uniform circular motion (constant speed circular motion), the total

acceleration vector points directly towards the center of the circle. Note: if

there was no upwards component of force from the seat, the person’s weight

would be unbalanced and the system would have tangential acceleration.

c) The FBD is shown at right. The force equations are Σ�\º :� = � Σ�b̂:�� = ��b

d) Mass is unnecessary info. Use �b = ±S¬ to give 5 = ²£�b ≈ 4.5ms

e) For the right rider, �a = �−260­̂ + 650Ê�̂N or 700 N pointing about 68° above the negative f-direction.

f) For the left rider, �a = �260­̂ + 650Ê�̂N or 700 N pointing about 68° above the positive f-direction.

6.55

a) The FBD is shown at right. The bottom of the bucket exerts normal force �. � + � = ��b

b) Think: if the water falls out of the bucket, it is losing contact with the bucket.

If losing contact, � ≈ 0.

We may set � = 0 in the force equation to learn about the physical scenario of the

water just barely able to stay in contact.

Any slower the water falls out.

c) At minimum speed �b = giving 5*+, = ²£

d) The period is given by ½ = �¾¬± = 2¼Ë¬T.

e) At the minimum speed the orbit takes a maximum amount of time…this is the maximum period.

6.56

a) When the rotating object is not balanced it will always tend to speed up on the way down and slow down

on the way up. For a balanced object, gravitational force acts at the pivot in the center and causes no

torque. We learn about torque in chapter 10…

b) Strictly speaking, unbalanced objects travelling in vertical circles slow down on the way up and speed up

on the way down. If you swing it very quickly there is very little time to speed up on the way down (or

slow down on the way up).

Another way to think about it: the largest value of �ik, is occurs midway between the top and bottom.

If �b ≫ we know that �b ≫ �ik,.

This implies the object is accelerating, essentially, towards the middle of the circle.

Notice, if you substitute �b = ±S¬ , we see �ik, is negligible whenever 5 ≫ ²£ .

�

��

�

�b ¸̂

9¶

� + � �b

43

6.57

a) The FBD is shown at right. The force equations are Σ�Í̂: = � Σ�b̂:� = ��b

On the verge of slipping (max period) use = ���.

½ = 2¼£5 = 2¼p��£

Notice the mass doesn’t matter! Of course, the type of clothing will affect ��.

b) The net force has magnitude �,�i = ²�� + � = � ²1 + ���

c) The normal force will quadruple (proportional to speed squared). The frictional force remains � (else person would magically slide upwards?!?!?). I suppose if you wanted to match the wording of the question you could say the frictional force increases by factor 1…which means the same thing as no change.

6.58

a) Gradual changes in speed imply �ik, ≈ 0. b) The force equations are Σ�Í̂:� = � Σ�b̂: = ��b

On the verge of slipping (max period) use = ���.

½ = 2¼£5 = 2¼p £��

c) This equation is independent of mass. They should all fly off at the same time.

d) �� = ξS¬½ST

6.59 The FBD and force equations are identical to those in 6.58. One finds £*+, = ±SVWT.

6.60

a) The FBD and force equations.

Σ�b:� sin 9 = ��b = � 5�¦ Σ�Í:� cos 9 = �

b) Solving for v in the force equations gives 5+��kl = ²¦ tan 9. As 9

increases the turn requires more speed. I made a plot of 5+��klvs 9

using ¦ = 200m shown above and right. Notice I used miles per hour

for the units. It is good practice to make these plots yourself after class.

c) The normal force is � = � ±SÏ []^ \ = � LYZ[\.

Notice as 9 increases cos 9 decreases and � increases.

A plot of � vs 9 using � = 900kg is shown at right.

d) For speeds less than 5+��kl , the car slides to the inside of the track!

� sin 9

� cos 9 9

�

¸

Ð �b

�

�

�b ¸̂

Ð̂

�

�

�b ¸̂

Ð̂

0

5

10

15

0 10 20 30 40 50 60

n(k

N)

θ (deg)

n vs θ

0

50

100

150

0 10 20 30 40 50 60

vid

eal(m

ph)

θ (deg)

videal vs θ

44

6.61

a) I chose to show two styles of FBDs. Notice friction points down the plane at high speeds.

Style 1 Style 2 Σ�Ñ:� − � cos 9 = ��b sin 9 Σ�b:� sin 9 + cos 9 = ��b Σ�∥: + � sin 9 = ��b cos 9 Σ�Í:� cos 9 − sin 9 = �

Style 1 is nice for determining , �, or �. Style 2 is nice for finding5, �, or ¦. Trick: if on the verge of

slipping, Style 2 works great for both. Replace all ’s with ���. Take a ratio & all �’s cancel!

b) The minimum value of �� required is given by

��*+, = 5�¦ − tan 91 + 5�¦ tan 9

c) For slow speeds friction flips direction in the above FBDs and

flips the minus signs on all ’s in the force equations. You

might recognize this as being the same direction but use a

negative value for �! In any event we find

��*+, = tan 9 − 5�¦ 1 + 5�¦ tan 9

d) Finding the ideal angle is similar to finding the ideal speed.

We find 9+��kl = tanuL �±SÏT� ≈ 24.7°. e) The plot is shown at right. Notice that beyond 70° the coefficient exceeds 1. The use of negative lift or

downforce is required to ensure the car stays on the road. Side note: most race tracks do not exceed 25°.

6.62

a) To determine ideal speed do the problem with no friction (see 6.60 parts a & b).

Find 5+��kl = ²¦ tan 9 = 22.9m/s.

b) Since the speed of the car is less than the ideal speed, friction points up the plane.

c) Using an FBD similar to the last problem (but with friction pointing up the plane) I found � = � cos 9 + ��b sin 9 ≈ 8.99kN

d) Using an FBD similar to the last problem (but with friction pointing up the plane) I found = � sin 9 − ��b cos 9 ≈ 544N

e) Notice the car’s speed is close to the ideal speed. At the ideal speed no frictional force is required. We

expect the car is not on the verge of slipping. The maximum possible frictional force is given by = ���.

Notice ≈ 544Nwhile ��� = 0.15�8.99kN� ≈ 1350N! Clearly ) ��� which seems reasonable.

� sin 9

�

9 � cos9

Ó

∥

�b

9 �b cos9 �b sin9 9

Ð

¸

�b

9

� cos9

cos9

� sin 9

sin9

�

Style 1 Style 2

0.00

0.50

1.00

1.50

2.00

2.50

0 10 20 30 40 50 60 70 80 90

µm

in

θ (deg)

µmin vs θ

45

6.63

a) I used Style 2 (see previous problem) and found

5*k� = p¦ �� + tan 91 − ��tan 9

b) Plot is the top figure at right.

The graph has a vertical asymptote at about 51° c) Plot is the bottom figure at right.

The graph has a vertical asymptote at about 81°. d) The equation in part a) assumed the car was on the verge of

slipping to the outside. As the angle of the incline increases the

car is in danger of sliding to the inside of the turn instead. At this

point we should really be considering a minimum speed instead of

a maximum speed! The vertical asymptotes in the previous

graphs indicate the angle at which ) ��� regardless of the

speed. The car will never slide to the outside of the turn

beyond the asymptote angle. It makes sense, then, for the car to

have a smaller asymptote angle for a higher coefficient.

6.64

a) In general, the radius of circular motion must always go from the center of the circle to the center of mass

of the rotating object.

b) � = � Ô − Õ±S¥Ï Ö c) The block loses contact when � = 0. This gives 5*k� = Ë¥Õ ¦ . As the radius increases the hill becomes

more flat. We expect vmax should increase and the formula agrees with that.

d) In this problem the normal force points radially outwards. In the bucket of water problem the normal force

pointed radially inwards. That is the main difference.

6.65

a) Force equations from the FBD at right:

Take a ratio to eliminate K&�. Sub in �b = 5�/£. Notice in the original

figure £ = h sin 9. Solve for 5. 5 = ²h sin 9 tan 9

b) Increase. Note: both sin 9 & tan 9 increase as θ increases. Therefore

increasing 9 will require a greater speed. The formula makes sense in this

respect.

c) I rearranged the result to find ±S»T = []^S \YZ[\ . Then I use

cos� 9 + 5�h cos 9 − 1 = 0

The quadratic formula gives

cos 9 = − 5�h � pN5�h P� + 42

I found 9 = 51.8°.

Σ�b:K sin 9 = ��b Σ�Í:K cos 9 = �

0

100

200

300

0 10 20 30 40

v max

(mp

h)

θ (deg)

vmax vs θ for µ = 0.8 & R = 200

0

100

200

300

0 10 20 30 40 50 60 70

v max

(mp

h)

θ (deg)

vmax vs θ for µ = 0.15 & R = 200

K sin 9

K cos 9 9

�

¸

Ð �b

46

6.66

a) 5 = ²�h sin 9 + ¦� tan 9

b) The solution is independent of mass! People that design these types of rides do need to set weight limits on

rides based on the maximum tension allowable in the cables. They do not need to worry about riders of

different masses swinging out to different angles and impacting each other.

c) I first found ¦ = ±ST ��^ \ − h sin 9. Taking the derivative with respect to 9 gives ³¦³9 = −5� csc� 9 − h cos 9

Since both terms are negative increasing 9 decreases R. The converse is also true.

d) I started by rearranging the equation to show 0 = 5� − h sin 9 tan 9 − ¦ tan 9.

I then created the function �9� = 5� − h sin 9 tan 9 − ¦ tan 9. I substituted 5 = �¦ + h�¿. The

solution occurs when �9� = 0. Making a quick table or plot shows the solution occurs when 9 = 55.0°.

6.67

a) �b = £¿�

b) The force equations are × sin 9 = �f¿� ×cos 9 = � c) Take a ratio, Horatio… tan 9 = f¿�

d) The derivative of a curve is the tangent line to the curve.

Literally, tan 9 = ³Ð³f = f¿�

Separate the variables to find ³Ð = ¿� f³f

Ø ³ÐÍÙÍÚ = Ø ¿� f³f�Ù

�Ú

Ðm − Ð+ = ¿�2 fm� − ¿�2 f+�

Now shift f+ → 0 (the center of rotation is position zero).

Also shift fm → f (distance horizontally from the center of rotation is f).

This implies Ðm → Ð (the height of the water arbitrary distance from the center).

It also implies Ð+ = n (the height of the water at the center of rotation). Ðm = ¿�2 f� + n

which is a parabola where n is the depth of water at the center of the turntable.

e) If you spin it too quickly, there is no water in the center so the parabolic equation is no longer valid.

× sin 9

× cos9 9

�

¸

Ð �b

47

6.68

a) At the bottom.

b) When 5 = 17.3 ~[ ≈ 39 ~]ÛÃ.

6.69

a) ÜÝÞßÜàÞÝ = áSâãuLáSâãUL

b) As 5 → ∞ the ratio goes to 1 (the tensions are equal).

Think: as 5 → ∞, � is insignificant compared to ��b.

The tension in the string must be massive to produce adequate force towards the center of circular motion.

This massive tension implies � is negligible (compared to tension).

When � is negligible, it seems reasonable to have roughly equal tension at the top & bottom of the loop.

6.70

a) � = *S*XU*S

b) 9 = tanuL � *S*XU*S�.

c) The hanging ball moves forwards (to the right) relative to the earth.

The hanging ball moves backwards (to the left) relative to the little dude.

6.71

a) Forward (relative to you).

View the acceleration of the car as a fictitious force… think of it as artificial gravity in the −­ ̂direction.

While accelerating, if we use fictitious forces, gravity effectively pulls downwards and back.

The helium balloon is less dense than air and is pushed by a buoyant force opposite the direction of gravity.

Another way to think of it, when you accelerate forwards the air in the van will tend to slightly compress at

the back and be less dense by the windshield.

b) Backward (relative to you). Opposite reasoning to part a.

c) In the little dude problem (6.70) the ball hangs in the direction of the gravitational force.

If one adds fictitious gravity, the ball aligns in the direction of net gravitational force (artificial + real).

In this problem, the balloon tends to align opposite gravity due to the nature of the buoyant force (discussed

in Chapter 14). This is a tough one to answer. I put it in here in an attempt to make you think…that’s all.

6.72

a) The car must hold you up and press you towards the center of the circle.

If the car didn’t do this, you would tend to move in a straight line and leave the circular path!

If the car stays in a circular path but you travel in a straight line, you are no longer in the car…OUCH!

b) You accelerate towards the center of the circle.

c) You perceive that you are pressed to the outside by a fictitious force called the “centrifugal force”.

When you thought you were sliding towards the outside of the turn, really the car was accelerating towards

the center of the circle without enough friction to bring you with it.

Once you press outwards against the door, the door presses inwards on you.

The door provides the required force to push you towards the center and keep you accelerating towards the

center of the circle.

48

6.73

a) The floats go to the center. Similar to the balloon in a van (6.71).

When rotating, there is centripetal acceleration towards the center.

We perceive this acceleration as a fictitious gravitational force of gravity away from the center.

Dense objects “sink” to the outside in the same direction as the artificial gravity.

Objects less dense than water “float” opposite the direction of artificial gravity (towards the center).

b) Centrifuge.

6.74

a) Do an FBD at the top of a vertical circle.

Normal force and � both point downwards (towards the center of the circle).

Don’t forget these forces sum to ��b at the top of the circle!

Recall, if � = 0, that implies the rider is about to lose contact with the circular path.

Set � = 0 and solve for 5. 5 = ²¦b�,i�¬jm*k�� = Ë�2.50m� �9.8 ms�� ≈ 4.95ms = 11.1mihr

b) In this problem, at the top of a vertical circle, both normal force and � point radially inwards.

In the ball on a string problem, at the top of a vertical circle, tension and � point radially inwards.

In the block sliding over a hill, at the top, normal force points radially outwards (� radially inwards).

This problem and the ball on the string both have a minimum required speed at the top.

The block sliding over the hill has a maximum speed at the top.

This problem is more like a ball on a string in my opinion.

c) Probably not.

Note: if $3,000,000 in cash shows up in my office, everyone gets an A!

6.75 For the first circle we know � − �L = *±SÏ .

Remember, for the first circle � points towards the center and �L points away from the center.

For the second we know �� − � = *±SÏ .

Remember, for the second circle � points away from the center and �� points towards the center.

The problem statement tells us �L = LÎ � .

From there use algebra to show �� = æÎ � .

49

6.76

a) Pythagorean theorem (consider upper figure at right) shows the radius is £� + �0.6h�� = h� £� = 0.64h� £ = 0.8h

Rather than determining the angle directly, use sin 9 = 0.6hh = 0.6 cos 9 = £h = 0.8 tan 9 = sin 9cos 9 = 0.75 WATCH OUT! The upper string has more tension than the lower string. If not, there is

no possible way to support the weight of the ball! �b:KL cos 9 + K� cos 9 = ��b

KL + K� = �£¿�cos 9

Σ�Í:KL sin 9 = K� sin 9 + �

KL − K� = � sin 9

Adding the equations gives 2KL = �£¿�cos 9 + � sin 9

KL = �£¿�2 cos 9 + � 2 sin 9

KL = 0.625�£¿� + 0.833�

b) Notice we can rearrange KL − K� = *T[]^ \ to get K� = KL − � sin 9

K� = �£¿�2 cos 9 + � 2 sin 9 − � sin 9

K� = �£¿�2 cos 9 − � 2 sin 9

K� = 0.625�£¿� − 0.833�

c) The ratio of tensions (upper to lower) is KLK� = �£¿�2 cos 9 + � 2 sin 9�£¿�2 cos 9 − � 2 sin 9

KLK� = £¿�cos 9 + sin 9£¿�cos 9 − sin 9

KLK� = £¿� tan 9 + 1£¿� tan 9 − 1

d) Notice this ratio is greater than 1. We expect upper tension to be larger than lower tension. Seems reasonable.

e) Mass drops out! Surprised me a bit.

f) As rotation rate increases, the ratio goes to 1. When things rotate very rapidly, the weight of the ball is negligible

compared to tension. We expect the two tensions to be approximately equal. Seems to make sense.

¸ Ð �b

KL

K� �

£ h 0.6h 9

9 9

50

6.77 Not typed up yet. This problem can be found online but the typical solution method is beyond the scope of our

course. Come by office hours to discuss if you are curious.

6.78

a) Use a system FBD to show � = �*XU*S cos 9

b) Use an FBD on block 2 to find �L� = ���. Plug in previous result to find �L� = *S*XU*S � cos 9

c) In these limits

i. 9 → 90° implies � = 0 and �L� = 0. This seems reasonable since you are pushing down on 1.

ii. 9 → 0° and �L → 0 gives � = �*S and �L� = �. This could be thought of as 1 being a piece of

paper next to the block. When thinking about it this way, it makes sense that all of the force you

apply to the paper is transmitted through to block 2. Also, the acceleration makes since because

all of the force is essentially pushing only on m2.

iii. 9 → 0° and �� → 0 gives � = �*X and �L� = 0. Now we are pushing a block with a piece of paper

on the back side. Again the mass of the paper shouldn’t affect the acceleration. Also, since the

paper is very low mass, it takes almost no force to cause it to accelerate; the normal force makes

sense.

6.79

a) � = 3� �[]^ \V − cos 9� = 11.5� . I used a system FBD to get the result. As an intermediate check you

should find the normal force from the incline on 2m is � = � + 3� cos 9.

b) = 3� sin 9 = 2.60� . The frictional force from the incline must balance the component of weight of

the entire system down the plane.

c) = ���. You could’ve used this equation for part b. Notice you find = ���� + 3� cos 9� = 0.2�11.49� + 3� cos 60°� = 2.60�

d) L� = � sin 9 = 0.866� . Again, this frictional force must balance the component of mg.

e) I expect L� ) ��L��L� since there is a higher coefficient of static friction compared to the incline. First

find the normal force between 1 & 2 is �L� = � + � cos 9 = 11.49� + � cos 60° = 11.99� . Now

compute ��L��L� = �0.5�11.99� = 6.00� . We see clearly that L� ) ��L��L�.

Beware: the trick used to find the frictional force in parts b and d works fine UNLESS there is a force other than �

and parallel to the plane! For example, this trick won’t work if there is tension parallel to the plane.

51

6.80

a) The force equations on 2� are � + 2� sin 9 = 2�� and � = �L� + 2� cos 9. When looking for the

maximum hanging mass ¨ we know 2� is on the verge of slipping. Use � = ���L�. Notice we must be

careful to use the normal force between the blocks and not from the incline on 2�. An FBD on � reveals

that �L� = � cos 9. It is worth double checking to see � = 3� cos 9 as we expect. Using all this crap

to solve for � gives � = �sin 9 + ��2 cos 9�

Notice � is now known, in theory, since this equation contains only known parameters!

A system FBD gives the force equation ¨ + 3� sin 9 = �¨ + 3���. It is convenient to rearrange and

solve for ¨ before plugging equation for �. One finds

¨ = 3� � − sin 9 − � = 3� ��2 cos 91 − sin 9 − ��2 cos 9 = 0.75� cos 91 − sin 9 − 0.25 cos 9

b) Consider the denominator 1 − sin 9 − 0.25 cos 9. If this denominator is negative the mass equation no

longer makes sense. Setting the denominator equal to zero gives 1 − sin 9 = 0.25 cos 9

Squaring both sides gives 1 − 2 sin 9 + sin� 9 = 0.0625 cos� 9 1 − 2 sin 9 + sin� 9 = 0.0625�1 − sin� 9� 1.0625sin� 9 − 2 sin 9 + 0.9375 = 0

sin 9 = −�−2� � ²4 − 4�1.0625��0.9375�2�1.0625� = 1 � ²1 − �1.0625��0.9375�1.0625 = 0.8824

I ignored the positive root which gives 9 = 90° which will be discussed in a moment. Using this angle

gives 9 = 61.93°. c) If 9 ≥ 61.93° the denominator for M is negative. This implies something in our model isn’t correct. My

suspicion is at that large of an angle it is no longer possible for the blocks to be on the verge of slipping.

Notice this logic invalidates the usage of the root 9 = 90°. I guess you could consider at 9 = 90° all three

blocks are in freefall together since the normal forces (and string tension) drop to zero. I suspect when

doing the experiment in the real world the blocks would separate slightly. Perhaps air resistance is no

longer negligible due to the different amounts of drag on each block which might contribute a tiny amount

of string tension? I have no clue really…