CHAPTER 4

NEWTON’S LAWS

• Little bit of history

• Forces

• Newton’s 1st Law

• Newton’s 2nd Law ! mass and weight

• Newton’s 3rd Law ! action reaction pairs ! free body diagrams ! the normal force ! analysis of day-to-day situations

Two books laid the foundations for our understanding of forces and motion ...

Forces

A force is best described as an action or influence that when acting on an object, will change its motion.

There are two main types of forces:• contact forces (push, pull, friction, etc.)

⇒ direct force• non-contact (or field) forces (gravity, magnetic, electric, etc.)

⇒ “action” at a distance and long range... but they are treated in exactly the same way.

Dimension

[M][L][T]2

(show later)

Unit: Newton (N)1N ⇒ weight of a medium size apple, i.e., the downward force exerted on your hand when you hold it.

Force is a vector, i.e., with both direction and magnitude and so to find the resultant or the net force when several forces act on a object, they must be added vectorially.

The net force is ! F R =

! F ii∑ =! F 1 +! F 2 + ...

The components of the net force are:

FRx = Fx∑ and FRy = Fy∑ .

! F

! F 1

! F 1

! F 2

! F 2

! F R

! F R

! F 2

! F 1

! F 3

! F 1

! F 2

! F 3

! F R

! F

θ

Fx =! F cosθ

Fy =! F sinθ

y

x

Question 4.1: An object on a horizontal table is acted

on by three forces, ! F 1 = 3 N,

! F 2 = 4 N and

! F 3 = 3 N, as

shown below. What is the net force acting on the object?

30"

60" ! F 3

! F 1

! F 2

30!

60! " F 3 = 3 N

y

x

" F 1 = 3 N

" F 2 = 4 N

Define x and y axes. Then, the net force in the x direction is Fx = Fxii∑ and

the net force in the y direction is Fy = Fyii∑ .

∴Fx = F1 cos30! + F2 cos60! − F3

= 2.60 N + 2.00 N − 3.00 N = 1.60 N,

and Fy = F1 sin 30! + F2 sin 60!

= 1.50 N − 3.46 N = −1.96 N.Thus, the net force is

" F R = (1.60ˆ i −1.96ˆ j ) N,

with magnitude

" F R = (Fx)2 + (Fy)2

= (1.60 N)2 + (−1.96 N)2 = 2.53 N.

The angle θ = tan−1 Fy

Fx

⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1(1.225) = 50.8!.

Fxˆ i

Fy ˆ j θ

" F R

Example of forces:Two forces: the weight of the ball downward (due to the attraction of the Earth) is balanced by the “normal” force of the ground on the ball.

No net force ⇒ no motion

Two forces: the weight of the ball downward and the force of the foot on the ball.

Net resultant force ⇒ change in motion

acceleration

One force: the weight of the ball downward.

Net force ⇒ change in motion

acceleration

Important concept ...

... we cannot neglect the effect of the Earth ... it always produces a non-contact force!

! W

! N

! W

! F

! W

! F R

Analysis of moving objects led to ...

Newton’s 1st Law:A body at rest, or moving with constant velocity, will remain at rest, or moving with constant velocity, unless it is acted on by a net force. (NB: the reference frame is the Earth.)

Sometimes called the Law of inertia ...• Tendancy for a body that’s moving to keep moving.• Tendancy for a body at rest to remain at rest.

∴ If the sum of forces ! F i∑ , i.e., the net force, on an

object is zero it remains at rest or continues moving with constant velocity. ( Fx∑ = 0 and Fy∑ = 0.)

This is called EQUILIBRIUM.

Implication ⇒ to change motion we have to apply a net force ... Newton’s 2nd Law gives us the relationship between the force and the change in motion.

A “simple” example ....

1 The force on you due to the Earth, i.e., your weight,

(downward) is balanced by the force of the seat of the chair on you (upward). No net force.

2 The force on the ice cream cone due to the Earth

(downward) is balanced by the force of your hand on the cone (upward). No net force.

3 The total weight of you, the ice cream cone and the

chair (downward) is balanced by the force of the ground on the chair legs (upward). No net force.

1 2

3

But don’t get fooled!!some common misconceptions ....

Are you “thrown” backwards when a car accelerates from rest? NO!! There is no “throwing” force; you tend to remain at rest as the seat moves forward!

Are you “thrown” outwards when a car turns a corner? NO!! There is no “throwing” force; as the car turns, you tend to continue to move in a straight line; so the side of the car exerts an inwards force on you!

! F

DISCUSSION PROBLEM [4.1]:

Why does a wire stretched between two posts always sag in the middle no matter how tightly it is stretched?

So, to produce a change in motion, there must be a net force ...

... to move an object from rest, i.e., to change its velocity,

or slow down a puck or change the direction of a ball.

! v

! F

! v

! F

! F

! v 1

! v 2

! F

frictional force opposing motion

! v

Newton’s 2nd Law:In the Principia Newton stated that

The change in motion is proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

Today, we write the 2nd Law as

! F =

d! p dt

,

where p = m! v , which today is called the momentum; Newton called it the quantity of motion.

∴! F =

d! p dt

=d(m! v )

dt= m

d! v dt

+ ! v dmdt

= m! a + ! v dmdt

.

However, most textbooks ignore the full Newtonian expression and assume that m is constant. So, you will find Newton’s 2nd Law usually stated this way:

If a net force acts on an object it accelerates (i.e., there is a change in motion). The direction of the acceleration is the same as the net force,

i.e., ! F i∑ i = m! a .

And in component form,

Fx∑ = ma x : Fy∑ = ma y : Fz∑ = ma z

where m is a constant. Remember, however, these expressions are only true if the mass remains constant. So, this form of the Law does not apply to a rocket that is losing mass by burning fuel.

NOTE:• The net force has to be an external force, i.e., not

inside the object or system, e.g., you can’t lift yourself!

• The product m! a is not itself a force ... it is the net

force that is put equal to m! a .

• The acceleration ! a is parallel to the net force

! F .

The dimension of force is the same as

mass × acceleration ⇒ [M] [L]

[T]2=[M][L][T]2

.

The unit of force is the Newton (N).

Question 4.2: An object on a horizontal table is acted on by three forces, F1 = 3 N, F2 = 4 N and F3 = 3 N, as shown

above. If the mass of the object is 2.0 kg, and it is initially at rest at the origin,

(a) what is its acceleration?(b) What is its velocity after 4.0 s?(c) What is its displacement after 4.0 s?

30!

60! " F 3

" F 1

" F 2

i

j

30!

60! " F 3 = 3 N

" F 1 = 3 N

" F 2 = 4 N

y

x

(a) From before (Q4.1) we found the resultant force was:

" F R = (1.60ˆ i −1.96ˆ j ) N.

Using Newton’s 2nd Law " F R = m" a .

∴ " a =

" F Rm

=(1.60ˆ i −1.96ˆ j ) N

2.0 kg

= (0.80ˆ i − 0.98ˆ j ) m/s2,

and the magnitude of the acceleration is

" a = (0.80 m/s2 )2 + (−0.98 m/s2 )2 = 1.27 m/s2

and it is parallel to " F R.

(b) " v = " v ! +

" a t = (0.80ˆ i − 0.98ˆ j ) m/s2 × 4.0 s

= (3.20ˆ i − 3.92ˆ j ) m/s.

∴" v = (3.20 m/s)2 + (−3.92 m/s)2 = 5.06 m/s.

(c) ! r − ! r " =

! v "t +12! a t2 =

12

(0.80ˆ i − 0.98ˆ j ) m/s2 × (4.0 s)2

= (6.40ˆ i − 7.84ˆ j ) m.

∴! r − ! r " = (6.40 m)2 + (−7.84 m)2 = 10.1 m.

The constant, m, in the expression

F = mais really a measure of how difficult it is to change the motion of an object. We call it the inertial mass, or simply, the mass. Note: mass is a scalar quantity.

To understand that concept, think about how you check the “weight” of something ...

... you check to see “how easy or difficult it is to make them move”! For the same acceleration ...

a = F1

m1= F2

m2,

i.e., m1

m2= F1

F2

So, in fact, you are comparing the masses !

m1 m2

DISCUSSION PROBLEM [4.2]:

We have just seen that the way we choose between the weights of two objects we move them up-and-down and determine how difficult it is to move them.

Will this idea work in a weightless environment?

m1 m2

OK ... so, a force changes motion. But when you throw a baseball, what keeps it moving to the right after it leaves your hand?

The answer is ... inertia ... remember Newton’s 1st Law. Neglecting air resistance, there are no horizontal forces to change its horizontal motion, so it keeps going to the right! Note, inertia is NOT a force.

There is a force downward (its weight), which causes the ball to “fall”. These effects combine so the ball follows a parabolic trajectory.

! w Question 4.3: An object of mass m is traveling at an

initial speed v! = 25.0 m/s. A net force frictional force of 15.0 N acting on the object in a direction opposing the motion, causes it to come to a stop after a distance of 62.5 m. (a) What is the mass of the object? (b) How long does it take to stop?

Define the positive direction to the right. First, we need to find the (negative) acceleration.

(a) We have: v2 = v!

2 + 2a(x − x!)

∴a =

v2 − v!2

2(x − x!)= −

(25.0 m)2

2 × 62.5 m= −5.00 m/s2.

Note that F opposes the motion, i.e., it acts in the −ve direction (same as the acceleration). Using Newton’s 2nd Law, F = ma,

i.e., m =

Fa=

−15.0 N−5.00 m/s2 = 3.00 kg.

(b) We have v = v! + at,

i.e., t =

v − v!a

=0 − 25.0 m−5.00 m/s2 = 5.00 s.

x! x

v! = 25.0 m/s v = 0

F = 15.0 N

(x − x!) = 62.5 m

+ve direction

Question 4.4: Two forces, F1 and F2, act on an object on

a frictionless surface, as shown. In which case is the acceleration greatest?

F1 = 8 N F2 = 2 N

D

2 kg

F1 = 8 N F2 = 4 N

B

4 kg

F1 = 8 N F2 = 2 N

C

3 kg

F1 = 12 N F2 = 3 N

A

6 kg

The net force on a block in each case is

F1 − F2 = F.

By Newton’s 2nd law, that force produces an

acceleration given by F = ma, i.e., a = Fm.

∴aA =

9 N6 kg

= 1.5 m/s2 : aB =

4 N4 kg

= 1 m/s2

aC =

6 N3 kg

= 2 m/s2 : aD =

6 N2 kg

= 3 m/s2

So, the answer is D.

F1 = 8 N F2 = 2 N

D

2 kg

F1 = 8 N F2 = 4 N

B

4 kg

F1 = 8 N F2 = 2 N

C

3 kg

F1 = 12 N F2 = 3 N

A

6 kg

What’s the difference between mass and weight ?

• MASS: inertial propertylarger mass ⇒ larger force required to accelerate.

• WEIGHT: the force exerted by the Earth on an object.

Example:Bowling ball ⇒ difficult to throw (mass)

⇒ difficult to lift (weight)

If you analyze a situation involving forces you’ll find that forces ALWAYS occur in pairs:

Books on table (down)Table on books (up)

! F FB

! F BF

(Spring)

! F BT

! F TB

A B

! F AB

! F BA

Pulling force A (to the left) is applied on B ! F AB( ).

Pulling force B (to the right) is applied on A ! F BA( ).

Pulling force A (to the left) is applied on the wall ! F AW( ).

The wall applies a force (to the right) on A ! F WA( ).

A

! F AW

! F WA

50

0

0 2 4 6 8 10 Time (s)

−50

FBA

FAB

FAB FBA

Who pulls harder, i.e., who applies the greater force, in the scenario shown here?

Ha! They both “pull” equal amounts ... the force exerted by A on B is equal and opposite to the force exerted by B on A, i.e.,

! F AB = −

! F BA and FAB = FBA

at all times!This leads us to ...

A B

! F AB

! F BA

Newton’s third Law:If an object A exerts a force on an object B (an action), object B exerts an equal and opposite force on A (the reaction), i.e.,

! F AB = −

! F BA

! F AB =

! F BA( )

Note:

• the two forces act on different objects, sometimes called an action/reaction pair.

• the Law applies whether the objects are at rest, moving with constant velocity or accelerating.

Examples of action/reaction pairs:

Books on table (down)Table on books (up)

! F BT =

! F TB

! F BT

! F TB

! F FB

! F FB =

! F BF

! F BF

That’s weird! How come it seems to hurt your eye morethan it hurts your finger?

But we have to be careful in identifying action/reaction pairs correctly (later).

! F FE

! F FE =

! F EF

! F EF

Often, Newton’s laws appear in combination look ...

First Law: “Every body perseveres in its state of rest, or of uniform motion in a right [straight] line, unless it is compelled to change that state by forces impressed thereon.”

Third Law: “To every action there is always opposed an equal reaction: or the mutual action of two bodies on each other are always equal, and directed to contrary parts.”

Principia Mathematica (1687).

Centripetal force(Action)

Car on you

Reaction(“Centrifugal

force”)You on car

So, if the Earth applies a force onan apple (its weight), the apple must apply an equal and opposite force on the Earth! Does that mean theEarth moves towards an apple whenwe drop the apple? Let’s see ...

We know ! F EA =

! F AE ,

and the 2nd Law tells us:

FEA = mAaA and FAE = mEaE.

∴mAaA = mEaE ⇒ aE =

mAaAmE

.

But mE ≈ 6 ×1024 kg, mA ≈ 0.1 kg and aA = g

∴aE ≈ 1.7 ×10−25 m/s2.

If the apple takes 1 s to reach Earth ...

ΔyA =

12

aAt2 ≈ 4.91 m.

ΔyE =

12

aEt2 ≈ 0.85 ×10−25m.

WOW ... that’s small!

! F EA

! F AE

Take something as simple as a book or cups on a table on the Earth (i.e., the floor) ...

~ Let’s identify ALL action-reaction pairs ~

• Forces acting ON the book: ! F TB and

! F EB(= ! w B)

These are not an action-reaction pairWhy ??

• Forces acting ON the table: ! F BT and

! F ET (= ! w T )

• Forces acting ON the Earth: ! F BE and

! F TE

For action-reaction pairs:

! F 12 =

! F 21 .

DISCUSSION PROBLEM [4.3]:

You are holding a baseball in your hand. The reaction force to the weight of the baseball is ....

A: the gravitational force of the Earth on the ball,

B: the gravitational force of the ball on the Earth,

C: the force of your hand on the ball,

D: the force of the ball on your hand, or

E: the gravitational force of the Earth on your hand.

Does Newton’s 3rd Law apply if objects are moving? You betcha!!

Identify all the forces associated with the rope:(Take L⇒ R as positive direction)

! F RS (rope on sled)* +

! F SR (sled on rope)* −

! F WR (woman on rope)# +

! F RW (rope on woman)# −

But ! F RS = −

! F SR and

! F WR = −

! F RW.

The net force on the rope ⇒! F WR −

! F SR .

So, if ! F WR >

! F SR

the rope and everything attached to it, moves to the right, but the equations above still hold!

S

RW

A free-body diagram (FBD) shows all of the forces acting ON an object. For example, the forces acting on a ball that’s been kicked ...

! W

! N

⇒

! F

! W

⇒

Believe me, dear, crucial tosolving “force problems” is

the free-body diagram.

You know all about the birdsand bees, but now it’s time to

learn about FBD’s

⇒ ! W

DISCUSSION PROBLEM [4.4]:

Which of the free-body diagrams below best represents the forces acting on an object sliding down a frictionless incline?

! v

A B C D

The Normal Force

In mechanics, the normal force FN (sometimes n or N) is

the component of the contact force exerted on an object , perpendicular to the surface of contact by, for example, the surface of a floor or incline, preventing the object from penetrating the surface. In the case of an object

resting on a horizontal surface (e.g., a table top, the ground), the normal force is equal and opposite to the weight of the

object (Newton’s 3rd Law), i.e., FN = mg .

When an object rests on an incline, the normal force is perpendicular to the plane the object rests on. In that case, the normal force is equal and opposite to the component of the weight perpendicular to the surface, i.e., FN = mg cosθ.

FN

mg

FN

mgθ

mgcosθ

mgsinθ

There is a force acting parallel to the incline, i.e., the component of the weight parallel to the surface, mg sin θ.

Therefore, when an object sits on a frictionless incline, the resultant (net) force,

FR = mg sinθproduces motion down the incline. By Newton’s 2nd Law the acceleration of the object down the slope is given by

a =

FRm

= gsinθ,

a result I showed in Chapter 2.

FN

mgθ

mgcosθ

mgsinθ

FN

mgθ

mgcosθ

mgsinθ FR

STRATEGY for analyzing “force” problems:

Separate the force components in two (perpendicular) directions.

Draw free body diagram(identify the forces acting ON an object)

Newton’s third Law

Equilibrium(i.e., at rest or constant velocity)

Newton’s first Law

Motion(i.e., accelerating)

Newton’s second Law

F1 = 8 N F2 = 2 N

D

2 kg

F1 = 8 N F2 = 4 N

B

4 kg

F1 = 8 N F2 = 2 N

C

3 kg

F1 = 12 N F2 = 3 N

A

6 kg

Question 4.5: Two forces, F1 and F2, act on an object on

a surface, as shown. If the objects are all moving with a constant velocity, in which case is the frictional force greatest?

Draw the FBD for the blocks. Since the blocks are not accelerating, Newton’s 2nd Law tells us the net force on each block is zero, i.e.,

F1 − F2 − f = 0

∴f = F1 − F2.

( F1 − F2) is largest for block A ( = 9 N to the right). Since

the net force must be zero, the frictional force of the ground on block A must be 9 N (to the left).

F1 F2 f

F1 = 8 N F2 = 2 N

D

2 kg

F1 = 8 N F2 = 4 N

B

4 kg

F1 = 8 N F2 = 2 N

C

3 kg

F1 = 12 N F2 = 3 N

A

6 kg f

DISCUSSION PROBLEM [4.5]:

You and a friend have identical cars. You want to rip an old pair of Levi’s apart; there are two possible arrangements, as shown above. Which is more likely to do the job, i.e., in which is the tension in the jeans greater?

A: A

B: B

C: They’re the same

A

B

Question 4.6: To prevent a box from sliding down a frictionless inclined plane, physics student Anna pushes on the box horizontally with just enough force so that the box is stationary. If the mass of the box is 2.00 kg and

the slope of the incline is 35!, what is the magnitude of

the force she exerts?

35!

Anna

Draw the free body diagram for the box:

The components of the weight force along x and along y are (−mg sin θ) and (−mg cosθ), respectively. The components of the force Anna applies (FA ) along x and y

are FA cosθ and −FA sinθ,

respectively. Summing the forces in the x-direction, we have

Fx∑ = 0

= FA cosθ − mg sin θ,

i.e., FA = mg tanθ.

∴FA = 2.00 kg( ) 9.81 m/s2( ) 0.700( ) = 13.7 N.

FN

mg

mgcosθ

mgsinθ

FA

θ

x y

θ

FAθ

FA sinθ

FA cosθ

T1 T2

T3

6 kg M

60! 60!

Question 4.7: The system shown below is in equilibrium. Find the three tensions T1, T2 and T3 in the string, and the

unknown mass, M.

First draw the free body diagram for the 6 kg mass; the two forces are the weight (mg) and

T2. Since the system is in equilibrium,

T2 = mg = 6.00 kg × 9.81 m/s2 = 58.9 N.

The tensions all pass through a single point (circled); draw the free body diagram at that point. Then, we have

Fh∑ = 0 and Fv∑ = 0.

(h) −T1 sin30! + T3 sin 30! = 0.

∴T1 = T3

(v) T1cos30! + T3cos30! − T2 = 0,

i.e., T1 + T3 =

T2cos30!

=58.9 N0.866

= 68.0 N.

∴T1 = T3 = 34.0 N.

mg

T2

T2

T1 T3 30! 30!

T1 T2

T3

6 kg M

60! 60!

T1 T2

T3

6 kg M

60! 60!

Since the pulley only changes the direction of T1, it does

not affect its magnitude.

∴T1 = Mg,

i.e., M =

T1g

=34.0 N

9.81 m/s2 = 3.47 kg.

Sometimes, we may not know the angles as in this problem ...

Question 4.8: Isobel Newton hangs motionless by one hand from a clothes line, as shown. If the clothes line is about to break, which side is most likely to break?

A: The left side.B: The right side.C: Equal chance of either side breaking.

Identify all of the forces acting through Isobel’s hand and draw the vector force diagram.

Since she is not moving, the net force on her hand is zero, i.e.,

! F net =

! F 1 +! F 2 +

! w = 0.

So the vector diagram shown on the right must be closed.

Clearly, ! F 1 >

! F 2 and so the part of the line on the right

has the greater tension and so is more likely to break!

Hence, even if the angles are not known, you can use the fact that the vector diagram of forces must be a closed figure to solve a problem.

! w

! F 1

! F 2

! F 1

! F 2

! w

Question 4.9: A 2 kg mass hangs from a spring scale, calibrated in Newtons, that hangs from the ceiling of an

elevator. What does the scale read when the elevator is (a) moving up with a constant velocity of 30 m/s, (b) moving down with a constant velocity of 30 m/s, and (c) is ascending at 20 m/s and gaining

speed at the rate of 3 m/s2?

(d) From t = 0 to t = 5 s, the elevator moves up at 10 m/s. Its velocity is then reduced uniformly to zero in the next 4 s, so that it comes to rest at

t = 9 s. What does the scale read during the time interval

0 < t < 9 s?

Draw the free body diagram for the 2 kg block. In parts (a) and (b) the block is not accelerating.

∴av = 0 so ! F v∑ = 0.

Take up direction as positive:∴ N +(−mg ) = 0,

i.e., N = mg = (2 kg)(9.81 m/s2)

= 19.8 N ... (2 kg).

(c) Since the block is accelerating upward we have:

! F v∑ = ma v ( av > 0).

Then N +(−mg ) = mav

i.e., N = m(g + av ) = (2 kg)(12.81 m/s2)

= 25.6 N ... (2.61 kg).

(d) When 0 < t ≤ 5 s, av = 0, so N = 19.8 N.

N (Upward force due to springbalance, i.e., apparent weight.

w = mg (Force due to Earth)

a v

When 5 s < t ≤ 9s, av =

ΔvΔt

=(0 −10 m/s)

4 s= −2.5 m/s2.

Since N +(−mg ) = mav ... but av < 0.

∴ N = m(g + av ) = (2 kg)(7.31 m/s2)

= 14.6 N ... (1.49 kg)

This is ME ... !Draw the free body diagram for the forces on me:

Fvv∑ = N −mg = ma v.

My apparent weight is the force I exert on the scale

⇒ N = m(g + av ).

So, if you know your true weight, you can find the acceleration of an elevator!

N (Upward force from scale)

w = mg (force due to Earth)

0

Apparent weight (pounds) 80 120 160 200 240

accelerationupward

acceleration downward

−2.0

−4.0

2.0

4.0

a v (m/s2)

Note that the apparent weight of someone in an elevator if the cable breaks is zero. Draw the FBD for the forces acting on the person ...

From the free-body diagram

N −mg = ma v,

so apparent weight N = m g + av( ) = m g + (−g)( )

= 0Conceptually what’s happening is ... the floor and weighing scales are falling at the same rate as the person inside the car so there’s no net force on the weighing scale.

This is what is popularly known as weightlessness except, of course, the person still has real weight!

Yikes!

N (Upward force from scale)

w = mg (force due to Earth)

Question 4.10: Two objects are connected by a massless string, as shown above. The incline and pulleys are frictionless. (a) Show that the acceleration of the objects is

a =

(m2 − m1 sinθ)g(m1 + m2)

.

(b) Find the acceleration and tension in the string when

m1 = m2 = 5 kg and θ = 30!.

(a) Treat each object separately. Note that T, the tension, is common to both free-body diagrams as the

pulley only changes the direction of T. Assume m2

descends (i.e., m1 ascends the slope).

[#1] along x: T + (−m1gsin θ) = m1a ... (i)

[#2] along y: T + (−m2g) = −m2a ... ... (ii)

Subtract (ii) from (i):

(m2 − m1 sin θ)g = (m1 + m2)a

∴a =

(m2 − m1 sinθ)g(m1 + m2 )

.

Note: if m2 > m1 sin θ then a > 0,

if m2 < m1 sin θ then a < 0,

if m2 = m1 sin θ then a = 0.

m2g

Tx

y

N T

m1g

xy

m2g

Tx

y

N T

m1g

xy

(b) With m1 = m2 = 5 kg and θ = 30!,

a =

m2 − m1sin θ( )gm1 + m2( )

=

5 kg − 5 kg × 0.50( )( ) 9.81 m/s2( )(5 kg + 5 kg)

= 2.45 m/s2.

Since a > 0, m1 moves up the slope.

Also, using equation (ii),

T = m2(g − a) = (5 kg)(7.36 m/s2 )

= 36.8 N.

Question 4.11: Two blocks are in contact on a

frictionless, horizontal surface. A horizontal force ! F is

applied to one of them, as shown. Find the acceleration

and the contact force for (a) general values of F, m1 and

m2, and (b) when F = 3.2 N, m1 = 2 kg and m2 = 6 kg.

Draw the free-body diagrams for each object. It is the contact force of m1 on m2 that causes m2 to accelerate.

(a) Consider only horizontal forces (why??) ...#1 ... F − F21 = m1a.

#2 ... F12 = m2a.

But F12 = F21. ∴a =

Fm1 + m2

.

The contact force is: F12 (= F21) =

m2m1 + m2

F.

(b) a =

Fm1 + m2

=3.2

2 + 6= 0.400 m/s2.

F12 =

62 + 6

× 3.2 = 2.4 N.

F21

m1g

F F12

m2g

N1 N2

Do the results change if the force ! F is applied from the

right? Conceptually, the magnitude of the acceleration

will be the same because the total mass (m1 + m2 ) and ! F

are the same. But, the contact force will be smaller since a smaller force will be required to produce the same acceleration for m1. We get the same result analytically.

We have

F − F12 = m2a,

and

F21 = m1a.

Eliminating a, and putting F12 = F21, we find

F21 =

m1(m1 + m2)

F,

i.e., the contact force of m2 on m1, F21, depends on the

mass of the second block, m1, which in this case is smaller

(2 kg) than in the problem (6 kg).

F21

m1g

F F12

m2g

N1 N2

! F

! F

(a) (b) m1 < m2