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Ch.3 PROBLEMS

Sections 3.1 and 3.2

3.1 (I) What net force must be exerted on a 7.0-kg sack of potatoes to give it an acceleration of 3.5 m/sec2? Solution. a = f

m∴ f = ma = 7kg 3.5 m

sec2⎛⎝⎜

⎞⎠⎟=24.5 kg •m

sec2= 24.5 Newtons

3.2. (I) A net force of 30 N is applied to an object, which is then observed to accelerate at 0.25 m/sec2. Calculate the mass of the object.

Solution. a=fm

∴m= fa=30 kg •m

sec2.25m/sec2

= 120kg

3.3. (I) Calculate the acceleration of a rocket of mass 1.2 x 106 kg if the net force on it is 2.0 x 106 N. Express in (a) meters per second-squared and (b) as a multiple of the acceleration of gravity, g.

Solution 3.3a. a=fm

=2x106N1.2x106kg

=1.67 msec2

Solution 3.3b.

1.67 ms2

ig

9.8 ms2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 0.170g

3.4. (I) (a) What is the weight in newtons of a 50-kg person on earth? (b) The acceleration due to gravity on the moon is 1.67 m/s2. What would this person's weight be on the moon?

Solution 3.4a. Weight = mg = 50kg( 9.8msec2 ) = 490 kgm

sec2 = 490 newtons

Solution 3.4b. Weight on the moon.

Weight = mg = mgmoon = 50kg(1.67msec2

) =83.5newtons

Selected solutions to Ch.3 Problems

3.5 (I) Calculate the mass in kilograms of a flea that has a weight of 5 x 10-6 N.

Solution 3.5. Wt = mg therefore, m =wtg

=5x10−6 kgm/sec2

9.8m/sec2 =5.1x10−7kg

TABLE 3.1 I COEFFICIENTS OF KINETIC AND STATIC FRICTIONa

µk µS Kinetic Friction Static Friction Rubber on dry concrete 0.7 1.0Rubber on wet concrete 0.5 0.7Wood on wood 0.3 0.5Waxed wood on wet snow 0.1 0.14Metal on wood 0.3 0.5Steel on steel (dry) 0.3 0.6Steel on steel (oiled) 0.03 0.05Teflon on steel 0.04 0.04Bone lubricated with synovial fluid 0.015 0.016 Shoes on wood 0.7 0.9Shoes on ice 0.05 0.1Ice on ice 0.03 0.1Steel on ice 0.02 0.4

"Values are approximate.

3.6 (I) How much kinetic friction will there be in a knee joint if the weight supported by the joint is 500 newtons. (Use table 3.1 for coefficient of friction). Solution 3.6. From table 3.1, the coefficient of friction, µ, of synovial fluid is 0.015. f= µFN = 0.015(500 N)= 7.5 newtons.

3.7. (I) If a steel spatula experiences a frictional force of 0.20 N when scraping against a Teflon frying pan, what is the normal force between the spatula and the pan? (Use table 3.1 for coefficient of friction).

Solution 3.7. friction = FN ⋅µ Therefore, FN =fµ=

0.2N0.04

=5.0newtons

3.8. (I) A frictional force of 300 N is observed between two moving pieces of steel when a normal force of 1000 N exists between them. Please refer to table 3.1 in order to determine if the steel oiled or dry?

Solution 3.8 friction = FNµ∴µ =fFN

=300n1000n

=0.30


Ch.3 PROBLEMS continuedSections 3.1 and 3.23.9. (I) What is the tension in a strand of spider thread when a spider of mass 1.0 x 10-4 kg hangs motionlessly from it? Hint: The tension will have the same magnitude as the spider’s weight.

Solution. 3.9. Tension = weight = mg= 1 x 10- 4 kg(9.8 m/sec2) = 9.8 x 10- 4 newtons

3.11. (II) An orderly exerts a horizontal force of 50 N on a gurney with a patient on it. The gurney and patient have a total mass of 90 kg. If the gurney and patient accelerate at 0.35 m/sec2, what is the magnitude of the frictional force opposing the motion?

Solution. 3.11.

a =fm

=50n + f90kg

=0.35 msec2

∴ f = 90 kg(.35 msec2

) − 50n =−18.5newtons

The negative sign indicates that the frictional force is backward. You should write your answer as friction = 18.5 newtons backward.


Tension

Weight

90 kgfriction

50 newtons

Ch.3 PROBLEMS continued Section 3.3

3.20. (II) Two movers push horizontally on a refrigerator. One pushes due north with a force of 150 N and the other pushes due east with a force of 200 N. Using trigonometric functions, determine the magnitude and direction of the resultant force on the refrigerator.

Solution 3.20a. We will use the Pythagorean theorem to find the magnitude of the resultant force.

Resultant magnitude = 1502 + 2002 =250 newtons Solution 3.20b. The direction of the resultant can be found by using the trig. rule for the tangent of an angle.

Tanθ = oppositeadjacent

=150200

=0.75∴ θ = 36.9

The direction of the force is 36.9˚ North of East. That is equivalent to a bearing of 53.1 degrees.

3.22. (I) Using trigonometric functions, determine the magnitudes of the two forces, F I and F 2, which add up to the total force shown in Figure 3.25.

Solution 3.22, F1

Sin35˚= opposite50N

opposite = Sin35˚i50N = 28.7Newtons

F1 = 28.7 Newtons

Solution 3.22, F2 Cos35˚=adjacent50N

; Adjacent = Cos35˚i50N = 41.0N


Fig. 3.25

150 N

200 N

Resultant force

3.23. (II) Figure 3.26 represents a sled viewed from above being pulled by two children and experiencing a frictional force. Find the acceleration of the 9.0-kg sled. Solution 3.23

Baby step 1. Determine the forward and left components of F1

Forward component of F1: Cos30˚i10N = 8.66N Left component of F1: Sin30˚i10N = 5.0N

Baby step 2. Determine the forward and right components of F2. Forward component of F2: Cos45˚i14.1N = 9.97N Right component of F2: Sin45˚i14.1N = 9.97N

Baby step 3. Total up the forward components and total up the left-right components. Assign forward as positive and backward as negative. Let right be positive and left be negative. Don’t forget the frictional force.

Total forward components = 8.63 Newtons forward. Total left-right components = 4.97 newtons to the right.


Forward-----Backward Right-----Left

8.66 -5

9.97 9.97

-10 0

8.63 4.97

Newtons

Total

friction

Baby step 4. Draw a diagram and determine the magnitude and direction of the resultant force due to the boys and friction.

Magnitude =

8.632 + 4.972 = 9.96Newtons

Direction = Tan−1 4.978.63

⎛⎝⎜

⎞⎠⎟=29.9˚

Baby step 5. Use Newton’s second law to determine the acceleration of the sled.

a =F∑

m=

9.96 kg ⋅ms2 27˚ to the right of forward( )

9kg

Ans. a = 1.11ms2 , 29.9˚ to the right of forward.

Section 3.4


8.63 N

4.97 N

ø

3.38. (II) Even when the head is held erect, its center of gravity is not directly over its major point of support, the atlanto-occipital joint. The splenius muscles in the back of the neck must therefore exert a force to keep the head erect. Calculate the force they must exert, using the information in Figure 3.36, if the /mass of the head is 5.0 kg.

The weight of an object can be viewed as acting in a straight line from the center of gravity of the object, toward the center of the earth.FM is the unknown force produced by the splenius muscle. The mass of the head ‘m’= 5 kg. We can solve this problem by using the concept that when a system is in rotational equilibrium, the clockwise and counterclockwise torques are equal. Solution 3.38. Clockwise torque = Counterclockwise torque

49N i2.5cm = FM i5cm

Ans.

FM =49N i2.5cm

5cm= 24.5N No wonder my neck aches.


center of gravity

2.5 cm

FM

5 cm

weight = mg =

5kgi9.8 ms2

= 49Newtons

3.39. (II) A man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure 3.37. Calculate the force in the Achilles tendon if he stands on one foot and has a mass of 80 kg. (Yes, this is another torque problem).Solution 3.39.Note: that 80 kg is his mass, not his weight. First determine his weight.

weight = mg = 80kgi9.8 ms2

= 784Newtons

Clockwise torque = Counterclockwise torque

FA i4.0cm = 784N i12.0cm

Ans.

FA =784N i12.0cm

4.0cm= 2,350 newtons

3.41. (II) What force must the woman in Figure 3.38 exert on the floor with her hands in order to do a pushup?

Solution 3.41

wt = mg = 50kgi9.8 ms2

= 490newtons

Clockwise torque = Counterclockwise torque 490N i0.90m = Freaction i1.50m

Ans.

Freaction =490N i0.90m1.50m

= 294newtons


Section 3.5

3.45. (II) What acceleration is experienced by materials 10 cm from the center of rotation of a centrifuge that spins at 4000 revolutions per minute? Express your answer in m/s2. Solution 3.45. Baby step 1. Convert 10 cm to meters.

10cm 1m100cm

= 0.10m

Baby step 2. Objects moving in a circular path, undergo centripetal acceleration. Write down the equation for centripetal acceleration.

ac =v2

r Note that we do not know the tangential velocity, v. We have been given the angular velocity in revolutions per minute. Baby step 3. Determine the tangential velocity in meters per second. Use the fact that distance traveled in one revolution is 2πr, (the circumference of a circle).

v = 4000 r evmin

i2π i0.10m

revi1min60sec =

= 41.9 ms

Baby step 4.

Ans. ac =v2

r=41.9 m

s⎛⎝⎜

⎞⎠⎟2

0.10m= 17,546 m

s2= 17,500 m

s2


3.46. (I) (a) How much sideways force must the wheels of a 950-kg car exert to cause it to round a corner of radius 200 m at a speed of 35 m/sec? (b) What acceleration does the driver experience? Express your answer in meters per second-squared and as a multiple of g. Solution 3.46b. Baby step 1. Let’s determine the centripetal acceleration first.

ac =v2

r=35 m

s⎛⎝⎜

⎞⎠⎟

200m= 6.13m

s2

Baby step 2. Convert from m/s2 to number of ‘g’s.

6.13ms2

i1g

9.8 ms2

= 0.625g

Solution 3.46a. Newton’s 2nd law tells us that F∑ = ma.

Fcentripetal = miac = 950kg∑ i6.13ms2

= 5,820Newtons

3.47. (I) (a) Calculate the acceleration of a 0.55-m-diameter car tire while the car is traveling at a constant speed of 25 m/sec. (b) Compare this with the acceleration experienced by a jet car tire of diameter 1.0 m when setting a land speed record of 310 m/sec.

Solution 3.47a. ac =v2

r=25 m

s⎛⎝⎜

⎞⎠⎟2

0.275m= 2,270 m

s2

Solution 3.47b. ac =v2

r=

( 310 msec

)2

.5 m=192,000 m

sec2

acceleration of jet tireacceleration of car tire

=192,000 m

s2

2270 ms2

= 84.6


3.48 (II) What centripetal acceleration (in meters per second squared) is experienced by the passengers in a jet airplane making a level turn of radius 1. 0 km at a speed of 400 km/hr? Solution 3.48. Baby step 1. Convert the speed to m/s.

400 kmhr

i1000m1km

i1hr3600s

= 111ms

Baby step 2. Convert 1.0 km to meters.

1kmi1000m1km

= 1000m

Baby step 3. Write down the centripetal acceleration equation and do the math.

Ans. ac =v2

r=(111 m

sec)2

1000m=12.3 m

sec2

3.49. (II) What centripetal force is exerted by the rope on a 1.2-kg tether ball swung in a 2.0-m-diameter circle at 45 revolutions per minute? Solution 3.49.

Baby step 1. ac =v2

r We don’t have the tangential velocity. Remember

that every revolution is once around a circle. The circumference of the circle is 2π ir . Let’s calculate v. Note that the radius is 1.0 meters.

45 rev1min

i2π i1.0m1rev

1min60s

= 4.71ms

Baby step 2. We are ready to use the centripetal acceleration equation.

ac =v2

r=4.71m

s⎛⎝⎜

⎞⎠⎟2

1.0m= 22.2 m

s2

Baby step 3. Determine the centripetal force.

Ans.

Fcentripetal = miac∑ = 1.2kgi22.2 ms2

= 26.6newtons


3.50. (II) At how many revolutions per minute must an astronaut-training centrifuge rotate to produce an acceleration of 10g if the radius of rotation is 25 m?

Solution 3.50. Baby step 1. Convert 10g to m/s/s.

10g9.8 m

s2g

= 98 ms2

Baby step 2. Determine the tangential velocity.

ac =v2

r therefore v= ac ir

v = 98 ms2

i25m = 49.5 ms

Baby step 3. Convert tangential velocity to angular velocity in revolutions per minute.

Ans.

ω = 49.5 ms

i1Rev

2π i25mi

60s1min

= 18.9 revmin


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