Problem 4C - Coefficients of Friction

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Holt Physics

Problem 4C COEFFICIENTS OF FRICTION PROBLEM

A cabinet initially at rest on a horizontal surface requires a 115 N horizontal force to set it in motion. If the coefficient of static friction between the cabinet and the floor is 0.38, what is the normal force exerted on the cabinet? What is the mass of the cabinet? SOLUTION

Given:

Fs,max = 115 N ms = 0.38 g = 9.81 m/s2

Unknown:

Fn = ?

m=?

Use the equation for the coefficient of static friction to find Fn. Fs, max  ms =  Fn Fs, max 115 N  =  = 3.0 × 102 N Fn =  ms 0.38 Fn = 3.0 × 102 N upward Use the definition for the normal force to find m. Fn = mg, for a horizontal surface F 3.0 × 102 N m = n =  g 9.81 m/s2

Copyright © by Holt, Rinehart and Winston. All rights reserved.

m = 31 kg

ADDITIONAL PRACTICE 1. A ship launched from a dry-dock slides into the water at a constant velocity. Suppose the force of gravity that pulls the ship downward along the dry-dock is 4.26 × 107 N. If the coefficient of kinetic friction between the ship’s hull and the dry-dock is 0.25, what is the magnitude of the normal force that the dry-dock exerts on the ship’s hull? 2. If the incline of the dry-dock in problem 1 is 10.0°, what is the ship’s mass? 3. A frictional force of 2400 N keeps a crate of machine parts from sliding down a ramp. If the coefficient of static friction between the box and the ramp is 0.20 and the incline of the ramp is 30.0°, what is the normal force of the ramp on the box? What is the mass of the crate of machine parts?

Problem 4C

Ch. 4–5

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4. A passenger with a mass of 60.0 kg is standing in a subway car that is accelerating at 3.70 m/s2. If the coefficient of static friction between the passenger’s shoes and the car floor is 0.455, will the passenger be able to stand without sliding? 5. A 90.0 kg skier glides down a slope with an incline of 17.0°. What frictional force is needed for the skier to move at a constant velocity? 6. A dogsled with a mass of 47 kg is loaded with 33 kg of supplies. If the coefficient of kinetic friction between the sled’s runners and the snow is 0.075, what is the magnitude of the frictional force on the sled as it moves across flat ground? What is the magnitude of the frictional force on the sled as it moves up a hill with a 15° incline? 7.

A car with a mass of 1.8 × 103 kg is parked on a hill in San Francisco. Suppose the hill makes a 15.0° incline with the horizontal. If the frictional force required to keep the car from sliding down the hill is 1.25 × 104 N, what is the coefficient of static friction between the pavement and the car’s tires?

8. A metal disk with a mass of 15.0 g slides along a metal sheet. Both the disk and sheet have been coated with a substance that reduces frictional forces. If the sheet needs to be tilted only 2.3° for the disk to slide down the sheet with a constant velocity, what is the coefficient of kinetic friction between the disk and sheet?

10. A snowboarder slides down a 5.0° slope at a constant speed. What is the coefficient of kinetic friction between the snow and the board?

Ch. 4–6

Holt Physics Problem Bank

Copyright © by Holt, Rinehart and Winston. All rights reserved.

9. In 1994, a commercial automobile accelerated from rest to 88.0 km/h in 3.07 s. Cars accelerate because of traction, which in turn depends on the force of static friction between the rubber of their tires and the road. If the force of acceleration is entirely provided by static friction between the tires and pavement (an overly simplified assumption), calculate the coefficient of static friction between the tires and the road.

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Givens

Print Solutions Fnet −65.0 N anet =  =  = m 0.145 kg

7. Fnet = −65.0 N m = 0.145 kg 8. m = 214 kg

−448 m/s2

Fnet = Fbuoyant − mg = 790 N − (214 kg)(9.81 m/s2)

Fbuoyant = 790 N

Fnet = 790 N − 2.10 × 103 N = −1310 N Fnet −1310 N anet =  =  = −6.12 m/s2 m 214 kg

g = 9.81 m/s2

9. m = 0.080 kg

Fnet = m anet = m g(sin q) anet = g(sin q) = (9.81 m/s2)(sin 37.0°) = 5.90 m/s2

q = 37.0° g = 9.81 m/s2

anet = 5.90 m/s2 down the incline (37.0° below horizontal)

Fnet = Fupward − m adownward = 1.40 N − (0.080 kg)(5.90 m/s2) = 1.40 N − 0.47 N = 0.93 N

10. m = 0.080 kg Fupward = 1.40 N adownward = 5.90 m/s2

Fnet = 0.93 N up the incline (37.0° above the horizontal) 0 .93 N F  =   = 12 m/s2 anet = net 0.0 80 kg m

Additional Practice 4C 1. Fdownward = 4.26 × 107 N mk = 0.25

Fnet = Fdownward − Fk = 0 Fk = mk Fn = Fdownward Fdownward 4.26 × 107 N Fn =  =  = 1.7 × 108 N 0.25 mk

q = 10.0° 2

g = 9.81 m/s

3. Fs,max = 2400 N ms = 0.20

Fn = mg (cos q) 1.7 × 108 N Fn m =  =  = 1.8 × 107 kg g (cos q) (9.81 m/s2)(cos 10.0°) Fs,max = ms Fn 2400 N Fs,max Fn =  =  = 1.2 × 104 N 0.20 ms

q = 30.0° g = 9.81 m/s2

Fn = 1.2 × 104 N perpendicular to and away from the incline Fn = mg(cos q ) Fn 1.2 × 104 N m =  =  = 1400 kg g (cos q) (9.81 m/s2)(cos 30.0°)

4. m = 60.0 kg

For the passenger to remain standing without sliding,

a = 3.70 m/s

Fs,max ≥ F = ma

ms = 0.455

Fs,max = ms Fn = ms mg

g = 9.81 m/s2

ms mg ≥ ma

2

V

ms g ≥ a (0.455)(9.81 m/s2) = 4.46 m/s2 > 3.70 m/s2 The passenger will be able to stand without sliding.

V Ch. 4–4

Holt Physics Solution Manual

Copyright © by Holt, Rinehart and Winston. All rights reserved.

2. Fn = 1.7 × 108 N

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Givens

Solutions Fnet = mg(sin q) − Fk = 0

5. m = 90.0 kg q = 17.0°

Fk = mg(sin q) = (90.0 kg)(9.81 m/s2)(sin 17.0°) = 258 N 2

g = 9.81 m/s

6. msled = 47 kg msupplies = 33 kg mk = 0.075 q = 15°

Fk = 258 N up the slope

Fk = mk Fn = mk(msled + msupplies)g = (0.075)(47 kg + 33kg)(9.81 m/s2) Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2) Fk = 59 N Fk = mk Fn = mk(msled + msupplies)g(cos q) = (0.075)(47 kg + 33 kg)(9.81 m/s2)(cos 15°) Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)(cos 15°) = 57 N

7. m = 1.8 × 103 kg q = 15.0° Fs,max = 1.25 × 104 N g = 9.81 m/s2

Fs,max = ms Fn = ms mg(cos q) 1.25 × 104 N Fs,max  =  ms =  (1.8 × 103 kg)(9.81 m/s2)(cos 15.0°) mg(cos q) ms = 0.73

8. m = 15.0 g

Fnet = mg(sin q) − Fk = 0

q = 2.3°

Fk = mk Fn = mk mg(cos q)

g = 9.81 m/s2

m g(sin q) mk =   = tan q = tan 2.3° mg (c os q) mk = 0.040

Copyright © by Holt, Rinehart and Winston. All rights reserved.

9. vf = 88.0 km/h

Fnet = Fapplied − Fs,max = 0

vi = 0 km/h

Fs,max = ms Fn = ms mg

∆t = 3.07 s

Fapplied = m a

g = 9.81 m/s2

m a = ms mg (88.0 km/h − 0 km/h)(103 m/km)(1 h/3600 s) a vf − vi ms =  =  =  (9.81 m/s2)(3.07 s) g g∆t ms = 0.812

10. q = 5.0°

Fnet = mg(sin q) − Fk = 0 Fk = mk Fn = mk mg(cos q) mg(sin q ) mk =  = tan q = tan 5.0° mg(cos q ) mk = 0.087

V

Section Five—Solution Manual

V Ch. 4–5