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NAME ______________________________________ DATE _______________ CLASS ____________________
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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NAME ______________________________________ DATE _______________ CLASS ____________________
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
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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