Problem#1
The person in Figure 1 weighs 170 lb. As seen from the front, each light crutch makes an angle of 22.0° with the vertical. Half of the person’s weight is supported by the crutches. The other half is supported by the vertical forces of the ground on his feet. Assuming the person is moving with constant velocity and the force exerted by the ground on the crutches acts along the crutches, determine (a) the smallest possible coefficient of friction between crutches and ground and (b) the magnitude of the compression force in each crutch.
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| Fig.1 |
Answer:
From the free-body diagram of the person
∑Fx = F1 sin 22.00 – F2 sin 22.00 = 0
which gives
F1 = F2 = F
Then,
∑Fy = 2F cos 22.00 + 85.0 lbs – 170 lbs = 0
yields F = 45.8 lb
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| Fig.2 |
(a) Now consider the free-body diagram of a crutch tip
∑Fx = f – (45.8 lb) sin 22.00 = 0
or
f = 17.2 lb
∑Fy = ntip – (45.8 lb) cos 22.00 = 0
which gives
ntip = 42.5 lb
For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so
f = fs,max = µsntip
and
µs = f/ntip = 17.2 lb/42.5 lb = 0.404
(b) As found above, the compression force in each crutch is
F1 = F2 = F = 45.8 lb
Problem#2
A 25.0-kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion. After it is in motion, a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find the coefficients of static and kinetic friction from this information.
Answer:
 |
| Fig.3 |
For equilibrium:
f = F and n = mg. Also, f = µn i.e.,
µ = f/n
µs = 75.0 N/(25.0 kg x 9.80 m/s2) = 0.306
and
µk = 60.0 N/(25.0 kg x 9.80 m/s2) = 0.245 N
Problem#3
A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? (b) What is the stopping distance when the surface is dry and µs = 0.600?
Answer:
∑Fy = may
+n – mg = 0
n = mg
and
fs = µsn = µsmg
This maximum magnitude of static friction acts so long as the tires roll without skidding.
∑Fx = max
–fs = max
The maximum acceleration is
ax = –µsg
The initial and final conditions are: xi = 0 , vi = = 50.0 mi/h = 22.4 m/s, vf = 0
vf2 = vi2 + 2a(xf - xi)
(a) then
0 = vi2 + 2(–µg)(xf - xi)
xf = vi2/2µg
xf = (22.4 m/s)2/(2 x 0.100 x 9.80 m/s2)
xf = 256 m
(b) xf = vi2/2µg
xf = (22.4 m/s)2/(2 x 0.600 x 9.80 m/s2)
xf = 42.7 m
Problem#4
Before 1960 it was believed that the maximum attainable coefficient of static friction for an automobile tire was less than 1. Then, about 1962, three companies independently developed racing tires with coefficients of 1.6. Since then, tires have improved, as illustrated in this problem. According to the 1990 Guinness Book of Records, the shortest time in which a piston-engine car initially at rest has covered a distance of one-quarter mile is 4.96 s. This record was set by Shirley Muldowney in September 1989. (a) Assume that, as in Figure 4, the rear wheels lifted the front wheels off the pavement. What minimum value of µs is necessary to achieve the record time? (b) Suppose Muldowney were able to double her engine power, keeping other things equal. How would this change affect the elapsed time?
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| Fig.4 |
Answer:
(a) ∑F = ma
µsmg = ma
a = µsg
but
Δx = ½ at2 = ½ µsgt2
So
µs = 2Δx/(gt2)
µs = 2(2.50 mi)(1609 m/mi)/{(9.80 m/s2)(4.96 s)}2
µs = 3.34
(b) Time would increase, as the wheels would skid and only kinetic friction would act; or perhaps the car would flip over.
Problem#5
To meet a U.S. Postal Service requirement, footwear must have a coefficient of static friction of 0.5 or more on a specified tile surface. A typical athletic shoe has a coefficient of 0.800. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.00 m on a tile surface if she is wearing (a) footwear meeting the Postal Service minimum? (b) a typical athletic shoe?
Answer:
(a) The person pushes backward on the floor. The floor pushes forward on the person with a force of friction. This is the only horizontal force on the person. If the person’s shoe is on the point of slipping the static friction force has its maximum value.
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| Fig.5 |
∑Fx = max
f = µsn = max
and
∑Fy = may
n – mg = 0
n = mg
then
max = µsmg
ax = µsg = 0.5(9.80 m/s2) = 4.90 m/s2
we use
xf = xi + vxit + ½ axt2
3 m = 0 + 0 + ½ (4.90 m/s2)t2
t = 1.11 s
(b) xf = xi + vxit + ½ axt2
xf = ½ µsgt2
3 m = ½ (0.8)(9.80 m/s2)t2
t = 0.875 s
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