A package of dishes (mass 55.0 kg) sits on the flatbed of a pickup truck with an
ID: 1491705 • Letter: A
Question
A package of dishes (mass 55.0 kg) sits on the flatbed of a pickup truck with an open tailgate. The coefficient of static friction between the package and the truck's flatbed is 0.330, and the coefficient of kinetic friction is 0.150.
(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the package does not slide relative to the truck bed?
(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the package sliding along its bed. What is the acceleration of the package relative to the ground?
(c) The driver cleans up the fragments of dishes and starts over again with an identical package at rest in the truck. The truck accelerates up a hill inclined at 9.0° with the horizontal. Now what is the maximum acceleration the truck can have such that the package does not slide relative to the flatbed?
(d) When the truck exceeds this acceleration, what is the acceleration of the package relative to the ground?
(e) For the truck parked at rest on a hill, what is the maximum slope the hill can have such that the package does not slide? (Answer in degrees)
Explanation / Answer
The crate will not move as long as the force exerted through the motion of the truck is less than the static friction between the crate and the flatbed. However, once this is overcome, the crate is no longer subject to static friction but to kinetic which is usually less than static. In other words it is more difficult to get something moving than it is to keep it moving.
(a) You need to balance the force from static friction with that provided on the crate by the forward motion of the truck.
Us = coefficient of static friction
Fs = static friction = (normal force)*Us = m*g*Us
Force pulling the crate = mass * acceleration = m*a
Equate the forces.
m*a = m*g*Us
a = g*Us = 9.81*0.33 = 3.2373 m/s^2
So the acceleration of the truck is 3.2373 m/s^2
(b) We now need to use the coefficient of kinetic friction in the above equation and then take the difference with the forward force to get the net acceleration. I will use the fact that the crate is still up to a maximu acceleration of g*Us (see above).
Uk = coefficient of kinetic friction
Fk = kinetic friction = m*g*Uk
g*Uk = acceleration due to friction
net acceleration = forward minus friction = g*Us - g*Uk = g(Us - Uk)
a = 9.81(0.33 - 0.15) = 9.81*0.18
a = 1.7658 m/s^2
So the acceleration of the crate relative to the ground is 1.7658 m/s^2 and it is in the same direction as the truck is moving.
c)
Fnet = ƒ = µ mg cos = ma
a = µ g cos = 0.33*9.81 cos 9 = 3.2373*0.994 = 3.197443465 m/s²
d)
net acceleration = forward minus friction = g*Us*cos 9 - g*Uk* cos 9 = g*cos 9 (Us - Uk) = 1.744060072 m/s^2
So the acceleration of the crate relative to the ground is 1.744060072 m/s^2 and it is in the same direction as the truck is moving
e)
As by free body diagram
Us*N = mg *cos
Us * mg * sin = mg * cos
So
tan = 1/Us
= 71.73711006 degrees
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