A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. T

Question

A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is 0.270, and the coefficient of kinetic friction is 0.240.
(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed?


m/s2

(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?



Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2

Explanation / Answer

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*0.270 = 2.7468 m/s^2 So the acceleration of the truck is 2.6487m/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(0.270 - 0.240) = 1.0791 a = 1.0791 m/s^2 So the acceleration of the crate relative to the ground is 1.0791 m/s^2 and it is in the same direction as the truck is moving.

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