A package of dishes (mass 60.0 kg) sits on the flatbed of a pickup truck with an

Question

A package of dishes (mass 60.0 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 0.340, and the coefficient of kinetic friction is 0.230.

(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?
3.332 m/s^2

(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 at5.0° 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?

Explanation / Answer

Mass of package = M = 60 kg

Static friction coefficient = 0.34 m/s2

kinetic friction coefficient = 0.23 m/s2

(a). Maximum acceleration of truck be 'a'

Psudo force on package = Ma. In opposite direction of motion of the truck.

This force will be balanced by static friction force .

Maximum Static friction force = fs = 0.34×Mg

Now for limiting case both the forces will be eqal in magnitude and opposite in direction .

0.34×Mg = Ma

Or. a = 0.34g = 3.332 m/s2

(b). When the packagace slides , the friction acting on it will be kinetic .

Hence net force on the block with respect to truck

Fnet = ma - 0.23×mg

Acceleretaion of package with respect to truck = a - 0.23g

Hence acceleration of block with respect to ground = a-(a-0.23g) = 0.23g = 2.254 m /s2

(c). Let the maximum acceleration be a.

a = g×Sin(5°) + 0.34×g×cos(5°) = 0.853 + 3.319 = 4.172 m/s2

(d). Acceleration of package with respect to ground = 0.23×g×cos(5°). - g×sin(5°) = 2.245 - 0.853 = 1.392 m/s2

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