A package of dishes (mass 55.0 kg) sits on the flatbed of a pickup truck with an

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 0.200.

(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?
°

(f) Is any piece of data unnecessary for the solution in all the parts of this problem?

The mass of the package

The coefficient of static friction

The coefficient of kinetic friction

Explanation / Answer

a)
maximum acceleration should be such that its force should not exceed force of static friction
m*a = mius*m*g
a = mius*g
a = 0.330*9.8
=3.232 m/s^2
Answer: 3.232 m/s^2

b)
Once its sliding, kinetic friction acts in backward and force equal to static friction acts forward
a = net force / mass
= (mius*m*g - miuk*m*g) / m
= (mius*g - miuk*g)
= (0.330*9.8 - 0.200*9.8)
= 1.274 m/s^2
Answer: 1.274 m/s^2

c)
force of truck must balance static friction - compoennet of weight along incline
net force = fs - m*g*sin thetha
m*a = miu*m*g*cos thetha - m*g*sin thetha
a = mius*g*cos thetha - g*sin thetha
a = 0.330*9.8*cos 9 - 9.8*sin 9
= 1.66 m/s^2
Answer: 1.66 m/s^2
d)
m*a = static friction + m*g*sin thetha - kinetic friction
m*a = mius*m*g*cos thetha + m*g*sin thetha - miuk*m*g*cos thetha
a = mius*g*cos thetha + g*sin thetha - miuk*g*cos thetha
= 0.330*9.8*cos 9 + 9.8*sin 9 - 0.200*9.8*cos 9
= 2.79 m/s^2
Answer: 2.79 m/s^2

only 4 parts at a time please

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