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 0.310, and the coefficient of kinetic friction is 0.150. 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? 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? 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 10.0 degree with the horizontal. Now what is the maximum acceleration the truck can have such that the package does not slide relative to the flatbed? When the truck exceeds this acceleration, what is the acceleration of the package relative to the ground? 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? Is any piece of data unnecessary for the solution in all the parts of this problem? The mass of the package The coefficient of kinetic friction The coefficient of static friction

Explanation / Answer

a)

let max acceleration = amax

so force on package = m * a max

static friction force = us * mg

m * a max = us * mg

amax = us*g = 0.31*9.8 = 3.038 m/s^2

b)

when truck is moving with above aceeleration, force on package = m*atruck - uk*mg   = m*apackage

3.038 - 0.15*9.8 = apackage

apackage = 1.568 m/s^2

a package = 1.568 m/s^2 with respect to truck...

so with respect to ground a = 3.038 - 1.568 = 1.47 m/s^2

c)

when truck is moving up hill,

the net force on the package = m*atruck + mgsin(10) - us*(mgcos(10))

for it not to slip it is 0.

so m*atruck + mgsin(10) - us*(mgcos(10)) = 0

atruck = 0.31*9.8*cos(10) - 9.8sin(10) = 1.29 m/s^2

d)

when truck is moving with above aceeleration, force on package = m*atruck + mgsin(10) - uk*mgcos(10) = m*apackage

apackage = 1.29 + 9.8*sin(10) - 0.15*9.8*cos(10)

                 = 1.544 m/s^2

a with respect to ground = 1.29-1.544 = -0.254 m/s^2 ..... '-' sign says the direction of acceleration is opposite to truck... value is 0.254 m/s^2

e) for truck at rest, force on package = mgsin(angle) - us*mgcos(angle)

so tan(angl;e) = us = 0.31

angle 17.2234

f)

mass is not required

In none of the calculation we require mass of the package.. we didnot use m= 60kg anywhere.. so it is not necessary

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