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 0.350, and the coefficient of kinetic friction is 0.170.

(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 8.0

Explanation / Answer

a) friction = m a

u m g = m a

a = u g = 0.35*9.81= 3.43 m/s^2

b)

now use kinetic friction

a = 0.17*9.81= 1.67 m/s^2

c)

sum perp to slope

N - m g cos theta = 0

N = m g cos theta

sum parallel to slope

friction - m g sin theta = m a

u m g cos theta - m g sin theta = m a

a = 0.35*9.81*cos(8 degrees) -9.81*sin(8 degrees)= 2.03 m/s^2

d) a = 0.17*9.81*cos(8 degrees) -9.81*sin(8 degrees) = 0.286 m/s^2

e) if doesnt slide

u m g cos hteta - m g sin theta = 0

u cos theta = sin theta

tan theta = u

theta = arctan(0.35)= 19.29 degrees

f) mass doesnt matter

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.