Question

General Physics with Calc 1

8. A package of dishes (mass 50.0 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 0.240. (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? Pick one below

The coefficient of kinetic friction

The mass of the package

The coefficient of static friction

Answer 1

8. a)

N - mg = 0

N = mg

maximum static friction =us N = 0.310mg

this is the maximum force on package.

and Fnet = ma

0.310mg = ma

a =0.310 x 9.8 = 3.04 m/s^2

b) when package is sliding then kinetic friction will work.

kinetic friction= uk N = 0.240mg

a = F/m = 0.240g = 0.240 x9.8 = 2.35 m/s^2

c) now N - mgcos9 = 0

N = mgcos9

maximum static friction =us N = 0.310mg cos9

this is the maximum force on package.

and Fnet = ma

0.310mgcos9 - mgsin9 = ma

a =9.8(0.310cos9 - sin9) = 1.47 m/s^2

d) when package is sliding then kinetic friction will work.

kinetic friction= uk N = 0.240mg cos9

0.240mgcos9 - mgsin9 = ma

a =9.8(0.240cos9 - sin9) = 0.79 m/s^2

e) at slope @,

N = mgcos@

f = 0.31mgcos@

then along the incline,

mgsin@ - 0.31mgcos@ = 0

tan@ = 0.31

@ = 17.22 deg

f) mass is not neessary.