A truck pulls boxes up an incline plane. The truck can exert a maximum force of 2,400 N. If the plane's incline is (2 pi )/3 and the coefficient of friction is 5/3 , what is the maximum mass that can be pulled up at one time?

1 Answer
May 10, 2018

The truck will be able to pull up to 144 kg.

Explanation:

This calculation is neglecting any friction or gravity acting on the truck, so it's net force is 2400 N up the slope. Also, I'll take theta as pi/3 or 60˚. This is because (2pi)/3 is 120˚ and takes you beyond a vertical slope and back around to 60˚.

The parallel component of the gravitational force on the box is given by

F_(g|\|)=mgsin(theta)

This is the force that gravity exerts on the box down the slope.

The normal force acting on the box is given by the perpendicular component of the gravitational force

N=mgcos(theta)

The frictional force is then given by

F_f=mu_sN=mu_smgcos(theta)

F_(g|\|) and F_f are the opposing forces so if we set the sum of these equal to 2400 N, that will allow us to solve for the maximum mass

F_(g|\|)+F_f=2400

rArrmgsin(theta)+mu_smgcos(theta)=2400

rArrm(gsin(theta)+mu_sgcos(theta))=2400

rArrm=2400/(gsin(theta)+mu_sgcos(theta))

m=2400/(8.487+8.167)=144" kg"

Where

g=9.80" m/s"

mu_s=5/3

theta=pi/3