Question

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?

Answer 1

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`