Question

A truck pulls boxes up an incline plane. The truck can exert a maximum force of `2,700 N`. If the plane's incline is `(2 pi )/3` and the coefficient of friction is `8/5`, what is the maximum mass that can be pulled up at one time?

Answer 1

`192.667\ kg`

Explanation:

Let `m` be the maximum mass that can be pulled by an applied force `F=2700\ N` up a plane inclined at an angle `\theta=\pi-{2\pi}/3=\pi/3` & having coefficient of friction `\mu =8/5`

The total force acting down the inclined plane

`=mg\sin\theta+mu mg\cos\theta`

for maximum mass to be pulled up the inclined plane, the applied force `F` up the plane must be equal to the total force down the inclined plane hence we have

`F=mg\sin\theta+mumg\cos\theta`

`F=mg(\sin\theta+mu\cos\theta)`

`m=\frac{F}{g(\sin\theta+mu\cos\theta)}`

setting the corresponding values in above equation, we get

`m=\frac{2700}{9.81(\sin(\pi/3)+9/8\cos(\pi/3))}`

`=192.667`

The truck can pull up a maximum mass of `192.667\ kg`