Question

A 100 kg box sits on an incline held at rest by a very thin rope attached to another mass hanging as shown. The force of friction between the box and the surface is 100 N and the hanging mass=50 kg. Determine both of the possible angles of the incline.

Answer 1

If friction is acting along the plane upwards

then in this case we will have

For equilibrium of 100 kg box on inclined plane we have

`mgsin\theta = F_f + T`

also for other side of hanging mass we have

`T = Mg = 50(9.8) = 490 N`

now we have

`100(9.8)sin\theta = 100 + 490`

`980sin\theta = 590`

`sin\theta = 0.602`

`\theta = 37 degree`

In other case we can assume that friction will act along the plane downwards

so now in that case we will have

`mgsin\theta + F_f = T`

also we have

`T = Mg = 50(9.8) N`

now we have

`100(9.8)sin\theta + 100 = 50(9.8)`

`980sin\theta + 100 = 490`

`980 sin\theta = 490 - 100`

`sin\theta = 0.397`

`\theta = 23.45 degree`

So the range of angle will be 23.45 degree to 37 degree