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