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algol [13]
3 years ago
7

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 frictio

n between the box and the surface is 100 N and the hanging mass=50 kg. Determine both of the possible angles of the incline.
Physics
1 answer:
MAVERICK [17]3 years ago
6 0

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

<em>So the range of angle will be 23.45 degree to 37 degree</em>

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