Question

1. What minimum force is required to keep a 5 kg block at rest on an inclined plane of inclination 30°? The coefficient of static friction between the block and the inclined plane is 0.25.

Answer 1

Given that Block should remain at rest, which means acceleration of block is zero, So from Newton's second law net force is also zero. Since

Fnet = m*a

Since a = 0, So

Fnet = 0

Now Fnet will be

Fnet = Weight of block - F_applied - F_friction = 0

Fnet = W - F_a - F_f = 0

F_a = W - F_f

F_f = uk*N

N = Normal Force on block on incline = m*g*cos theta

W = m*g*sin theta

So,

F_a = m*g*sin theta - uk*m*g*cos theta

Using given values:

m = 5 kg

theta = 30 deg

uk = 0.25

So,

F_a = 5*9.81*sin 30 deg - 0.25*5*9.81*cos 30 deg

F_a = 13.9 N

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