A block weighing 10.0 N rests on a 30° inclined plane. Find (a) the normal force

Question

A block weighing 10.0 N rests on a 30° inclined plane.
Find (a) the normal force exerted by the plane on the
block; (b) the frictional force exerted by the plane on
the block; (c) the magnitude of the total force exerted
by the plane on the block; (d) the normal force and the
frictional force exerted by the block on the plane.

Explanation / Answer

a) Normal force = 10cos30 = 8.66 N b)Since the block is at rest Frictional force - 10sin30 =0 => Frictional force 5N c) Total force = square root(Frictional force^2 + Normal force^2) = 10 d) Normal force and the the frictional force exerted by block on plane are same in magnitude as that exerted by plane on block but opposite in direction. so -8.66 and -5

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