Question

A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force exerted by the plane on the block is

Answer 1

Answer: N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Explanation:

See attached for a sketch.

From the attachment.

.

N = normal reaction force on block

W = weight of the block

theta = angle of the inclined plane to the horizontal

From the sketch, we can see that

N is equal in magnitude but opposite direction to Wy

N = Wy

And

Wy = Wcos(theta)

Wx = Wsin(theta)

Then,

N = Wy = Wcos(theta)

But W = mass × acceleration due to gravity = mg

N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Answer 2

Answer:

Mg Cos theta

Explanation:

Choosing a coordinate system parallel and perpendicular to the plane, since the acceleration perpendicular to the plane is 0, the net force perpendicular to the plane is also 0. Therefore, the magnitude of the normal force equals the magnitude of the y component of the weight.