# a block weighing 50 N rests on an inclined plane. its weight is a force directed vertically downward, a37 degree angel find the componants of the force parallel to the surface of the plane and perpendicular to it

## ghfgf

## The weight force components are:

50 sin 37 Newtons parallel to the surface of the plane, and

50 cos 37 Newtons perpendicular to the surface of the plane

## 40 Newtons

## To find the components of the weight force parallel and perpendicular to the inclined plane, we can use trigonometry.

First, let's identify the given information:

Weight of the block, W = 50 N

Angle of the inclined plane, θ = 37 degrees

Now, let's find the component of the weight force perpendicular to the inclined plane (W⊥). This component is equal to the weight force multiplied by the cosine of the angle.

W⊥ = W * cos(θ)

W⊥ = 50 N * cos(37 degrees)

W⊥ ≈ 50 N * 0.7986

W⊥ ≈ 39.93 N

So, the perpendicular component of the weight force is approximately 39.93 N.

Next, let's find the component of the weight force parallel to the inclined plane (W//). This component is equal to the weight force multiplied by the sine of the angle.

W// = W * sin(θ)

W// = 50 N * sin(37 degrees)

W// ≈ 50 N * 0.6018

W// ≈ 30.09 N

Therefore, the parallel component of the weight force is approximately 30.09 N.

To summarize:

The component of the weight force perpendicular to the inclined plane is approximately 39.93 N.

The component of the weight force parallel to the inclined plane is approximately 30.09 N.