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
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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.