Question

A 30N block is pulled at constant speed up a frictionless inclined plane by a weight of 10N hanging from a cord and passing over a frictionless pulley at the top of the plane. Find: a) the slope angle of the plane, b) the tension in the cord, and c) the normal force exerted on the block by the plane.

Answer 1

So the free body diagram appears on the right, and close up of what the forces of the 30N block are on the left. Mg=30N for the block.

Decompose the forces on the block to bite size chunks.I see that tension of the string is at an angle Theta to the plane and the direction of travel is up the plane, so I will take that as my line of reference. The arrows are pointing up down diagonally down (since i'm taking the tension as my horizontal line). We can do better by making the diagonal force (Mg) become only up and down left and right. You might need some trigonometry to convince yourself that the magnitudes are the ones shown in the diagram.

Since it is moving at a constant speed, F=ma = 0
Which means that MgSin θ=N=0 since the block isn't going into the inclined plane.
And MgCos θ -T=0
But the string doesn't stretch, neither does it have a netforce (or there will be an accerleration) so applying newton's thrid law on the weight we have mg-T=0=10-T=0 and so T=10N

Now putting T=10N back into MgCos θ-T=0 , Cos θ=1/3 = 70.5 degrees.

Taking MgSin θ=N=0, we find that the normal force N is 10N