A block weighing 9.02 N rests on a32° inclined plane. (a) Find thenormal force e

Question

A block weighing 9.02 N rests on a32° inclined plane. (a) Find thenormal force exerted by the plane on the block.
1 N out of theplane

(b) Find the frictional force exerted by the plane on theblock.
2 N up the slope

(c) Find the magnitude of the total force exerted by the plane onthe block
3 N

(d) Find the normal force and the frictional force exerted by theblock on the plane. normal force 4 N into theplane frictional friction 5 N down theslope normal force 4 N into theplane frictional friction 5 N down theslope

Explanation / Answer

F=mg cos(theta)   =(9.02 N)(cos 32)   =7.649 N
2) The frictional force be the force pulling up the plane.Using the same triangle you have drawn, you can find the frictionalforce.
F= mg sin(theta)    =(9.02 N)(sin 32)    = 4.78 N

3) I'm not sure :/
4) The forces are the same as in part 1. Whatever force theblock exerts on the plane, the plane exerts on the block (this iswhy the block does not just fall straight through the plane).
Normal Force: 7.649 Frictional Force: 4.78 N

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