Answer to Question #84785 in Mechanics | Relativity for Diosa Mamauag

Question

A 200N block rests on an inclined plane at 30° to the horizontal. A force parallel to and acting up the plane is applied to the body if coefficient is 0.20. Find the value of force

Answer 1

There are three forces that act on the block: the force of gravity (or weight)

$$mg$$

directed downward and can be resolved into two perpendicular components (

$$F_{||} = mgsin\alpha$$

and

$$F_{\perp} = mgcos\alpha$$

), the force of reaction

$$N$$

directed perpendicular to the surface, the applied force

$$F_p$$

directed parallel and upward to the plane. Let’s draw a free-body diagram and write all forces that act on a block:

There are three forces that act on the block: the force of gravity (or weight)

$$mg$$

directed downward and can be resolved into two perpendicular components (

$$F_{||} = mgsin\alpha$$

and

$$F_{\perp} = mgcos\alpha$$

), the force of reaction

$$N$$

directed perpendicular to the surface, the applied force

$$F_p$$

directed parallel and upward to the plane. Let’s draw a free-body diagram and write all forces that act on a block:

Since the block is at rest the applied force is equal to the rolling force:

$$F_p - F_r = 0,$$ $$F_p - mgsin\alpha = 0,$$ $$F_p = mgsin\alpha = Wsin\alpha.$$

From this formula, we can find the applied force:

$$F_p = Wsin\alpha = 200 N \cdot sin30^\circ = 100 N.$$

Answer:

$$F_p = 100 N$$

.