Question

A 75Kg block is placed on an inclined plane with an angle of 30o. Find the components of the weight of the block parallel and perpendicular to the incline plane

Answer 1

Answer:

There will be two components of the block i.e. 75gcos30°  direction downwards towards the block from the center of the block and 75gsin30° direction parallel to the incline from the center of the block. The downward force will balance the normal force and the other force will help the block to either accelerate or stay in its equilibrium position.

Thus, the component parallel to inclined is 75gsin30°= $75 \times 9.8 \times \frac{1}{2}$ = 367.5 N. So the value for parallel component is 367.5 N.

Answer 2

Answer:

There will be two components of the block i.e. 75gcos30° direction downwards towards the block from the center of the block and 75gsin30° direction parallel to the incline from the center of the block. The downward force will balance the normal force and the other force will help the block to either accelerate or stay in its equilibrium position.

Thus, the component parallel to inclined is 75gsin30°= 75 \times 9.8 \times \frac{1}{2}75×9.8×

2

1

= 367.5 N. So the value for parallel component is 367.5 N.