A 2.5 m long diving board weighs 120 N. It has two supports, one at the end and another at a distance of 1.0 m from that end. A 100 N diver stands at the other end of the board. What is the magnitude and direction of the force exerted by the support at the end of the diving board

46.67 N  Upwards (with a clockwise moment)

Explanation:

length of board = 2.5 m

weight of board = 120 N

the board has two supports,  say support A and support B

support A is at one end,

support B is at 100 m from the other end.

weight of diver = 100 N

diver stands on the other end of the board.

Magnitude of support A at the end of the board

To get the magnitude and force exerted by the support at the end of the board (support A, we take moment of the forces about support B

Moment of a force is the product of force and perpendicular distance of the force about a center.

The weight of the board acts at the center of the board (1.25 m from each end of the board). That is 2.5 m from the support B.

moment of board's weight about support B is  120 x 0.25 = 30 N-m

The moment due to the weight of the board acts anticlockwise.

Weight of the diver acts at the opposite side of the board, and it acts 1 m from support B.

Moment of diver about support B is 100 x 1 = 100 N-m

Th moment due to the diver acts clockwise.

The moment due to the reaction at support A acts at a distance 1.5 m from support B

If the reaction force on support A is Fa, then the reaction about support B is Fa x 1.5 = 1.5Fa.

The moment due to support A acts clockwise.

According to moment laws, the total clockwise movement must be equal to the total anticlockwise movement.

Total clockwise movements = 100 N-m + 1.5Fa

Total anticlockwise moment = 30 N-m

according to moment laws,

100 + 1.5Fa = 30

1.5 Fa = 30 - 100 = -70

Fa = -70/1.5 = -46.67 N

The magnitude of the force exerted at support A is equal but opposite to the reaction at support A and is equal to 46.67 N

the rearrangement of particles in a physical change. hope this