# A diving board length L = 5.20m that weighs 205N is supported by two pillars. One pillar is at the left end of the diving board, as shown below; the other is distance d = 1.70m away. Find the magnitude of the force exerted by the left pillar when a 90.0kg diver stands at the far end of the board.

## I can't conceptualize this, but draw a sketch, with forces both up and down (forces down: weight of board, and diver )forces up: forces at each pilar.

Of course, the sume of all vertical forces are zero.

then, sum moments about any point, I suggest the right pillar. the sum of moments is zero.

## To find the magnitude of the force exerted by the left pillar, we can use the principle of torque equilibrium. Torque is the rotational equivalent of force and is given by the equation:

τ = r x F

where τ is the torque, r is the perpendicular distance from the axis of rotation to the point of application of force, and F is the force applied.

In this case, the axis of rotation is at the right end of the diving board where the diver stands, and the left pillar is the point of application of force. We need to consider the forces acting on the diving board:

1. The weight of the diving board (205N) acting downward at the center of mass.
2. The weight of the diver (90.0 kg) acting downward at the far end of the board.

The torque due to the weight of the diver is given by:

τ_diver = r_diver x F_diver

where r_diver is the distance from the right end of the diving board (axis of rotation) to the far end of the board and F_diver is the force applied by the diver's weight.

Given:
L = 5.20 m (length of the diving board)
d = 1.70 m (distance between the two pillars)
g = 9.8 m/s² (acceleration due to gravity)

First, let's calculate r_diver:

r_diver = L - d

r_diver = 5.20 m - 1.70 m
r_diver = 3.50 m

Next, let's calculate F_diver:

F_diver = m x g

F_diver = 90.0 kg x 9.8 m/s²
F_diver = 882 N

Now, we can calculate the torque due to the weight of the diver:

τ_diver = r_diver x F_diver

τ_diver = 3.50 m x 882 N
τ_diver = 3087 N∙m

Since we are looking for the force exerted by the left pillar, we also need to consider the torque due to the weight of the diving board. The torque due to the weight of the diving board is given by:

τ_diving_board = (L/2) x F_diving_board

Given that the weight of the diving board is 205N, we can find F_diving_board:

F_diving_board = 205 N

Now, let's calculate τ_diving_board:

τ_diving_board = (L/2) x F_diving_board

τ_diving_board = (5.20 m/ 2) x 205 N
τ_diving_board = 533 N∙m

The net torque on the diving board must be zero for it to be in equilibrium. Therefore, the torque exerted by the left pillar is equal in magnitude and opposite in direction to the combined torque of the weight of the diving board and the weight of the diver:

τ_left_pillar = -(τ_diving_board + τ_diver)

τ_left_pillar = -(533 N∙m + 3087 N∙m)
τ_left_pillar = -3620 N∙m

Finally, we can find the magnitude of the force exerted by the left pillar using the equation:

τ = r x F

F = τ / r

F = -3620 N∙m / d

F = -3620 N∙m / 1.70 m
F ≈ -2135.29 N

Since we are interested in the magnitude of the force, we can take the absolute value:

|F| = 2135.29 N

Therefore, the magnitude of the force exerted by the left pillar when a 90.0kg diver stands at the far end of the board is approximately 2135.29 N.