Find the two forces F1, and F2.

Question

A diver of weight 1500 newtons stand at the end of a 4.0 meter diving board. Assume that the weight of the diving board is negligible. The board is attached to two pedestal that are 1.5 meters apart. Find the two forces F1, and F2.

Answer 1

I am assume that one pedestal is at one edge of the board, call it point P₁,

that the diver is at the opposite edge of the board, and

the board is stationary (is able to support the weight of the diver).

Call the point where the other pedestal is attached point P₂.

Let

r = 4.0m be the length of the board,

d = 1.5m be the distance between the pedestals,

F = -1500N be the weight of the diver,

F₁ (to be found) be the force of the pedestal at point P₁,

F₂ (to be found) be the force of the pedestal at point P₂.

We let upward acting forces to be positive, while downward acting forces are negative.

That is why the weight F is negative.

Similarly, distance vectors pointing in the direction P₁ to P₂ will be positive,

and distance vectors pointing in the direction P₂ to P₁ will be negative.

Since the board is stationary, the sum of all torques around P₁ as well as around P₂ must be 0.

1) Point P₁:

The torque of the diver is rF.

The torque of the pedestal acting at point P₂ is dF₂.

The sum of the torques is 0:

rF + dF₂ = 0

F₂ = -Fr/d

2) Point P₂:

The torque of the diver is (r-d)F.

The torque of the pedestal acting at point P₁ is -dF₁.

The sign of d is negative because the distance vector from P₂ to P₁ is in the negative direction.

Again the sum of the two torques is 0:

(r-d)F - dF₁ = 0

F₁ = F(r-d)/d

Substituting actual numbers:

F₁ = -1500(4.0-1.5)/1.5 = -2500N

F₂ = 1500 x 4/1.5 = 4000N

Note that F₁ is negative, i.e., pulling the board down, while

F₂ is positive, pushing the board up.