Vertical forces physics problem.?

Question

A uniform diving board, of length 5.0 m and mass 55 kg, is supported at two points; one support is located 3.4 m from the end of the board and the second is at 4.6 m from the end (see the figure below). What are the forces acting on the board due to the two supports when a diver of mass 65 kg stands at the end of the board over the water? Assume that these forces are vertical. [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]

Left support _____kN downward
Right support _____kN upward


This was a question on my homework and I need help!

Answer 1

For this one, you have to consider both the weight of the board and the weight of the diver in your torque equations. We can use the weight of the board as a point force in the middle of the board, so we'll have a force of (55 kg)(9.81 m/s^2) acting at 2.5 m from the end of the board. We'll need that for both of the following scenarios.

Since there's no picture here, I'm going to assume that the setup looks like this:

x----------c--1-----------2

where x is the diver, 1 is the support at 3.6 m, and 2 is the support at 4.6 m. If that's the layout, then the center of gravity is at c.

Let's start by looking at the end support and using the middle support (1) as our fulcrum. You'll have a torque from the weight of the board at 1.1 m from the fulcrum, and a torque from the diver at 1.4 m from the fulcrum. The support is 1 m away. From this, we can balance our forces (Fx refers to the force from the diver, Fc is the force due to the board, and F2 is the force on the opposite support):

Fxdx + Fcdc = F2d2
(65 kg)(9.81 m/s^2)(1.4 m) + (55kg)(9.81 m/s^2)(1.1 m) = F2(1 m)
892.71 Nm + 593.505 Nm = F2(1 m)
F2 = 1486.215 N

Since the two forces are pulling down on the fulcrum, the force on the support is acting up.

We can use the same process using the end support (2) as our fulcrum. The diver is 4.6 m away from the fulcrum, and the center is 2.1 m away, and (1) is again 1 m away:

Fxdx + Fcdc = F1d1
(65 kg)(9.81 m/s^2)(4.6 m) + (55kg)(9.81 m/s^2)(2.1 m) = F1(1 m)
2933.19 Nm + 1133.055 Nm = F1 (1 m)
F1 = 4066.245 N

And since the forces and the support are on the same side of the fulcrum, the force on this support is acting down.