# Diving board

1. Jun 7, 2005

### MAPgirl23

A diving board of length L = 3.00 m is supported at a point a distance 1.00 m from the end, and a diver weighing 490 N stands at the free end. The diving board is of uniform cross section and weighs 295 N.

Find the force at the support point.
** Now I know net torque = 0
Suppose F1 is the force applied at the support point and F2 is the force at the end that is held down.
(F1*x_f1) - (w_diver * L) - (w_board * x_point) = 0

x_f1 = 3.0 m * {490 N/(490N + 295N)} = 1.87 m ?

--> F1(1.87 m) - (490 N)(3.0 m) - (295 N)(1.0 m) = 0 solve for F1

F1 = {(490N)(3.0 m) + (295N)(1.0 m)}/1.87 m = 944 N which is wrong

how do I find x_f1? is my formula for F1 correct?

2. Jun 7, 2005

### learningphysics

Ok. So you're going to find the torque about the end of the board that is held fast right?

That's fine. You're getting the torque about the end that is held fast.

Not sure why you did this. The length of the board is 3m. The support point is 1m from the free end. So it is 2m from the unfree end. x_f1=2m (draw a picture if you haven't)

Also the weight of the board acts at the center of the board. So that is half the length of the board from the end. so x_point=1.5m

Now you should be able to solve for F1.

3. Jun 7, 2005

### Staff: Mentor

Only the original poster knows for sure, but I would have guessed that the support point was 1m from the fixed end.

4. Jun 7, 2005

### learningphysics

Ah... you're probably right. The OP used "end" for the support point then "free end" later. That would also make more sense I believe for a diving board.

5. Jun 7, 2005

### MAPgirl23

but if I make x_f1 = 1 m and F1 = {(490N)(3.0 m) + (295N)(1.0 m)}/1 m = 1765 N the answer is wrong.

6. Jun 7, 2005

### Staff: Mentor

The weight of the board acts at its middle, not at x = 1.0m.

7. Jun 7, 2005

### MAPgirl23

so it acts at 1.5 m (L/2 = 3.0 m/2 = 1.5 m)

is x_f1 = 1.5 therefore:
F1 = {(490N)(3.0 m) + (295N)(1.5 m)}/1 m = 1912.5 m

8. Jun 7, 2005

### MAPgirl23

thank you for all of your help.