# Homework Help: Reaction Forces

1. Sep 6, 2010

### ana111790

1. The problem statement, all variables and given/known data
An 80kg person is preparing to dive into a pool. The diving board is a uniform horizontal beam that is hinged to the ground at point A and supported by a frictionless roller at D. B is the point directly under the center of gravity of the person. The distance between A to B is l=6m, and the distance between A to D is d=2m. Note the 1/3 of the board is located between A and B and 2/3 between D and B. You can assume that the board consists of two parts with two different weights connected at D.

If the diving board has a total weight of 1500 N determine the reactions on the beam at A and D.

Picture:
[PLAIN]http://img837.imageshack.us/img837/3036/drawinge.jpg [Broken]

2. Relevant equations
$$\sum Fy$$=0
$$\sum Fx$$=0
$$\sum M$$=0

3. The attempt at a solution
I think the only equation that can be used here is $$\sum Fy$$=0.

Forces that act at point A is only 1/2 of 1/3 of the total weight of the board, (I multiplied by 1/2 since that 1/3 of the weight is shared by points A and D)

So for $$\sum$$= 1/2 * 1/3 * -Wboard + Ron A = 0
To solve for Ron A = 250 N (up)
This is however a wrong answer. The RA should be 2318 N (down) according to the answer in the book.

Forces that act at point D are 1/2 of the 1/3 of the Weight of the board, 2/3 of the weight of the board and the weight of the person.
So for $$\sum$$= (1/2 * 1/3 * Wboard) + (2/3 Wboard) + ( Wperson) + Ron D = 0
To solve for Ron D = - (-250 N - 1000 N - 784 N) = 2034 (up).
This answer is also wrong, correct answer should be 4602 N (up)

I am obviously missing something major here since the answers are so off. What am I doing?

Last edited by a moderator: May 4, 2017
2. Sep 6, 2010

### Staff: Mentor

Don't make any assumptions about how the weight of the board will be divided up between the supports.
(1) The weight of an object--such as the board--can be considered to act through its center of mass.
(2) Label all the forces acting on the board.
(3) Use the torque condition for equilibrium.

3. Sep 6, 2010

### ana111790

[PLAIN]http://img138.imageshack.us/img138/9051/drawing2n.jpg [Broken]

Last edited by a moderator: May 4, 2017
4. Sep 6, 2010

### Staff: Mentor

Your analysis of the weights is good. Your diagram needs to show the upward force on the board at D and the downward force at A. Call those forces FD and FA (or whatever you like--just give them names).

I don't understand your torque equations. What are you using as your pivot in each case?

(M stands for 'moment'--another term for torque--not momentum!)

5. Sep 6, 2010

### ana111790

(I meant to write moment not momentum, sorry for the confusion)

[PLAIN]http://img193.imageshack.us/img193/8467/drawing2t.jpg [Broken]

For the first equation (Sum of torque on A) I am using point A as a pivot (so the whole weight of the board is the force and the radius is the whole length of the board, same for weight of person)
For the second equation (Sum of torque on D) I am using point D as a pivot, so 1/3 weight of board with radius 2 m (cw) and then 2/3 of the weight of the board with radius 4, and weight of person with radius 4m (ccw).

Last edited by a moderator: May 4, 2017
6. Sep 6, 2010

### Staff: Mentor

Your diagram shows 5 forces. If you want to treat the weight of the board as a single force, that's fine--where would that force act?

Going by your diagram, that equation will show the torques about A for all 5 forces. (Some of the torques will be trivial, of course.)

Here I expect to see the torques about D for all forces. Careful that you use the proper distance from the pivot for each force. The W1 and W2 forces are in the middle of their sections.

7. Sep 6, 2010

### ana111790

Solved it! Thank you so much for your help.