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ana111790
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Homework Statement
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
Homework Equations
[tex]\sum Fy[/tex]=0
[tex]\sum Fx[/tex]=0
[tex]\sum M[/tex]=0
The Attempt at a Solution
I think the only equation that can be used here is [tex]\sum Fy[/tex]=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 [tex]\sum[/tex]= 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 [tex]\sum[/tex]= (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?
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