Physics help please.?

Question

A uniform diving board, of length 5.0 m and mass 54 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.]

Answer 1

I agree with the first answerer...where is the figure? But let's see if we can solve it just the same. We have to guess a little regarding the location of the supports.

Given: length of board=5.0m
mass of board=54kg
Hence, weight of board =54g N.
Place this at the middle of the board because it is a uniform board as given in the problem.

First support is 3.4m from the end, and the second is 4.6m form the end.

Diver stands at end of the board. He weighs 65g N.

Find: F1 at first support and F2 at second support.

Take torque about F1:

54g*(2.5-0.4)+65g*4.6=
F2*(1.6-.4)

54g*2.1+65g*4.6=1.2F2 Call this equation 1.
113.4g+299g=1.2F2
F2=(1111.3+2930.2)/1.2
=3368N upward.

Take torque about F2:

54g*(3.4-2.5)+65g*3.4=
F1*(1.6-.4)

54g*0.9+65g*3.4=1.2F1 Call this equation 2.

48.6g+221g=1.2F1
F1=2642/1.2
=2201.7N downward

Check your answer:

F1+F2=65*9.8+54*9.7
-2201.7+3368=637+529.2

3368=637+529.2+2201.7
=3367.9
0.1 difference is acceptable. so our solution is correct. (Assuming my assumptions as to locations of supports are same as in your figure.)

Answer 2

I drew the diagram from your description and found the supports on the right and the diver on the left. Then I adjusted the measurements and re-drew it with the supports on the left and the diver on the right.

Let force A be at the support located 5.0 - 4.6 = 0.4m from the left. Force B is at the support located 5.0 - 3.4 = 1.6m from the left. The weight of the board, 54g, is located at the center, 2.5m from the left; and the weight of the diver, 65g, is located at the right, 5.0m from the left.

Force A and the two weights will be directed downward, and force B will be upward. Figure the torques about point B (1.6m from the left). Here's what you get:

A(1.6 - 0.4) = 54g(2.5 - 1.6) + 65g(5.0 - 1.6)

Solve that for A. I get A = 224 2/3 g.

Now, B = A + 54g + 65g. I get B = 343 2/3 g.

Using g = 9.8 m/s^2, A = 2201.7N and B = 3367.9N.

Answer 3

Where is the figure???