A uniform diving board, of length 5.0 m and mass 55 kg, is supported at two poin

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 62 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.]

Explanation / Answer

........................................…
----------------------------------
...^........^.....^
...L.......R.....B

L is the left support, R is the right, B is the center of mass of th e board and P is a person with her tongue out. All forces are positive up. I chose counter clock wise to be positive for torques.

1) around point L:
Fr*(4.6-3.4) + Fp*4.6 + Fb*(4.6-2.5)= 0

2) around point R
-Fl*(4.6-3.4)+ Fp * 3.4 + Fb*(3.4-2.5) = 0

3) total forces:

Fl+Fr+Fp + Fb= 0

Fb= -55*9.8 [N] (you can use 10 if your teacher prefers) = -539 [N]
Fp = -62*9.8 [N] = -607.6[N]

three equations are too many. I wrote them just in case one is easier to use than the other. We have 2 unknown and so 2 equations are enough.
so now the equations look like:

Fr*1.2-607.6*3.4[N]-485[N] = 0
or Fr*1.2-2550.8[N] = 0
or Fr= 2125.7[N] (so indeed it points up)
= 2.125 kN (upward)

-Fl*(4.6-3.4) -607.6 * 3.4 -539*(3.4-2.5) = 0
-Fl*1.2 - 2550.94 = 0
Fl = -2125.78 N (it is pointing downwards)
= 2.125 kN (downward)

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