A uniform diving board, of length 5.00 m and mass 53.0 kg, is supported at two p

Question

A uniform diving board, of length 5.00 m and mass 53.0 kg, is supported at two points; one support is located 3.40 m from the end of the board and the second is at 4.60 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 63.0 kg stands at the end of the board over the water? Assume that these forces are vertical

Answer:

Left support
? kN (select)downward,upward,right,left

Right support
kN (select)downward, upward, right, left

Explanation / Answer

As the board is in equilibrium, The torque about Cnter of Mass = 0
Let the forces acting on the board due to the Left & Right support, be Fl & Fr respectively.

Fl * (5/2 - (5.0 - 3.4 -1.2) ) + Fr * (5.0/2 - 1.6) = 63.0 * 2.5 * 9.8 --------1

Fl + Fr = (53.0 + 63.0) * 9.8
Fl + Fr = 1136.8 N ------2

Solving eq 1 & 2,

Fl = 433.65 N
Fr = 703.15 N

Left Support, = 0.434 kN Upward.
Right Support, = 0.703 kN Upward.

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