A 75 kg diver stands at the end of the 1.5 meter long, 10 kg diving board. The d

Question

A 75 kg diver stands at the end of the 1.5 meter long, 10 kg diving board. The diving board is nailed to two supports: one at the opposite end of the board from the diver, and another 30 cm away from that. If the diver leaps straight up into the air with an acceleration of 3.0 m/s^2, determine: a. What force the diving board must be exerting on him as he leaps. b. The forces that must be exerted on the diving board by the two supports while he is leaping. (Please note that the answer would be different if he were just standing still!)

Explanation / Answer

mass of diver =75 kg

mass of board =10 kg

! O----------------------------------!B-------0.30 m-----------|A

<--------------------------------1.5 m------------------------>

<----------------1.2 m--------->

Mid of board distance from O is 0.75m so distance to B is0.45 m

Let reaction at A is F (downward)

Consider rotational equilibrium at B

F*0.30 - 75*9.8*1.2 - 10*9.8*0.45=0

so F=(75*9.8*1.2 + 10*9.8*0.45)/0.30=3087 N

again Now reation at B, =(75+10)*9.8 + 3087= 3920 N (upward)

So reaction at the end of the board=3087 downward and at another support 3920 N upward

As you can see diver and board do not weigh much, the system of lever put great force at A.
so the board should be firmly bolted down.

a) Diving board must be exerting force on him= mass*acceleration=75kg*3=225 N

   3920-3087-10*9.8=

b)The two support exerting forces on the diving board is = 3920-3087=833 N

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