Question

Question 1: A 70 kg diver stands at the right end of a uniform 30 kg diving board of length 3.0 m. The board is hinged at the left end and rests on a bar at the 1.0 m mark as shown. Find the vertical force exerted by the bar?

Answer 1

3m 0.5m Mg

Take moment about the bar

$\sum M=0$

$F_{y}*1m-mg*0.5m-Mg*2m=0$

$F_{y}*1m-30kg*9.81m/s^{2}*0.5m-70kg*9.81m/s^{2}*2m=0$

$F_{y}*1-30*9.81*0.5-70*9.81*2=0$

$F_{y}*1=30*9.81*0.5+70*9.81*2$

ANSWER: ${\color{Red} F_{y}=1520.55N}$

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