Static Equilibrium EX 3

We will consider the board and the diver as one object. Then the forces exerted on this system are the gravitational force on the diver (the magnitude is 60 N), $F_{G}$ and the forces exerted by the two supports, $F_{1}$ and $F_{2}$. The force $F_{1}$ acts on the system at the pivot, the force $F_{2}$ acts at a distance 0.5 m from the pivot and the gravitational force acts at the distance of 3 m from the pivot. The diving board is in equilibrium and therefore

$\Sigma F_{y}=0$ $\Sigma \tau =0$

The diagram below shows the free body diagram and it also shows the force vectors tail-to-tail with the lever arm vectors.

We will compute the torques. Note that the torque exerted by the force $F_{1}$ is zero:

Forces	Direction	Torque
$F_{G}$	CW	$\tau =-(3)(600)\sin {90^{\circ }}=-1800Nm$
$F_{2}$	CCW	$\tau =(0.5)F_{2}\sin {90^{\circ }}=0.5F_{2}$
$\Sigma \tau$		0

Solving the equation

$F_{2}={1800 \over 0.5}=3600N$

The net force must be zero:

$F_{1}+F_{2}-600=0$

$F_{1}+3600-600=0$

$F_{1}=-3000N$

The negative sign indicates that the force $F_{1}$ points downward. It means that the diving board has to be "clamped" to the supporting pilon. The clamps that exert the downward force on the diving board.