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

$\mathrm{\Sigma }{F}_y=0$ $\mathrm{\Sigma }\tau =0$

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	$\tau =(0.5)F_2\sin 90^{\circ }=0.5F_2$
$\mathrm{\Sigma }\tau$		0

Solving the equation

$F_2=\frac{1800}{0.5}=3600N$

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.