PRACTICE PROBLEM

Find the forces exerted by the pillars when the diver is $1\mathrm{.}00$ m to the left of the right end.

$[$Answer: $F_{1,y}=-1470N,F_{2,y}=2350N]$

How do we find these answers? Use the picture below to answer

EXAMPLE $11-8$ TAKING THE PLUNGE

A $5\mathrm{.}00-$m$-$long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board, as shown in the sketch; the other is $1\mathrm{.}50$ m away. Find the forces exerted by the pillars when a $90\mathrm{.}0-$ kg diver stands at the far end of the board.

PICTURE THE PROBLEM

We choose upward to be the positive direction for the forces. When calculating torques, we use the left end of the diving board as the axis of rotation, which means that vec$(F{)}_1$ causes zero torque. In addition, vec$(F{)}_2$ would cause a counterclockwise rotation if it acted alone, soits torque is positive. On the other hand, mvec$(g)$ would cause a clockwise rotation, so its torque is negative. Finally, vec$(F{)}_2$ acts atadistance a from the axis of rotation, and mvec$(g)$ acts at a distance $L.$

REASONING AND STRATEGY

As usual in static equilibrium problems, we use the conditions of $($i$)$ zero net force and $($ii$)$ zero net torque to determine the unknown forces, vec$(F{)}_1$ and vec$(F{)}_2.$ In this system, all forces act in the positive or negative $y$ direction; thus we need only set the net $y$ component of force equal to zero.

Known Length of diving board, $L=5\mathrm{.}00m$; distance of second pillar, $d=1\mathrm{.}50m$; mass of diver, $m=90\mathrm{.}0kg.$

Unknown Force exerted by pillar $1,F_{1,y}=?$; force exerted by pillar $2,F_{2,y}=?$

SOLUTion

Set the net $y$ component of force acting on the diving board equal to zero:

Calculate the torque due to each force, using the left end of the board as the axis of rotation. Notice that each force is at right angles to the radius and that vec$(F{)}_1$ goes directly through the axis of rotation:

${\displaystyle \underset{?}{\overset{?}{}}}{F}_y=F_{1,y}+F_{2,y}-mg=0$

${}_1=F_{1,y}(0)=0$

${}_2=F_{2,y}(d)$

${}_3=-mg(L)$