91Ó°ÊÓ

The condition for translational equilibrium can be expressed as: $\underset{}{∑}{F}_{ext}=0$

And that for rotational equilibrium can be expressed as: $\underset{}{∑}{\mathrm{τ}}_{ext}=0$. The sum of all the forces acting on the body will be zero.

Given that the diver standing at one end of the diving board weighs . The weight and length of the board, respectively, are 280 N and 3.00 m.

Let the whole given setup be illustrated as a free body diagram for forces on the board, as shown in the figure below:

Now, for the force , considering the criteria for rotational equilibrium, we have

$\begin{array}{r}∑\mathrm{τ}=0\\ F_1\left(1.00\right)=\left(1.5\right)\left(280\right)+\left(3.00\right)\left(500\right)\\ F_1=1920\mathrm{N}\end{array}$

Thus, the force at the support point is 1920 N.

Similarly, for the force $F_2$, considering the criteria for rotational equilibrium at the other end, we have

$\begin{array}{r}∑\mathrm{τ}=0\\ F_2\left(1.00\right)=(0.50)\left(280\right)+(2.00)\left(500\right)\\ F_2=1140\mathrm{N}\end{array}$

Thus, the force at the left-hand end of the board is 1140 N.