91Ó°ÊÓ

The condition for translational equilibrium can be expressed as:$\mathbf{∑}{\mathit{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ .

And that for rotational equilibrium can be expressed as:$\mathbf{∑}{\mathit{τ}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ . 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 500 N. 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 $F_1$, considering the criteria for rotational equilibrium, we have

$$\begin{gathered} ∑\mathrm{τ}=0 \\ {F}_1\left(1.00\right)=\left(1.5\right)\left(280\right)+\left(3.00\right)\left(500\right) \\ {F}_1=1920\text{N} \end{gathered}$$

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:

$$\begin{gathered} ∑\mathrm{τ}=0 \\ {F}_2\left(1.00\right)=\left(0.50\right)\left(280\right)+\left(2.00\right)\left(500\right) \\ {F}_2=1140\text{N} \end{gathered}$$

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