91Ó°ÊÓ

The condition for translational equilibrium is: $\underset{}{\mathbf{∑}{\mathbf{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}}$.

And that for rotational equilibrium is: $\underset{}{\mathbf{∑}{\mathbf{π}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}}$.

The force$60N$is applied at one end of the board weighing$160N$.The length of the board is $3.00m$.Since the board is uniform, the center of gravity will be at $1.50m$

Let the whole set up illustrated as a free body diagram for forces shown in the figure as:

Considering the upward force to be positive and applying the condition for translational equilibrium, we have:

$$\begin{gathered} \underset{}{∑F_y=F_1+F_{2-}160=0} \\ {F}_2=160-F_1 \\ {F}_2=160-60 \\ {F}_2=100N \end{gathered}$$

Again, considering the anticlockwise rotation to be positive and applying the condition for rotational equilibrium, we have:

$$\begin{gathered} \underset{}{∑{\mathrm{π}}_{ext}=0} \\ {\mathrm{F}}_{2.}\mathrm{X}-160.\left(\frac{\mathrm{L}}{2}\right)=0 \\ \mathrm{X}=\frac{160×1.50}{100} \\ \mathrm{X}=2.40\mathrm{m} \end{gathered}$$

Thus, the person needs a force of 100 newtons to lift the board at a distance of 2.40 m from the end of the board.