Two people are carrying a uniform wooden board that is $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\text{m}}$long and weighs $\mathbf{160}\mathbf{\text{N}}$. If one person applies an upward force equal to $\mathbf{60}\mathbf{\text{N}}$at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

The condition for translational equilibrium is: $\sum F_{ext}=0$.

And that for rotational equilibrium is:$\sum {\tau }_{ext}=0$

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

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:

$\begin{array}{rcl}\sum F_y& =& F_1+F_2-160=0\\ F_2& =& 160-F_1\\ F_2& =& 160-60\\ F_2& =& 100\text{N}\end{array}$

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

$\begin{array}{rcl}\sum {\tau }_{ext}& =& 0\\ F_2·x-160·\left(\frac{L}{2}\right)& =& 0\\ x& =& \frac{160\times 1.50}{100}\\ x& =& 2.40\text{m}\end{array}$

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.