Q6 E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

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.

Step 1: Equilibrium

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

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

Step 2: Find the Force

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

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:

$$\begin{gathered} \underset{}{\sum F_y=F_1+F_{2-}160=0} \\ F_2=160-F_1 \\ F_2=160-60 \\ F_2=100 N \end{gathered}$$

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

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

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.