Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 N.

(a) What is the weight of the motor, and where along the board is its center of gravity located?

(b) Suppose the board is not light but weighs 200 N, with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

The condition for translational equilibrium is: $\sum {\mathrm{F}}_{\mathrm{ext}}=0$

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

(a)

Given the forces applied on both the ends of the board are $400\mathrm{N}\mathrm{and}600\hspace{0.33em}\mathrm{N}$. The length of the board is $2\mathrm{m}$.

Considering the given system to be in equilibrium, the equivalent weight of the motor will be:

$$\begin{gathered} \mathrm{W}={\mathrm{F}}_1+{\mathrm{F}}_2 \\ =400+600 \\ =1000 \\ =1\mathrm{kN} \end{gathered}$$

Thus, the motor weighs $1000\mathrm{N}$.

(b)

Let the center of gravity be located at $xm$from force of $400\mathrm{N}$.

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

$$\begin{gathered} \underset{}{\sum {\mathrm{\tau }}_{\mathrm{ext}}={\mathrm{F}}_1\mathrm{x}-{\mathrm{F}}_2\left(2-\mathrm{x}\right)}=0 \\ 400\mathrm{x}=1200-600\mathrm{x} \\ 1000\mathrm{x}=1200 \\ \mathrm{x}=1.2\mathrm{m} \end{gathered}$$

$\begin{array}{rcl}\sum {\tau }_{ext}& =& F_1·x-F_2·\left(2-x\right)=0\\ 400x& =& 1200-600x\\ 1000x& =& 1200\\ x& =& 1.2\text{m}\end{array}$

Thus, the motor weighs 1-kilo newton with the center of gravity at a distance of 1.20 m from the end of the board, having a force of 400 N.