91Ó°ÊÓ

A force is an external factor that can change the rest or motion of a body. It has a size and a general direction.

The diagram for the forearm is shown below:

(a)

Applying the condition of equilibrium under torque we can write,

$$\begin{gathered} F_r×1.50=W×0.9 \\ {F}_r=\frac{W×0.9}{1.50} \end{gathered}$$

For $W=50×9.8\text{N}$we get,

$\begin{array}{rcl}{F}_r& =& \frac{50×9.8×0.9}{1.50}\\ & =& 294\text{N}\end{array}$

The force is the same for both the hands so the force on each hand is,

$\begin{array}{rcl}\frac{F_r}{2}& =& \frac{294}{2}\\ & =& 147\text{N}\end{array}$

Hence, the force is 147N.

(b)

For the equilibrium under the torques,

$\begin{array}{rcl}{F}_t×1.75& =& \frac{F_r}{2}×20\\ F_t& =& \frac{147×20}{1.75}\\ & =& 1680\text{N}\end{array}$

The ratio of this force to her weight is,

$\begin{array}{rcl}\frac{2F_t}{W}& =& \frac{2×1680}{50×9.8}\\ & =& 6.86\end{array}$

Hence, the force is 1680N with ratio 6.86.

(c)

The work is,

$\begin{array}{rcl}work& =& W×0.240\\ & =& 50×9.8×0.240\\ & =& 117.6\text{J}\end{array}$

Hence, the work is 117.6 J.

(d)

The power is,

$\begin{array}{rcl}\frac{Totalwork}{t}& =& \frac{25×117.6}{60}\\ & =& 49\text{W}\end{array}$

Hence, the power is 49W.