Question

Verify that the force in the elbow joint inExample$9.4$is $407\mathrm{N}$, as stated in the text.

Short answer

The force in the elbow joint is $407\mathrm{N}$.

Step-by-step solution

Force

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

Free-body diagram of the given system

The diagram is given below:

Calculation of the force

The torque around a point of equilibrium is zero. So, we write,

$\begin{array}{rl}{F}_E{r}_1& =w_a\left(r_2-r_1\right)+w_b\left(r_3-r_1\right)\\ F_E& =\frac{w_a\left(r_2-r_1\right)+w_b\left(r_3-r_1\right)}{r_1}\\ F_E& =w_a\left(\frac{r_2}{r_1}-1\right)+w_b\left(\frac{r_3}{r_1}-1\right)\end{array}$

Here, the forearm's weight is,

$\begin{array}{rl}{w}_a& =2.5\times 9.8\\ & =24.5N\end{array}$

The load's total weight is,

$\begin{array}{rl}{w}_b& =4\times 9.8\\ & =39.2N\end{array}$

The distance between the biceps and the elbow where they exert force is$r_1=4\mathrm{c}\mathrm{m}$.

The force exerted by the forearm's weight on the elbow is$r_2=16\mathrm{c}\mathrm{m}$.

The load and elbow are separated by a distance of$r_3=38\mathrm{c}\mathrm{m}$.

For all of these reasons,

$\begin{array}{rl}{F}_E& =24.5\times \left(\frac{16}{4}-1\right)+39.2\times \left(\frac{38}{4}-1\right)\\ \approx 407N\end{array}$

Hence, it is proved.