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Q26 PE

Expert-verified
Found in: Page 317

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Verify that the force in the elbow joint in Example $${\rm{9}}{\rm{.4}}$$ is $${\rm{407}}\;{\rm{N}}$$, as stated in the text.

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

See the step by step solution

## Step 1: 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.

## Step 2: Free-body diagram of the given system

The diagram is given below:

## Step 3: Calculation of the force

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

\begin{align}{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{align}

Here, the forearm's weight is,

\begin{align}{w_a} &= 2.5 \times 9.8\\ &= 24.5\;N\end{align}

The load's total weight is,

\begin{align}{w_b} &= 4 \times 9.8\\ &= 39.2\;N\end{align}

The distance between the biceps and the elbow where they exert force is $${r_1} = 4\;{\rm{cm}}$$.

The force exerted by the forearm's weight on the elbow is $${r_2} = 16\;{\rm{cm}}$$.

The load and elbow are separated by a distance of $${r_3} = 38\;{\rm{cm}}$$.

For all of these reasons,

\begin{align}{F_E} &= 24.5 \times \left( {\frac{{16}}{4} - 1} \right) + 39.2 \times \left( {\frac{{38}}{4} - 1} \right)\\ \approx 407\;N\end{align}

Hence, it is proved.