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Q 29PE

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Found in: Page 318

### College Physics (Urone)

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

# Question: A device for exercising the upper leg muscle is shown in Figure , together with a schematic representation of an equivalent lever system. Calculate the force exerted by the upper leg muscle to lift the mass at a constant speed. Explicitly show how you follow the steps in the Problem- Solving Strategy for static equilibrium in Applications of Statistics, Including Problem-Solving Strategies.

The force exerted by the upper leg muscle is 1715 N.

See the step by step solution

## Step 1: Definition of the 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: Understanding the free body diagram

The diagram for the leg muscles is shown below:

The perpendicular distance of the weight from the pivot point is ${\text{r}}_{\perp }=\text{35}\text{cm}$.

The perpendicular distance of the applied force from the pivot point is ${{\text{r}}_{\perp }}^{\text{'}}=\text{2.0}\text{cm}$.

## Step 3: Calculation of the force

As the upper leg muscle lifts the mass at a constant speed, the net force is zero. So, we can write,

$\begin{array}{rcl}T& =& mg\\ & =& 10×9.8\\ & =& 98\text{N}\end{array}$

The torque about the pivot point is zero. So, we can write,

$\begin{array}{rcl}{F}_{M}{r\text{'}}_{\perp }-T{r}_{\perp }& =& 0\\ {F}_{M}& =& \frac{T{r}_{\perp }}{{r\text{'}}_{\perp }}\\ & =& \frac{98×35}{2}\\ & =& 1715\text{N}\end{array}$

Hence, the force is 1715 N.