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Q 37PE

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Found in: Page 319

### College Physics (Urone)

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

# Question: (a) What force should the woman in Figure 9.45 exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75cm, and she exerts force on the floor at a horizontal distance of 20.0cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) How much work does she do if her center of mass rises 0.240m? (d) What is her useful power output if she does 25 pushups in one minute?

The force exerted on the floor is 147N.

The force on the triceps muscle is 1680N and the ratio of the force to her weight is 6.86.

The work is 117.6J.

The power output is 49W.

See the step by step solution

## Step 1: Define 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: Free-body diagram

The diagram for the forearm is shown below:

## Step 3: Calculation of the force exerted on the floor

(a)

Applying the condition of equilibrium under torque we can write,

${F}_{r}×1.50=W×0.9\phantom{\rule{0ex}{0ex}}{F}_{r}=\frac{W×0.9}{1.50}$

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.

## Step 4: Calculation of the force on the triceps muscle

(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,

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

Hence, the force is 1680N with ratio 6.86.

## Step 5: Calculation of the work

(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.

## Step 6: Calculation of the power output

(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.