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Q 37PE

Expert-verifiedFound in: Page 319

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.

**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,

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

For $W=50\times 9.8\text{N}$ we get,

$\begin{array}{rcl}{F}_{r}& =& \frac{50\times 9.8\times 0.9}{1.50}\\ & =& 294\text{N}\end{array}$

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,

$\begin{array}{rcl}{F}_{t}\times 1.75& =& \frac{{F}_{r}}{2}\times 20\\ {F}_{t}& =& \frac{147\times 20}{1.75}\\ & =& 1680\text{N}\end{array}$

The ratio of this force to her weight is,

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

Hence, the force is 1680N with ratio 6.86.

(c)

The work is,

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

Hence, the work is 117.6 J.

(d)

The power is,

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

Hence, the power is 49W.

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