Chapter 9, Problem 37PE

To determine

(a)

The force exerted on the floor by each hand to do pushups.

Answer to Problem 37PE

The force exerted on the floor by each hand to do pushups is $147\text{N}$.

Explanation of Solution

Given:

The mass of the woman is $m_1=50\text{kg}$.

The acceleration due to gravity is $g=9.8\text{m}/{\text{s}}^{\text{2}}$.

Formula used:

The free body diagram is as shown below.

  

  Figure (1)

Apply second condition of equilibrium; the net torque around the pivot is zero.The force is same on each arm.

The relation between weight and mass of the woman is

  $w=mg$

By second condition of equilibrium,

  $w_1{x}_1=F_{\text{reaction}}{x}_2$

The force on each arm is

  $F_{\text{arm}}=\frac{F_{\text{reaction}}}{2}$

Calculation:

The weight of the woman is calculated as

  $\begin{array}{c}w=mg\\ =\left(50\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =490\text{N}\end{array}$

The reaction force is calculated as

  $\begin{array}{c}{w}_1{x}_1=F_{\text{reaction}}{x}_2\\ \left(490\text{N}\right)\left(0.9\text{m}\right)=F_{\text{reaction}}(1.5\text{m})\\ F_{\text{reaction}}=\left(490\text{N}\right)\left(\frac{0.9\text{m}}{1.5\text{m}}\right)\\ =294\text{N}\end{array}$

The force on each arm is calculated as

  $\begin{array}{c}{F}_{\text{arm}}=\frac{F_{\text{reaction}}}{2}\\ =\frac{294\text{N}}{2}\\ =147\text{N}\end{array}$

Conclusion:

The force exerted on the floor by each hand to do the pushups is $147\text{N}$.

To determine

(b)

The magnitude of force in each triceps muscle and the ratio of forces between the triceps muscle to the weight.

Answer to Problem 37PE

The magnitude of force in each triceps muscle is $1680\text{N}$ and the ratio of forces between the triceps muscle to the weight is $3.4$.

Explanation of Solution

Given:

The reaction force is $F_{\text{reaction}}=147\text{N}$.

Formula used:

The torque around the pivot point gives the relation between the forces in triceps muscles $F_{\text{t}}$ and the reaction force.

  $F_{\text{t}}\left(1.75\text{cm}\right)=F_{\text{reaction}}\left(20\text{cm}\right)$

The ratio between the forces in the triceps muscle to the weight is

  $R=\frac{F_{\text{t}}}{w}$

Calculation:

The force in each triceps muscle is calculated as

  $\begin{array}{c}{F}_{\text{t}}\left(1.75\text{cm}\right)=F_{\text{reaction}}\left(20\text{cm}\right)\\ F_{\text{t}}\left(1.75\text{cm}\right)=\left(147\text{N}\right)\left(20\text{cm}\right)\\ F_{\text{t}}=\left(147\text{N}\right)\left(\frac{20\text{cm}}{1.75\text{cm}}\right)\\ =1680\text{N}\end{array}$

The ratio is calculated as

  $\begin{array}{c}R=\frac{F_{\text{t}}}{w}\\ =\frac{1680\text{N}}{490\text{N}}\\ =3.4\end{array}$

Conclusion:

The magnitude of force in each triceps muscle is $1680\text{N}$ and the ratio of forces between the triceps muscle to the weight is $3.4$.

To determine

(c)

The work done by the women.

Answer to Problem 37PE

The work done by the women is $118\text{J}$.

Explanation of Solution

Given:

The height from the floor when the centre of mass rises her body is $h=0.24\text{m}$.

Formula used:

The work done by the woman to raise her centre of mass is given by

  $W=mgh$

Calculation:

The work done by the woman is calculated as

  $\begin{array}{c}W=mgh\\ =\left(50\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0.24\text{m}\right)\\ =117.6\text{J}\\ \approx 118\text{J}\end{array}$

Conclusion:

The work done by the women is $118\text{J}$.

To determine

(d)

The useful power of the women.

Answer to Problem 37PE

The useful power output of the women is $49\text{W}$.

Explanation of Solution

Given:

The work done by the woman is $W=118\text{J}$.

The number pushups the woman do in one minute is $n=25$.

Formula used:

The time taken by the woman to do one pushup is

  $t=\frac{1\text{min}}{n}$

The relation between the work done, the time taken to complete the work and the power is

  $P=\frac{W}{t}$

Calculation:

The time taken by the woman to do one pushup is calculated as

  $\begin{array}{c}t=\frac{1\text{min}}{n}\\ =\frac{1\text{min}}{25}\\ =\left(0.04\min \times \frac{60\text{s}}{1\text{min}}\right)\\ =2.4\text{s}\end{array}$

The power output is calculated as

  $\begin{array}{c}P=\frac{W}{t}\\ =\frac{118\text{J}}{2.4\text{s}}\\ =49\text{W}\end{array}$

Conclusion:

The useful power output of the women is $49\text{W}$.