Chapter 5, Problem 41P

(a)

To determine

The force exerted by the chin-up bar on the person’s body at $t=0$.

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at $t=0$ is $\text{646 N up}$

Explanation of Solution

The slope of the speed versus time graph gives the magnitude of the acceleration. Consider positive slope of speed versus time graph as positive acceleration and negative slope of speed versus time graph gives negative acceleration. When the slope is zero the acceleration is zero.

According to Newton’s second law,

  $m{a}_y=F_{\text{bar}}-mg$

Here, $F_{\text{bar}}$ is the force exerted by the chin-up bar on the body, $m$ is the mass of the person, $g$ is the acceleration due to gravity, and $a_y$ is the vertical acceleration.

Rewrite the above equation in terms of $F_{\text{bar}}$

  $\begin{array}{c}{F}_{\text{bar}}=mg+m{a}_y\\ =m\left(g+a_y\right)\end{array}$                                                                                                  (I)

Write the expression to calculate the acceleration.

    $a_y=\frac{s_f-s_i}{t_f-t_i}$                                                                                                    (II)

Here, $s_i$ is the initial speed, $s_f$ is the final speed, $t_i$ is the initial time, and $t_f$ is the final time.

From the graph the slope at time $t=0$.

$s_f=s_2$, $s_i=s_1$, $t_f=t_2$, and $t_i=t_1$

Conclusion:

Substitute $6\text{cm}/\text{s}$ for $s_i$, $18\text{cm}/\text{s}$ for $s_f$, $0.2\text{s}$ for $t_i$, $0.6\text{s}$ for $t_f$ in (II) to find $a_y$.

    $\begin{array}{c}{a}_y=\frac{18\text{cm}/\text{s}-6\text{cm}/\text{s}}{0.6\text{s}-0.2\text{s}}\\ =\frac{12\text{cm}/\text{s}}{0.4\text{s}}\\ =30{\text{cm/s}}^2\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.3{\text{ m/s}}^{\text{2}}\end{array}$ s

Substitute $0.3{\text{ m/s}}^{\text{2}}$ for $a_y$ , $9.80{\text{ m/s}}^{\text{2}}$ for $g$ , and $64.0\text{kg}$ for $m$ in (I) to find $F_{\text{bar}}$.

    $\begin{array}{c}{F}_{\text{bar}}=\left(64.0\text{ kg}\right)\left(9.80{\text{ m/s}}^{\text{2}}+0.3{\text{ m/s}}^{\text{2}}\right)\\ =\text{646}\text{.4 N up}\\ \approx \text{646 N up}\end{array}$

Therefore, the force exerted by the chin-up bar on the person’s body at $t=0$ is $\text{646 N up}$

(b)

To determine

The force exerted by the chin-up bar on the person’s body at $t=0.5\text{s}$.

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at $t=0.5\text{s}$ is $\text{646 N up}$

Explanation of Solution

The slope of the speed versus time graph gives the magnitude of the acceleration. The slope of the graph at 0.5 s is same as that of 0 s. That is the graph is a straight line at these moment. If the graph is a straight line, the acceleration is same.

The mass of the person does not change and the acceleration is same as that at $t=0\text{s}$

Conclusion:

Therefore, the force exerted by the chin-up bar on the person’s body at $t=0.5\text{s}$ is $\text{646 N up}$

(c)

To determine

The force exerted by the chin-up bar on the person’s body at $t=1.1\text{s}$.

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at $t=1.1\text{s}$ is $627\text{N}\text{up}$

Explanation of Solution

The graph below is the speed versus time graph

From the graph it is evident that at $t=1.1\text{s}$, the slope is zero. Thus, the acceleration is zero.

Conclusion:

Substitute $0{\text{ m/s}}^{\text{2}}$ for $a_y$ , $9.80{\text{ m/s}}^{\text{2}}$ for $g$ , and $64.0\text{kg}$ for $m$ in (I) to find $F_{\text{bar}}$.

    $\begin{array}{c}{F}_{\text{bar}}=\left(64.0\text{ kg}\right)\left(9.80{\text{ m/s}}^{\text{2}}+0{\text{ m/s}}^{\text{2}}\right)\\ =627.2\text{ N}\text{up}\\ \approx 627\text{ N}\text{up}\end{array}$

Therefore, the force exerted by the chin-up bar on the person’s body at $t=1.1\text{s}$ is $627\text{ N}\text{up}$

(d)

To determine

The force exerted by the chin-up bar on the person’s body at $t=1.6\text{s}$.

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at $t=1.6\text{s}$ is $593\text{ N up}$

Explanation of Solution

The graph below is the speed versus time graph.

From the graph the slope at time $t=1.6\text{s}$.

$s_f=s_4$, $s_i=s_3$, $t_f=t_4$, and $t_i=t_3$

Conclusion:

Substitute $11\text{cm}/\text{s}$ for $s_i$, $2\text{cm}/\text{s}$ for $s_f$, $1.5\text{s}$ for $t_i$, $1.67\text{s}$ for $t_f$ in (II) to find $a_y$.

    $\begin{array}{c}{a}_y=\frac{2\text{cm}/\text{s}-11\text{cm}/\text{s}}{1.67\text{s}-1.5\text{s}}\\ =\frac{-9\text{cm}/\text{s}}{0.17\text{s}}\\ =-53{\text{cm/s}}^2\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =-0.53{\text{ m/s}}^{\text{2}}\end{array}$

Substitute $-0.53{\text{ m/s}}^{\text{2}}$ for $a_y$ , $9.80{\text{ m/s}}^{\text{2}}$ for $g$ , and $64.0\text{kg}$ for $m$ in (I) to find $F_{\text{bar}}$.

    $\begin{array}{c}{F}_{\text{bar}}=\left(64.0\text{ kg}\right)\left(\text{9}{\text{.80 m/s}}^{\text{2}}-0.53{\text{ m/s}}^{\text{2}}\right)\\ =593\text{ N up}\end{array}$

Therefore, the force exerted by the chin-up bar on the person’s body at $t=1.6\text{s}$ is $593\text{ N up}$