Chapter 12.1, Problem 12.29P

To determine

(a)

The acceleration of the panel and the tension in the cord after the system released from rest.

Answer to Problem 12.29P

Acceleration of the panel $a=8.944ft/s^2$

Tension in the cord $T=18.056lb$

Explanation of Solution

Given information:

Weight of panel $=40lb$

Weight of the counterweight $=25lb$

First, we draw the free body diagram and kinetic diagram of the panel:

F = force exerted by counterweight

Weight $W=mg$

$m=\frac{W}{g}$

Now, force $F=ma$

$F=\frac{W}{g}\times a$

$\sum F_x=ma$

$T-F=ma$

$T-F=\frac{40}{g}a\_\_\_\_\_\_\_(1)$

Now, computing the forces for counterweight A,

For this we draw the free body and kinetic diagram of the counterweight A;

Hence, from the above diagram it is clear that the acceleration component of counterweight A has two components.

Now, $\begin{array}{l}{a}_A=a_P+a_{A/P}\\ a_A=a\to +a\downarrow \end{array}$

$\sum F_x=m{a}_x$

$F=\frac{25}{g}a\_\_\_\_\_\_(2)$

$\sum F_g=m{a}_g$

$\begin{array}{l}W-T=\frac{25}{g}a\\ 25-T=\frac{25}{g}a\_\_\_\_\_\_\_\_(3)\end{array}$

Now, adding equation (1), (2) and (3);

$\begin{array}{l}T-F+F+25-T=\frac{40+25+25}{g}a\\ \Rightarrow 25=\frac{90}{g}a\\ \Rightarrow a=\frac{25}{90}g\\ \Rightarrow a=\frac{25}{90}\left(32.2\right)\\ \Rightarrow a=8.944ft/s^2\end{array}$

Now, Tension in the cord from equation (3),

$25-T=\frac{25}{g}a$

$\begin{array}{l}\Rightarrow 25-T=\frac{25}{32.2}\times 8.944\\ \Rightarrow 25-T=6.944\\ \Rightarrow T=18.056lb\end{array}$

To determine

(b)

The acceleration of the panel and the tension in the cord after the system released from rest.

Answer to Problem 12.29P

Acceleration of the panel $a=12.38ft/s^2$

Tension in the cord $T=15.38lb$

Explanation of Solution

Given information:

Weight of panel $=40lb$

Weight of the counterweight $=25lb$

First, we draw the free body diagram and kinetic diagram of the panel:

F = force exerted by counterweight.

Weight $W=mg$

$m=\frac{W}{g}$

Now, force $F=ma$

$F=\frac{W}{g}\times a$

$\sum F_y=ma$

$T=ma$

$T=\frac{40}{g}a\_\_\_\_\_\_\_(1)$

Now, computing the forces for counterweight A,

For this we draw the free body and kinetic diagram of the counterweight A;

$\sum F_y=ma$

$25-T=\frac{25}{g}a\_\_\_\_\_\_(2)$

Now, adding equation (1) and (2);

$\begin{array}{l}T+25-T=\frac{40+25}{g}a\\ \Rightarrow 25=\frac{65}{g}a\\ \Rightarrow a=\frac{25}{65}g\\ \Rightarrow a=\frac{25}{65}\left(32.2\right)\\ \Rightarrow a=12.38ft/s^2\end{array}$

Now, Tension in the cord from equation (1),

$T=\frac{40}{g}a$

$\begin{array}{l}\Rightarrow T=\frac{40}{32.2}\times 12.38\\ \Rightarrow T=15.38lb\end{array}$

To determine

(c)

The acceleration of the panel and the tension in the cord after the system released from rest.

Answer to Problem 12.29P

Acceleration of the panel $a=12.38ft/s^2$

Tension in the cord $T=15.38lb$

Explanation of Solution

Given information:

Weight of panel $=40lb$

Weight of the counterweight $=25lb$

First, we draw the free body diagram and kinetic diagram of the panel:

F = force exerted by counterweight

Weight $W=mg$

$m=\frac{W}{g}$

Now, force $F=ma$

$F=\frac{W}{g}\times a$

Since, panel is accelerated to the left, there is no force exerted by the panel on counterweight hence;

$\sum F_x=ma$

$T=ma$

$T=\frac{40}{g}a\_\_\_\_\_\_\_(1)$

Now, computing the forces for counterweight A,

For this, we draw the free body and kinetic diagram of the counterweight A;

$\sum F_y=ma$

$25-T=\frac{25}{g}a\_\_\_\_\_\_(2)$

Now, adding equation (1) and (2);

$\begin{array}{l}T+25-T=\frac{40+25}{g}a\\ \Rightarrow 25=\frac{65}{g}a\\ \Rightarrow a=\frac{25}{65}g\\ \Rightarrow a=\frac{25}{65}\left(32.2\right)\\ \Rightarrow a=12.38ft/s^2\end{array}$

Now, Tension in the cord from equation (1),

$T=\frac{40}{g}a$

$\begin{array}{l}\Rightarrow T=\frac{40}{32.2}\times 12.38\\ \Rightarrow T=15.38lb\end{array}$