Chapter 12.1, Problem 12.15P

(a)

To determine

Find the acceleration of block A for each system.

Answer to Problem 12.15P

The acceleration of block A for system 1 is $\underset{\_}{10.73{\text{ft/s}}^2}$.

The acceleration of block A for system 2 is $\underset{\_}{16.10{\text{ft/s}}^2}$.

The acceleration of block A for system 1 is $\underset{\_}{0.740{\text{ft/s}}^2}$.

Explanation of Solution

Calculation:

Sketch the general diagram of systems as shown in Figure (1).

Write total length of cable connecting block A and block B.

$y_A+y_B=\text{constant}$ (1)

Here,$y_A$ is the length of cable connecting block A and $y_B$ is the length of cable connecting block B.

Differentiate Equation (1) with respect to t to write velocity of the blocks.

$v_A+v_B=\text{0}$ (2)

Here, $v_A$ is the velocity of the block A and $v_B$ is the velocity of the block B.

Differentiate Equation (2) with respect to t to write acceleration of the blocks.

$\begin{array}{c}{a}_A+a_B=\text{0}\\ a_B=-a_A\end{array}$ (3)

First of all check the required static friction with static friction to maintain equilibrium.

Sketch the free body diagram and kinetic diagram of block A as shown in Figure (2).

Refer Figure (2).

Consider downward direction as positive.

Apply Newton’s law of motion along y-axis.

$\begin{array}{l}\Sigma F_y=m_y{a}_A\\ W_A-T=\frac{W_A}{g}{a}_A\\ T=W_A-\frac{W_A}{g}{a}_A\end{array}$ (4)

Here, T is the tension in the cable, $W_A$ is the weight of cable A, g is the acceleration due to gravity, and $a_A$ acceleration of block A.

Sketch the free body diagram and kinetic diagram of block B as shown in Figure (3).

Refer Figure (3).

Consider downward direction as positive.

Apply Newton’s law of motion along y-axis.

Find the equation of acceleration of block A.

$\begin{array}{l}\Sigma F_y=m_y{a}_B\\ P+W_B-T=\frac{W_B}{g}{a}_B\end{array}$ (5)

Here, T is the tension in the cable, $W_B$ is the weight of cable B, and $a_B$ acceleration of block B.

Substitute $W_A-\frac{W_A}{g}{a}_A$ for T and $-a_A$ for $a_B$

$\begin{array}{l}P+W_B-W_A+\frac{W_A}{g}{a}_A=\frac{W_B}{g}\left(-a_A\right)\\ \frac{Pg+W_{B}g-W_{A}g+W_A{a}_A}{g}=-\frac{W_B{a}_A}{g}\\ Pg+W_{B}g-W_{A}g+W_A{a}_A=-W_B{a}_A\\ W_A{a}_A+W_B{a}_A=\left(W_A-W_B-P\right)g\end{array}$

$\begin{array}{l}{a}_A\left(W_A+W_B\right)=\left(W_A-W_B-P\right)g\\ a_A=\frac{\left(W_A-W_B-P\right)g}{\left(W_A+W_B\right)}\end{array}$ (6)

The initial velocity of block A is zero.

Find the equation of velocity of block A using kinematics:

$v_A^2-{\left(v_A\right)}_0^2=2a_A\left[y_A-{\left(y_A\right)}_0\right]$

Here, ${\left(v_A\right)}_0$ is the initial velocity of block A and ${\left(y_A\right)}_0$ is the length of movement of block A at 0 second.

Substitute At $0$ for ${\left(v_A\right)}_0$ and 0 for At ${\left(y_A\right)}_0$.

$\begin{array}{l}{v}_A^2-{\left(v_A\right)}_0^2=2a_A\left[y_A-{\left(y_A\right)}_0\right]\\ v_A^2=2a_A{y}_A\\ v_A=\sqrt{2a_A{y}_A}\end{array}$ (7)

Find the equation of time required for block A to reach any velocity.

$\begin{array}{l}{v}_A=a_{A}t\\ t=\frac{v_A}{a_A}\end{array}$ (8)

Find the acceleration of block A $\left[{\left(a_A\right)}_1\right]$ for system 1 using equation 6:

${\left(a_A\right)}_1=\frac{\left(W_A+W_B-P\right)g}{\left(W_A+W_B\right)}$

Substitute 200 lb for $W_A$, 100 lb for $W_B$, 0 for P, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}{\left(a_A\right)}_1=\frac{\left(200-100-0\right)\left(32.2\right)}{\left(200+100\right)}\\ =10.73{\text{ft/s}}^2\end{array}$

Therefore, the acceleration of block A for system 1 is $\underset{\_}{10.73{\text{ft/s}}^2}$.

Find the acceleration of block A $\left[{\left(a_A\right)}_2\right]$ for system 2 using equation 6:

${\left(a_A\right)}_2=\frac{\left(W_A+W_B-P\right)g}{\left(W_A+W_B\right)}$

Substitute 200 lb for $W_A$, 0 for $W_B$, 100 lb for P, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}{\left(a_A\right)}_2=\frac{\left(200-0-100\right)\left(32.2\right)}{\left(200+0\right)}\\ =16.10{\text{ft/s}}^2\end{array}$

Therefore, the acceleration of block A for system 2 is $\underset{\_}{16.10{\text{ft/s}}^2}$.

Find the acceleration of block A $\left[{\left(a_A\right)}_3\right]$ for system 2 using equation 6:

${\left(a_A\right)}_3=\frac{\left(W_A+W_B-P\right)g}{\left(W_A+W_B\right)}$

Substitute 2200 lb for $W_A$, 2100 for $W_B$, 0 for P, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}{\left(a_A\right)}_3=\frac{\left(2200-2100-0\right)\left(32.2\right)}{\left(2200+2100\right)}\\ =0.749{\text{ft/s}}^2\end{array}$

Therefore, the acceleration of block A for system 2 is $\underset{\_}{0.749{\text{ft/s}}^2}$.

(b)

To determine

Find the velocity of block A for each system after it has moved through 10 ft

Answer to Problem 12.15P

The velocity of block A for system 1 after it has moved through 10 ft is $\underset{\_}{14.65\text{ft/s}}$.

The velocity of block A for system 2 after it has moved through 10 ft is $\underset{\_}{17.94\text{ft/s}}$.

The velocity of block A for system 3 after it has moved through 10 ft is $\underset{\_}{3.87\text{ft/s}}$.

Explanation of Solution

Calculation:

Find the velocity of block A for system 1 $\left[{\left(v_A\right)}_1\right]$ after it has moved through 10 ft using Equation (7):

${\left(v_A\right)}_1=\sqrt{2{\left(a_A\right)}_1{y}_A}$

Substitute $10.73{\text{ft/s}}^2$ for ${\left(a_A\right)}_1$ and 10 ft for $y_A$

$\begin{array}{c}{\left(v_A\right)}_1=\sqrt{2\times 10.73\times 10}\\ =14.65\text{ft/s}\end{array}$

Thus, the velocity of block A for system 1 after it has moved through 10 ft is $\underset{\_}{14.65\text{ft/s}}$.

Find the velocity of block A for system 2 $\left[{\left(v_A\right)}_2\right]$ after it has moved through 10 ft using Equation (7):

${\left(v_A\right)}_2=\sqrt{2{\left(a_A\right)}_2{y}_A}$

Substitute $16.10{\text{ft/s}}^2$ for ${\left(a_A\right)}_1$ and 10 ft for $y_A$

$\begin{array}{c}{\left(v_A\right)}_2=\sqrt{2\times 16.10\times 10}\\ =17.94{\text{ft/s}}^2\end{array}$

Thus, the velocity of block A for system 2 after it has moved through 10 ft is $\underset{\_}{17.94\text{ft/s}}$.

Find the velocity of block A for system 3 $\left[{\left(v_A\right)}_3\right]$ after it has moved through 10 ft using Equation (7):

${\left(v_A\right)}_3=\sqrt{2{\left(a_A\right)}_1{y}_A}$

Substitute $0.749{\text{ft/s}}^2$ for ${\left(a_A\right)}_1$ and 10 ft for $y_A$

$\begin{array}{c}{\left(v_A\right)}_1=\sqrt{2\times 0.749\times 10}\\ =3.87\text{ft/s}\end{array}$

Thus, the velocity of block A for system 3 after it has moved through 10 ft is $\underset{\_}{3.87\text{ft/s}}$.

(c)

To determine

Find the time required for block A to reach a velocity of 20 ft/s

Answer to Problem 12.15P

The time required for block A for system 1 to reach a velocity of 20 ft/s is $\underset{\_}{1.864\text{s}}$.

The time required for block A for system 2 to reach a velocity of 20 ft/s is $\underset{\_}{1.242\text{s}}$.

The time required for block A for system 3 to reach a velocity of 20 ft/s is $\underset{\_}{26.7\text{s}}$.

Explanation of Solution

Calculation:

Find the time required for block A for system 1 $\left[{\left(t_A\right)}_1\right]$ to reach a velocity of 20 ft/s using Equation (8):

${\left(t_A\right)}_1=\frac{v_A}{{\left(a_A\right)}_1}$

Substitute $20\text{ft/s}$ for $v_A$ and $10.73{\text{ft/s}}^2$ for ${\left(a_A\right)}_1$.

$\begin{array}{c}{\left(t_A\right)}_1=\frac{20}{10.73}\\ =1.864\text{s}\end{array}$

Thus, the time of required for block A for system 1 to reach a velocity of 20 ft/s is $\underset{\_}{1.864\text{s}}$..

Find the time required for block A for system 2 $\left[{\left(t_A\right)}_2\right]$ to reach a velocity of 20 ft/s using Equation (8):

${\left(t_A\right)}_2=\frac{v_A}{{\left(a_A\right)}_2}$

Substitute $20\text{ft/s}$ for $v_A$ and $16.10{\text{ft/s}}^2$ for ${\left(a_A\right)}_2$.

$\begin{array}{c}{\left(t_A\right)}_2=\frac{20}{16.10}\\ =1.242\text{s}\end{array}$

Thus, the time of required for block A for system 2 to reach a velocity of 20 ft/s is $\underset{\_}{1.242\text{s}}$..

Find the time required for block A for system 3 $\left[{\left(t_A\right)}_3\right]$ to reach a velocity of 20 ft/s using Equation (8):

${\left(t_A\right)}_3=\frac{v_A}{{\left(a_A\right)}_3}$

Substitute $20\text{ft/s}$ for $v_A$ and $0.749{\text{ft/s}}^2$ for ${\left(a_A\right)}_3$.

$\begin{array}{c}{\left(t_A\right)}_1=\frac{20}{0.749}\\ =26.7\text{s}\end{array}$

Thus, the time of required for block A for system 3 to reach a velocity of 20 ft/s is $\underset{\_}{26.7\text{s}}$..