Chapter 13.1, Problem 13.23P

(a)

To determine

The velocity of block B $\left(v_B\right)$ after block A has moved 2 m.

Answer to Problem 13.23P

The velocity of block B $\left(v_B\right)$ after block A has moved 2 m is $\underset{\_}{1.218\text{m}/\text{s}(\leftarrow )}$.

Explanation of Solution

Given information:

The mass of the block A $\left(m_A\right)$ is 30 kg.

The mass of the block B $\left(m_B\right)$ is 25 kg.

The force (P) acting at block A is 250 N.

The coefficient of static friction between block A and horizontal surface $\left({\mu }_s\right)$ is 0.25.

The coefficient of kinetic friction between block A and horizontal surface $\left({\mu }_k\right)$ is 0.20.

Calculation:

Show the system with the distance as in Figure (1).

When the block A moves one unit left, block B moves 3 units upwards,

Write the expression for the constraint of the cable from Figure (1) as follows:

$\begin{array}{l}{x}_A+3y_B=0\\ \Delta x_A+3\Delta y_B=0\\ v_A+3v_B=0\end{array}$

Here, $x_A$ and $y_B$ are the distances, $\Delta x_A$ and $\Delta y_B$ are change in distances, $v_A$ is the velocity of block A and $v_B$ is the velocity of block B.

Show the free body diagram of block B with all the forces acting on it as in Figure (2).

Check the equilibrium position of the blocks to verify whether the blocks move.

From Figure (2), for block B to remain in equilibrium, the net resultant force acting on the block B should be zero.

$\begin{array}{l}\text{Net force}\text{on }B\text{=0}\\ 3F-m_{B}g=0\\ F=\frac{m_{B}g}{3}\end{array}$

Here, g is the acceleration due to gravity and F is the tension in the cable.

Substitute 25 kg for $m_B$, and 9.81 $\text{m}/{\text{s}}^2$ for g.

$\begin{array}{c}F=\frac{\left(25\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)}{3}\\ =81.75\text{N}\end{array}$

Show the free body diagram of block B with all the forces acting on it as in Figure (3).

Calculate the net forces acting on the block A in Y-axis direction using the relation:

$\begin{array}{l}{N}_A-m_{A}g=0\\ N_A=m_{A}g\end{array}$

Here, $N_A$ is the normal reaction acting on block A.

Substitute 30 kg for $m_A$ and $9.81\text{m}/{\text{s}}^2$ for g.

$\begin{array}{c}{N}_A=\left(30\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =294.3\text{N}\end{array}$

Calculate the net forces acting on block A in X-axis direction using the relation:

$\text{}{\left(\text{Net force}\right)}_X=P-F$

Substitute 250 N for $P$ and $81.75$ N for F.

$\begin{array}{c}{\left(\text{Net force}\right)}_X=250\text{N}-81.75\text{N}\\ =168.25\text{N}\end{array}$

Calculate the available static friction acting on block A $\left(F_S\right)$ using the formula:

$F_S={\mu }_s{N}_A$

Substitute 0.25 for ${\mu }_s$ and $294.3$ N for $N_A$.

$\begin{array}{c}{F}_S=\left(0.25\right)\left(294.3\text{N}\right)\\ =73.57\text{N}\end{array}$

Since ${\left(\text{Net force}\right)}_X>F_S$, the blocks move. Hence, the kinetic friction comes into action.

Show the free body diagram of block B with kinetic frictional force acting on it as in Figure (4).

Consider the block A.

The kinetic energy of block A ${\left(T_1\right)}_A$ is zero as it rest initially.

The expression for the final kinetic energy of the block A $\left[{\left(T_2\right)}_A\right]$ after A has moved through a distance of 2 m as follows;

${\left(T_2\right)}_A=\frac{1}{2}{m}_A{v}_A^2$

Write the expression for the kinetic frictional force acting on the block A $\left(F_A\right)$ during sliding as follows:

$F_A={\mu }_k{N}_A$

Write the expression for the work done by the block A $\left[{\left(U_{1-2}\right)}_A\right]$ in moving through a distance of $\Delta x_A$ as follows:

${\left(U_{1-2}\right)}_A=\left(P-F_A-F\right)\left(\Delta x_A\right)$

Apply the principle of work and energy to block A.

Work and energy principle states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Write the expression for the work and energy principle as follows:

${\left(T_1\right)}_A+{\left(U_{1-2}\right)}_A={\left(T_2\right)}_A$

Substitute 0 for ${\left(T_1\right)}_A$, $\left(P-F_A-F\right)\left(\Delta x_A\right)$ for ${\left(U_{1-2}\right)}_A$, and $\frac{1}{2}{m}_A{v}_A^2$ for ${\left(T_2\right)}_A$.

$\begin{array}{l}{\left(T_1\right)}_A+{\left(U_{1-2}\right)}_A={\left(T_2\right)}_A\\ 0+\left(P-F_A-F\right)\left(\Delta x_A\right)=\frac{1}{2}{m}_A{v}_A^2\\ \left(P-F_A-F\right)\left(\Delta x_A\right)=\frac{1}{2}{m}_A{v}_A^2\end{array}$

Substitute 250 N for $P$, 30 kg for $m_A$, 2 m for $\Delta x_A$, 58.86 N for $F_A$, and $3v_B$ for $v_A$.

$\begin{array}{l}\left(250-58.86-F\right)\left(2\right)=\frac{1}{2}\left(30\right){\left(3v_B\right)}^2\\ 382.28-2F=135v_B^2\end{array}$ (1)

Consider the block B.

The kinetic energy of block B ${\left(T_1\right)}_B$ is zero as it rest initially.

The expression for the final kinetic energy $\left[{\left(T_2\right)}_B\right]$ of the block B after A has moved through a distance of 2 m.

${\left(T_2\right)}_B=\frac{1}{2}{m}_B{v}_B^2$

Write the expression the work done by the block B in moving through a distance of $\Delta y_B$ as follows:

$\begin{array}{c}{\left(U_{1-2}\right)}_B=\left\{\begin{array}{l}\left(\text{Net force acting on }B\right)\\ \left(\text{distance through which }B\text{ moves}\right)\end{array}\right\}\\ =\left(3F-W_B\right)\left(-\Delta y_B\right)\end{array}$

Apply the principle of work and energy to the block B.

According to the work and energy principle,

${\left(T_1\right)}_B+{\left(U_{1-2}\right)}_B={\left(T_2\right)}_B$

Substitute 0 for ${\left(T_1\right)}_B$, $\frac{1}{2}{m}_B{v}_B^2$ for ${\left(T_2\right)}_B$, and $\left(3F-W_B\right)\left(-\Delta y_B\right)$ for ${\left(U_{1-2}\right)}_B$.

$\begin{array}{l}{\left(T_1\right)}_B+{\left(U_{1-2}\right)}_B={\left(T_2\right)}_B\\ 0+\left(3F-W_B\right)\left(-\Delta y_B\right)=\frac{1}{2}{m}_B{v}_B^2\\ -\left(3F-m_{B}g\right)\left(\Delta y_B\right)=\frac{1}{2}{m}_B{v}_B^2\end{array}$

Substitute 25 kg for $m_B$, 9.81 $\text{m}/{\text{s}}^2$ for $g$, and 2/3 for $\Delta y_B$.

$\begin{array}{l}\left(3F-\left(25\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\right)\left(\frac{2}{3}\right)=\frac{1}{2}\left(25\text{kg}\right)v_B^2\\ \left(3F-245.25\right)\left(\frac{2}{3}\right)=12.5v_B^2\\ 2F-163.5=12.5v_B^2\end{array}$ (2)

Add equation (2) and equation (4) to eliminate $F$.

$\begin{array}{l}382.28-163.5=147.5v_B^2\\ v_B^2=\frac{218.78}{147.5}\\ v_B=\sqrt{1.4833}\\ v_B=1.218\text{m}/\text{s}(\leftarrow )\end{array}$

Therefore, the velocity of block B $\left(v_B\right)$ after block A has moved 2 m is $\underset{\_}{1.218\text{m}/\text{s}(\leftarrow )}$.

(b)

To determine

The tension (F) in the cable.

Answer to Problem 13.23P

The tension (F) in the cable is $\underset{\_}{91.0\text{N}}$.

Explanation of Solution

Given information:

The mass of the block A $\left(m_A\right)$ is 30 kg.

The mass of the block B $\left(m_B\right)$ is 25 kg.

The force (P) acting at block A is 250 N.

The coefficient of static friction between block A and horizontal surface $\left({\mu }_s\right)$ is 0.25.

The coefficient of kinetic friction between block A and horizontal surface $\left({\mu }_k\right)$ is 0.20.

Calculation:

Calculate the tension (F) in the cable:

Substitute $1.218\text{m}/\text{s}$ for $v_B$ in equation (2).

$\begin{array}{l}2F-163.5=12.5v_B^2\\ 2F-163.5=\left(12.5\right){\left(1.218\right)}^2\\ 2F=18.541+163.5\\ F=91.0\text{N}\end{array}$

Therefore, the tension (F) in the cable is $\underset{\_}{91.0\text{N}}$.