Chapter 12.1, Problem 12.28P

Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.

Fig. P12.28

To determine

Find the magnitude of the force P.

Answer to Problem 12.28P

The magnitude of the force P is $\underset{\_}{10.00\text{N}}$.

Explanation of Solution

Given information:

The mass of block A $\left(m_A\right)$ is 10 kg.

The mass of blocks B $\left(m_B\right)$ is 5 kg.

The mass of blocks C $\left(m_C\right)$ is 5 kg.

The initially block B $\left[{\left(y_B\right)}_0=0\right]$ is at rest.

The distance of movement for block B $\left(y_B\right)$ is 3 m.

The time taken by block B to move 3 m (t) is 2 s.

Calculation:

Sketch the system with position of blocks as shown in Figure 1.

Write the general equation of weight (W):

$W=mg$

Here, m is the mass, g is the acceleration due to gravity.

Refer Figure (1).

Consider the position of y be positive downward.

Consider the constraint of cord AD.

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

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

Here, $y_A$ is the length of cord connecting block A and $y_D$ is the length of cord connecting block D.

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

$v_A+v_D=\text{constant}$ (2)

Here, $v_A$ is the velocity of the block A and $v_D$ is the velocity of the block D.

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

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

Here, $a_A$ is the acceleration of block A and $a_D$ acceleration of block D.

Consider the constraint of cord BC.

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

$\begin{array}{l}\left(y_B-y_D\right)+\left(y_C-y_D\right)=\text{constant}\\ y_B+y_C-2y_D=\text{constant}\end{array}$ (4)

Here,$y_B-y_D$ is the length of cord connecting block B and $y_C-y_D$ is the length of cord connecting block C.

Differentiate Equation (4) with respect to t to write velocity of the blocks.

$v_B+v_C-2v_D=\text{constant}$ (5)

Here, $v_B$ is the velocity of the block B and $v_C$ is the velocity of the block C.

Differentiate Equation (5) with respect to t to write acceleration of the blocks.

$a_B+a_C-2a_D=\text{0}$ (6)

Here, $a_B$ is the acceleration of block B and $a_C$ acceleration of block C.

Substitute $-a_A$ for $a_D$.

$\begin{array}{l}{a}_B+a_C-2\left(-a_A\right)=\text{0}\\ 2a_A+a_B+a_C=0\end{array}$ (7)

The motion of blocks is uniform.

Find the acceleration of block B $\left(a_B\right)$ using general kinematic equation:

$y_B={\left(y_B\right)}_0+{\left(v_B\right)}_{0}t+\frac{1}{2}{a}_B{t}^2$

Here, ${\left(v_B\right)}_0$ is the initial velocity of block B.

Substitute $0$ for ${\left(v_B\right)}_0$.

$\begin{array}{l}{y}_B={\left(y_B\right)}_0+0+\frac{1}{2}{a}_B{t}^2\\ \frac{1}{2}{a}_B{t}^2=y_B-{\left(y_B\right)}_0\\ a_B=\frac{2\left[y_B-{\left(y_B\right)}_0\right]}{t^2}\end{array}$

Substitute 3 m for $y_B$, 0 for ${\left(y_B\right)}_0$, and 2 s for t.

$\begin{array}{c}{a}_B=\frac{2\left[3-0\right]}{2^2}\\ =1.5{\text{m/s}}^2\end{array}$

Sketch the free body diagram of pulley D as shown in Figure 2.

Refer Figure (2),

Consider equilibrium along y-axis.

$\begin{array}{l}\Sigma F_y=0\\ 2T_{BC}-T_{AD}=0\\ T_{AD}=2T_{BC}\end{array}$ (8)

Here, $T_{BC}$ is the tension in the cord BC and $T_{AD}$ is the tension in the cord AD.

Sketch the free body diagram of block A as shown in Figure (3).

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

$\begin{array}{l}\Sigma F_y=m_A{a}_A\\ W_A-T_{AD}=m_A{a}_A\\ a_A=\frac{W_A-T_{AD}}{m_A}\end{array}$

Here, $W_A$ is the weight of block A.

Substitute $m_{A}g$ for $W_A$ and $2T_{BC}$ for $T_{AD}$.

$a_A=\frac{m_{A}g-2T_{BC}}{m_A}$ (9)

Sketch the free body diagram of block C as shown in Figure (4).

Refer Figure (4),

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

$\begin{array}{l}\Sigma F_y=m_C{a}_C\\ W_C-T_{BC}=m_C{a}_C\\ a_C=\frac{W_C-T_{BC}}{m_C}\end{array}$

Here, $W_C$ is the weight of block C.

Substitute $m_{C}g$ for $W_C$.

$a_C=\frac{m_{C}g-T_{BC}}{m_C}$ (10)

Find the tension in the cord BC using Equation (7).

$2a_A+a_B+a_C=0$

Substitute $\frac{m_{A}g-2T_{BC}}{m_A}$ for $a_A$, $1.5{\text{m/s}}^2$ for $a_B$, and $\frac{m_{C}g-T_{BC}}{m_C}$ for $a_C$.

$2\left(\frac{m_{A}g-2T_{BC}}{m_A}\right)+1.5+\frac{m_{C}g-T_{BC}}{m_C}=0$

Substitute 10 kg for $m_A$, $9.81{\text{m/s}}^2$ for g, and 5 kg for $m_C$.

$\begin{array}{l}2\left(\frac{10\times 9.81-2T_{BC}}{10}\right)+1.5+\frac{5\times 9.81-T_{BC}}{5}=0\\ \frac{98.1-2T_{BC}}{5}+\frac{49.05-T_{BC}}{5}=-1.5\\ 98.1-2T_{BC}+49.05-T_{BC}=-7.5\\ T_{BC}=51.55\text{N}\end{array}$

Sketch the free body diagram of block B as shown in Figure (5).

Refer Figure (5),

Find the magnitude of the force P.

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

$\begin{array}{l}\Sigma F_y=m_B{a}_B\\ W_B+P-T_{BC}=m_B{a}_B\end{array}$

Here, $W_B$ is the weight of block B.

Substitute $m_{B}g$ for $W_B$.

$m_{B}g+P-T_{BC}=m_B{a}_B$

Substitute 5 kg for $m_B$, $9.81{\text{m/s}}^2$ for g, 51.55 N for P, and $1.5{\text{m/s}}^2$ for $a_B$.

$\begin{array}{l}5\times 9.81+P-51.55=5\times 1.5\\ P=10.00\text{N}\end{array}$

Thus, the magnitude of the force P is $\underset{\_}{10.00\text{N}}$.

To determine

Find the tension in the cord AD.

Answer to Problem 12.28P

The tension in the cord AD is $\underset{\_}{103.1\text{N}}$.

Explanation of Solution

Calculation:

Find the tension in the cord AD using Equation (8).

$T_{AD}=2T_{BC}$

Substitute $51.55\text{N}$ for $T_{BC}$.

$\begin{array}{c}{T}_{AD}=2\left(51.55\right)\\ =103.1\text{N}\end{array}$

Thus, the tension in the cord is $\underset{\_}{103.1\text{N}}$.