Chapter 12.1, Problem 12.18P

Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μ_s = 0.20 and μ_k = 0.15. If P = 0, determine (a) the acceleration of block B, (b) the tension in the cord.

Fig. P12.18 and P12.19

To determine

Find the acceleration of block B.

Answer to Problem 12.18P

The acceleration of block B is $\underset{\_}{0.986{\text{m/s}}^2}$.

Explanation of Solution

Given information:

The mass of block A $\left(m_A\right)$ is 40 kg.

The mass of block B $\left(m_B\right)$ is 8 kg.

The coefficient of static friction between all surfaces of contact $\left({\mu }_s\right)$ is 0.20.

The coefficient of kinetic friction between all surfaces of contact $\left({\mu }_k\right)$ is 0.15.

The horizontal load (P) is zero.

Calculation:

Write the general equation of weight  (W):

$W=mg$

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

Consider the constraint of cord.

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

$2x_A+x_{B/A}=\text{constant}$ (1)

Here,$x_A$ is the length of cord connecting block A and $x_{B/A}$ is the length of cord connecting block B relative to block A.

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

$2v_A+v_{B/A}=\text{constant}$ (2)

Here, $v_A$ is the velocity of the block A and $v_{B/A}$ is the velocity of the block B relative to block A.

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

$\begin{array}{l}2a_A+a_{B/A}=\text{0}\\ a_{B/A}=-2a_A\end{array}$ (3)

Here, $a_A$ is the acceleration of block A and $a_{B/A}$ acceleration of block B relative to block A.

Find the equation of acceleration of block B in terms of acceleration of block A.

$a_B=a_A+a_{B/A}$

Here, $a_B$ is the acceleration of block B.

Substitute $-2a_A$ for $a_{B/A}$.

$\begin{array}{c}{a}_B=a_A+\left(-2a_A\right)\\ =-a_A\end{array}$ (4)

First of all determine whether the blocks will move for the given value of $\theta \left(25°\right)$.

Sketch the free body diagram of block B as shown in Figure (1).

Refer Figure (1).

Consider equilibrium along y-axis .

$\begin{array}{l}\Sigma F_y=0\\ N_{AB}-W_B\cos \theta =0\\ N_{AB}=W_B\cos \theta \end{array}$

Here, $N_{AB}$ is the normal force on block B from block A.

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

$N_{AB}=m_{B}g\cos \theta$

Write the equation of frictional force $\left(F_{AB}\right)$ on the block B from A.

$F_{AB}={\mu }_s{N}_{AB}$

Substitute 0.20 for ${\mu }_s$ and $m_{B}g\cos \theta$ for $N_{AB}$.

$F_{AB}=0.20m_{B}g\cos \theta$

Consider equilibrium along x-axis.

$\begin{array}{l}\Sigma F_x=0\\ -T+F_{AB}+W_B\sin \theta =0\\ T=F_{AB}+W_B\sin \theta \end{array}$

Substitute $0.20m_{B}g\cos \theta$ for $F_{AB}$ and $m_{B}g$ for $W_B$.

$\begin{array}{c}T=0.20m_{B}g\cos \theta +m_{B}g\sin \theta \\ =m_{B}g\left(0.20\cos \theta +\sin \theta \right)\end{array}$ (5)

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

Refer Figure (2).

Consider equilibrium along y-axis.

$\begin{array}{l}\Sigma F_y=0\\ N_A-N_{AB}-W_A\cos \theta =0\\ N_A=N_{AB}+W_A\cos \theta \end{array}$

Here, $N_A$ is the normal force on block A.

Substitute $m_{B}g\cos \theta$ for $N_{AB}$ and $m_{A}g$ for $W_A$.

$\begin{array}{c}{N}_A=m_{B}g\cos \theta +m_{A}g\cos \theta \\ =g\cos \theta \left(m_A+m_B\right)\end{array}$

Write the equation of frictional force $F_A$ on the block A.

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

Substitute 0.20 for ${\mu }_s$ and $g\cos \theta \left(m_A+m_B\right)$ for $N_A$.

$F_A=0.20g\cos \theta \left(m_A+m_B\right)$

Consider equilibrium along x-axis.

$\begin{array}{l}\Sigma F_x=0\\ -T-F_A-F_{AB}+W_A\sin \theta =0\\ T=W_A\sin \theta -F_A-F_{AB}\end{array}$

Substitute $0.20g\cos \theta \left(m_A+m_B\right)$ for $F_A$, $0.20m_{B}g\cos \theta$ for $F_{AB}$, and $m_{A}g$ for $W_A$.

$\begin{array}{c}T=m_{A}g\sin \theta -0.20g\cos \theta \left(m_A+m_B\right)-0.20m_{B}g\cos \theta \\ =m_{A}g\sin \theta -0.20g\cos \theta \left(m_A+2m_B\right)\\ =g\left[m_A\sin \theta -0.20\cos \theta \left(m_A+2m_B\right)\right]\end{array}$ (6)

Find the angle $\theta$ for impending motion.

Equate Equation (5) and (6).

$\begin{array}{l}{m}_{B}g\left(0.20\cos \theta +\sin \theta \right)=g\left[m_A\sin \theta -0.20\cos \theta \left(m_A+2m_B\right)\right]\\ m_B\left(0.20\cos \theta +\sin \theta \right)=\left[m_A\sin \theta -0.20\cos \theta \left(m_A+2m_B\right)\right]\end{array}$ (7).

Divide Equation (3) by $\cos \theta$.

$\begin{array}{l}{m}_B\left(0.20+\frac{\sin \theta }{\cos \theta }\right)=\left[m_A\frac{\sin \theta }{\cos \theta }-0.20\left(m_A+2m_B\right)\right]\\ m_B\left(0.20+\tan \theta \right)=\left[m_A\tan \theta -0.20\left(m_A+2m_B\right)\right]\end{array}$

Substitute 40 kg for $m_A$ and 8 kg for $m_B$.

$\begin{array}{l}\left(8\right)\left(0.20+\tan \theta \right)=\left[\left(40\right)\tan \theta -0.20\left(40+2\times 8\right)\right]\\ 1.6+8\tan \theta =40\tan \theta -11.2\\ 32\tan \theta =12.8\\ \theta =21.8°\end{array}$

The angle $\theta$ of impending motion is less than the $25°$, the blocks will move. Therefore apply Newton’s law of motion.

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

Refer Figure 3.

Consider equilibrium along y-axis .

$\begin{array}{l}\Sigma F_y=0\\ N_{AB}-W_B\cos 25°=0\\ N_{AB}=W_B\cos 25°\end{array}$

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

$N_{AB}=m_{B}g\cos 25°$

Write the equation of frictional force $\left(F_{AB}\right)$ on the block B from A.

$F_{AB}={\mu }_k{N}_{AB}$

Substitute 0.15 for ${\mu }_k$ and $m_{B}g\cos 25°$ for $N_{AB}$.

$F_{AB}=0.15m_{B}g\cos 25°$

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

$\begin{array}{l}\Sigma F_x=m_B{a}_B\\ -T+F_{AB}+W_B\sin 25°=m_B{a}_B\\ T=F_{AB}+W_B\sin 25°-m_B{a}_B\end{array}$

Substitute $0.15m_{B}g\cos 25°$ for $F_{AB}$ and $m_{B}g$ for $W_B$.

$\begin{array}{c}T=0.15m_{B}g\cos 25°+m_{B}g\sin 25°-m_B{a}_B\\ =m_B\left(0.15g\cos 25°+g\sin 25°-a_B\right)\end{array}$

Substitute 8 kg for $m_B$ and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}T=m_B\left(0.15g\cos 25°+g\sin 25°-a_B\right)\\ =\left(8\right)\left(0.15\times 9.81\cos 25°+9.81\sin 25°-a_B\right)\\ =8\left(5.47952-a_B\right)\end{array}$ (8)

Sketch the free body diagram and kinetic diagram of block A as shown in Figure 4.

Refer Figure (4).

Consider equilibrium along y-axis.

$\begin{array}{l}\Sigma F_y=0\\ N_A-N_{AB}-W_A\cos 25°=0\\ N_A=N_{AB}+W_A\cos 25°\end{array}$

Substitute $m_{B}g\cos 25°$ for $N_{AB}$ and $m_{A}g$ for $W_A$.

$\begin{array}{c}{N}_A=m_{B}g\cos 25°+m_{A}g\cos \theta \\ =g\cos 25°\left(m_A+m_B\right)\end{array}$

Write the equation of frictional force $F_A$ on the block A.

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

Substitute 0.15 for ${\mu }_k$ and $g\cos 25°\left(m_A+m_B\right)$ for $N_A$.

$F_A=0.15g\cos 25°\left(m_A+m_B\right)$

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

$\begin{array}{l}\Sigma F_x=m_A{a}_A\\ -T-F_A-F_{AB}+W_A\sin 25°=0\\ T=W_A\sin 25°-F_A-F_{AB}-m_A{a}_A\end{array}$

Substitute $0.15g\cos 25°\left(m_A+m_B\right)$ for $F_A$, $0.15m_{B}g\cos 25°$ for $F_{AB}$, $m_{A}g$ for $W_A$ and $-a_B$ for $a_A$.

$\begin{array}{c}T=m_{A}g\sin 25°-0.15g\cos 25°\left(m_A+m_B\right)-0.15m_{B}g\cos 25°-m_A\left(-a_B\right)\\ =m_{A}g\sin 25°-0.15g\cos \theta \left(m_A+2m_B\right)+m_A{a}_B\\ =g\left[m_A\sin 25°-0.15\cos 25°\left(m_A+2m_B\right)\right]+m_A{a}_B\end{array}$

Substitute 40 kg for $m_A$, 8 kg for $m_B$ and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}T=g\left[m_A\sin 25°-0.15\cos 25°\left(m_A+2m_B\right)\right]+m_A{a}_B\\ =9.81\left[40\sin 25°-0.15\cos 25°\left(40+2\times 8\right)\right]+40a_B\\ =91.15202+40a_B\end{array}$ (9)

Find the acceleration of block B $\left(a_B\right)$.

Equate equation (8) and (9).

$\begin{array}{l}8\left(5.47952-a_B\right)=91.15202+40a_B\\ 43.83616-8a_B=91.15202+40a_B\\ -48a_B=47.31586\\ a_B=-0.986{\text{m/s}}^2\end{array}$

Negative sign indicates the motion of block B in opposite to x-axis.

Thus, the acceleration of block B is $\underset{\_}{0.986{\text{m/s}}^2}$.

To determine

Find the tension in the cord.

Answer to Problem 12.18P

The tension in the cord is $\underset{\_}{51.7\text{N}}$.

Explanation of Solution

Calculation:

Find the tension in the cord using Equation (9).

$T=91.15202+40a_B$

Substitute $-0.986{\text{m/s}}^2$ for $a_B$.

$\begin{array}{c}T=91.15202+40\left(-0.986\right)\\ =51.7\text{N}\end{array}$

Thus, the tension in the cord is $\underset{\_}{51.7\text{N}}$.