Chapter 12.1, Problem 12.31P

A 10-lb block B rests as shown on a 20-lb bracket A. The coefficients of friction are μ_s = 0.30 and μ_k = 0.25 between block B and bracket A, and there is no friction in the pulley or between the bracket and the horizontal surface. (a) Determine the maximum weight of block C if block B is not to slide on bracket A. (b) If the weight of block C is 10 percent larger than the answer found in a, determine the accelerations of A, B, and C.

Fig. P12.31

To determine

Find the maximum weight of block C if block B is not to slide on bracket A.

Answer to Problem 12.31P

The maximum weight of block C if block B is not to slide on bracket A is $\underset{\_}{2.43\text{lbs}}$.

Explanation of Solution

Given information:

The weight of block B $\left(W_B\right)$ is 10 lb.

The weight of bracket A $\left(W_A\right)$ is 20 lb.

The coefficient of static friction between block B and bracket A $\left({\mu }_s\right)$ is 0.30.

The coefficient of kinetic friction between block B and bracket A $\left({\mu }_k\right)$ is 0.25.

Calculation:

Let as consider the horizontal coordinate of A is $x_A$ from the fixed vertical line to the

left of A.

Let as consider the horizontal coordinate of B is $x_B$ from the fixed vertical line to the

left of A.

Let as consider $y_c$ is the distance that block C is below the pulley.

Sketch the system with coordinates points as shown in Figure (1).

Write the general equation of mass (m):

$\begin{array}{c}W=mg\\ m=\frac{W}{g}\end{array}$

Here, W is the weight, g is the acceleration due to gravity.

Consider the constraint of cord.

Write the total length of cable length (L).

$L=\left(x_B-x_B\right)+\left(x_P-x_A\right)+y_C+\text{constant}$ (1)

Here, $x_P$ is the distance that pulley P from the fixed vertical line to the left of A.

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

$\begin{array}{l}0=\left(v_B-v_A\right)+\left(0-v_A\right)+v_C+0\\ v_B-2v_A+v_C=0\end{array}$ (2)

Here, $v_A$ is the velocity of the block B and $v_A$ is the velocity of the bracket A, $v_C$ is the velocity of the block C.

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

$-2a_A+a_B+a_C=0$ (3)

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

No slip between bracket A and block B. Therefore, the acceleration of bracket A is equal to acceleration of block B.

$a_A=a_B$ (4)

Substitute $a_A$ for $a_B$ in Equation (3).

$\begin{array}{l}-2a_A+a_A+a_C=0\\ -a_A+a_C=0\\ a_A=a_C\end{array}$ (5)

From Equation (4) and (5), consider

$a_A=a_B=a_C=a$ (6)

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

Refer Figure (2).

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

$\begin{array}{l}\Sigma F_x=m{a}_x\\ 2T-F_{AB}=m_A{a}_A\end{array}$

Here, T is the tension in the cable, $F_{AB}$ is the friction force between bracket A and block B, and $m_A$ is the mass of the bracket A.

Substitute $\frac{W_A}{g}$ for $m_A$.

$2T-F_{AB}=\frac{W_A}{g}{a}_A$ (7)

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

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

Refer Figure (3).

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

$\begin{array}{l}\Sigma F_x=m{a}_x\\ F_{AB}-T=m_B{a}_B\end{array}$

Substitute $\frac{W_B}{g}$ for $m_B$.

$F_{AB}-T=\frac{W_B}{g}{a}_B$ (8)

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

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

$\begin{array}{l}\Sigma F_y=m{a}_y\\ N_{AB}-W_B=0\\ N_{AB}=W_B\end{array}$ (9)

Here, $N_{AB}$ is the normal force between bracket A and block B and $W_B$ is the weight of block B.

Write the equation of frictional force $\left(F_{AB}\right)$.

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

Substitute $W_B$ for $N_{AB}$.

$F_{AB}={\mu }_s{W}_B$ . (10)

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

Refer Figure (4).

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

$\Sigma F_y=m_C{a}_y$

Substitute $\frac{W_C}{g}$ for $m_C$.

$\begin{array}{l}\left(\frac{W_C}{g}\right)g-T=\frac{W_C}{g}{a}_C\\ W_C-T=\frac{W_C}{g}{a}_C\end{array}$ (11)

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

Adding Equation (5), (6), and (9).

$\begin{array}{l}2T-F_{AB}+F_{AB}-T+W_C-T=\frac{W_A}{g}{a}_A+\frac{W_B}{g}{a}_B+\frac{W_C}{g}{a}_C\\ \frac{W_A}{g}{a}_A+\frac{W_B}{g}{a}_B+\frac{W_C}{g}{a}_C=W_C\end{array}$ (12)

Substitute Equation (6) in Equation (12).

$\begin{array}{l}\frac{W_A}{g}\left(a\right)+\frac{W_B}{g}\left(a\right)+\frac{W_C}{g}\left(a\right)=W_C\\ a=\frac{W_{C}g}{W_A+W_B+W_C}\end{array}$ (13)

Subtract Equation (6) and (9).

$\begin{array}{l}{F}_{AB}-T-W_C+T=\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C\\ F_{AB}-W_C=\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C\end{array}$ (14)

Substitute Equation (6) and (10) in Equation (14).

$\begin{array}{l}\left({\mu }_s{W}_B\right)-W_C=\frac{W_B}{g}\left(a\right)-\frac{W_C}{g}\left(a\right)\\ {\mu }_s{W}_B-W_C=\frac{a}{g}\left(W_B-W_C\right)\end{array}$ (15)

Subtracting Equation (13) and (9).

$\begin{array}{l}{\mu }_s{W}_B-W_C=\frac{\frac{W_{C}g}{W_A+W_B+W_C}}{g}\left(W_B-W_C\right)\\ {\mu }_s{W}_B-W_C=\frac{W_{C}g}{\left(W_A+W_B+W_C\right)g}\left(W_B-W_C\right)\\ {\mu }_s{W}_B-W_C=\frac{W_C}{\left(W_A+W_B+W_C\right)}\left(W_B-W_C\right)\end{array}$ (16)

Substitute 0.30 for ${\mu }_s$, 10 lb for $W_B$, and 20 lb for $W_A$.

$\begin{array}{l}0.30\times 10-W_C=\frac{W_C}{\left(20+10+W_C\right)}\left(10-W_C\right)\\ 3-W_C=\frac{W_C}{30+W_C}\left(10-W_C\right)\\ 90-30W_C+3W_C-W_C^2=10W_C-W_C^2\\ 10W_C+30W_C-3W_C=90\end{array}$

$\begin{array}{l}37W_C=90\\ W_C=2.43\text{lbs}\end{array}$

Thus, the maximum weight of block C if block B is not to slide on bracket A is $\underset{\_}{2.43\text{lbs}}$.

To determine

Find the accelerations of bracket A, block B and block C.

Answer to Problem 12.31P

The accelerations of bracket A is $\underset{\_}{3.14{\text{ft/s}}^2}$.

The accelerations of block B is $\underset{\_}{0.881{\text{ft/s}}^2}$.

The accelerations of block C is $\underset{\_}{5.41{\text{ft/s}}^2}$.

Explanation of Solution

Given information:

The weight of block C is 10 % larger than the answer found in part (a).

Calculation:

Find the weight of block C.

$\begin{array}{c}{W}_C=1.1\times 2.43\\ =2.673\text{lb}\end{array}$

The slip is occurring. Therefore consider the kinetic acceleration.

Write the equation of frictional force $\left(F_{AB}\right)$.

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

Substitute $W_B$ for $N_{AB}$.

$F_{AB}={\mu }_k{W}_B$ . (17)

Subtract Equation (6) and (9).

$\begin{array}{l}{F}_{AB}-T-W_C+T=\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C\\ F_{AB}-W_C=\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C\end{array}$ (18)

Substitute Equation (17) in Equation (18).

$\begin{array}{l}{\mu }_k{W}_B-W_C=\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C\\ W_B{a}_B-W_C{a}_C=g\left({\mu }_k{W}_B-W_C\right)\end{array}$ (19)

Substitute 0.25 for ${\mu }_k$, 10 lb for $W_B$, 20 lb for $W_A$, 2.673 lb for $W_C$, and $32.2{\text{ft/s}}^2$ for $g$.

$\begin{array}{l}10a_B-2.673a_C=\left(32.2\right)\left[\left(0.25\right)\left(10\right)-\left(2.673\right)\right]\\ 10a_B-2.673a_C=-5.5706\end{array}$ (20)

Rewrite the Equation (5).

$\begin{array}{l}\frac{W_A}{g}{a}_A+\frac{W_B}{g}{a}_B+\frac{W_C}{g}{a}_C=W_C\\ W_A{a}_A+W_B{a}_B+W_C{a}_C=W_{C}g\end{array}$

Substitute 10 lb for $W_B$, 20 lb for $W_A$, 2.673 lb for $W_C$, and $32.2{\text{ft/s}}^2$ for $g$.

$\begin{array}{l}20a_A+10a_B+2.673a_C=2.673\times 32.2\\ 20a_A+10a_B+2.673a_C=86.0706\end{array}$ (21)

Find the accelerations of bracket A, block B and block C.

Solve Equation (3), (20), and (21).

$\begin{array}{l}{a}_A=3.14{\text{ft/s}}^2\\ a_B=0.881{\text{ft/s}}^2\\ a_C=5.41{\text{ft/s}}^2\end{array}$

Thus, the accelerations of bracket A is $\underset{\_}{3.14{\text{ft/s}}^2}$.

Thus, the accelerations of block B is $\underset{\_}{0.881{\text{ft/s}}^2}$.

Thus, the accelerations of block C is $\underset{\_}{5.41{\text{ft/s}}^2}$.