Chapter 12.1, Problem 12.31P

A 10-Ib block B rests as shown on a 20-1b bracket A. The coefficients of friction are ${\mu }_s=0.30$ and ${\mu }_k=0.25$ 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.

  

To determine

(a)

Maximum weight of block C if block B is not to slide on bracket A.

Answer to Problem 12.31P

Maximum weight of block C is $W_C=2.43lbs.$

Explanation of Solution

Given:

${\mu }_s=0.30$

${\mu }_k=0.25$

$W_A=20-\text{lb}$

$W_B=10-\text{lb}$

Concept used:

Kinematics :

Let $x_A$ and $x_B$ be the horizontal coordinates of A and B measured from a fixed vertical line to the left of A and B.

Let $y_C$ be the distance that block C is below the pulley.

Note that $y_C$ increases when C moves downward.

The cable length is fixed.

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

Differentiating and noting that $x_P=0$

$v_B-2v_A+v_C=0$

$-2a_A+a_B+a_C=0$ … (1)

Here. $a_A$ and $a_B$ are positive to the right and $a_C$ is positive downward.

Kinetics :

Let $T$ be the tension in the cable and $F_{AB}$ be the friction force between blocks A and B.

The free body diagrams are shown here:

Bracket A:

Block B:

Block C:

Bracket A:

$\stackrel{+}{\to }\sum F_x=m{a}_x$ :

$2T-F_{AB}=\frac{W_A}{g}{a}_A$ … (2)

Block B:

$\stackrel{+}{\to }\sum F_x=m{a}_x$

$F_{AB}-T=\frac{W_B}{g}{a}_B$ … (3)

$+\uparrow \sum F_y=m{a}_y$ :

$N_{AB}-W_B=0$

$N_{AB}=W_B$

Block C:

$+\downarrow \sum F_y=m{a}_y$ :

$m_C-T=\frac{W_C}{g}{a}_C$ … (4)

Adding Eqs. (2). (3) and (4) and transposing.

$\frac{W_A}{g}{a}_A+\frac{W_B}{g}{a}_B+\frac{W_C}{g}{a}_C=W_C$ … (5)

Subtracting Eq. (4) from Eq. (3) and transposing.

$\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C=F_{AB}-W_C$ … (6)

No slip between A and B.

$a_B=a_A$

From Eq. (1).

$a_A=a_B=a_C=a$

From Eq. (5).

$a=\frac{W_{C}g}{W_A+W_B+W_C}$

For impeding slip.

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

Substituting into Eq. (6).

$\frac{\left(W_B-W_C\right)}{W_A+W_B+W_C}={\mu }_s{W}_B-W_C$

Solving for $W_C$.

$W_C=\frac{{\mu }_s{W}_B\left(W_A+W_B\right)}{W_A+2W_B-{\mu }_s{W}_B}$

$W_C=\frac{\left(0.30\right)\left(10\right)\left(20+10\right)}{20+2\left(10\right)-\left(0.30\right)\left(10\right)}$

$W_C=2.43\text{lb}$

To determine

(b)

Accelerations of A, B, and C, if the weight of block C is 10 percent larger than the result of part a.

Answer to Problem 12.31P

Accelerations of A. B and C:

$a_A=3.14\text{ft}/s^2\to$

$a_B=0.881\text{ft}/s^2\to$

$a_C=5.41\text{ft}/s^2\downarrow$

Explanation of Solution

Given:

$W_C$ increased by 10%.

$W_B=2.6757\text{lb}$ (increased by 10%)

Since slip is occurring.

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

Eq. (6) becomes.

$\frac{W_B}{g}{a}_B-\frac{W_C}{g}{a}_C=F_{AB}-W_C$

Or:

$10a_B-2.6757a_C=\left[\left(0.25\right)\left(10\right)-2.6757\right]\left(32.2\right)$ … (7)

With given data. Eq. (5) becomes

$20a_C+10a_B+2.6757a_C=\left(2.6757\right)\left(32.2\right)$ … (8)

Solving Eqs. (1). (7). and (8) gives

$a_A=3.144\text{ft}/s^2,a_B=0.881\text{ft}/s^2,a_C=5.407\text{ft}/s^2$

So,

$a_A=3.14\text{ft}/s^2\to$

And,

$a_B=0.881\text{ft}/s^2\to$

And

$a_C=5.41\text{ft}/s^2\downarrow$