Chapter 8.4, Problem 8.119P

(a)

To determine

Find the smallest combined mass m of the bucket for the block C will remain at rest.

Answer to Problem 8.119P

The smallest combined mass m of the bucket is $\underset{\_}{9.46\text{kg}}$.

Explanation of Solution

Given information:

The mass of the block C is $m_C=100\text{kg}$.

The coefficient of static friction is ${\mu }_s=0.35$.

The coefficient of kinetic friction is ${\mu }_k=0.25$.

The drum B is frozen and cannot rotate.

Calculation:

Show the free-body diagram of the drum B as in Figure 1.

Find the angle of the belt wounded around the drum as follows;

$\begin{array}{c}\beta =120°\times \frac{\pi }{180°}\\ =\frac{2\pi }{3}\text{rad}\end{array}$

Find the tension $T_2$ in terms of tension $T_1$ using the equation.

$\frac{T_2}{T_1}=e^{\mu \beta }$

Substitute mg for $T_1$.

$\begin{array}{l}\frac{T_2}{mg}=e^{\mu \beta }\\ T_2=mg{e}^{\mu \beta }\end{array}$ (1)

Here, the acceleration due to gravity is g.

Consider the value of acceleration due to gravity is $g=9.81{\text{m/s}}^2$.

Show the free-body diagram of the block C as in Figure 2.

At rest, the cable slips on the drum. The motion impending is along the x-axis. $(\searrow )$.

$\mu ={\mu }_s$

Substitute $9.81{\text{m/s}}^2$ for g, 0.23 for $\mu$ and $\frac{2\pi }{3}$ for $\beta$ in Equation (1).

$\begin{array}{c}{T}_2=m\times 9.81\times e^{0.35\times \frac{2\pi }{3}}\\ =20.41848m\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\nearrow {\displaystyle \sum F_x=0}\\ N-m_{C}g\cos 30°=0\end{array}$

Substitute 100 kg for $m_C$ and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{l}N-100\times 9.81\times \cos 30°=0\\ N=849.571\text{N}\end{array}$

Find the friction force (F) using the relation.

$F={\mu }_{s}N$

Substitute 0.35 for ${\mu }_s$ and 849.571 N for N.

$\begin{array}{c}F=0.35\times 849.571\\ =297.35\text{N}\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}+\nwarrow {\displaystyle \sum F_y=0}\\ T_2+F-m_{C}g\sin 30°=0\end{array}$

Substitute $20.41848m$ for $T_2$, 297.35 N for F, and 100 kg for $m_C$, and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{l}20.41848m+297.35-100\times 9.81\times \sin 30°=0\\ 20.41848m+297.35-490.5=0\\ m=9.46\text{kg}\end{array}$

Therefore, the smallest combined mass m of the bucket is $\underset{\_}{9.46\text{kg}}$.

(b)

To determine

Find the smallest combined mass m of the bucket for the block C start moving up the incline.

Answer to Problem 8.119P

The smallest combined mass m of the bucket is $\underset{\_}{167.2\text{kg}}$.

Explanation of Solution

Given information:

The mass of the block C is $m_C=100\text{kg}$.

The coefficient of static friction is ${\mu }_s=0.35$.

The coefficient of kinetic friction is ${\mu }_k=0.25$.

The drum B is frozen and cannot rotate.

Calculation:

Show the free-body diagram of the drum B as in Figure 3.

Find the angle of the belt wounded around the drum as follows;

$\begin{array}{c}\beta =120°\times \frac{\pi }{180°}\\ =\frac{2\pi }{3}\text{rad}\end{array}$

Find the tension $T_2$ in terms of tension $T_1$ using the equation.

$\frac{T_2}{T_1}=e^{\mu \beta }$

Substitute mg for $T_2$.

$\begin{array}{l}\frac{mg}{T_1}=e^{\mu \beta }\\ T_1=\frac{mg}{e^{\mu \beta }}\end{array}$ (2)

Show the free-body diagram of the block C as in Figure 4.

When the block start moving up the incline; $\mu ={\mu }_s$.

No slipping occurs at block and drum. The motion impending is against the x-axis. $(\nwarrow )$.

Substitute $9.81{\text{m/s}}^2$ for g, 0.35 for $\mu$ and $\frac{2\pi }{3}$ for $\beta$ in Equation (2).

$\begin{array}{c}{T}_1=\frac{m\times 9.81}{e^{0.35\times \frac{2\pi }{3}}}\\ =4.71319m\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\nearrow {\displaystyle \sum F_x=0}\\ N-m_{C}g\cos 30°=0\end{array}$

Substitute 100 kg for $m_C$ and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{l}N-100\times 9.81\times \cos 30°=0\\ N=849.571\text{N}\end{array}$

Find the friction force (F) using the relation.

$F={\mu }_{s}N$

Substitute 0.35 for ${\mu }_s$ and 849.571 N for N.

$\begin{array}{c}F=0.35\times 849.571\\ =297.35\text{N}\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}+\nwarrow {\displaystyle \sum F_y=0}\\ T_1-F-m_{C}g\sin 30°=0\end{array}$

Substitute $4.71319m$ for $T_1$, 297.35 N for F, and 100 kg for $m_C$, and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{l}4.71319m-297.35-100\times 9.81\times \sin 30°=0\\ 4.71319m-297.35-490.5=0\\ m=167.2\text{kg}\end{array}$

Therefore, the smallest combined mass m of the bucket is $\underset{\_}{167.2\text{kg}}$.

(c)

To determine

Find the smallest combined mass m of the bucket for the block C continue moving up the incline at constant speed.

Answer to Problem 8.119P

The smallest combined mass m of the bucket is $\underset{\_}{121\text{kg}}$.

Explanation of Solution

Given information:

The mass of the block C is $m_C=100\text{kg}$.

The coefficient of static friction is ${\mu }_s=0.35$.

The coefficient of kinetic friction is ${\mu }_k=0.25$.

The drum B is frozen and cannot rotate.

Calculation:

Show the free-body diagram of the drum B as in Figure 5.

Find the angle of the belt wounded around the drum as follows;

$\begin{array}{c}\beta =120°\times \frac{\pi }{180°}\\ =\frac{2\pi }{3}\text{rad}\end{array}$

Find the tension $T_2$ in terms of tension $T_1$ using the equation.

$\frac{T_2}{T_1}=e^{\mu \beta }$

Substitute mg for $T_2$.

$\begin{array}{l}\frac{mg}{T_1}=e^{\mu \beta }\\ T_1=\frac{mg}{e^{\mu \beta }}\end{array}$ (3)

Show the free-body diagram of the block C as in Figure 6.

When the block start moving up the incline; $\mu ={\mu }_k$.

No slipping occurs at block and drum. The motion impending is against the x-axis. $(\nwarrow )$.

Substitute $9.81{\text{m/s}}^2$ for g, 0.25 for $\mu$ and $\frac{2\pi }{3}$ for $\beta$ in Equation (3).

$\begin{array}{c}{T}_1=\frac{m\times 9.81}{e^{0.25\times \frac{2\pi }{3}}}\\ =5.81130m\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\nearrow {\displaystyle \sum F_x=0}\\ N-m_{C}g\cos 30°=0\end{array}$

Substitute 100 kg for $m_C$ and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{l}N-100\times 9.81\times \cos 30°=0\\ N=849.571\text{N}\end{array}$

Find the friction force (F) using the relation.

$F={\mu }_{k}N$

Substitute 0.25 for ${\mu }_s$ and 849.571 N for N.

$\begin{array}{c}F=0.25\times 849.571\\ =212.39275\text{N}\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}+\nwarrow {\displaystyle \sum F_y=0}\\ T_1-F-m_{C}g\sin 30°=0\end{array}$

Substitute $5.81130m$ for $T_1$, 212.39275 N for F, and 100 kg for $m_C$, and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{l}5.81130m-212.39275-100\times 9.81\times \sin 30°=0\\ 5.81130m-212.39275-490.5=0\\ m=121\text{kg}\end{array}$

Therefore, the smallest combined mass m of the bucket is $\underset{\_}{121\text{kg}}$.