Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00\text{lb}$.

The weight of the cylinder A $\left(W_A\right)$ is $5\text{lb}$.

The weight of the cylinder B $\left(W_B\right)$ is $20\text{lb}$.

The coefficient of the static friction $\left({\mu }_s\right)$ is $0.50$.

The coefficient of the kinetic friction $\left({\mu }_k\right)$ is $0.40$.

The radius of the cylinder A $\left(r_A\right)$ is $4\text{in}\text{.}$.

The radius of the cylinder B $\left(r_B\right)$ is $8\text{in}\text{.}$.

Calculation:

Consider the acceleration due to gravity (g) as $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Convert the unit of the radius of the cylinder A $\left(r_A\right)$:

$\begin{array}{c}{r}_A=4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{1}{3}\text{ft}\end{array}$

Convert the unit of the radius of the cylinder B $\left(r_B\right)$:

$\begin{array}{c}{r}_B=8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{2}{3}\text{ft}\end{array}$

Consider that no slipping occurs.

Calculate the acceleration of the belt $\left(a_{belt}\right)$:

$\begin{array}{l}{a}_{belt}=r_A{\alpha }_A=r_B{\alpha }_B\\ a_{belt}=\frac{{\alpha }_A}{3}=\frac{2{\alpha }_B}{3}\end{array}$

$\begin{array}{l}\frac{{\alpha }_A}{3}=\frac{2{\alpha }_B}{3}\\ {\alpha }_A=2{\alpha }_B\\ {\alpha }_B=\frac{{\alpha }_A}{2}\end{array}$

Calculate the mass of the cylinder A $\left(m_A\right)$:

$m_A=\frac{W_A}{g}$

Substitute $5\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$m_A=\frac{5}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$

Calculate the mass of the cylinder B $\left(m_B\right)$:

$m_B=\frac{W_B}{g}$

Substitute $20\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$m_B=\frac{20}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$

Calculate the mass moment of inertia of the cylinder A $\left({\overline{I}}_A\right)$:

${\overline{I}}_A=\frac{1}{2}{m}_A{r}_A^2$

Substitute $\frac{5}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_A$ and $\frac{1}{3}\text{ft}$ for $r_A$.

$\begin{array}{c}{\overline{I}}_A=\frac{1}{2}\times \frac{5}{32.2}\times {\left(\frac{1}{3}\right)}^2\\ =8.6266\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the mass moment of inertia of the cylinder B $\left({\overline{I}}_B\right)$:

${\overline{I}}_B=\frac{1}{2}{m}_B{r}_B^2$

Substitute $\frac{20}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_B$ and $\frac{2}{3}\text{ft}$ for $r_B$.

$\begin{array}{c}{\overline{I}}_B=\frac{1}{2}\times \frac{20}{32.2}\times {\left(\frac{2}{3}\right)}^2\\ =138.0262\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Show the free body diagram of the cylinder A as in Figure 1.

Here, $F_A$ is the horizontal force of the cylinder A and ${\alpha }_A$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum M_G}=I_G\alpha \\ \left(F_A{r}_A\right)={\overline{I}}_A{\alpha }_A\end{array}$

Substitute $\frac{1}{3}\text{ft}$ for $r_A$ and $8.6266\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for ${\overline{I}}_A$.

$\begin{array}{l}\left(F_A\times \frac{1}{3}\right)=8.6266\times {10}^{-3}{\alpha }_A\\ \frac{1}{3}{F}_A=8.6266\times {10}^{-3}{\alpha }_A\\ F_A=8.6266\times {10}^{-3}{\alpha }_A\times 3\\ F_A=25.8798\times {10}^{-3}{\alpha }_A\end{array}$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Here, $F_B$ is the horizontal force of the cylinder B and ${\alpha }_B$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum M_G}=I_G\alpha \\ \left(F_B{r}_B\right)={\overline{I}}_B{\alpha }_B\end{array}$

Substitute $\frac{2}{3}\text{ft}$ for $r_B$, $\frac{{\alpha }_A}{2}$ for ${\alpha }_B$, and $138.0262\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for ${\overline{I}}_B$.

$\begin{array}{l}\left(F_B\times \frac{2}{3}\right)=138.0262\times {10}^{-3}\times \frac{{\alpha }_A}{2}\\ \frac{2}{3}{F}_B=69.0131\times {10}^{-3}{\alpha }_A\\ F_B=69.0131\times {10}^{-3}{\alpha }_A\times \frac{3}{2}\\ F_B=103.51965\times {10}^{-3}{\alpha }_A\end{array}$ (2)

Show the free body diagram of the belt as in Figure 3.

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ 2.00-F_A-F_B=0\\ F_A+F_B=2.00\end{array}$ (3)

Calculate the angular acceleration of the cylinder A $\left({\alpha }_A\right)$:

Substitute $25.8798\times {10}^{-3}{\alpha }_A$ for $F_A$ and $103.51965\times {10}^{-3}{\alpha }_A$ for $F_B$ in Equation (3)

$\begin{array}{l}25.8798\times {10}^{-3}{\alpha }_A+103.51965\times {10}^{-3}{\alpha }_A=2.00\\ 129.4125\times {10}^{-3}{\alpha }_A=2.00\\ {\alpha }_A=\frac{2.00}{129.4125\times {10}^{-3}}\\ {\alpha }_A=15.46\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the horizontal force of the cylinder A $\left(F_A\right)$:

Substitute $15.46\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$ in Equation (1).

$\begin{array}{c}{F}_A=25.8798\times {10}^{-3}\times 15.46\\ =0.4\text{lb}\end{array}$

Calculate the horizontal force of the cylinder B $\left(F_B\right)$:

Substitute $15.46\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$ in Equation (2).

$\begin{array}{c}{F}_B=103.51965\times {10}^{-3}\times 15.46\\ =1.6\text{lb}\end{array}$

Calculate the magnitude of the friction force $\left(F_m\right)$:

$F_m={\mu }_s{W}_A$

Substitute $5\text{lb}$ for $W_A$ and $0.50$ for ${\mu }_s$.

$\begin{array}{c}{F}_m=0.50\times 5\\ =2.5\text{lb}\end{array}$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $\left(F_A\text{and}{F}_B<F_m\right)$.

Therefore, there is no slipping occurs between cylinders and belt.

(b)

To determine

Find the angular acceleration of each cylinder $\left({\alpha }_A\text{and}{\alpha }_B\right)$.

Answer to Problem 16.40P

The angular acceleration of each cylinder $\left({\alpha }_A\text{and}{\alpha }_B\right)$ are $\underset{\_}{15.46\text{rad}/{\text{s}}^{\text{2}}}$ and $\underset{\_}{7.73\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6\text{lb}$.

The weight of the cylinder A $\left(W_A\right)$ is $5\text{lb}$.

The weight of the cylinder B $\left(W_B\right)$ is $20\text{lb}$.

The coefficient of the static friction $\left({\mu }_s\right)$ is $0.50$.

The coefficient of the kinetic friction $\left({\mu }_k\right)$ is $0.40$.

The radius of the cylinder A $\left(r_A\right)$ is $4\text{in}\text{.}$.

The radius of the cylinder B $\left(r_B\right)$ is $8\text{in}\text{.}$.

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is ${\alpha }_B=\frac{1}{2}{\alpha }_A$

Calculate the angular acceleration of the cylinder A $\left({\alpha }_A\right)$:

Substitute $1.6\text{lb}$ for $F_B$ in Equation (2).

$\begin{array}{l}1.6=103.51965\times {10}^{-3}{\alpha }_A\\ {\alpha }_A=\frac{1.6}{103.51965\times {10}^{-3}}\\ {\alpha }_A=15.46\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the angular acceleration of the cylinder B $\left({\alpha }_B\right)$:

${\alpha }_B=\frac{1}{2}{\alpha }_A$

Substitute $15.46\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$.

$\begin{array}{c}{\alpha }_B=\frac{1}{2}\times 15.46\\ =7.73\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of each cylinder $\left({\alpha }_A\text{and}{\alpha }_B\right)$ are $\underset{\_}{15.46\text{rad}/{\text{s}}^{\text{2}}}$ and $\underset{\_}{7.73\text{rad}/{\text{s}}^{\text{2}}}$.