Chapter 16.1, Problem 16.39P

A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are μ_s = 0.50 and μ_k = 0.40. For P = 3.6 lb, determine (a) whether slipping occurs between the belt and either cylinder, (b) the angular acceleration of each cylinder.

Fig. P16.39

To determine

Find whether slipping occurs between the belt and either cylinder.

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:

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.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ 3.6-F_A-F_B=0\\ F_A+F_B=3.6\end{array}$ (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)

$\begin{array}{l}25.8798\times {10}^{-3}{\alpha }_A+103.51965\times {10}^{-3}{\alpha }_A=3.6\\ 129.4125\times {10}^{-3}{\alpha }_A=3.6\\ {\alpha }_A=\frac{3.6}{129.4125\times {10}^{-3}}\\ {\alpha }_A=27.82\text{rad}/{\text{s}}^{\text{2}}\end{array}$

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

Substitute $27.82\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$ in Equation (1).

$\begin{array}{c}{F}_A=25.8798\times {10}^{-3}\times 27.82\\ =0.72\text{lb}\end{array}$

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

Substitute $27.82\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$ in Equation (2).

$\begin{array}{c}{F}_B=103.51965\times {10}^{-3}\times 27.82\\ =2.88\text{lb}\end{array}$

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 force of the cylinder B is greater than the magnitude of the friction force $\left(F_B>F_m\right)$.

Therefore, the slipping occurs between cylinder B and the belt and the slipping not occur between cylinder A and the belt.

To determine

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

Answer to Problem 16.39P

The angular acceleration of each cylinder $\left({\alpha }_A\text{and}{\alpha }_B\right)$ are $\underset{\_}{61.8\text{rad}/{\text{s}}^{\text{2}}}$ and $\underset{\_}{9.66\text{rad}/{\text{s}}^{\text{2}}}$.

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 slipping occurs at cylinder B.

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

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

$F_B={\mu }_k{W}_A$

Substitute $5\text{lb}$ for $W_A$ and $0.40$ for ${\mu }_k$.

$\begin{array}{c}{F}_B=0.40\times 5\\ =2\text{lb}\end{array}$

Show the free body diagram of the cylinder B as in Figure 4.

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 4.

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

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

$\begin{array}{l}\left(2\times \frac{2}{3}\right)=138.0262\times {10}^{-3}\times {\alpha }_B\\ \frac{4}{3}=138.0262\times {10}^{-3}{\alpha }_B\\ {\alpha }_B=\frac{4}{3\times 138.0262\times {10}^{-3}}\\ {\alpha }_B=9.66\text{rad}/{\text{s}}^{\text{2}}\end{array}$

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

Substitute $2\text{lb}$ for $F_B$ in Equation (3).

$\begin{array}{c}{F}_A+2=3.6\\ F_A=3.6-2\\ =1.6\text{lb}\end{array}$

The horizontal force of the cylinder A is less than the force of the cylinder B due to the static friction $\left(F_A<F_m\right)$.

There is no slipping between the cylinder A and the belt.

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

Substitute $1.6\text{lb}$ for $F_A$ in Equation (1).

$\begin{array}{c}1.6=25.8798\times {10}^{-3}{\alpha }_A\\ {\alpha }_A=\frac{1.6}{25.8798\times {10}^{-3}}\\ =61.8\text{lb}\end{array}$

Hence, the angular acceleration of each cylinder $\left({\alpha }_A\text{and}{\alpha }_B\right)$ are $\underset{\_}{61.8\text{rad}/{\text{s}}^{\text{2}}}$ and $\underset{\_}{9.66\text{rad}/{\text{s}}^{\text{2}}}$.