Chapter 16.1, Problem 16.37P

Gear A weighs 1 lb and has a radius of gyration of 1.3 in.; gear B weighs 6 lb and has a radius of gyration of 3 in.; gear C weighs 9 lb and has a radius of gyration of 4.3 in. Knowing a couple M of constant magnitude of 40 lb·in. is applied to gear A, determine (a) the angular acceleration of gear C, (b) the tangential force that gear B exerts on gear C.

Fig. P16.37

To determine

Find the angular acceleration of the gear C $\left({\alpha }_C\right)$.

Answer to Problem 16.37P

The angular acceleration of the gear C $\left({\alpha }_C\right)$ is $\underset{\_}{130\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

The weight of the gear A $\left(W_A\right)$ is $1\text{lb}$.

The weight of the gear B $\left(W_B\right)$ is $6\text{lb}$.

The weight of the gear C $\left(W_C\right)$ is $9\text{lb}$.

The radius of gyration of the gear A $\left({\overline{k}}_A\right)$ is $1.3\text{in}\text{.}$.

The radius of gyration of the gear B $\left({\overline{k}}_B\right)$ is $3\text{in}\text{.}$.

The radius of gyration of the gear C $\left({\overline{k}}_C\right)$ is $4.3\text{in}\text{.}$.

The couple of the constant magnitude applied to gear A (M) is $40\text{lb}\cdot \text{in}\text{.}$.

The radius of the gear A $\left(r_A\right)$ is $2\text{in}\text{.}$.

The radius of the outer gear B $\left(R_B\right)$ is $4\text{in}\text{.}$.

The radius of the inner gear B $\left(r_B\right)$ is $2\text{in}\text{.}$.

The radius of the gear C $\left(r_B\right)$ is $6\text{in}\text{.}$.

The angular acceleration of the gear A is ${\alpha }_A$.

The angular acceleration of the gear B is ${\alpha }_B$.

The angular acceleration of the gear C is ${\alpha }_C$.

Calculation:

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

Convert the unit of the couple (M):

$\begin{array}{c}M=40\text{lb}\cdot \text{in}\text{.}\times \frac{1\text{lb}\cdot \text{ft}}{12\text{lb}\cdot \text{in}\text{.}}\\ =\frac{10}{3}\text{lb}\cdot \text{ft}\end{array}$

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

$\begin{array}{c}{r}_A=2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{1}{6}\text{ft}\end{array}$

Convert the unit of the radius of the outer gear B $\left(R_B\right)$:

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

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

$\begin{array}{c}{r}_B=2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{1}{6}\text{ft}\end{array}$

Convert the unit of the radius of the gear C $\left(r_C\right)$:

$\begin{array}{c}{r}_C=6\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =0.5\text{ft}\end{array}$

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

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

Substitute $1\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_A=\frac{1}{32.2}\\ =0.031056\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

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

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

Substitute $6\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_B=\frac{6}{32.2}\\ =0.18634\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Calculate the mass of the gear C $\left(m_C\right)$:

$m_C=\frac{W_C}{g}$

Substitute $9\text{lb}$ for $W_C$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_C=\frac{9}{32.2}\\ =0.2795\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Calculate the mass moment of inertia of the gear A $\left(I_A\right)$:

$I_A=m_A{\overline{k}}_A^2$

Substitute $0.031056\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_A$ and $1.3\text{in}\text{.}$ for ${\overline{k}}_A$.

$\begin{array}{c}{I}_A=0.031056\times {\left(1.3\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =0.36448\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the mass moment of inertia of the gear B $\left(I_B\right)$:

$I_B=m_B{\overline{k}}_B^2$

Substitute $0.18634\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_B$ and $3\text{in}\text{.}$ for ${\overline{k}}_B$.

$\begin{array}{c}{I}_B=0.18634\times {\left(3\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =11.646\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the mass moment of inertia of the gear C $\left(I_C\right)$:

$I_C=m_C{\overline{k}}_C^2$

Substitute $0.2795\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_C$ and $4.3\text{in}\text{.}$ for ${\overline{k}}_C$.

$\begin{array}{c}{I}_C=0.2795\times {\left(4.3\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =35.889\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

The point of contact between A and B:

$\begin{array}{l}{r}_A{\alpha }_A=R_B{\alpha }_B\\ {\alpha }_A=\frac{R_B{\alpha }_B}{r_A}\end{array}$

Substitute $\frac{1}{6}\text{ft}$ for $r_A$, and $\frac{1}{3}\text{ft}$ for $R_B$

$\begin{array}{c}{\alpha }_A=\frac{\frac{1}{3}\times {\alpha }_B}{\frac{1}{6}}\\ =2{\alpha }_B\end{array}$

The point of contact between B and C:

$\begin{array}{l}{r}_B{\alpha }_B=r_C{\alpha }_C\\ {\alpha }_B=\frac{r_C{\alpha }_C}{r_B}\end{array}$

Substitute $\frac{1}{6}\text{ft}$ for $r_B$, and $0.5\text{ft}$ for $r_C$

$\begin{array}{c}{\alpha }_B=\frac{0.5\times {\alpha }_C}{\frac{1}{6}}\\ =3{\alpha }_C\end{array}$

Therefore, the angular acceleration of the gear A is ${\alpha }_A=2{\alpha }_B=6{\alpha }_C$

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

Here, $A_x$ is the horizontal force of the gear A, $A_y$ is the vertical force of the gear A, and $F_{AB}$ is the tangential force which gear A exerts on gear B.

Refer to Figure 1.

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

$\begin{array}{l}{\displaystyle \sum M_A}=I_A{\alpha }_A\\ M-F_{AB}{r}_A=I_A{\alpha }_A\\ F_{AB}{r}_A=M-I_A{\alpha }_A\\ F_{AB}=\frac{M-I_A{\alpha }_A}{r_A}\end{array}$

Substitute $\frac{10}{3}\text{lb}\cdot \text{ft}$ for M, $6{\alpha }_C$ for ${\alpha }_A$, $\frac{1}{6}\text{ft}$ for $r_A$, and $0.36448\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_A$.

$\begin{array}{c}{F}_{AB}=\frac{\frac{10}{3}-\left(0.36448\times {10}^{-3}\times 6{\alpha }_C\right)}{\frac{1}{6}}\\ =20-0.013121{\alpha }_C\end{array}$ (1)

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

Here, $B_x$ is the horizontal force of the gear B, $B_y$ is the vertical force of the gear B, and $F_{BC}$ is the tangential force which gear B exerts on gear C.

Refer to Figure 2.

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

$\begin{array}{l}{\displaystyle \sum M_B}=I_B{\alpha }_B\\ F_{AB}{R}_B-F_{BC}{r}_B=I_B{\alpha }_B\\ F_{BC}{r}_B=F_{AB}{R}_B-I_B{\alpha }_B\\ F_{BC}=\frac{F_{AB}{R}_B-I_B{\alpha }_B}{r_B}\end{array}$

Substitute $\left(20-0.013121{\alpha }_C\right)$ for $F_{AB}$, $3{\alpha }_C$ for ${\alpha }_B$, $\frac{1}{3}\text{ft}$ for $R_B$, $\frac{1}{6}\text{ft}$ for $r_B$, and $11.646\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_B$.

$\begin{array}{c}{F}_{BC}=\frac{\left[\left(20-0.013121{\alpha }_C\right)\times \frac{1}{3}\right]-\left(11.646\times {10}^{-3}\times 3{\alpha }_C\right)}{\frac{1}{6}}\\ =\frac{\left(\frac{20}{3}-4.37367\times {10}^{-3}{\alpha }_C-34.938\times {10}^{-3}{\alpha }_C\right)}{\frac{1}{6}}\\ =40-0.23587{\alpha }_C\end{array}$ (2)

Show the free body diagram of the gear C as in Figure 3.

Here, $C_x$ is the horizontal force of the gear C, and $C_y$ is the vertical force of the gear C.

Refer to Figure 3.

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

$\begin{array}{l}{\displaystyle \sum M_C}=I_C{\alpha }_C\\ F_{BC}{r}_C=I_C{\alpha }_C\end{array}$ (3)

Calculate the angular acceleration of the gear C $\left({\alpha }_C\right)$:

Substitute $\left(40-0.23587{\alpha }_C\right)$ for $F_{AC}$, $0.5\text{ft}$ for $r_C$, and $35.889\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_C$ in Equation (3).

$\begin{array}{l}{F}_{BC}{r}_C=I_C{\alpha }_C\\ \left(40-0.23587{\alpha }_C\right)\times 0.5=35.889\times {10}^{-3}{\alpha }_C\\ 20=0.117935{\alpha }_C+35.889\times {10}^{-3}{\alpha }_C\end{array}$

$\begin{array}{l}0.153834{\alpha }_C=20\\ {\alpha }_C=\frac{20}{0.153834}\\ {\alpha }_C=130\text{rad}/{\text{s}}^2\end{array}$

Hence, the angular acceleration of the gear C $\left({\alpha }_C\right)$ is $\underset{\_}{130\text{rad}/{\text{s}}^{\text{2}}}$.

To determine

Find the tangential force which gear B exerts on gear C.$\left(F_{BC}\right)$.

Answer to Problem 16.37P

The tangential force which gear B exerts on gear C $\left(F_{BC}\right)$ is $\underset{\_}{9.33\text{lb}\uparrow }$.

Explanation of Solution

The weight of the gear A $\left(W_A\right)$ is $1\text{lb}$.

The weight of the gear B $\left(W_B\right)$ is $6\text{lb}$.

The weight of the gear C $\left(W_C\right)$ is $9\text{lb}$.

The radius of gyration of the gear A $\left({\overline{k}}_A\right)$ is $1.3\text{in}\text{.}$.

The radius of gyration of the gear B $\left({\overline{k}}_B\right)$ is $3\text{in}\text{.}$.

The radius of gyration of the gear C $\left({\overline{k}}_C\right)$ is $4.3\text{in}\text{.}$.

The couple of the constant magnitude applied to gear A (M) is $40\text{lb}\cdot \text{in}\text{.}$.

The radius of the gear A $\left(r_A\right)$ is $2\text{in}\text{.}$.

The radius of the outer gear B $\left(R_B\right)$ is $4\text{in}\text{.}$.

The radius of the inner gear B $\left(r_B\right)$ is $2\text{in}\text{.}$.

The radius of the gear C $\left(r_B\right)$ is $6\text{in}\text{.}$.

The angular acceleration of the gear A is ${\alpha }_A$.

The angular acceleration of the gear B is ${\alpha }_B$.

The angular acceleration of the gear C is ${\alpha }_C$.

Calculation:

Refer the part (a).

Calculate the tangential force which gear B exerts on gear C $\left(F_{BC}\right)$:

Substitute $130\text{rad}/{\text{s}}^2$ for ${\alpha }_C$ in Equation (3).

$\begin{array}{c}{F}_{BC}=40-\left(0.23587\times 130\right)\\ =9.33\text{lb}\uparrow \end{array}$

Hence, the tangential force which gear B exerts on gear C $\left(F_{BC}\right)$ is $\underset{\_}{9.33\text{lb}\uparrow }$.