Chapter 13, Problem 50P

The figure shows a double-reduction helical gearset. Pinion 2 is the driver, and it receives a torque of 1200 lbf · in from its shaft in the direction shown. Pinion 2 has a normal diametral pitch of 8 teeth/in, 14 teeth, and a normal pressure angle of 20° and is cut right-handed with a helix angle of 30°. The mating gear 3 on shaft b has 36 teeth. Gear 4, which is the driver for the second pair of gears in the train, has a normal diametral pitch of 5 teeth/in, 15 teeth, and a normal pressure angle of 20° and is cut left-handed with a helix angle of 15°. Mating gear 5 has 45 teeth. Find the magnitude and direction of the force exerted by the bearings C and D on shaft b if bearing C can take only a radial load while bearing D is mounted to take both radial and thrust loads.

Problem 13–50

Dimensions in inches.

To determine

The magnitude and direction of the force exerted by the bearing $C$.

The magnitude and direction of the force exerted by the bearing $D$.

Answer to Problem 50P

The magnitude and direction of the force exerted by the bearing $C$ is ${\overrightarrow{F}}_C=\left(1564\text{lbf}\right)\widehat{i}+\left(674\text{lbf}\right)\widehat{j}$.

The magnitude and direction of the force exerted by the bearing $D$ is ${\overrightarrow{F}}_D=\left(1610\text{lbf}\right)\widehat{i}-\left(425\text{lbf}\right)\widehat{j}+\left(154\text{lbf}\right)\widehat{k}$.

Explanation of Solution

The figure below shows the forces acting on the assembly of gears on the shaft $b$.

Figure-(1)

The center of the gear 3 is $O_3$ and the centre of the gear 4 is $O_4$. The force acting tangential, radial and axial to the gear 3 are $F_{23}^t$, $F_{23}^r$ and $F_{23}^a$ respectively. The forces in x, y and z directions on the bearing $D$ are $F_D^x$, $F_D^y$ and $F_D^z$ respectively. The force acting tangential, radial and axial to the gear 4 are $F_{54}^t$, $F_{54}^r$ and $F_{54}^a$ respectively. The forces in the x and y directions on the bearing $C$ are $F_C^x$ and $F_C^y$.

Write the expression for the diameter of the gear 2.

    $d_2=\frac{N_2}{P_{n2}\cos {\psi }_2}$                                                           (I)

Here, the number of teeth on the gear 2 is $N_2$, the normal pitch of gear 2 is $P_{n2}$ and the helix angle on gear 2 is ${\psi }_2$.

Write the expression for the diameter of the gear 4.

    $d_4=\frac{N_4}{P_n\cos {\psi }_4}$                                                                          (II)

Here, the number of teeth on the gear 2 is $N_2$, the normal pitch of gear 4 is $P_{n4}$ and the helix angle on gear 4 is ${\psi }_4$.

Write the expression for the diameter of the gear 3.

    $d_3=\frac{N_3}{P_{n3}\cos {\psi }_3}$                                                                         (III)

Here, the number of teeth on the gear 3 is $N_3$, the normal pitch of gear 3 is $P_{n3}$ and the helix angle on gear 3 is ${\psi }_3$.

Write the expression for the diameter of the gear 5.

    $d_5=\frac{N_5}{P_{n5}\cos {\psi }_5}$                                                                         (IV)

Here, the number of teeth on the gear 5 is $N_5$, the normal pitch of gear 5 is $P_{n5}$ and the helix angle on gear 5 is ${\psi }_5$.

Write the expression for the transverse pressure angle for gears 2.

    ${\varphi }_{t2}={\tan }^{-1}\left(\frac{\tan {\varphi }_n}{\cos {\psi }_2}\right)$                                                                    (V)

Here, the normal pressure angle is ${\varphi }_n$.

The transverse pressure angle for gears 2 and 3 are same.

Write the expression for the transverse pressure angle for gears 4.

    ${\varphi }_{t4}={\tan }^{-1}\left(\frac{\tan {\varphi }_n}{\cos {\psi }_4}\right)$                                                                     (VI)

Here, the normal pressure angle is ${\varphi }_n$.

Write the expression for the tangential force on the gear 3.

    $F_{23}^t=\frac{T_2}{\left(\frac{d_2}{2}\right)}$                                                                                       (VII)

Here, the torque on the gear 2 is $T_2$.

Write the expression for the radial force on the gear 3.

    $F_{23}^r=F_{23}^t\tan {\varphi }_{t2}$                                                                                (VIII)

Write the expression for the axial force on the gear 3.

    $F_{23}^a=F_{23}^t\tan {\psi }_2$                                                                                (IX)

Calculate the tangential load on the gear 4.

    $\begin{array}{l}{F}_{54}^t\left(d_4\right)=F_{23}^t\left(d_3\right)\\ F_{54}^t=\frac{F_{23}^t\left(d_3\right)}{d_4}\end{array}$                                                                            (X)

Write the expression for the radial force on the gear 4.

    $F_{54}^r=F_{54}^t\tan {\varphi }_{t4}$                                                                                (XI)

Write the expression for the axial force on the gear 4.

    $F_{54}^a=F_{54}^t\tan {\psi }_4$                                                                                (XII)

Write the expression for the position vector for $CG$.

    $\begin{array}{c}{\overrightarrow{R}}_{CG}=\left(\frac{d_4}{2}\right)\widehat{j}+O_{4}C\left(-\widehat{k}\right)\\ =\left(\frac{d_4}{2}\right)\widehat{j}-O_{4}C\left(\widehat{k}\right)\end{array}$

Here, the length of $O_4$ and $C$ is $O_{4}C$.

Write the expression for the position vector for $CH$.

    $\begin{array}{c}{\overrightarrow{R}}_{CH}=\left(\frac{d_3}{2}\right)\left(-\widehat{j}\right)+C{O}_3\left(-\widehat{k}\right)\\ =-\left(\frac{d_3}{2}\right)\widehat{j}-C{O}_3\left(\widehat{k}\right)\end{array}$

Here, the length of $O_3$ and $C$ is $C{O}_3$.

Write the expression for the position vector for $CD$.

    $\begin{array}{c}{\overrightarrow{R}}_{CD}=CD\left(-\widehat{k}\right)\\ =-CD\widehat{k}\end{array}$

Here, the length of $D$ and $C$ is $CD$.

Write the expression for the force $F_{54}$ in vector form.

    ${\overrightarrow{F}}_{54}=F_{54}^t\left(-\widehat{i}\right)+F_{54}^r\left(-\widehat{j}\right)+F_{54}^a\left(\widehat{k}\right)$                                                         (XIII)

Write the expression for the force $F_{23}$ in vector form.

    ${\overrightarrow{F}}_{23}=F_{23}^t\left(-\widehat{i}\right)+F_{23}^r\left(\widehat{j}\right)+F_{23}^a\left(-\widehat{k}\right)$                                                          (XIV)

Write the expression for the force on the bearing $C$.

    ${\overrightarrow{F}}_C=F_C^x\widehat{i}+F_C^y\widehat{j}$                                                                                       (XV)

Write the expression for the force on the bearing $D$.

    $F_D=F_D^x\widehat{i}+F_D^y\widehat{j}+F_D^z\widehat{k}$                                                                              (XVI)

Write the expression for the moment at $C$.

    ${\overrightarrow{R}}_{CG}\times {\overrightarrow{F}}_{54}+{\overrightarrow{R}}_{CH}\times {\overrightarrow{F}}_{23}+{\overrightarrow{R}}_{CD}\times {\overrightarrow{F}}_D=0$                                                     (XVII)

Substitute $\left(\frac{d_4}{2}\right)\widehat{j}-O_{4}C\left(\widehat{k}\right)$ for ${\overrightarrow{R}}_{CG}$, $F_{54}^t\left(-\widehat{i}\right)+F_{54}^r\left(-\widehat{j}\right)+F_{54}^a\left(\widehat{k}\right)$ for ${\overrightarrow{F}}_{54}$, $-\left(\frac{d_3}{2}\right)\widehat{j}-C{O}_3\left(\widehat{k}\right)$ for ${\overrightarrow{R}}_{CH}$, $F_{23}^t\left(-\widehat{i}\right)+F_{23}^r\left(\widehat{j}\right)+F_{23}^a\left(-\widehat{k}\right)$ for ${\overrightarrow{F}}_{23}$, $F_D^x\widehat{i}+F_D^y\widehat{j}+F_D^z\widehat{k}$ for ${\overrightarrow{F}}_D$ and $-CD\widehat{k}$ for ${\overrightarrow{R}}_{CD}$ in the Equation (XXI).

    $\begin{array}{l}\left\{\begin{array}{l}\left[\left(\frac{d_4}{2}\right)\widehat{j}-O_{4}C\left(\widehat{k}\right)\right]\times \\ \left[\begin{array}{l}{F}_{54}^t\left(-\widehat{i}\right)+F_{54}^r\left(-\widehat{j}\right)\\ +F_{54}^a\left(\widehat{k}\right)\end{array}\right]\end{array}\right\}+\left\{\begin{array}{l}\left[\begin{array}{l}-\left(\frac{d_3}{2}\right)\widehat{j}\\ -C{O}_3\left(\widehat{k}\right)\end{array}\right]\times \\ \left[\begin{array}{l}{F}_{23}^t\left(-\widehat{i}\right)\\ +F_{23}^r\left(\widehat{j}\right)\\ +F_{23}^a\left(-\widehat{k}\right)\end{array}\right]\end{array}\right\}+\left\{\begin{array}{l}\left[-CD\widehat{k}\right]\times \\ \left[F_D^x\widehat{i}+F_D^y\widehat{j}+F_D^z\widehat{k}\right]\end{array}\right\}=0\\ \left[\begin{array}{l}\left(\frac{d_4}{2}\right)F_{54}^t\left(\widehat{k}\right)+\left(\frac{d_4}{2}\right)F_{54}^a\widehat{i}\\ +\left(O_{4}C\right)F_{54}^t\widehat{j}\\ -O_{4}C\left(F_{54}^r\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}\frac{d_3}{2}{F}_{23}^t\left(-\widehat{k}\right)+F_{23}^a\frac{d_3}{2}\widehat{i}\\ +C{O}_3\left(F_{23}^t\right)\widehat{j}\\ +C{O}_3\left(F_{23}^r\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}-CD\left(F_D^x\right)\widehat{j}\\ +CD\left(F_D^y\right)\widehat{i}\end{array}\right]=0\end{array}$ . (XVIII)

Write the expression for the force balance equation.

    ${\overrightarrow{F}}_C+{\overrightarrow{F}}_{54}+{\overrightarrow{F}}_{23}+{\overrightarrow{F}}_D=0$                                                                        (XIX)

Conclusion:

Substitute $14$ for $N_2$, $8{\text{in}}^{-1}$ for $P_{n2}$ and $30°$ for ${\psi }_2$ in equation (I).

    $\begin{array}{c}{d}_2=\frac{14}{\left(8{\text{in}}^{-1}\right)\cos 30°}\\ =\frac{14}{6.928}\text{in}\\ \approx \text{2}\text{.021}\text{in}\end{array}$

Substitute $15$ for $N_4$, $5{\text{in}}^{-1}$ for $P_{n4}$ and $15°$ for ${\psi }_4$ in equation (II).

    $\begin{array}{c}{d}_4=\frac{15}{\left(5{\text{in}}^{-1}\right)\cos 15°}\\ =\frac{15}{4.829}\text{in}\\ \approx \text{3}\text{.106}\text{in}\end{array}$

Substitute $36$ for $N_3$, $8{\text{in}}^{-1}$ for $P_{n3}$ and $30°$ for ${\psi }_2$ in equation (III).

    $\begin{array}{c}{d}_3=\frac{36}{\left(8{\text{in}}^{-1}\right)\cos 30°}\\ =\frac{36}{6.928}\text{in}\\ \approx \text{5}\text{.196}\text{in}\end{array}$

Substitute $45$ for $N_5$, $5{\text{in}}^{-1}$ for $P_{n5}$ and $15°$ for ${\psi }_5$ in Equation (IV).

    $\begin{array}{c}{d}_5=\frac{45}{\left(5{\text{in}}^{-1}\right)\cos 15°}\\ =\frac{45}{4.829}\text{in}\\ \approx \text{9}\text{.317}\text{in}\end{array}$

Substitute $20°$ for ${\varphi }_n$ and $30°$ for ${\psi }_2$ in Equation (V).

    $\begin{array}{c}{\varphi }_{t2}={\tan }^{-1}\left(\frac{\tan 20°}{\cos 30°}\right)\\ ={\tan }^{-1}\left(\frac{0.3639}{0.8660}\right)\\ ={\tan }^{-1}0.420\\ \approx 22.8°\end{array}$

Substitute $20°$ for ${\varphi }_n$ and $15°$ for ${\psi }_4$ in Equation (VI).

    $\begin{array}{c}{\varphi }_{t4}={\tan }^{-1}\left(\frac{\tan 20°}{\cos 15°}\right)\\ ={\tan }^{-1}\left(\frac{0.3639}{0.9659}\right)\\ ={\tan }^{-1}0.376\\ \approx 20.6°\end{array}$

Substitute $1200\text{lbf}\cdot \text{in}$ for $T_2$ and $2.021\text{in}$ for $d_2$ in Equation (VII).

    $\begin{array}{c}{F}_{23}^t=\frac{1200\text{lbf}\cdot \text{in}}{\left(\frac{2.021\text{in}}{2}\right)}\\ =\frac{1200\text{lbf}\cdot \text{in}}{1.0105\text{in}}\\ =1187.53\text{lbf}\\ \approx \text{1188}\text{lbf}\end{array}$

Substitute $\text{1188}\text{lbf}$ for $F_{23}^t$ and $22.8°$ for ${\varphi }_{t2}$ in the Equation (VIII).

    $\begin{array}{c}{F}_{23}^r=\left(\text{1188}\text{lbf}\right)\tan 22.8°\\ =\left(\text{1188}\text{lbf}\right)0.42036\\ =499\text{lbf}\end{array}$

Substitute $\text{1188}\text{lbf}$ for $F_{23}^t$ and $30°$ for ${\psi }_2$ in Equation (IX).

    $\begin{array}{c}{F}_{23}^a=\left(\text{1188}\text{lbf}\right)\tan 30°\\ =\left(\text{1188}\text{lbf}\right)0.57735\\ \approx 686\text{lbf}\end{array}$

Substitute $\text{1188}\text{lbf}$ for $F_{23}^t$, $\text{5}\text{.196}\text{in}$ for $d_3$ and $3.106\text{in}$ for $d_4$ in Equation (X).

    $\begin{array}{c}{F}_{54}^t=\frac{\left(\text{1188}\text{lbf}\right)\left(\text{5}\text{.196}\text{in}\right)}{3.106\text{in}}\\ =\text{1188}\text{lbf}\times 1.1673\\ \approx 1987\text{lbf}\end{array}$

Substitute $1987\text{lbf}$ for $F_{54}^t$ and $20.6°$ for ${\varphi }_{t4}$ in Equation (XI).

    $\begin{array}{c}{F}_{54}^r=\left(1987\text{lbf}\right)\tan 20.6°\\ =\left(1987\text{lbf}\right)\times 0.3758\\ =746\text{lbf}\end{array}$

Substitute $1987\text{lbf}$ for $F_{54}^t$ and $15°$ for ${\psi }_4$ in Equation (XII).

    $\begin{array}{c}{F}_{54}^a=\left(1987\text{lbf}\right)\tan 15°\\ =1987\text{lbf}\times \text{0}\text{.2679}\\ \approx \text{532}\text{lbf}\end{array}$

Substitute $\text{5}\text{.196}\text{in}$ for $d_3$, $\text{532}\text{lbf}$ for $F_{54}^a$, $746\text{lbf}$ for $F_{54}^r$, $1987\text{lbf}$ for $F_{54}^t$, $686\text{lbf}$ for $F_{23}^a$, $499\text{lbf}$ for $F_{23}^r$, $\text{1188}\text{lbf}$ for $F_{23}^t$, $3\text{in}$ for $O_{4}C$, $3.106\text{in}$ for $d_4$, $6.5\text{in}$ for $C{O}_3$, $8.5\text{in}$ for $CD$ in Equation (XVIII).

$\begin{array}{l}\left[\begin{array}{l}\left(\frac{3.106\text{in}}{2}\right)\left(1987\text{lbf}\right)\left(\widehat{k}\right)\\ +\left(\frac{3.106\text{in}}{2}\right)\left(\text{532}\text{lbf}\right)\widehat{i}\\ +\left(3\text{in}\right)\left(1987\text{lbf}\right)\widehat{j}-3\text{in}\left(746\text{lbf}\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}\frac{\text{5}\text{.196}\text{in}}{2}\text{1188}\text{lbf}\left(-\widehat{k}\right)\\ +\left(686\text{lbf}\right)\frac{\text{5}\text{.196}\text{in}}{2}\widehat{i}\\ +6.5\text{in}\left(\text{1188}\text{lbf}\right)\widehat{j}\\ +6.5\text{in}\left(499\text{lbf}\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}-8.5\text{in}\left(F_D^x\right)\widehat{j}\\ +8.5\text{in}\left(F_D^y\right)\widehat{i}\end{array}\right]=0\\ \left[\begin{array}{l}\left(3085.811\text{lbf}\cdot \text{in}\right)\left(\widehat{k}\right)\\ +\left(826.196\text{lbf}\cdot \text{in}\right)\widehat{i}\\ +\left(5961\text{lbf}\cdot \text{in}\right)\widehat{j}-\left(2238\text{lbf}\cdot \text{in}\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}\text{3086}\text{.424}\text{lbf}\cdot \text{in}\left(-\widehat{k}\right)\\ +\left(1782.228\text{lbf}\cdot \text{in}\right)\widehat{i}\\ +\left(\text{7722}\text{lbf}\cdot \text{in}\right)\widehat{j}\\ +\left(3243.5\text{lbf}\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}-8.5\text{in}\left(F_D^x\right)\widehat{j}\\ +8.5\text{in}\left(F_D^y\right)\widehat{i}\end{array}\right]=0\\ \left[\begin{array}{l}\left(3085.811\text{lbf}\cdot \text{in}\right)\left(\widehat{k}\right)\\ +\left(826.196\text{lbf}\cdot \text{in}\right)\widehat{i}\\ +\left(5961\text{lbf}\cdot \text{in}\right)\widehat{j}-\left(2238\text{lbf}\cdot \text{in}\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}\text{3086}\text{.424}\text{lbf}\cdot \text{in}\left(-\widehat{k}\right)\\ +\left(1782.228\text{lbf}\cdot \text{in}\right)\widehat{i}\\ +\left(\text{7722}\text{lbf}\cdot \text{in}\right)\widehat{j}\\ +\left(3243.5\text{lbf}\right)\widehat{i}\end{array}\right]+\left[\begin{array}{l}-8.5\text{in}\left(F_D^x\right)\widehat{j}\\ +8.5\text{in}\left(F_D^y\right)\widehat{i}\end{array}\right]=0\end{array}$

Further simplification,

    $\left\{\left(3613.924+8.5F_D^y\right)\text{lbf}\cdot \text{in}\right\}\left(\widehat{i}\right)+\left\{\left(13683-8.5F_D^x\right)\text{lbf}\cdot \text{in}\right\}\left(\widehat{j}\right)+\left(0\right)\widehat{k}=0$ (XX)

Compare the $\widehat{i}$ coordinates of Equation (XX).

    $\begin{array}{l}\left(3613.924+8.5F_D^y\right)\text{lbf}\cdot \text{in}=0\\ 3613.924\text{lbf}\cdot \text{in}=-8.5F_D^y\\ F_D^y=\frac{3613.924\text{lbf}\cdot \text{in}}{-8.5\text{in}}\\ F_D^y\approx -425\text{lbf}\end{array}$

Compare the $\widehat{j}$ coordinates of Equation (XX).

    $\begin{array}{l}13683\text{lbf}\cdot \text{in}-\left(8.5\text{in}\right)F_D^x=0\\ F_D^x=\frac{13683\text{lbf}\cdot \text{in}}{8.5\text{in}}\\ F_D^x\approx 1610\text{lbf}\end{array}$

Substitute $1610\text{lbf}$ for $F_D^x$ and $-425\text{lbf}$ for $F_D^y$ in Equation (XV).

    $\begin{array}{c}{\overrightarrow{F}}_D=\left(1610\text{lbf}\right)\widehat{i}+\left(-425\text{lbf}\right)\widehat{j}+F_D^z\widehat{k}\\ =\left(1610\text{lbf}\right)\widehat{i}-\left(425\text{lbf}\right)\widehat{j}+F_D^z\widehat{k}\end{array}$ (XX)

Substitute $\text{532}\text{lbf}$ for $F_{54}^a$, $746\text{lbf}$ for $F_{54}^r$ and $1987\text{lbf}$ for $F_{54}^t$, in Equation (XIII).

    ${\overrightarrow{F}}_{54}=\left(\left(-1986\right)\left(\widehat{i}\right)-748\widehat{j}+532\widehat{k}\right)\text{lbf}$

Substitute $686\text{lbf}$ for $F_{23}^a$, $499\text{lbf}$ for $F_{23}^r$ and $\text{1188}\text{lbf}$ for $F_{23}^t$ in Equation (XIV).

    $\begin{array}{c}{\overrightarrow{F}}_{23}=\left(\text{1188}\text{lbf}\right)\left(-\widehat{i}\right)+\left(499\text{lbf}\right)\left(\widehat{j}\right)+\left(686\text{lbf}\right)\left(-\widehat{k}\right)\\ =\left(-1188\widehat{i}+499\widehat{j}-686\widehat{k}\right)\text{lbf}\end{array}$

Substitute $\left(-1188\widehat{i}+499\widehat{j}-686\widehat{k}\right)\text{lbf}$ for ${\overrightarrow{F}}_{23}$, $\left(\left(-1986\right)\left(\widehat{i}\right)-748\widehat{j}+532\widehat{k}\right)\text{lbf}$ for ${\overrightarrow{F}}_{54}$, $\left(1610\text{lbf}\right)\widehat{i}+\left(-425\text{lbf}\right)\widehat{j}$ for ${\overrightarrow{F}}_D$ and $F_C^x\widehat{i}+F_C^y\widehat{j}$ for ${\overrightarrow{F}}_C$ in Equation (XIX).

    $\begin{array}{l}\left\{\begin{array}{l}\left[F_C^x\widehat{i}+F_C^y\widehat{j}\right]+\left[\left(\left(-1986\right)\left(\widehat{i}\right)-748\widehat{j}+532\widehat{k}\right)\text{lbf}\right]\\ +\left[\left(-1188\widehat{i}+499\widehat{j}-686\widehat{k}\right)\text{lbf}\right]+\left[\left(1610\text{lbf}\right)\widehat{i}+\left(-425\text{lbf}\right)\widehat{j}+F_D^z\widehat{k}\right]\end{array}\right\}=0\\ \left(F_C^x-1986-1188+1610\right)\widehat{i}+\left(F_C^y-748+499-425\right)\widehat{j}+\left(532-686+F_D^z\right)\widehat{k}=0\end{array}$

Further simplification.

    $\left(F_C^x-1564\right)\widehat{i}+\left(F_C^y-674\right)\widehat{j}+\left(-154+F_D^z\right)\widehat{k}=0$         (XXI)

Compare the $\widehat{i}$ coordinate of Equation (XXI).

    $\begin{array}{l}{F}_C^x-1564\text{lbf}=0\\ F_C^x=1564\text{lbf}\end{array}$

Compare the $\widehat{j}$ coordinate of Equation (XXI).

    $\begin{array}{l}{F}_C^y-674\text{lbf}=0\\ F_C^y=674\text{lbf}\end{array}$

Compare the $\widehat{k}$ coordinate of Equation (XXI).

    $\begin{array}{l}-154\text{lbf}+F_D^z=0\\ F_D^z=154\text{lbf}\end{array}$

Substitute $154\text{lbf}$ for $F_D^z$ in Equation (XX).

    ${\overrightarrow{F}}_D=\left(1610\text{lbf}\right)\widehat{i}-\left(425\text{lbf}\right)\widehat{j}+\left(154\text{lbf}\right)\widehat{k}$

Thus, the force in the bearing $D$ is ${\overrightarrow{F}}_D=\left(1610\text{lbf}\right)\widehat{i}-\left(425\text{lbf}\right)\widehat{j}+\left(154\text{lbf}\right)\widehat{k}$.

Substitute $674\text{lbf}$ for $F_C^y$ and $1564\text{lbf}$ for $F_C^x$ in Equation (XV).

    ${\overrightarrow{F}}_C=\left(1564\text{lbf}\right)\widehat{i}+\left(674\text{lbf}\right)\widehat{j}$

Thus, the magnitude and direction of the force exerted by the bearing $C$ is ${\overrightarrow{F}}_C=\left(1564\text{lbf}\right)\widehat{i}+\left(674\text{lbf}\right)\widehat{j}$.