The bearing consists of rollers, symmetrically confined within the housing. The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft. Determine the normal reactions N_B and N_C on the bearing at its contact points B and C for equilibrium.

Solution:

Free-body diagram of the roller:

Equations of Equilibrium:

Note that if we take the sum of forces in the x-direction, there are two unknown forces involve, but if we take the sum of forces in the y-direction, there is only one unknown force involve.

Summation of forces in the y-direction:

\begin{aligned}
+\uparrow \sum F_y & =0& & & & & \\
125- N_C \cos 40 \degree &=0  & & & & &\\
N_C &=\dfrac{125}{\cos 40 \degree} & & & & &  \\
N_C & =163.1759 \ \text{N} \\
\end{aligned}

Summation of forces in the x-direction:

\begin{aligned}
\xrightarrow{+} \sum F_x & =0 \\
N_B - 163.1759\ \sin 40 \degree &=0 \\
N_B &=163.1759 \sin 40\degree \\
N_B & = 104.8874 \ \text{N}
\end{aligned}

Therefore, the normal reactions N_B and N_C on the bearing at its contact points B and C for equilibrium are 163.1759 N and 104.8874 N, respectively.