Step 1

We are given the weight of a machine $W = 50\;{\rm{lb}}$.

We are asked to determine the vertical reaction at the wheel $C$ and $B$ and the smooth contact point $A$.

 
Step 2

We will draw a free body diagram of the machine.

Here, ${R_A}$ is the reaction force at point $A$, ${R_B}$ is the reaction force at point $B$,and ${R_C}$ is the reaction force at point $C$.

The length $OC$ is $OC = 1.25\;{\rm{ft}}$.

The length $OB$ is $OB = 1.25\;{\rm{ft}}$.

The length $OA$ is $OA = 3.5\;{\rm{ft}}$.

The length $OD$ is $OD = 2\;{\rm{ft}}$.

 
Step 3

We will take moment about the $y$-axis to find the reaction force at point $A$.

\[\begin{array}{c} \sum {{M_y}} = 0\\ W \times OD - {R_A} \times OA = 0\\ {R_A} \times OA = W \times OD \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {R_A} \times \left( {{\rm{3}}{\rm{.5}}\;{\rm{ft}}} \right) = \left( {{\rm{50}}\;{\rm{lb}}} \right) \times \left( {2\;{\rm{ft}}} \right)\\ {R_A} = \frac{{\left( {{\rm{50}}\;{\rm{lb}}} \right) \times \left( {2\;{\rm{ft}}} \right)}}{{\left( {{\rm{3}}{\rm{.5}}\;{\rm{ft}}} \right)}}\\ \approx 28.6\;{\rm{lb}} \end{array}\]
 
Step 4

We will take moment about the $x$-axis.

\[\begin{array}{c} \sum {{M_x}} = 0\\ {R_B} \times OB - {R_C} \times OC = 0\\ {R_B} \times OB = {R_C} \times OC \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {R_B} \times \left( {{\rm{1}}{\rm{.25}}\;{\rm{ft}}} \right) = {R_C} \times \left( {{\rm{1}}{\rm{.25}}\;{\rm{ft}}} \right)\\ {R_B} = {R_C}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 1 \right) \end{array}\]
 
Step 5

We will resolve the forces in vertical direction.

\[\begin{array}{c} \sum {{F_z}} = 0\\ {R_A} + {R_B} + {R_C} - W = 0\\ {R_A} + {R_B} + {R_C} = W\;\;\;\;\;\;\;...\left( 2 \right) \end{array}\]

Substitute the given value in equation (2).

\[\begin{array}{c} 28.6\;{\rm{lb}} + {R_C} + {R_C} = 50\;{\rm{lb}}\\ 2{R_C} = 50\;{\rm{lb}} - 28.6\;{\rm{lb}}\\ {R_C} = \frac{{50\;{\rm{lb}} - 28.6\;{\rm{lb}}}}{2}\\ {R_C} = 10.7\;{\rm{lb}} \end{array}\]
 
Step 6

Substitute the given value in the equation (2) to find the reaction force at point $B$.

\[\begin{array}{c} {R_B} = {R_C}\\ = 10.7\;{\rm{lb}} \end{array}\]