This can be done considering the mechanical equilibrium of the rod.

On the rod there are three forces acting. Two of them are the reactions from supports C and D. The other one is the weight of the rod. The weight of the rod is balanced by the support reactions.

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

This can be done considering the mechanical equilibrium of the rod.

On the rod there are three forces acting. Two of them are the reactions from supports C and D. The other one is the weight of the rod. The weight of the rod is balanced by the support reactions.

Considering the equilibrium of the rod AB;

`R_C+R_D = M`

`R_C+56.7 = 7.5xx9.81`

`R_C = 16.875N`

*So the support reaction at point C is 16.875N.*

**Further Reading**