A non-uniform rod AB, of mass 7·5 kg and length 8 m, rests horizontally in equilibrium on two smooth supports at C and D, where AC = 1·5 m and AD = 5·0 m. The reaction of the support at D on the rod is 56·7 N.

Determine the reaction of the support at C on the rod.

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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.*

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