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.

Calculate the distance of the centre of gravity of the rod from C

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Please refer the attached image with the explanations.

Let us sat the center of gravity of the rod is at point E. Since the rod is in equilibrium point E lies between C and D.

According to the given data;

`AB = 8m`

`AC = 1.5m`

`AD = 5m`

`M = 7.5kg`

`R_D = 56.7N`

Assume that centre of gravity is Xm away from point C.

Let us take moments for the rod from the point C;

`M_c rarr R_Dxx(5-1.5) = MgX`

`M_c rarr 56.7(5-1.5) = 7.5xx9.81X`

`M_c rarr X = 2.7m`

*So the centre of gravity of the rod is 2.7m from the point C.*

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This is the referred image.

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