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...

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

Expert Answers
jeew-m eNotes educator| Certified Educator

Please refer to the attached image with the explanations.

Let us say 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.

jeew-m eNotes educator| Certified Educator

This is the referred image.

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