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.
This is the referred image.
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