A nonuniform, horizontal bar of mass 3.43 is supported by two massless wires aga

Question

A nonuniform, horizontal bar of mass 3.43 is supported by two massless wires against gravity. The left wire makes an angle 36.6 , with the horizontal, and the right wire makes an angle 62.9 . The bar has length 1.55 . Find the position of the center of mass of the bar, , measured from the bar's left end.

Explanation / Answer

It may help you a lot Let F1 and F2 be the forces exerted by lift and right wire on the bar. Both forces act upward in direction of the wires. So you can split the forces into x- and y-component. F1x = -F1 · sin(f1) F1y = F1 · cos(f1) F2x = F2 · sin(f2) F2y = F2 · cos(f2) As always for such a plane system in equilibrium you can balances horizontal forces(x) , vertical forces (y) and the torques about an arbitrary fulcrum. Set up balance of forces acting on the bar in x-direction F1x + F2x = 0 -F1 · sin(f1) + F2 · sin(f2) = 0 F1/F2 = sin(f2)/sin(f1) From the balance in y-direction you could calculate the magnitude of F1 and F2 but this is not the focus, so skip this step. As fulcrum for torque balance choose the point directly above the center of mass at the upper edge of the bar. By this choice there are no torques exerted by the F1x and F2x nor the weight of the bar. - F1y · x + F2y · (L-x) = 0 => x = L · F2y / (F1y + F2y) x = L / ( (F1y/F2y ) + 1) express forces in terms of magnitude an angle x = L / ( (F1/F2)·(cos(f1)/cos(f2)) + 1) use ratio derive from x-direction balance x = L / ( (sin(f2)/sin(f1))·(cos(f1)/cos(f2)) + 1) x = L / ( (tan(f2)/tan(f1) + 1)

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