Part A Find the position of the center of mass of the bar, x, measured from the

Question

Part A
Find the position of the center of mass of the bar, x, measured from the bar's left end.

Part B
What is the tension in the wire in the right side of the bar (T2)?

T2 = ___ N

Part C
What is the tension in the wire in the left side of the bar (T1)?

T1 = ___ N

A nonuniform, horizontal bar of mass 4.37 kg is supported by two massless wires against gravity. The left wire makes an angle 30.2 degrees, with the horizontal, and the right wire makes an angle 60 degrees. The bar has length 1.29 m. uploaded image Part A Find the position of the center of mass of the bar, x, measured from the bar's left end. Part B What is the tension in the wire in the right side of the bar (T2)? Part C What is the tension in the wire in the left side of the bar (T1)?

Explanation / Answer

given that    mass m = 4.37kg 1=30.20 2 = 600    length L = 1.29m    if T1 and T2 are torque at both ends     then from the equlibrium condition    T1 sin1 ( L/2) - T2 sin2 (L/2) = 0 ( along vertical)     T1 cos1 = T2 cos2   ( alomg horizontal)           T1 =  T2 cos2/cos1                =0.578T2               at certain position algebric sum of torque become zero       T1 sin1 (x) = T2 sin2 (L-x)       0.578T2 sin30.20 x = T2 sin60 ( L-x)           0.290x = 0.5 ( L-x)                       = 0.5L - 0.5x                 0.79x = 0.5L                  0.79x = 0.5 * 1.29                       x = 0.81               at certain position algebric sum of torque become zero       T1 sin1 (x) = T2 sin2 (L-x)       0.578T2 sin30.20 x = T2 sin60 ( L-x)           0.290x = 0.5 ( L-x)                       = 0.5L - 0.5x                 0.79x = 0.5L                  0.79x = 0.5 * 1.29                       x = 0.81

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