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