Can you explain how you got the answer, step by step, instead of giving straight
ID: 1659886 • Letter: C
Question
Can you explain how you got the answer, step by step, instead of giving straight calculations plz ty
A uniform bar of mass m is suspended horizontally. A rope of tension Ti, which is secured around its center of mass, pulls vertically to hold it up. To the right, at a distance l1 from the center of the bar is a mass M. On the left at a distance l2 from the center of the bar another rope pulls the bar downward. The tension in this second rope is T2. TI 12 T2 If m-2.5kg, M=1.2kg, l,-1.9m, and 12-2.1m, what it Ti? Answer in Newtons (N).Explanation / Answer
As bar is in equilibrium, net force and the net torque on the Bar must be zero.
Net torque on Bar will be zero about any point. To get rid of T2, we take torque about the point where second rope is connected to bar, that is l2 to left from center of Bar.
There ar 4 force acting on the Bar
it's weight, mg at the the center, in downward direction
Weight of mass M, Mg at l1 from center on the right side, in the downward direction
Tension in the first rope T1 at the center, in the upward direction
tension in the second rope T2 at l2 from center, in dowanward direction.
Corresponding torques due to these forces about the point where rope 2 connects bar are
mg l2 ( clockwise) = 51.45 Nm
Mg (l1 + l2) ( clockwise) = 47.04 Nm
T1 (l2) ( anticlockwise) = 2.1 T1 Nm
T2 ( 0 ) = 0
Net torque = 0 => sum of the torques in clockwise direction = sum of torques in anticlockwise direction
51.45 + 47.04 = 1.9 T1
T1 = 46.9 N
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