A rigid, uniform horizontal bar of mass 55 kg and length 5.0 m is supported by t

Question

A rigid, uniform horizontal bar of mass 55 kg and length 5.0 m is supported by two vertical massless strings. String A is attached at a distance of 1.3 m from the left edge of the bar and is connected to the ceiling. String B is attached to the left edge of the bar and is connected to the floor. An object of mass 25 kg is supported by the bar at a distance of 4.1 m from its left edge.

Let the axis of rotation be at the mass m2 throughout the problem.

If the system is perfectly balanced then calculate the magnitude of the tension in both Strings A and B.

I got 1809 N as answer for tension in String A (which is right) and 1025 N as answer for String B (which was wrong); it being wrong has to do with the axis of rotation being at the mass m2.

Please help.

Explanation / Answer

Here ,

balancing the torque about string A ,

TB * 1.3 - 25 * 9.8 *(4.1 - 1.3) - 9.8 * 55 *(2.5 - 1.3) = 0

TB = 1025.2 N

the tension in string B is 1160.9 N

Now, balancing vertical forces ,

TA = TB + m3

TA = 1025.3 + 55 * 9.8 + 25 * 9.8

TA = 1809.3 N

both tension in A and B is correct

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