The figure shows a torque bar similar to the one you used in the lab. The torque

Question

The figure shows a torque bar similar to the one you used in the lab. The torque bar is 80.0 cm long and has a mass of 0.55kg. The two calibrated springs hold the torque bar horizontal. if alpha=70 degrees and beta=120 degrees, determine the tension in the left spring, the tension in the right spring, and the location of the center of mass of the bar with respect to the left end of the bar. Note that point A is 20.0 cm from the left end of the bar, and point B is 60.0 cm from the left end of the bar.

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^ figure

Explanation / Answer

The Force equilibrium will give us :

T1cos20 + T2cos 30 = mg (Along Y-axis or vertical direction)    (mg = 0.55 * 9.81 = 5.4 N)

T1cos20 + T2cos 30 = 5.4

T1cos70 = T2cos 60 (Along X-axis or Horizontal direction)

Taking torque at centre of mass we get (Here we take Tension perpendicular to length of bar)

T1cos20 * x - T2cos 30 * (40 - x)=0 ( Since system is in equilibrium)

Solving the equations we get x is approximately 15.46 cm from A

So Location of Centre of mass w.r.t to left end of bar is 35.46 cm approx.

T1 is approx. 3.52 N and T2 is approx. 2.41 N.

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