Torque = (r)(F)(sin (theta)). R is the magnitude/absolute value of the rotation

Question

Torque = (r)(F)(sin (theta)). R is the magnitude/absolute value of the rotation axis to the force. F is the magnitude/ absolute value of that force. Theta should be 90 degrees --> sin 90 = 1 --> Net torque in the picture below is: -0.5L (2/3)w sin 90 + (1/4)(L) (w) sin 90 + (1/4)(L)(1/3)(w) sin 90. Please help me to show that net torque is also zero if we choose a rotation axis at the right end of the beam.

Note: X is the random point.

Torque = (r)(F)(sin (theta)). R is the magnitude/absolute value of the rotation axis to the force. F is the magnitude/ absolute value of that force. Theta should be 90 degrees --> sin 90 = 1 --> Net torque in the picture below is: -0.5L (2/3)w sin 90 + (1/4)(L) (w) sin 90 + (1/4)(L)(1/3)(w) sin 90. Please help me to show that net torque is also zero if we choose a rotation axis at the right end of the beam. Note: X is the random point.

Explanation / Answer

T= 1/3wx1/4L + w x1/4L - 2/3wx1/2L =0

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