Calculate a torque from a pressure

  • #26
Dale
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It is not so difficult than that
Then why are you getting results that you know are wrong? I recommend either analyzing this device with a simpler method or analyzing a simpler device using this method.

If you don't want to listen to my recommendations, then I can only wish you good luck.
 
  • #27
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Then why are you getting results that you know are wrong?
Because if I rotate the device of an angle, for example 0.1 rd, then the sum of the energy is not at 0. I studied some others examples, easier, and I always found the sum of energy at 0. For example, if I take the center of rotation of the black arm in the red dot, then the sum of energy is 0.
I found the sum of the forces at 0. I found the sum of the torques around the red dot (without the green dot) at 0. I need a little help for the work of the black arm.
 
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  • #28
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I change the expression of the force F3 because I divided by 4 but it's because I knew the result (sum of forces =0). So the integrals are:

##F_3x = \int_{-\pi/2}^{\pi/2} \int_{0}^{0.5} x \frac{\cos(\arctan(\frac{1.5+x\sin(y)}{x\cos(y)}))}{(x\cos(y))^2 + (1.5+x\sin(y))^2} dxdy = 0.0264##

##F_3y = \int_{-\pi/2}^{\pi/2} \int_{0}^{0.5} x \frac{\sin(\arctan(\frac{1.5+x\sin(y)}{x\cos(y)}))}{(x\cos(y))^2 + (1.5+x\sin(y))^2} dxdy = 0.182##

The results are the same. Note this is polar coordinates, 'x' is 'r' and 'y' is the angle.
 

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