Trying to determine the resultant force F with the setup pictured below, given that L1, L2, and theta can vary. Not sure how to translate the torque applied by the weight to the force at L1. Thanks in advance!

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  • $\begingroup$ You haven't specified how $L_2$ is positioned with respect to the horizontal. $\endgroup$ – PiKindOfGuy Feb 27 at 4:21
  • $\begingroup$ If $\theta$ were $\pi/2$ then I would naturally assume that $L_2$ is horizontal, but as it stands, I don't know. $\endgroup$ – PiKindOfGuy Feb 27 at 4:22
  • $\begingroup$ What way do you want to find the force, does the question demand that the system is in equilibrium? $\endgroup$ – think__tech Feb 27 at 15:57

If there's no resistance, then there's no force either (you can't push air with a force of 10 pounds). So we can assume there's a resistance, and one that's sufficient to stop the hinge rotating.

If so, then the torque from the weight and the torque from the resistance must sum to zero. $\tau_1 + \tau_2 = 0$

Do you know how to figure the torque from those two elements? Remember that torque is the perpendicular distance from the axis based on the direction of the force. The force on the L1 arm is not directed downward, it will be directed to the left. So the angle between the force and the vector from the axis would be $\frac{\pi}{2}$.

The torque contribution from weight on the L1 arm depends on the horizontal angle of that arm, not on the angle between the two arms. Because the weight force is always downward.

  • $\begingroup$ You're right, this is assuming that F1 is pushing against an object that doesn't move, thus I'd like to know the force F1 imparts on the object. $\endgroup$ – hp43 Feb 26 at 21:13
  • $\begingroup$ I'd guess τ1 would be L2*W*sin(theta)... and T1=L1*F1*sin(zero), but that doesn't seem to be right as T1 would be zero. $\endgroup$ – hp43 Feb 26 at 21:15

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