compound bow

[reburbia]

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Problem attached. According to EK, the lever arm for the string at A in greater than at B in position 1. This is reverse in position 2. Lever arm is stationary so torques must equal zero, where lever arm is greater, tension force must be less.


I understand the torque idea, but I just can't see how the lever arm for A is ever greater. Where is the arm measured. Can someone please drop a little more detail on this...

 

[tsujita annex]

I dont know... i was looking at this for a day.

 

[ELTURC0]

Anybody who has an answer for this?
I am not sure where they are measuring the lever arm from.

 

[sazerac]

The lever arm is the minimum distance between the point of rotation and the line of force.

In this problem it is the distance between the DIT on the pulley, and (roughly) where the string touches the pulley.

If the pulley is not spinning, the torques must be equal, therefore F times Lever Arm must be equal.

How do the lever arms of A and B compare in Position 1? What does this day about the tension in the string A and string B in position 1?

 

[ELTURC0]

Thank you! 🙂 I really appreciate it.

So based on what you are saying, in position 1, lever arm for A is longer than lever arm for B, hence the F for A must be less than F for point B? Am I correct?

One more question, for clarification. Are you measuring the lever arm (say for position 1- point A) from point A to the dot on the pulley OR from where the string touches the pulley to the dot on pulley? This is where I got confused.

 

[sazerac]

Thank you! 🙂 I really appreciate it.

So based on what you are saying, in position 1, lever arm for A is longer than lever arm for B, hence the F for A must be less than F for point B? Am I correct?

One more question, for clarification. Are you measuring the lever arm (say for position 1- point A) from point A to the dot on the pulley OR from where the string touches the pulley to the dot on pulley? This is where I got confused.

Technically the torque is the force, times the displacement from the center, times the sine of the angle between them. You can put the force vector anywhere along the string (at point A, at the pulley, etc) and it won't make a difference. As you get closer and closer to the pulley, the displacement vector R gets smaller, but the angle between R and F approaches 90 degrees and Sin(theta) goes up so the torque is the same.

Generally the math is easiest if the angle between R and F is 90 degrees, and the messy sin(theta) term drops out.

So to answer your question, I used the point where the string touches the pulley, because there the angle between the force vector and the displacement (from pulley rotation point to force vector) is approximately 90 degrees. When you examine where the string touches the pulley from A and from B, it is obvious which has the larger moment arm (and therefore smallest force) in positions 1 and 2.