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Free body diagram help 
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#1
Nov908, 01:48 PM

#2
Nov908, 01:52 PM

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#3
Nov908, 02:04 PM

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I was going to start with the 3.0Kg weight at the end of the bar. The image shows that it is 60 degrees less than the horizontal plane of the bar. But, what doesn't make sense, is what is that in relation to? The weight is acting vertically, but it is hanging off the end of the bar at 60 degrees, so the torque applied to the bar is modified by that angle somehow. The equation that is in the book is dFsinTheta, but the example in the book shown with that equation has the angle in the positive above the x axis.
Wow, I'm running around in circles... :) Is the torque caused by the weight going to be calculated by dFsin(60) ? Thanks! 


#4
Nov908, 09:25 PM

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Free body diagram help
The alternate way to find a torque is multiply the force times the perpendicular distance from the line of action of that force to the pivot point. Doing it this way, the torque created by the mass is mg(.40)(sin60), or mg(.40)(cos 30). It all leads to the same result. Now find the torque created by the tensile force, and solve for the tension by summing torques = 0 . 


#5
Nov1008, 11:31 AM

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Thank you very much for your reply. The first method you document is using the crossproduct of the two vectors, which I understand. But, where does the 120 degrees come from in "mg(.40)(sin 120)"?
Thanks again!  Evan 


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Nov1008, 12:14 PM

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Nov1008, 12:39 PM

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Nov1008, 01:20 PM

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#9
Nov1008, 01:31 PM

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I personally tend NOT to recommend using the angle formulation of torque (especially if you are having difficulty with choosing what angle to plug into a trig formula, and the problems can give you either angle!). What I recommend instead is to draw the free body diagram for the bar then replace the forces that that act at odd angles with their parallel and perpendicular components (relative to the bar).... Using this method, the torque due to each force is the perpendicular component of the force times the lever arm length. This way, after all, strikes at the true physical meaning of torque.
The method also sets up things nicely for if you need the net force equations (to determine multiple unknown parameters through a set of equations: You can easily set up two additional equations setting net parallel force to zero and net perpendicular force to zero (using the bar coordinates as opposed to using vertical and horizontal coordinates). 


#10
Nov1008, 01:40 PM

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#11
Nov1008, 01:46 PM

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This is a fundamental problem I've been running into all semester... :( Thanks for your help!  Evan 


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Nov1008, 02:44 PM

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#13
Nov1008, 03:22 PM

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Well I can't thank you all enough, you have been a huge help with this!  Evan 


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