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Free body diagram help

by emoseman
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emoseman
#1
Nov9-08, 01:48 PM
P: 13
1. The problem statement, all variables and given/known data


I'm trying to setup a free body diagram for this problem, and I'm not even sure where to start with the angles. That is basically my question, when do I use cos and when do I use sin. It has never made sense...

Links to tutorials, or an explanation on setting up a free body diagram would be enormously helpful.

Thanks!
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asleight
#2
Nov9-08, 01:52 PM
P: 154
Quote Quote by emoseman View Post
1. The problem statement, all variables and given/known data


I'm trying to setup a free body diagram for this problem, and I'm not even sure where to start with the angles. That is basically my question, when do I use cos and when do I use sin. It has never made sense...

Links to tutorials, or an explanation on setting up a free body diagram would be enormously helpful.

Thanks!
Set-up your axis along the regular x/y coordinates and then shift the forces away from the axes and compare the angles with one another. What are you drawing a FBD for right now?
emoseman
#3
Nov9-08, 02:04 PM
P: 13
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!

PhanthomJay
#4
Nov9-08, 09:25 PM
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Free body diagram help

Quote Quote by emoseman View Post
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!
Yes, as long as you correctly identify the value of d and F. There are 2 ways to calculate torque. One way is the vector cross product of the force vector times the position vector. The result is Torque= Fd sin theta, where theta is the angle between the 2 vectors. The sign of the torque, positive or negative, is determined by whether it tends to rotate the rod clockwise of counterclockwise. In this example , looking at torques about the pivot point, the torque created by the mass is mg(.40)(sin 120).

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 .
emoseman
#5
Nov10-08, 11:31 AM
P: 13
Thank you very much for your reply. The first method you document is using the cross-product of the two vectors, which I understand. But, where does the 120 degrees come from in "mg(.40)(sin 120)"?

Thanks again!

--
Evan
PhanthomJay
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Nov10-08, 12:14 PM
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Quote Quote by emoseman View Post
Thank you very much for your reply. The first method you document is using the cross-product of the two vectors, which I understand. But, where does the 120 degrees come from in "mg(.40)(sin 120)"?

Thanks again!

--
Evan
120 degrees is the angle between the force vector , mg, and the position vector, d. It comes from the fact that supplementary angles add up to 180 degrees (a straight line), (180-60=120).
emoseman
#7
Nov10-08, 12:39 PM
P: 13
Quote Quote by PhanthomJay View Post
120 degrees is the angle between the force vector , mg, and the position vector, d. It comes from the fact that supplementary angles add up to 180 degrees (a straight line), (180-60=120).
Sorry for being elementary with this, but this is a fundamental concept I'm trying to understand better because it is causing me a lot of headaches. The position vector d is the bar coming off the right of the pivot point and at the end hangs the weight. And the force vector mg is the weight hanging on the end of the rod. Why isn't the angle between the two vectors 60 degrees? I realize that sin(60) == sin(120), but I'm more interested in the choice of angles. I would have chosen 60 degrees because that represents the angle between the two vectors as well.
PhanthomJay
#8
Nov10-08, 01:20 PM
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Quote Quote by emoseman View Post
Sorry for being elementary with this, but this is a fundamental concept I'm trying to understand better because it is causing me a lot of headaches. The position vector d is the bar coming off the right of the pivot point and at the end hangs the weight. And the force vector mg is the weight hanging on the end of the rod. Why isn't the angle between the two vectors 60 degrees? I realize that sin(60) == sin(120), but I'm more interested in the choice of angles. I would have chosen 60 degrees because that represents the angle between the two vectors as well.
Since sin theta = sin(180 - theta), it's really the same thing, but the definition of the torque cross product is (the magnitude of the force) times the (magnitude of the position vector) times the sine of the included angle between the two. It may be helpful to have each vector emanating from a common point, for example, at the point where the hanging mass and the rod meet, with the force vector pointing down, and the position vector pointing up and to the left, along the axis of the rod. This is definitely helpful when determining the vector direction of the torque along the positive or negative z axis perpendicular to the plane in which the force and position vectors lie.
physics girl phd
#9
Nov10-08, 01:31 PM
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P: 936
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).
emoseman
#10
Nov10-08, 01:40 PM
P: 13
Quote Quote by PhanthomJay View Post
Since sin theta = sin(180 - theta), it's really the same thing, but the definition of the torque cross product is (the magnitude of the force) times the (magnitude of the position vector) times the sine of the included angle between the two. It may be helpful to have each vector emanating from a common point, for example, at the point where the hanging mass and the rod meet, with the force vector pointing down, and the position vector pointing up and to the left, along the axis of the rod. This is definitely helpful when determining the vector direction of the torque along the positive or negative z axis perpendicular to the plane in which the force and position vectors lie.
Okay I think I see it now, thanks!
emoseman
#11
Nov10-08, 01:46 PM
P: 13
Quote Quote by physics girl phd View Post
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).
This brings up my next question which is in regards to the free body diagram for the other side of this bar which is the wire attached to the rod at 110 degrees. The method of splitting up the force into its x and y axis components makes sense and does provide extra equations when trying to solve for unknowns. But, when I set these up I don't know how to figure out the angles. When do I use sin and when do I use cos?

This is a fundamental problem I've been running into all semester... :(

Thanks for your help!

--
Evan
PhanthomJay
#12
Nov10-08, 02:44 PM
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Quote Quote by emoseman View Post
This brings up my next question which is in regards to the free body diagram for the other side of this bar which is the wire attached to the rod at 110 degrees. The method of splitting up the force into its x and y axis components makes sense and does provide extra equations when trying to solve for unknowns. But, when I set these up I don't know how to figure out the angles. When do I use sin and when do I use cos?

This is a fundamental problem I've been running into all semester... :(

Thanks for your help!

--
Evan
Generally I prefer the 2nd method of determining torques, which is the force times perpendicular distance method alluded to above. In this particular problem, the first method is simpler. For example, the torque caused by the tension force, F, is just (F)(.40)(sin70). If you want to use the 2nd method, which I use 90% of the time, remember that torque is equal to the force times the perpendicular distance from the line of action of the force to the pivot point. So for the torque caused by the wire tension, draw a line between the wire and pivot point that is perpendicular to the wire. That line represents the perpendicular distance. Now using a little basic geometry and the trigonometric properties of a right triangle, you should see that this distance is (.40)(cos20), or (.40)(sin70), if you wish. Alternatively, as Physics Girl suggests, you could break up the tension force into its x and y components, and sum the torques of each component times its perpendicular distance to the pivot, but the geometry and trig becomes a bit more difficult.
emoseman
#13
Nov10-08, 03:22 PM
P: 13
Quote Quote by PhanthomJay View Post
Generally I prefer the 2nd method of determining torques, which is the force times perpendicular distance method alluded to above. In this particular problem, the first method is simpler. For example, the torque caused by the tension force, F, is just (F)(.40)(sin70). If you want to use the 2nd method, which I use 90% of the time, remember that torque is equal to the force times the perpendicular distance from the line of action of the force to the pivot point. So for the torque caused by the wire tension, draw a line between the wire and pivot point that is perpendicular to the wire. That line represents the perpendicular distance. Now using a little basic geometry and the trigonometric properties of a right triangle, you should see that this distance is (.40)(cos20), or (.40)(sin70), if you wish. Alternatively, as Physics Girl suggests, you could break up the tension force into its x and y components, and sum the torques of each component times its perpendicular distance to the pivot, but the geometry and trig becomes a bit more difficult.
Wow I think I finally am starting to see it. I'm not sure why this of all things this is giving me the most difficulty. So in using the method of getting the proper distance from a perpendicular line from the wire I can see now how to construct the triangle. And, I would use cosine to find the adjacent side of the triangle, because that is the distance in question.

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

--
Evan


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