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Torque Angles

  1. Mar 12, 2008 #1
    The problem

    I know the process to solve the problem but I don't understand what I'm supposed to do with the 16 angle. Aren't the other angles already measured to the horizontal plane of the axis of rotation? If this is the case, there is no need to subtract 16 from the 80 and add 16 to the 60. However, why would they give me 16 then?

    Thank you for helping in advance!
     
  2. jcsd
  3. Mar 12, 2008 #2

    Dick

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    You don't need the 16 degree angle to balance the vertical forces on the bar. You do need it to balance the torques on the bar. The torque exerted by each tension is F*r*sin(theta) where theta is the angle of the force relative to the axis of the bar, correct?
     
  4. Mar 12, 2008 #3
    I was going to find the tensions using the fact that the sum of Fx and sum of Fy are 0 N in equilibrium. So I wanted to solve for the right tension in terms of the left.

    But I guess it would be easier to set the sum of the torques equal to 0 instead. Is 16 the angle of the gravitational force relative to the axis of the bar?
     
  5. Mar 12, 2008 #4

    Dick

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    Don't assume that the tensions are equal. They aren't. Since you have two unknown tensions you need two equations to solve for them. The sum of the vertical components of all of the forces AND the sum of the torques BOTH have to be zero.
     
  6. Mar 12, 2008 #5
    I'm sorry but I don't quite understand. The tensions aren't equal and yet I need to equate them? Plus, how can I know the vertical force without having the tension?

    If the bar isn't moving then there are no forces at work so why aren't the tensions equal at least in their x and y components? My two equations would be:

    (1) sum of Fx = FL*cos(thetaL) - FR*cos(thetaR)
    FR = [FL*cos(thetaL)] / cos(thetaR)

    (2) sum of Fy = FL*sin(thetaL) + FR*sin(thetaR) - mg
    FL = [-FR*sin(thetaL) + mg)] / sin(thetaL)

    Plug in the expression for FL into that of FR to get:

    FR = ( [-FR*sin(thetaL) + mg)] / sin(thetaL) )*cos(thetaL) / cos(thetaR)

    I did a problem similar to this in this way and got the correct answer so I guess I'm just really puzzled as to how that happened if this isn't the right way to do it.
     
  7. Mar 12, 2008 #6

    Dick

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    Apologies. You titled the problem with 'Torque angles' and I was thinking torques. I think your way of solving it should also work out correctly. And yes, in that case you don't need the 16 degree angle measure. You could also do it by balancing torques. I've been trying to work it out both ways to make sure they agree. Which actually means to make sure I understand the problem correctly.
     
  8. Mar 12, 2008 #7
    I'm sorry too! I titled it this way because this is a part of a lab involving torque. I'm confused as to why they put in a problem that has little to do with the hardcore concept of torque itself too. Just out of curiosity, could you explain to me how I would do it with torques? The entire concept is rather new to me despite the fact that I'm in my third year of physics.

    I know that equilibrium the sum of the torques must be equal. Therefore I should have an equation like this:

    0 = FR*r*sin(80+16) + FL*r*sin(60-16)

    r would be 0.4 m because the center of mass, the point of rotation, is in the middle of the bar? Or am I supposed to do one equation where the point of rotation is where the string on the right is connected and another where the string on the left is connected?
     
  9. Mar 12, 2008 #8

    Dick

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    No, you are on the right track. Except that the torque on the left is trying to rotate the object clockwise and the one on the right counter clockwise. So they should be subtracted to equal zero. And I've got to confess the angles are all swimming in my head. Maybe I can straighten this out tomorrow if you are still having problems?
     
  10. Mar 12, 2008 #9
    Like you said before, I need another equation to solve for FL and FR. I think I can use the summation of Fx or that of Fy, correct?

    If it's no trouble to you to explain to me about the angles after you've figured it out, I would really appreciate that.

    Thank you so much for everything so far. You've really helped me a lot with this devil of a problem.
     
  11. Mar 12, 2008 #10

    Dick

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    Ok, I think the angle on the left should be 84 degrees (180-(80+16)) and the angle on the right should be 44 degrees (I agree with you on that one).
     
    Last edited: Mar 12, 2008
  12. Mar 12, 2008 #11

    Dick

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    FL and FR are just the magnitudes of the two tensions. They aren't components.
     
  13. Mar 12, 2008 #12
    I can't do this?

    sum of Fx = FL*cos(thetaL) - FR*cos(thetaR)
    FR = [FL*cos(thetaL)] / cos(thetaR)

    Plug into

    0 = FR*r*sin(84) + FL*r*sin(44)
     
  14. Mar 12, 2008 #13

    Dick

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    The theta's are different for the Fx calculation versus the torque calculation, aren't they? You might be as tired as I am. Give it another thought in the morning.
     
  15. Mar 13, 2008 #14

    Dick

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    I've changed my mind. You can't solve it your way. Know why? Because the rod won't hang at rest in the position described. If you try to do what you just did and solve the F_x and torque equations simultanteously you'll see (and you probably already did) that the equations have no solution (except for FR=FL=0). In other words, you can adjust the tensions in such a way that there is no y-acceleration and either i) no rotation or ii) no x-acceleration but not both simultaneously. This is a badly set up problem. You'd better do it their way if you want to get their answer. No wonder it was so confusing.
     
    Last edited: Mar 13, 2008
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