1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum tension

  1. Apr 9, 2008 #1
    [​IMG]
    The maximum force of the block, T, must be determined. The 2nd rope that the pulley is attached to has the same maximum tension force as the rope of the pulley which is 10kN.
    Theta also needs to be determined.
     
  2. jcsd
  3. Apr 9, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Could you possibly write the sum of all the forces on the pulley and set it equal to zero? There are two unknown tensions in the problem, T and the unnamed tension between the pulley and the support point. Balancing the forces should give you two equations in two unknowns.
     
  4. Apr 10, 2008 #3
    is this what you mean?
    make T = 10 since it will undergo the highest tension.
    change in y = 10cos0-w=0
    how do i do the x component?
     
  5. Apr 10, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm not even completely sure what your original question means. I was hoping if you started solving it I could figure that out, but I guess I was wrong. Why T=10? y=10cos0-w=0 looks kind of like a force balance equation for the weight. Is that what it's supposed to be? How about balancing the forces on the pulley?
     
  6. Apr 10, 2008 #5

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The way I interpret it is the following: There are three unknowns (the two tensions and the angle) and the constraint that neither tension must be over 10 kN.

    The approach I would suggest would be to first assume that the tension T is the maximum value, 10 kN. And then solve for the other tension and the angle theta. If the other tension is less than 10 KN then that's it, that's the final answer to the problem.
    If the other tension is above 10 kN, one has to start all over again by setting that other tension equal to 10kN and finding T and theta.
     
  7. Apr 10, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ah! That makes sense, thanks.
     
  8. Apr 10, 2008 #7
    this is what I had in mind but i dont know how to solve for the other tension. Can you show me how to do that?
     
  9. Apr 10, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You have a direction (angle) and magnitude of each of the three forces acting on the pulley. Split each into x and y components. You'll need to use a trig function of the angles.
     
    Last edited: Apr 10, 2008
  10. Apr 10, 2008 #9
    This is what I came up with
    sum of y components = 10cos0 +Fw + bcos30
    sum of x components = bsin30 + 10sin0
    I cant solve it because there's 3 unknowns
     
  11. Apr 10, 2008 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, I'm guessing you are using b as the unknown tension and the symbol '0' to represent 'theta' (that's not a very good choice - just write e.g. cos(theta)). If that's the case then you are missing some signs. For example, one of the x forces should be positive and the other negative. They are pointing in opposite directions. Not all of the y forces are pointing in the same direction either. As for the other unknown 'Fw', the tensions on the two sides of the pulley are equal. So Fw isn't unknown, it's the same as T. Now there are two unknowns, b and theta.
     
  12. Apr 10, 2008 #11
    yes b = unknown tension
    Assuming T and Fw are both negative forces then
    -10cos(theta)-10+bcos30=0
    bsin30-10sin(theta)=0
    Is this correct?
    The answer I got was 60 degrees.
     
    Last edited: Apr 10, 2008
  13. Apr 10, 2008 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, the equations are correct. And, yes, I seem to get 60 degrees as well.
     
  14. Apr 10, 2008 #13
    No matter what value I have for T the answer is always 60 degrees. Should this matter?
     
  15. Apr 10, 2008 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, you finally wind up with an equation for theta in which the T cancels out. This is a good thing. Is b>10kN? If so you will have to resolve the problem with b=10kN and then figure out what T is. If everything varies in proportion, this will save you a lot of time.
     
  16. Apr 11, 2008 #15
    Is there a way to solve for b without having to make theta the subject of the equation.
     
  17. Apr 11, 2008 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Solve one equation for sin(theta) and the other for cos(theta). Square and add them. sin(theta)^2+cos(theta)^2=1. Poof, no more theta.
     
  18. Apr 12, 2008 #17
    -10cos(theta)-10+bcos30=0
    bsin30-10sin(theta)=0

    Im beginning to doubt that these equations are correct.
    I think that either the -10 or the -10cos(theta) shouldn't be in there because they are the same force. Pulleys just change the direction of the force.
    Does anyone agree with me?
     
  19. Apr 12, 2008 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I don't. Those two terms are vertical components of two 10kN forces pointing in different directions. You were almost done with this problem. What's bothering you?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Maximum tension
  1. Maximum deflection (Replies: 1)

  2. Maximum Hieght (Replies: 3)

  3. Maximum stress (Replies: 1)

  4. Maximum Temperature (Replies: 1)

Loading...