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Range of values of force for string to remain in tension

  1. Jul 4, 2009 #1
    The question is to find the range of values of force p such that the two strings in the picture remain in tension

    2. Relevant equations

    3. The attempt at a solution
    I tried to solve using moments about point C such that i can cancel out the tension in BC but it doesn't seem to work as the terms do not cancel each other out neither can i factorise out the terms.

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  2. jcsd
  3. Jul 4, 2009 #2


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    Moments ? This is a force problem

    Basically the 2 forces have to result in a force that has a vector with an angle of 0' to 50' relative to C.

    How can you do that?
  4. Jul 4, 2009 #3
    i don't quite seem to get you. which 2 forces are you talking about? you mean p and the 960N?
  5. Jul 4, 2009 #4


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    In statics you want the sum of the forces to net to 0.

    This means in both x and y.

    So you have the string exerting a force upward along both sides and the vertical components of those forces must add to 960N. The horizontal components must balance.

    Since they give you the angles for each then you can write 2 equations for the two tensions necessary to satisfy the picture.

    2 equations. 2 unknown tensions. 1 of them is the answer to your problem.
  6. Jul 4, 2009 #5
    ok i got it my ans is p<-1804 or p>1920 is it possible? thanks
  7. Jul 4, 2009 #6


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    Seems awfully high.

    Isn't in the x direction

    T1*Cos30 = T2*Cos50

    and in the y ...

    T1*sin30 + T2*Sin50 = 960 ?

    Solving for T1 shouldn't you get a number that's not all that different from the weight?
  8. Jul 4, 2009 #7


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    One question though.

    Can the angle of P vary in any way? Are the strings really at AB and AC? Is this why you are asking for a range? Your drawing is unclear on this point. If it is, my comments are not exactly correct. I thought it was a string pointing at P.
  9. Jul 4, 2009 #8
    AB and BC are 2 strings that are connected at the point where the 960N acts and force P is not a string but rather just a force applied at the angle of 30degrees. But from your equation, it seems that you totally ignored the tension in BC, shouldnt it be also included into the x direction?
  10. Jul 5, 2009 #9


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    I did indeed ignore it because I didn't understand from your drawing that BC was a string at all. It looked to me like a horizontal reference. I thought that P was the second string of interest.

    Is the angle of P then fixed at 30°? Or can the force P vary in angle with the horizontal in any way?
  11. Jul 5, 2009 #10
    yes it is fixed at 30 degrees. By the way i just solved for another more realistic ans for 672N<p<1920N
  12. Jul 5, 2009 #11


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    That looks more like it then. Because basically you solve both situations, i.e. the situation where one string has 0 tension, but not slack, and then the other.

    My equations solved the first condition, where BC was 0.
    I will trust that you similarly solved for when AB was 0.
  13. Jul 6, 2009 #12
    thanks pion i have solved it using your way and the answers tally with my results earlier on, thanks alot.

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