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Homework Help: Tension help?

  1. Nov 7, 2008 #1
    Tension help???

    Consider the 59 N weight held by two cables. The left-hand cable is horizontal.

    What is the tension in the cable slanted
    at an angle of 51◦? Answer in units of N.

    What is the tension in the horizontal cable?
    Answer in units of N
     
  2. jcsd
  3. Nov 8, 2008 #2

    Dick

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    Re: Tension help???

    You have three force vectors pointed out from the point where the cables meet. The sum must be zero. Now split the forces into x-y components and equate the horizontal and vertical forces. Now stop double posting. Now start showing an attempt at solving the problem if you expect significant help.
     
  4. Nov 8, 2008 #3
    Re: Tension help???

    Sorry for double posting but, I was absent today in school when we learned this, and my teacher expects me to know how to do this. My grandmother died and i had to go to her funeral. What do you mean by splitting them into x-y components, and isn't their only 2 vectors pointed out
     
  5. Nov 8, 2008 #4

    Dick

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    Re: Tension help???

    There are three forces. One to the left that's horizontal. One the the right that's upwards at a 51 degree angle. And one that's downwards with a magnitude of 59N. The vertical forces have to cancel first. Only two of them have a vertical component. Please don't say you don't know how to split a force into vertical and horizontal components. Use sin(51 degrees) and cos(51 degrees), ok? Sorry about your grandmother.
     
  6. Nov 8, 2008 #5
    Re: Tension help???

    Nah it's fine and thanks for helping me though.


    51sin= .777145 51cos= .6293

    Now after i split them what do i have to do with the numbers?
     
  7. Nov 8, 2008 #6

    Dick

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    Re: Tension help???

    Call the magnitude of the 51 degree force F. Then the upward part of F is F*sin(51 degrees), assuming 51 degrees is the angle from the horizonal, isn't it? So isn't F*sin(51 degrees)=59N? Since the vertical components have to balance? Solve for F. Now balance the horizontal components. Sorry, I've got to zzzzz now.
     
  8. Nov 8, 2008 #7

    tiny-tim

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    welcome to pf!

    hi bamb3ry! welcome to pf! :smile:
    i'll say this quietly so as not to wake Dick :zzz: …

    in the class you missed, the teacher probably said that forces are vectors, and so when you add forces, you must add them like vectors ("vector addition") …

    you can add vectors either by using a vector triangle

    or by taking the x and y components, and adding them separately …

    the three forces (two unknown tensions and the known weight) must add to zero, so the x and y components will give you two equations wiht two unknowns, which you should be able to solve :smile:

    shhh … :blushing:

     
  9. Nov 8, 2008 #8

    Dick

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    Re: Tension help???

    Thanks for keeping the volume down! I didn't hear a thing. Yaawwwn.
     
  10. Nov 8, 2008 #9

    tiny-tim

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    oooh … can we make a NOISE now? :biggrin:

    :rolleyes: so long as we don't fight!​
     
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