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Urgent help on mechanics!

  1. May 26, 2005 #1
    I am revising for a test which is next week. I came across this question and I keep getting the wrong answer. Someone please help!

    A particle has mass 2kg. It is attached at B to the end of two light inextensible strings AB and BC. When the particle is in equilibirum, AB makes an angle of 30 degrees with the vertical. The magnitude of the tension in BC is twice the magnitude of the tension in AB.

    Find, in degress to one decimal place, the size of the angle that BC makes with the vertical.
    :confused:
     
  2. jcsd
  3. May 26, 2005 #2

    HallsofIvy

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    Do this in terms of "components". Let T be the tension is the string AB. Then the x component of force for the AB string is Tcos(30) and the y component is Tsin(30).
    We don't know the angle the BC string makes with the vertical so call it θ. We do know that the tension in the BC string is "is twice the magnitude of the tension in AB." so it is 2T. The x component of force for the BC string is -2T sin(θ) (notice the negative- the two string pull in opposite directions horizontally) and the y component is 2Tcos(θ) (positive: both string pull upwards).
    Since the object is not moving horizontally, the two x components must "offset":
    T sin(30)- 2T sin(θ)= 0 or T sin(30)= 2T sin(θ).
    Since the object is not moving vertically, the two y components, together, must equal the weight of the object: Tcos(30)+ 2Tcos(&theta)= 2g.
    Can you solve those two equations for T and θ?
     
  4. May 26, 2005 #3
    I worked out the value of theta to be 90.4 degrees. I dont know where I am going wrong!
     
  5. May 26, 2005 #4
    I think you did a lilttle mistake. The X component of the AB string should be Tcos 60 and the y components is Tcos 30.
     
  6. May 26, 2005 #5
    oh sorry.my bad. Well I try again and get back to you!. Thanx
     
  7. May 26, 2005 #6
    The introductio of sin into the equation doesnt let me continue any further???
     
  8. May 26, 2005 #7
    Post your work, so we can help more!
     
  9. May 26, 2005 #8

    Doc Al

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    Hint: All you need consider are the horizontal components of the forces on the mass. Start there and show your work
     
  10. May 26, 2005 #9
    ok. I got two equations. 2Tcosθ + Tcos30 = 2g............1
    Tcos60 = 2Tcos(90-θ) ............2
    I have two simultanous equations. I rearrange equation 2 to make T the subject, and then sub T into equation 1.

    Am I going the right way?
     
  11. May 26, 2005 #10

    Doc Al

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    The second equation is all you need.
     
  12. May 26, 2005 #11
    how? that doesnt make sense.
    Tcos60 = 2Tcos (90-θ)
    since cos 90 = 0, the equation simplifies to: Tcos60 = -2Tcos θ
    cancel one T on each side>>>> cos60 = -tcosθ Now??
     
  13. May 26, 2005 #12

    Doc Al

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    FYI: In general, [itex]\cos (A - B) \ne \cos A - \cos B[/itex] !!!

    Instead, cancel the T and get:
    cos60 = 2cos (90-θ), and then
    cos (90-θ) = (cos60)/2

    Use your calculator.
     
  14. May 26, 2005 #13
    Oh no!!! Did I just make the mistake of the century? This is soo embarassing. OBVIOuSLY 2T/T is not T!!...SHOOT ME...and yes, what you said is right. How could I forget these simple rules??? all right..let me try again
     
  15. May 26, 2005 #14
    theta = 14.5 degrees!
     
  16. May 26, 2005 #15

    Doc Al

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    You got it.
     
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