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Need to find the tension at T in left-hand cable

  1. Oct 6, 2007 #1
    Hello. I'm new here and in need of some help.

    I have a problem here in which two cables are supporting a weight of 631 N. The right-hand cable has a tension of 680 N and and makes an angle of 32 degrees with the ceiling. The left-hand cable has a tension of T and makes an angle of X at the wall.

    I need to find the tension at T in left-hand cable in respect to the wall.

    I also need to find the angle that the left-hand cable makes with respect to the wall. I thought it should be 58 degrees, but apparently that is not correct.
     
  2. jcsd
  3. Oct 6, 2007 #2

    learningphysics

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    Hi bikerkid. try to write out these equations:

    [tex]\Sigma{F_x} = ma_x[/tex] which becomes [tex]\Sigma{F_x} = 0[/tex]

    and [tex]\Sigma{F_y} = ma_y[/tex] which becomes [tex]\Sigma{F_y}=0[/tex].

    So if T is the tension of the unknown then the vertical component is Tsin(X). Horizontal component is Tcos(X).
     
  4. Oct 6, 2007 #3
    Thank you. I got the first part right. I just don't know what to do for the second part.
     
  5. Oct 6, 2007 #4

    learningphysics

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    cool. can you show how you got the first part? that way it'll be easier to show what you need to do next.
     
  6. Oct 7, 2007 #5
    I found the horizontal, vertical, and hypotenuse of the known side. Then, I knew that the unknown horizontal was equivalent to the known horizontal. I knew the forces had to add up to zero, due to it being in equilibrium, so I solved for the unknown tension and found it to be 637.029 N. I solved for the other unknown perpendicular just do so and found that to be 270.6557549 N. I thought this angle would be arc tan, but that is not working. I don't know what to do now.
     
  7. Oct 7, 2007 #6
    Can anyone help me?
     
  8. Oct 8, 2007 #7

    learningphysics

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    it's arctan(opposite/adjacent)... or arcsin(opposite/hypotenesue) = arcsin(270.65575/637.029) = 25.1 degrees. Is this what you got?
     
  9. Oct 8, 2007 #8
    Yeah, that's what I got, but it says the answer is incorrect, even though it says that the force I got was correct. It makes no sense.
     
  10. Oct 8, 2007 #9

    learningphysics

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    Oh... it asks for the angle with the wall... I thought the second cable was connected to the ceiling also... I think then the answer is supposed to be 90-25.1 = 64.9 degrees
     
  11. Oct 8, 2007 #10
    Hooray. I was going to try that days ago to see if that could be right, but didn't want to bother wasting one of my guesses on something that probably wouldn't be right. Thanks for your help. I'll probably need some more help tomorrow. Haha.
     
  12. Oct 8, 2007 #11
    There is another problem that I don't really understand.

    An 8kg block is released from rest on an inclined plane and moves 2.2m during the next 3.4s. The acceleration of gravity is 9.8m/s^2.

    What is the coefficient of kinetic friction for the incline?
     
    Last edited: Oct 8, 2007
  13. Oct 8, 2007 #12
    If someone can help, that would be really nice. I need the answer to this question sometime before 12:00 tonight (it's 8:00 now) to get credit for it.
     
  14. Oct 8, 2007 #13

    learningphysics

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    Did you find the acceleration? What are the forces acting on the block along the plane?
     
  15. Oct 8, 2007 #14
    a = .3806228374 m/s^2

    There are no other forces acting on it.
     
  16. Oct 8, 2007 #15

    learningphysics

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    yes, that's the correct acceleration.

    There are 2 forces... what are they? 1 is friction... what's the other?
     
  17. Oct 8, 2007 #16
    Weight? The block has a mass of 8kg. It doesn't say any other forces.
     
  18. Oct 8, 2007 #17

    learningphysics

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    Yes... divide the weight into 2 components... what is the component along the plane... what is the component perpendicular to the plane...

    Get the equations:

    [tex]\Sigma{F_x} = ma_x[/tex]

    [tex]\Sigma{F_y} = ma_y[/tex]

    you already got [tex]a_x = .3806228374[/tex]. and ay = 0.
     
  19. Oct 8, 2007 #18
    72.59N in the Y and 29.3692N in the X ?
     
    Last edited: Oct 8, 2007
  20. Oct 8, 2007 #19

    learningphysics

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    What is exactly 72.58 and what exactly is 29.3692N? can you write the equations out...
     
  21. Oct 8, 2007 #20
    W = mg
    W = 8(9.8)
    W = 78.4

    I did F perpendicular as 78.4cos22 and found that to be 72.6912.

    I then did 78.4sin22 and found that to be 29.3692.
     
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