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Tension of 2 Ropes Find the Angle

  1. Sep 24, 2008 #1
    Tension of 2 Ropes are known angle and mass unknown help please

    1. The problem statement, all variables and given/known data
    A block mass M hangs in equilibrium. A horizontal rope attached to the block has a tension of 30N. Another rope attached to the cieling with a tension of 90N, and makes an angle with the ceiling. What is the angle

    2. Relevant equations
    I was thinking the inverse tangent
    tan = y/x being y= 90 and x = 30

    3. The attempt at a solution
    inverse tan = 30/90 = 0.3333 18 degrees

    i'm probably doing something wrong cause i think gravity here is important i just dont see how to put all this together
    Any suggestions or corrections would be helpful
    Last edited: Sep 24, 2008
  2. jcsd
  3. Sep 24, 2008 #2
    Draw a picture of the situation.

    The rope making an angle with the ceiling MUST have a certain tension in the horizontal direction. What is that tension and why must it have that magnitude?

    Think on that.
  4. Sep 24, 2008 #3
    The rope with tension 90N makes an angle with the ceiling

    I kind of thought of drawing in this triangle to help me out

    R = sqrrt{(30)^2 + (90)^2}
    sorry made a mistake in the triangle C is 95 N confused with another problem
    using this to help me find the angles
    what do u guys think?
    Last edited: Sep 24, 2008
  5. Sep 24, 2008 #4
    So what is the horizontal component of the angled rope? Remember, the object is in equilibrium. What does that mean about all of the forces in the problem?

  6. Sep 24, 2008 #5
    So for M to be in equilibrium it must be acted on by no forces? the net force is zero i think so if it is zero then what?
  7. Sep 24, 2008 #6
    Yes, the net force must be zero. That means the Net force in the x direction must be zero. So what is the horizontal component of the angled rope?
  8. Sep 24, 2008 #7
    I hit the wall with that one, sorry i'm
    still trying to figure out
  9. Sep 24, 2008 #8
    Ty = T sin([tex]\oslash[/tex]). would it be this
  10. Sep 24, 2008 #9
    for M in equilibrium each component of the net force must be zero so
    [tex]\Sigma[/tex]Fx = 0 and [tex]\Sigma[/tex]Fy = 0
  11. Sep 25, 2008 #10
    That's correct. So if the sum of the x forces is zero, then what would be T Sin Theta?
  12. Oct 9, 2008 #11
    Then it would be the inverse of Sine theta = 30/90 = 0.33 which is 19 degrees
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