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What must the resultant of the two tension forces in the ropes be?

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    A) What must the resultant of the two tension forces in the ropes be?
    B)What is the magnitude and direction of the tension in each of the two ropes?

    2. Relevant equations



    3. The attempt at a solution

    For A .......... shouldn't the tensions in the ropes equal 500N?? cause they need to oppose the force of gravity on the load.........
    and i have no idea where to start for B anyone know ??
     
  2. jcsd
  3. Jan 22, 2008 #2
    for A, because it has 2 ropes, and they are not equal , so it's not 500N
     
  4. Jan 22, 2008 #3
    ahhhh but what if it was one rope ???
    also the rope connects to a central point which is the hook .......
    so wouldn't the tension remain constant throughout the whole rope???
     
  5. Jan 22, 2008 #4
    dont you see the hook is at the connecting point of 2 ropes?
    the 2 ropes added up to 500N.
    but then one rope is longer, and one is shorter, so the force they contain is different too
     
  6. Jan 22, 2008 #5
    alright ........... the tensions are not equal ....... but they should add up to 500 N right??

    Edit : Also for each tension ....... is it the overall tension ....... or the Y direction of the tensions??
     
  7. Jan 22, 2008 #6
    bump.........
     
  8. Jan 22, 2008 #7
    The tensions' y-components add up to 500N.

    Draw a free-body diagram for the mass and write out Newton's 2nd law in the x and y directions.
     
  9. Jan 22, 2008 #8
    the y-components are not equal though right??
     
  10. Jan 22, 2008 #9
    No, because the angles are different.
     
  11. Jan 22, 2008 #10
    huh............ i said " they are NOT EQUAL " .......
    since they are at different angles they are not the same right??
    but where would i start at this problem??
     
  12. Jan 22, 2008 #11
    Yeah, I'm saying you're right, they aren't equal.

    Read what I said earlier, 'Draw a free-body diagram for the mass and write out Newton's 2nd law in the x and y directions.'
     
  13. Jan 22, 2008 #12
    alright i got the FBD ......... but for newton's 2nd law ........
    are u talking about the Fnet = ma in the x and y direction or a different one??
     
  14. Jan 22, 2008 #13
    Yes, write out the net forces in the x and y direction. Not sure what you're asking.
     
  15. Jan 22, 2008 #14
    alright is it
    Fnetx = fax + fbx = 0N
    Fnety = fay + fby + fg = 0N
    cause it's in equilibirum right??
     
  16. Jan 22, 2008 #15
    Yep.
     
  17. Jan 22, 2008 #16
    still doesn't make any sense to me ...... :S i don't have an Fa nor Fb
    so i can't use trig ....... what do i do next?
     
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