1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Forces Question

  1. Jun 18, 2010 #1
    1. The problem statement, all variables and given/known data
    This will probabaly seem like a very simple question and maybe I'm wayy over thinking it.
    This goes back to forces on an anchor system.. lets say I have two pieces of rope holding up a 70 kg load.. each rope is attached to a seperate anchor.. the two ropes are perfectly parallel so there is no angle between the two..
    how do i know thats the load will be split between the two.. whats the proof.

    I have T1 and T2 acting up in the same direction... (not sure on magnitude).. and mg acting down
    so
    T1 + T2 = mg = 686 N

    but how do i know T1 = T2 which it clearly does...


    2. Relevant equations
     

    Attached Files:

  2. jcsd
  3. Jun 18, 2010 #2
    No, not that obvious.
    If the problem says nothing more, then the most we can deduce is T1+T2 = mg. Only under some certain other conditions (which means the problem must provide more info), such as the system is symmetrical, can we state T1=T2.
     
  4. Jun 18, 2010 #3
    it just seems odd to me that if i were to take that exact same question and instead of having each rope parallel to each other i make an angle of 140 degrees between the two.. i can tell everything. Since it adds a horizontal component and the system is at equilibrium TIx must equal T2x and since the load has only Y components and no horizontal.
    T1x = T2x therefore T1 = T2

    but i take that same situation and remove the angle between the two ropes and now we can no longer deduce that T1 = T2 ? something seems off
     
  5. Jun 18, 2010 #4
    How do you deduce this?
     
  6. Jun 18, 2010 #5
    As per the diagrams attached on this reply.
    Situation a.) shows what i am trying to figure out.. and situation b.) shows a more complex situation that its clear to see that the two horizontal components must equal each other since the object is not accelerating the sum of forces on the x access must = 0(bad example here because it is perfectly symetrical) but regardless of the angles there .. if the load is hanging straight down.. it has no horizontal component so the forces created by the tension in the x axis must cancel each other out.. thus making them equal.

    its not quite as saying T1x = T2x off the start .. in a non symetrical system youd end up with T1 = T2cos(25) / cos(45) then you can sub into the Y access and solve.

    this is more complex but yet way easyier for me to find T1 and T2 but for a set up like a.) whats the math behind proving T1 = T2 ... it must.. even if the rope for T2 is 10ft and T1 is 2ft they should still be splitting that load 50/50 but i cant prove it...
     

    Attached Files:

  7. Jun 18, 2010 #6
    If you flip the original system right to left, can you tell the difference between rope 1 and rope 2? Assuming you did not see it flip?

    If your answer is no, that means you can deduce by symmetry that the two split the load evenly, since there is no reason why any other possibility should be plausible.
     
  8. Jun 18, 2010 #7
    well say for example rope1 was is 10ft and rope 2 is 4 feet.. so they are hung from diffrent heights but in the end they attach to the load at exactly the same point parrallel to each other.. the lengths are different so you can definitely see a difference between rope 1 and rope 2 but .. the force must be the same.. i can see it in my head.. doesnt mean its right but im 99% sure it is.. i just dont know how to create a proof on paper that can deduce...
    T1 = T2
     
  9. Jun 18, 2010 #8
    There's a difference as far as the rest of the system goes. The way the load is distributed among the ropes will be very different, but if you focus your attention on the box and its contacts with the two ropes, can you tell a difference?
     
  10. Jun 18, 2010 #9
    if you focus in right at the box no you cannot both ropes would leave the connection at the box from roughly the same point perfectly parallel..
    so i can see the force would have the exact same direction. but how do i know the magnitudes would be equal
     
  11. Jun 19, 2010 #10
    Oh so x denotes horizontal direction. I thought it's vertical.
    So how do you deduce this: T1x = T2x therefore T1 = T2? (post #3)
    Anyway we can never know whether the magnitudes are equal if no further information is provided.
     
  12. Jun 19, 2010 #11
    yes third post was a bad example i was looking at a symmetrical system at that point..

    so what dictates a perfectly symmetrical system in a situation where the two ropes a perfectly parallel with each other... if one rope is longer than the other.. yet they are parallel the force should still be split equally but it doesnt look symmetrical to me..
     
  13. Jun 19, 2010 #12
    There has to be a way that physics can solve this sort of a question short of sticking a load cell on the ropes to determine that the load would in fact be split equally.... I'm definitely missing something here. maybe its calculus I have no idea. but if i take those two ropes that form an angle of 140 degrees and now I make the angle of 0.00000001 degreess then you could easily determine that the resultant vectors created by T1 and T2 would be essentially half the weight of the load each(assuming a symmetrical set up like you said earlier).... now if i make the angle 0 degrees why is this so much more complex... this should be simple to explain. or maybe i just answered my own question... if the system is symetrical rope length doesnt not matter because the freebody diagram is only concerened with the connection point at the load and the direction of the forces, and it is the magnitude of these forces that will dictate the length of these vectors (but we do not know the magnitude to start so the lengths are arbitrary to begin with)?
    I would like some more input please
     
    Last edited: Jun 19, 2010
  14. Jun 20, 2010 #13
    do the length of ropes affect the magnitude?
    i don't think so
    because i think they have experienced the same force..
    what have made T1 and T2 unequal???
    can please list out the factors??
     
  15. Jun 20, 2010 #14
    thats also a good way of looking at it. what makes T1 and T2 unequal... well im not sure in this case.

    the two ropes are parallel with each other... and attached to the load at the same spot. one rope is longer than the other. the forces are only in the Y axis.

    I cannot think of any other factors?
     
  16. Jun 20, 2010 #15

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You started out with the 2 parallel ropes attached at different points, but since then you have been focusing on 2 ropes attached to the same point. Assuming inextensible cords, the tensions T1 and T2 are the same in both cases, if the loading is symettrical in case 1 and 2. You can show this by summing torques about any point for case 1, and by joint equilibrium of forces for case 2.

    In case one, for 2 parallel ropes attached at different points and not symmetrical, then you must use moments, T1(d1) =T2(d2), where d1 and d2 are the respective perpendicular rope distances from the center of mass of the block, which must be given. For example , if T1 was attached directly above the center of the block, and a parallel rope T2 was attached at the right corner of the block, then T1 would be equal to the blocks weight, and T2 would be zero. For the second symmetrical case, T1cos theta = T2 costheta, T1 = T2 (you run into dividing by 0 when theta is 90 degrees, but in the limit, T1 still equals T2).

    For inextensible rigid ropes, the length of the ropes make no difference.
     
  17. Jun 20, 2010 #16
    very good answer thanks so much for looking at my problem... There was some clarification issues on my behalf but you covered each situation perfectly. The initial question I posted "attached to seperate anchors" was misleading because i meant seperate anchors for connection to a wall or whatever surface one above the other and connected to the load at the same point..in the diagram i seperated T1 and T2 for clarification that there were two different forces but I think it lead to confusion.

    I will play around with your answer thanks !

    EDIT: after playing around with this im more concerned with case 2 and the method of joint equilibrium. can I get pointed in the right direction on joint equilibrium? I did some googling and found it to be a common concept for determining forces on trusses .. is this the same method?
     
    Last edited: Jun 20, 2010
  18. Jun 21, 2010 #17

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes.
     
  19. Jun 21, 2010 #18
    is this not a joint sum of forces (note T1 and T2 are seperated for clarity, you can think of them as attaching to the same point)? sorry to keep bugging you on this one. I really want to make sense of it and have not heard the term joint equilibrium before but from what I gathered on what I was reading this is all it was a summation of forces at each joint... I'm haveing trouble yielding T1 = T2 .
     

    Attached Files:

  20. Jun 21, 2010 #19

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This is a special case of symmetry where theta, the angle between each force and the horizontal, is 90 degrees, so you lose the identity of the triangle as in your '140 degree' example above. In general, for equal angles (the symmetrc case) from sum of forces in x direction = 0, then T1cos theta = T2 cos theta, ot T1=T2. So pretend that theta is 89.999999 degrees instead of 90 degrees, for the parallel ropes case, so you don't have to divide by 0 , and then T1 = T2.
     
  21. Jun 21, 2010 #20
    okay nice! thats logic I've been messing around with as well and it seemed everything was going good and then I thought...well if T1 and T2 are 90 degrees and have 0 horizontal component then we assume 89.999999 degrees etc... can the same not be said for the load pulling downward giving it a slight horiztonal component as well?
    so F_x= T2cos(theta) - T2cos(theta) + T3 Cos(theta)
    or because the force acting on the load is gravity we know it will always be perpendicular in this example?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple Forces Question
  1. Simple force question (Replies: 7)

  2. Simple Force question (Replies: 3)

Loading...