Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can tension be generated perpendicular to a rope?

  1. Feb 3, 2010 #1
    For example, if one end of a rope is attached to a wall, and the other end is attached to the floor (at an angle), if I pull straight up on the rope, would that exert a force on the wall?
  2. jcsd
  3. Feb 3, 2010 #2


    User Avatar

    Staff: Mentor

    No, you can't generating tension perpendicular. What happens when you try to pull straight up is you break the rope into two segments that are no longer parallel and each provides a regular tension force to counteract the force you've added.
  4. Feb 3, 2010 #3
    What a coincidence: I just came on to post a very similar question, but someone else had already posted a very similar problem.

    I figured I wasn't going to make a brand new thread, so I'll just post in here.

    My question is conceptual. I am having some difficulty understanding how a rope under tension can create a force against a pulley.

    I think I understand tension for a straight rope. A rope is pulled, and to prevent it from accelerating, the opposite end is pulled by an equal force. The force is transmitted directly through the rope: one "chunk" of atoms that makes up the rope is pulled, then it pulls an adjacent chunk, and so on.

    Let us reduce the case of a pulley to a rope being slightly "bent" by a pin.

    I understand that in this case the balance of forces in the x direction is:

    Fx= F*cos(a) - F*cos(a)

    This happens because the tension in the rope now has y and x components.

    The balance of forces in the y direction is now:

    Fy=-F*sin(a) - F*sin(a) + Fp

    Fp is the reaction force exerted by the pulley on the rope in order for the rope to not accelerate.

    But how can a rope create the y components of the force?

    If the rope were a solid rod, I would understand. The rod would be welded, or otherwise attached to the pulley. The rod would then exert this force through the attachment point on the pulley.

    But, the rope is not like that! The rope can't "tug" on the pulley, since there is no point of attachment. Friction is also not a factor.

    So, what is creating this force, physically?

    What seems intuitive to me is that the rope can be thought of as a spring, with Δx being the vertical displacement. The force applied by the rope on the pulley is in fact proportional to x by a constant.

    To summarize my question, what physically explains the rope creating a force on the pulley?

    Attached Files:

  5. Feb 3, 2010 #4
    Hmm...It seems even if you had a rope that would not stretch and secured at points that would not stretch there would still be a force. Adding the vectors of each half of rope doesn't give you the Fapp on the pin?

    And I would also think that it is supplied by the portion of rope directly touching the pin being pulled by the pieces on either side making it accelerate toward the pin and pushing it(Fapp).

    However I guess it would be proportional to x since with a set length of rope as the x increases the angle between the rope and a vertical axis through the pin would get proprtionatly smaller right, and since a smaller angle = a greater force component, the force would be proportional to x...

    Okay, just figured it out I think. the force on the pin(or pully) would be


    which is obvious since cos(theta) = A/H
    where A is x and H is L/2 so...
    2x/L = cos(theta) which you would use to add the vectors heh...

    And if the ropes are of uneven lengths you could use:


    [EDIT] I guess i didn't pay attention, you already knew the math.....
    Last edited: Feb 3, 2010
  6. Feb 3, 2010 #5
    And to be technical, you could generate a force perpendicular to the length of the rope if it was taught and you pulled one side of the rope away from the center of the rope, hard to do with a small rope but there are some BIG ropes that would be easy to pull one side...anyway wouldn't the fibers exert tension in every anlge radiating outward from where it was attached, one of which would be perpendicular to the length of the rope
  7. Feb 3, 2010 #6
    You'll greatly increase the force on the wall. The (vertical) normal force you apply to a short segment of the rope must (since it isn't accelerating) be balanced by (the vertical component of) the tension forces attached to the sides of that same short rope segment (and directed parallel with the rope's extension). This is the same tension that will be connected to the wall. But since those tensions are almost perpendicular to the normal force, the magnitude of the tensions must be very large (compared to the applied normal force) in order for their (relatively small) non-perpendicular components to cancel it.

    That said, the tension applied to the wall is always exactly in the direction of the rope; again only a fraction of the tension (or about half the force of your upward pull) is perpendicular to the original direction of the rope.
    Last edited: Feb 3, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook