1. Limited time only! Sign up for a free 30min personal 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!

Tension throughout a rope/wire

  1. Jul 31, 2006 #1
    I'm working on the following question:

    "A painting of mass 3.20 kg hangs on a wall. Two thin pieces of wire, each 0.250 m long, connect the painting's center to two hooks in the wall. The hooks are at the same height and are 0.330 m apart. When the painting hangs straight on the wall, how much tension is in each piece of wire? (It is assumed that the wire is massless)"

    I found the angle between one of the angles and the horizontal line connected the two hooks (48.7 degrees). Then, I used the weight (mg = 31.36) and the angle to calculate the tension in the rope (41.7 N).

    I know that the answer is half of that, but at what point should I divide by two in order to find the tension in each piece? My guess is that I divide the final answer by two. I guess I'm unsure of whether tension is the same throughout the whole wire.
  2. jcsd
  3. Jul 31, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Since they specifically refer to two pieces of wire, that's how you should treat it. You know (by symmetry) that each piece of wire has the same tension. Since two wires pull up on the painting, that force that you calculated is really the combined tension of both wires.
  4. Jul 31, 2006 #3
    So I can solve for the combined tension and then divide that answer by 2 (since they're symmetrical), or should I divide earlier in the problem? (I have another problem that involves a non-symmetrical arrangement, so I'm trying to figure out when to make the distinction.)
  5. Jul 31, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Since you've solved for the combined tension, just divide by 2. Mathematically, what you are doing is this:
    [tex]T\sin\theta + T\sin\theta = mg[/tex]

    [tex]2T\sin\theta = mg[/tex]
  6. Jul 31, 2006 #5
    Okay thanks, that makes sense. I was just getting confused about why T is multiplied by [tex]sin\theta[/tex]. I was solving it a different way and didn't see that. Thanks :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Tension throughout a rope/wire
  1. Tension of a rope (Replies: 3)

  2. Tension on a rope (Replies: 6)

  3. Tension on rope (Replies: 2)

  4. Tension in rope? (Replies: 2)