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

Statics problem, determinate without moments, indeterminate with?

  1. Feb 21, 2012 #1
    Hey everyone. First time poster here.

    I was helping a friend do some Statics homework (I completed the course last Fall), and I ran into a really unsettling problem.

    I have done this problem over several times with the same result.

    When I do the problem only summing forces in the X and then Y direction, I get an answer that "checks out."

    When I do the same problem, but FIRST solve for tension using sum of moments about point R, I get a result for tension that does NOT check out.

    I have tried doing this with simple algebra methods, and using cross products to determine the magnitude of tension. EVERY TIME I USE MOMENTS I get the same answer.

    Here's the problem:

    http://img829.imageshack.us/img829/3564/stuffcw.png [Broken]

    I just sketched this up. R is just a reaction force. In the actual problem it's a rocker end that is butted up against a floor and a wall. Tension is a wire that is connected to the same wall.

    When I sum forces in X and then Y, I get

    T = 167.2 N
    R = 192.9 N

    When I sum moments I get

    T = 83.56 N
    then sum moments in Y direction gives me R = 96.42

    Just looking for anyone to try to figure it out with moments and see if they can get it to work.

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 22, 2012 #2
    To me this means that the reaction 'R' as you have it is actually 2 components Rx and Ry. Try to solve it like that. The way you have it there are more equations than unknowns which could be why you are getting that issue. (more than one solution..which is not quite right). Opposite of statically indeterminent :)
  4. Feb 22, 2012 #3
    Hey, thanks for the reply.

    You are right. But I didn't draw that in the problem just for simplicity sake. If you break R and T into components, when you sum forces you get:

    →ƩFx = 0 = -12/13 T + 4/5 R

    Solve for T = 13/15 R

    ^ƩFy = 0 = +5/13 T + 3/5 R - 180

    180 = 14/15 R

    R = 192.9 newtons
    T = 167.2 newtons

    Doing the beginning of the same problem with moments (dont have much time to go through this):

    (ccw+) ƩMr = 0 = +T(12/13)(0.36 meters) + T(5/13)(0.48 meters) - 180(0.24 meters)

    180(0.24) = 0.517 T
    T = 83.57 newtons ????????????????

    edit: I just realized when I do moments, the magnitude is half of when I do forces. Any insight?
    Last edited: Feb 22, 2012
  5. Feb 22, 2012 #4
    For what it's worth, I showed this problem to a few colleages and we all get the same result. T when found only with forces is double the magnitude of T found starting with moments.
  6. Feb 22, 2012 #5
    I realized my error. Since it's a 3 force member you cannot assume that the reaction force is colinear to the member's length.
  7. Feb 22, 2012 #6
    You can't assume that R is in line with the beam.

    Edit: Ah, a minute too late.
  8. Feb 22, 2012 #7
    Thanks Travis, I still really appreciate your post. It's always something trivial, isn't it?
  9. Feb 22, 2012 #8
    Sure is, I did it twice looking at that diagram and I was scratching my head saying, "What the heck is going on with this?"

    It's hard to see those mistakes once you've got it drawn in. All better now though
  10. Feb 22, 2012 #9
    Glad you figured it out :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook