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!

Resolving forces

  1. Aug 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi guys, i hope this is posted in the right area, i have been struggling on the following question. Sorry about the diagram but i dont have a scanner

    I have a pin frame BCFGD and need to resolve the forces. It is attached to a wall at B and F. There is a negative force of 20KN acting vertically at C and at D a positive force of 10KN acting horizontally.


    2. Relevant equations



    3. The attempt at a solution

    I have worked out the following BC=20KN, CF=28.28KN, FG=10KN, CG=40KN GD=10KN

    I think i am on the right line but I just cant get it to add up. I started by working out the forces at the supports which i got as 20KN for B and 10KN for F acting horizontally and a vertical force of 20KN acting at F.

    Any help would be appreciated
     
  2. jcsd
  3. Aug 18, 2008 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Geoffishere, welcome to PF!
    Without a sketch, it's pretty tough for us to help. However, if you're not getting your numbers to add up, a likely cause might be in the incorrect use of plus and minus signs and incorrect directions for reaction forces. If you can't attach a sketch, perhaps you can describe the frame with words, giving dimensions between joints, angles, etc.
     
  4. Aug 18, 2008 #3
    Hi, thanks for your reply, i was struggling to attach the file last night. All horizontal and vertical lengths are 2m.
     

    Attached Files:

  5. Aug 18, 2008 #4
    The attachment wont work for me. Are you using method of joints of sections to try and solve the problem?
     
  6. Aug 18, 2008 #5
    I think i am usin joints of sections. I will try and discribe the frame if the attachment isnt working. Starting at B, it is attached to wall by a hinge on a roller. from this is a horizontal bar joining to C. From C are three further bars one vertically down to G, one going south west to F and one going south east to D. D and G are attached horizontally as are FG. F is attached to the wall by a hinge.

    All angles are 45 or 90 and all horizontal and vertical lengths are 2m. The forces are acting at C, -20KN vertically and at D +10KN horizontally
     
  7. Aug 18, 2008 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I was able to open your attachment. Be sure to specify the direction of the reaction forces, and the direction of the member forces (compression or tension). Otherwise, you have an error in the value of CG, and it's probably because you got careless with your signage at joint C. Isolate joint G to solve for the force in CG. It should bounce right out at you. What did you get for the force in DC?
     
  8. Aug 18, 2008 #7
    I got my reaction force at B as +20KN and at F -10KN. I was struggling because once i had worked these out I found I had a -40kN force which needed to be taken by CG and CD and i couldnt get these to be in equilibrium at G and D.

    I think it must be something wrong with what i am doing. I will try and explain what my thinking is.

    Firstly because of the -20KN force at C, I deduced there must be a reactant force of +20KN acting vertically at F. I the deduced that the moments at B and F are 20Knm however because of the 10KN force at D and to keep everything in equilibrium, F is +10Kn and B -20KN.

    I then used equilibrium of joints to try and work out the rest and thats where i got stuck
     
  9. Aug 18, 2008 #8

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You mean the reaction force at B is -20 (acting left) and at F it is +10 (acting right), which you seem to have correctly noted below.
    You are messing up your plus and minus signs. You must keep them straight, or else you'll be scratching your head forever. At Joint C, BC exerts a horizontal force of 20 to the left, and the horizontal component of FC acting on joint C is 20 to the right. They cancel, not add.
    This result is correct.
    Mind your plus and minus signs, they can wear you down.
     
  10. Aug 18, 2008 #9
    I think i get it in equilibrium, but CG and CD are zero, is this right???
     
  11. Aug 18, 2008 #10

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, correct, they are both zero. I think you've licked it now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Resolving forces
  1. Resolving forces (Replies: 1)

  2. Resolving Forces (Replies: 2)

  3. Resolving Forces (Replies: 1)

  4. Resolving force (Replies: 7)

  5. Resolving Forces (Replies: 11)

Loading...