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!

Homework Help: Pin-jointed structure - Mechanics of Solids

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    "A force F of magnitude 90 kN is applied at point C at an angle of 45°. The pin-joint B may be assumed to be resting on frictionless rollers. Determine the reaction forces at A and B"

    Image of the diagram here: http://i.imgur.com/jUL0rJe.jpg?1

    2. Relevant equations
    two force principle
    three force principle
    sum of moments = 0
    sum of X components = 0
    sum of Y components = 0

    3. The attempt at a solution

    First I started to try and fill in the free body diagram and considered AB. B is on frictionless rollers, so to stay in equilibrium I thought the reaction force must only be able to act vertically. At A I don't know the direction of the reaction force so I just drew in arrows for the X and Y components, At D I again had it perpendicular to AB vertically up ( not sure if this was correct ). I then resolved horizontally and vertically; Y: Ya+Rd+Rb=0 X: Xa = 0. Then I took moments about A : Rd(4.5)+Rb(9)=0. There's too many unknowns to solve so i know i have to resolve somewhere else but i'm not sure about the directions of the forces acting on C to try and resolve AC or CB.. Maybe i'm going about it completely wrong, as already it feels wrong having the X component of A being 0 if I look at the rest of the diagram.

    Any help will be greatly appreaciated
  2. jcsd
  3. Oct 20, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your very first free body diagram should look at the entire truss with a force F applied at C as shown, and unknown external forces at the supports.
    yes, that is correct, the roller cannot support any force parallel to its 'wheels'.
    yes, and it is a good idea to indicate ther direction if possible , which is simply determined in ths case.
    there is no external support at D, so don't put one in there.
    there is no Rd when looking at the system in a FBD. And what happened to the y component of F?
    No-o, what happened to the x component of F
    Find support reactions first , in terms of F, after first breaking F into its x and y components. After you get support reactions by summing moments = 0, then you can get internal member forces using 'method of joints'. Note also: what is the force in CD?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted