Pin-jointed structure - Mechanics of Solids

[furv]

Homework Statement

"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

Homework Equations

two force principle
three force principle
sum of moments = 0
sum of X components = 0
sum of Y components = 0

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

 

[PhanthomJay]

Homework Statement

"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

Homework Equations

two force principle
three force principle
sum of moments = 0
sum of X components = 0
sum of Y components = 0

The Attempt at a Solution

First I started to try and fill in the free body diagram and considered AB.

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.

B is on frictionless rollers, so to stay in equilibrium I thought the reaction force must only be able to act vertically.

yes, that is correct, the roller cannot support any force parallel to its 'wheels'.

At A I don't know the direction of the reaction force so I just drew in arrows for the X and Y components.

yes, and it is a good idea to indicate ther direction if possible , which is simply determined in ths case.

At D I again had it perpendicular to AB vertically up ( not sure if this was correct ).

there is no external support at D, so don't put one in there.

I then resolved horizontally and vertically; Y: Ya+Rd+Rb=0

there is no Rd when looking at the system in a FBD. And what happened to the y component of F?

X: Xa = 0.

No-o, what happened to the x component of F

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

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?