How to resolve this vector

Note from the original diagram that there is a right-angle sign between the force arrow and the beam AC, and just to make Melody's reply clearer, here is the diagram with the 'X' that Melody used marked on it.

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The angle HAB is given by tan^-1(CG/AG) so the horizontal (x) component of force is 2*sin(HAB) kN, and the vertical (y) component is -2*cos(HAB) kN  (assuming y is positive upwards).

Can you take it from there?

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I don't get it ! how do you use the angle HAB for this vector ?

can you draw it ! 

I am interested too Alan.  Could you please discuss this one some more ?

Does this help:

 Remember that cos(90-α) is the same as sin(α)  (I've used α for angle HAB)

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Yes that helps - thanks Alan.

So the downward force at B is 1.6kN

the inward horizonal force at B is 1.2kN

I guess it is the same at D

There is the downward force at G of 1.5kN

How does all this get put together to answer the rest of the questions?

Alan, how did you know that this angle is alpha  ( I put a blue line )  . 

Hi 315,

Consider the right angled triangle ABX   Where BX is the vertical line.

<ABX = 90- alpha

<ABX+ <XBH= 90

so angle XBH = alpha

The angle that you are questioning is vertically opposite this angle so it is also alpha.   

Thanks Alan :)

Ok , what if I want to take the x-axis along the beam AC ! ? 

The component of the 2kN force along AC is zero, as it is perpendicular to AC!

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And the y components will be as it is , right ? 

If your y-direction is perpendicular to AC then the component of the 2kN force in that direction will be either 2kN or -2kN depending on which direction you choose as positive.

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