How can I resolve the vector at point B ?
I want to find the x and y components of this force ?
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?
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)
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 ) .
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.
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.
The component of the 2kN force along AC is zero, as it is perpendicular to AC!