How can I resolve the vector at point B ?

I want to find the x and y components of this force ?

xvxvxv Jul 25, 2015

#1**+15 **

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?

.

Alan Jul 26, 2015

#2**0 **

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

can you draw it !

xvxvxv Jul 26, 2015

#4**+5 **

Does this help:

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

.

Alan Jul 26, 2015

#5**0 **

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?

Melody Jul 26, 2015

#7**+10 **

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.

Melody Jul 27, 2015

#8**+15 **

Best Answer

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.

.

Alan Jul 27, 2015