+0

How to resolve this vector

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746
13

How can I resolve the vector at point B ?

I want to find the x and y components of this force ? Jul 25, 2015

#8
+15

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. .

Jul 27, 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?

.

Jul 26, 2015
#2
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I don't get it ! how do you use the angle HAB for this vector ?

can you draw it !

Jul 26, 2015
#3
+5

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

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)

.

Jul 26, 2015
#5
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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?

Jul 26, 2015
#6
+5

Alan, how did you know that this angle is alpha  ( I put a blue line )  . 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. Jul 27, 2015
#8
+15

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
#9
0

Thanks Alan :)

Jul 27, 2015
#10
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Ok , what if I want to take the x-axis along the beam AC ! ?

Jul 28, 2015
#12
+10

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

.

Jul 29, 2015
#13
0

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

Jul 29, 2015
#14
+5

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.

.

Jul 29, 2015