+0  
 
0
336
13
avatar+1828 

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

Best Answer 

 #8
avatar+26329 
+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.

 Angles:

.

Alan  Jul 27, 2015
Sort: 

13+0 Answers

 #1
avatar+26329 
+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
avatar+1828 
0

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

can you draw it ! 

xvxvxv  Jul 26, 2015
 #3
avatar+91039 
+5

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

Melody  Jul 26, 2015
 #4
avatar+26329 
+5

Does this help:

 

 Angles:

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

.

Alan  Jul 26, 2015
 #5
avatar+91039 
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
 #6
avatar+1828 
+5

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

 

xvxvxv  Jul 26, 2015
 #7
avatar+91039 
+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
avatar+26329 
+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.

 Angles:

.

Alan  Jul 27, 2015
 #9
avatar+91039 
0

Thanks Alan :)

Melody  Jul 27, 2015
 #10
avatar+1828 
0

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

xvxvxv  Jul 28, 2015
 #12
avatar+26329 
+10

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

.

Alan  Jul 29, 2015
 #13
avatar+1828 
0

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

xvxvxv  Jul 29, 2015
 #14
avatar+26329 
+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.

.

Alan  Jul 29, 2015

18 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details