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Does collinear mean that two vectors start at the same point? 
 

Also when defining R = A + B and R' = A + (-B), is there further rules to define which is A or B?

Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

Stu  Jul 24, 2014

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 #4
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I did define collinear however the diagram mentioning it had two vectors from the same point of origin. Hence my question. Is that defining them as collinear or it would have been something else? 

No, they're not collinear just because they start at the same point.  None of the forces in your diagrams that start at the same point are collinear.  They would all have to lie along the same straight line (not necessarily in the same direction) to be collinear. 

...I have been resolving the " hypotenues of the triangle for example to get the resulting force of Fxi and Fyj etc. But not adding them and just giving the answer. 

Resolving a force on to two perpendicular axes just gives you the components of that force.  If you do this separately  with two forces, you would need to add the corresponding components to find the coordinates of the resultant force.

Alan  Jul 25, 2014
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 #1
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Is FR in the below image "resultant force"? 
Can some one please verbally interprut the image with more classification.


Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

How is the bellow question in the image worked out? 

Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

Stu  Jul 24, 2014
 #2
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"collinear" means along the same line; or if it refers to points, then it means the points lie along a straight line.

 

If you sum vectors A and B you get the same result as if you sum vectors B and A.

 

The picture with the chains means that the forces, F1 and F2, in two of the chains are added (vectorially) to give a resultant vector (F1+F2).  This is then added (vectorially) to the force F3 to give an overall resultant force FR.

Adding vectorially means you add the x-component values together, and add the y-component values together.

 

In the s***w eye you need to resolve F1 into it's components (F1*cos(theta1), F1sin(theta1)) and F2 into its components (F2cos(theta2), F2sin(theta2)), (where the thetas are angles between the vectors and the x-axis) and get the components of the resultant from

(FRx, FRy) = (F1*cos(theta1)+F2cos(theta2), F1sin(theta1)+F2sin(theta2))

The magnitude of the resultant is given by √( FRx2 + FRy2) and its angle to the x-axis by tan-1(FRy/FRx)

Alan  Jul 24, 2014
 #3
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I did define collinear however the diagram mentioning it had two vectors from the same point of origin. Hence my question. Is that defining them as collinear or it would have been something else? 

With regards to resultant force, I asked a lecturer today. When working a problem I have been resolving the " hypotenues of the triangle for example to get the resulting force of Fxi and Fyj etc. But not adding them and just giving the answer. Further example is 5sin45 = ... is the Fr (and times D is torque). Is this correct? The lecturer said the answer was correct, but it sounds like i'm only solving for force, not resulting force. Even today working with a parellelogram and multiple forces this was all I did. Or am I missing the definition of two different things here? 


Stu  Jul 25, 2014
 #4
avatar+26329 
+5
Best Answer

I did define collinear however the diagram mentioning it had two vectors from the same point of origin. Hence my question. Is that defining them as collinear or it would have been something else? 

No, they're not collinear just because they start at the same point.  None of the forces in your diagrams that start at the same point are collinear.  They would all have to lie along the same straight line (not necessarily in the same direction) to be collinear. 

...I have been resolving the " hypotenues of the triangle for example to get the resulting force of Fxi and Fyj etc. But not adding them and just giving the answer. 

Resolving a force on to two perpendicular axes just gives you the components of that force.  If you do this separately  with two forces, you would need to add the corresponding components to find the coordinates of the resultant force.

Alan  Jul 25, 2014
 #5
avatar+1314 
0

Thank you.

Stu  Jul 25, 2014

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