Dealing with vectors

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  • #1
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in normal situation, for convenience , we usually decomposed a vector into two components - horizontal and vertical.and setting up a series of equation to find out the answer...
but I am doubt if we can count up all the vectors to give a resultant vector that point to the direction of motion...(e.g. net force)
when do in practical, I found that this is impossible... but why ...

On the other hands, when dealing with more than three forces... I am confused in calculation because sometimes I am not sure which vector should be "decomposed"....

any hints/strategies when tackling these problems? Thanks :smile:
 

Answers and Replies

  • #2
dst
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For net force, decompose every vector and sum their components up (whichever directions you've chosen as negative, make sure all components in that direction are kept negative). You then get the components of the resultant and from there it's quite simple to get everything else.
 
  • #3
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For net force, decompose every vector and sum their components up (whichever directions you've chosen as negative, make sure all components in that direction are kept negative). You then get the components of the resultant and from there it's quite simple to get everything else.

1372736669290863e85b52d6578627487ec32be.jpg
 
  • #4
dst
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Can you draw a bigger diagram please? It's hard to see what's going on. But it doesn't look like you should be having problems with that example either.
 
  • #5
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Can you draw a bigger diagram please? It's hard to see what's going on. But it doesn't look like you should be having problems with that example either.
okay.

http://www.uploadgeek.com/uploads456/0/bigpic.PNG" [Broken]
updated .
thanks for your help.
 
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  • #6
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anybody can help me?
it is urgent. since I am going to attend the exam :(
 
  • #7
Doc Al
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I'm not clear as to what your question is. Since the system is in equilibrium, the net force must be zero. You can choose your axes any way you want, but vertical and horizontal work just fine.

Seems like you get the same answer either way, so what's your question?
 
  • #8
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Your answers are the same for both decompositions. Note that y=90-x, so that
A/B=sin(x)/sin(90-x) but sin(90-x)=cos(-x) and cos(-x)=cos(x) so that A/B=tan(x).

As for tricks with vectors the best thing to do is just practice, the more you do the better you will get and you will be able see shortcuts. I don't think there is any universal tricks. Young and Freedman university physics has some good questions on vectors.
 
  • #9
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Your answers are the same for both decompositions. Note that y=90-x, so that
A/B=sin(x)/sin(90-x) but sin(90-x)=cos(-x) and cos(-x)=cos(x) so that A/B=tan(x).

As for tricks with vectors the best thing to do is just practice, the more you do the better you will get and you will be able see shortcuts. I don't think there is any universal tricks. Young and Freedman university physics has some good questions on vectors.

but the point is. can x+y =/= 90' ?

and for different cases, if you cannot practise the experiment, I am confused and 'afraid' to do these questions....
 
  • #10
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I'm not clear as to what your question is. Since the system is in equilibrium, the net force must be zero. You can choose your axes any way you want, but vertical and horizontal work just fine.

Seems like you get the same answer either way, so what's your question?


perhaps my question is not right clear.
Let say when you have a small bob attached to the ceiling on one hand by a string...and then move it an angle to the vertical...

in this situation , we should decompose mg or tension of string on the bob ?
Tsinx =F
Tcosx=mg
tanx = F/mg

but if mg is decomposed
mgcosx =T
mgsinx=F ...

The situation is quite different if the direction of acceleration is undetermined...
 
  • #11
Doc Al
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but the point is. can x+y =/= 90' ?
I assume "=/=" means "equals"? That's up to you--it's your diagram!

Regardless of what x and y are, you can always take components about any pair of perpendicular axes. And you'll get the same answer.

and for different cases, if you cannot practise the experiment, I am confused and 'afraid' to do these questions....
Dig up just about any equilibrium problem and try to solve it using different axes. Of course, in most problems there's an "obvious" choice of axes that make the solution easier.
 
  • #12
Doc Al
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perhaps my question is not right clear.
Let say when you have a small bob attached to the ceiling on one hand by a string...and then move it an angle to the vertical...
OK. You exert a horizontal force F that produces equilibrium at some angle.

in this situation , we should decompose mg or tension of string on the bob ?
Tsinx =F
Tcosx=mg
tanx = F/mg
OK.

but if mg is decomposed
mgcosx =T
mgsinx=F ...
I get:
Fsinx + mgcosx = T
Fcosx = mgsinx

Same thing: tanx = F/mg

As to which axes to use: Doesn't matter! By habit, I almost always use vertical and horizontal unless there's an obvious advantage in using some other set of axes.
 
  • #13
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I get:
Fsinx + mgcosx = T
Fcosx = mgsinx

Same thing: tanx = F/mg

As to which axes to use: Doesn't matter! By habit, I almost always use vertical and horizontal unless there's an obvious advantage in using some other set of axes.
Haha...in fact, I usually represent "not equal to" as =/=
sorry for my poor presentation :(

for my decomposition of mg
I got mgcosx=T and mgsinx=F
if the question does not offer me the situation... it is possiblefor me to decompose like that
and then finally results will be tanx =F/T but not F/mg
T =/= mg
and I am worried about that
 
  • #14
Doc Al
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for my decomposition of mg
I got mgcosx=T and mgsinx=F
Which is incorrect. See my decomposition above.

You must find components of all forces along your chosen axes. F, mg, and T all have components parallel to T.
 
  • #15
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Which is incorrect. See my decomposition above.

You must find components of all forces along your chosen axes. F, mg, and T all have components parallel to T.
I think it should be related to direction of F as well ?
for example, when I put off my hand, the direction of acceleration... I am doubt...

for my way ,why T cannot be balanced by component of mg , leaving mgsinx which is further balanced by my applied force ?

I am frustrated....:(
 
  • #16
Doc Al
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for my way ,why T cannot be balanced by component of mg , leaving mgsinx which is further balanced by my applied force ?
Because you can't just forget about the component of F! You chose to use force compoents parallel to the string, so find the components of all the forces in that direction.

The bottom line: Since there's equilibrium, the sum of force components (all of them!) in any direction must add to zero.
 
  • #17
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Because you can't just forget about the component of F! You chose to use force compoents parallel to the string, so find the components of all the forces in that direction.

The bottom line: Since there's equilibrium, the sum of force components (all of them!) in any direction must add to zero.

oic ! I've got it
Thanks 10000 times
 

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