Solving Vector Problems: Tips & Strategies

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Discussion Overview

The discussion revolves around strategies for solving vector problems, particularly in the context of forces and equilibrium. Participants explore methods for decomposing vectors into components and the challenges associated with calculating net forces in various scenarios, including practical applications and exam preparation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses doubt about the feasibility of summing all vectors to find a resultant vector that indicates the direction of motion, especially when dealing with multiple forces.
  • Another participant suggests decomposing every vector and summing their components, emphasizing the importance of maintaining consistent sign conventions for directions.
  • Some participants request clearer diagrams to better understand the problem context, indicating that visual representation may aid in resolving confusion.
  • There is a discussion about the conditions under which the net force must be zero in equilibrium situations, with some participants questioning the choice of axes for decomposition.
  • One participant raises a concern about the possibility of angles not summing to 90 degrees in certain decompositions, leading to confusion about the relationships between forces.
  • Several participants discuss the decomposition of gravitational force and tension in a scenario involving a bob on a string, with differing opinions on which forces should be decomposed and how to approach the problem mathematically.
  • There is a mention of practicing various equilibrium problems to gain confidence in solving vector-related questions, with some participants noting that no universal tricks exist for these problems.
  • One participant expresses frustration over their understanding of force components and equilibrium, seeking clarification on how to properly balance forces in their calculations.
  • Another participant emphasizes the necessity of considering all force components in any direction when analyzing equilibrium, reiterating that the sum must equal zero.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to vector decomposition and the handling of forces in equilibrium situations. Multiple competing views and uncertainties remain regarding the correct methods and interpretations of the problems discussed.

Contextual Notes

Some participants express confusion about the application of vector decomposition in different scenarios, indicating that the choice of axes and the relationships between forces can significantly impact the calculations. There are unresolved questions about the implications of different decompositions on the results.

mysqlpress
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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:
 
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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.
 
dst said:
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
 
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.
 
dst said:
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"
updated .
thanks for your help.
 
Last edited by a moderator:
anybody can help me?
it is urgent. since I am going to attend the exam :(
 
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?
 
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.
 
Vuldoraq said:
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
Doc Al said:
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
mysqlpress said:
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
mysqlpress said:
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
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
mysqlpress said:
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
Doc Al said:
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
mysqlpress said:
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
Doc Al said:
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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