# Solving Vector Problems: Tips & Strategies

• mysqlpress
In summary: 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.
mysqlpress
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

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.

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.

updated .

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...

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...

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.

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.

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

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.

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...:(

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.

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

## 1. What is a vector and how is it represented in problem solving?

A vector is a mathematical quantity that has both magnitude and direction. It is represented by an arrow with its length representing the magnitude and its direction showing the direction. In problem solving, vectors are often represented using coordinates or components, such as x and y, to indicate their direction and magnitude.

## 2. What are some common strategies for solving vector problems?

Some common strategies for solving vector problems include breaking down the vector into its components, using trigonometric functions to find the magnitude and direction, and using the Pythagorean theorem to find the resultant vector. It is also helpful to draw a diagram to visually understand the problem and identify the given information.

## 3. How do I determine the direction of a vector in a problem?

The direction of a vector can be determined by finding the angle it makes with the positive x-axis. This can be done by using trigonometric functions, such as tangent, or by using inverse trigonometric functions, such as arctangent.

## 4. What is the difference between a displacement vector and a velocity vector?

A displacement vector represents the change in position of an object, while a velocity vector represents the change in position over time. Displacement is a vector quantity, while velocity is a vector quantity with both magnitude and direction. In problem solving, displacement vectors are often represented with a delta symbol (∆), while velocity vectors are represented with the letter v.

## 5. How do I find the resultant vector in a problem with multiple vectors?

To find the resultant vector, you can use the Pythagorean theorem to find the magnitude and trigonometric functions to find the direction. First, break down each vector into its components and then add the corresponding components together. Finally, use the Pythagorean theorem to find the magnitude of the resultant vector and use trigonometric functions to find its direction. Alternatively, you can also use the parallelogram method or the head-to-tail method to visually represent and find the resultant vector.

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