How to obtain the resultant and its modulus from a group of vectors in a circle?

Jun 2017
352
6
Lima, Peru
The problem is as follows:

Find the modulus of the resultant from the vectors shown in the picture from below:



The alternatives given on my book are:

$\begin{array}{ll}
1.&10\,\textrm{inch}\\
2.&20\,\textrm{inch}\\
3.&10\sqrt{3}\,\textrm{inch}\\
4.&20\sqrt{3}\,\textrm{inch}\\
5.&60\,\textrm{inch}\\
\end{array}$

For this problem the only thing I could come up with is described in my attempt seen in the figure from below:



I thought that it was easy to form a closed polygon and from there I could obtain a sum like this, hence the resultant:

$\vec{r} = \vec{u} + \vec{v} + \vec{w} + \vec{x} + \vec{y} + \vec{z}$

$\vec{x}+\vec{y}+\vec{z}=\vec{w}$

$\vec{w}+\vec{v}=\vec{u}$

$\vec{r}=\vec{u}+\vec{u}+\vec{w}$

For the sake of brevity, I'm omitting units. Since it is known the radius is $\textrm{10 inch}$ then:

$\vec{w}=20$

and from there it can be inferred that:

$\vec{u}=10$

Then:

$\vec{r}=10+10+20=40$

But my answer doesn't appear in the alternatives. Am I doing something wrong? Could it be that Am I misinterpreting the concepts?. I'd like somebody could take a look into this as I'm confused with vectors. Since I believe an auxiliary drawing can be required to aid understanding of the answer I hope that somebody could help me if there is some sort of geometrical manipulation which can be done to solve this problem. I'd like to note that apparently $\textrm{O}$ is the center of the circle. :)
 
Last edited by a moderator:

skipjack

Forum Staff
Dec 2006
21,393
2,412
$\vec{r}=10+10+20=40$
That's incorrect. The appropriate conclusion is that the modulus of $\vec{r}$ is twice the length of a median of an equilateral triangle with sides of length 20 inch, which is 20√3 inch.
 
Jun 2017
352
6
Lima, Peru
Uh I'm still confused

That's incorrect. The appropriate conclusion is that the modulus of $\vec{r}$ is twice the length of a median of an equilateral triangle with sides of length 20 inch, which is 20√3 inch.
Sorry skipjack. But I had to take an exam (on this subject) so I couldn't reply early. Anyways, can you please tell me exactly what step I did wrong?. I don't know the graphical justification for your answer. Perhaps if you could show me this part I can "see" where's the resultant. Can you help me with that because I'm slow at picturing these things on my head solely reading words. ;)
 

skipjack

Forum Staff
Dec 2006
21,393
2,412
You first went wrong with $\vec{w}=20$, as it's the vector's modulus that is 20.

Your $\vec{r}=\vec{u}+\vec{u}+\vec{w}$ was correct, but $\vec{r}=10+10+20=40$ was wrong.

Hence $\vec{r} = 2\vec{u} + \vec{w}$ would be correct.

As $2\vec{u}$ and $\vec{w}$ are vectors of equal modulus, but there's an angle of 60 degrees between them, adding them by use of the parallelogram method produces a resultant that is twice a median of a certain triangle, which is equilateral in this case. See the diagram below for the addition of any two vectors by the parallelogram method. As the diagonals of a parallelogram bisect each other, PR = 2PT and PT is a median of triangle PQS. For the current problem, the median's length can be found by using Pythagoras. In the general case, trigonometry can be used.
ParallelogramMethod.PNG
 
  • Like
Reactions: 1 person
Jun 2017
352
6
Lima, Peru
You first went wrong with $\vec{w}=20$, as it's the vector's modulus that is 20.

Your $\vec{r}=\vec{u}+\vec{u}+\vec{w}$ was correct, but $\vec{r}=10+10+20=40$ was wrong.

Hence $\vec{r} = 2\vec{u} + \vec{w}$ would be correct.

As $2\vec{u}$ and $\vec{w}$ are vectors of equal modulus, but there's an angle of 60 degrees between them, adding them by use of the parallelogram method produces a resultant that is twice a median of a certain triangle, which is equilateral in this case. See the diagram below for the addition of any two vectors by the parallelogram method. As the diagonals of a parallelogram bisect each other, PR = 2PT and PT is a median of triangle PQS. For the current problem, the median's length can be found by using Pythagoras. In the general case, trigonometry can be used.
View attachment 10595
My bad I totally overlooked the fact that you mentioned. I did used the cosines law because that is what (I could use) to obtain the sum of the vectors which I obtained (am I right with this?) so by redoing all these steps I ended with the same result you got which is $20\sqrt 3$, I knew about how to obtain the median length but I had to recall the identity. Btw, the drawing you made was very helpful. ;)
 
Jun 2017
352
6
Lima, Peru
I'd like to make an update of how I obtained this answer because it resembles a similar problem where I'm struggling with:

The resultant is:

$\vec{R}=\vec{x}+\vec{y}+\vec{z}+\vec{w}+\vec{v}+\vec{u}$

$\vec{x}+\vec{y}+\vec{z}=\vec{w}$

$\vec{R}=\vec{w}+\vec{w}+\vec{v}+\vec{u}$

$\vec{R}=2\vec{w}+\vec{v}+\vec{u}$

Now the problem is what to do with $\vec{u}$

This can be expressed as the sum of the other two vectors as follows:

$\vec{u}=\vec{w}+\vec{v}$

This can be inserted in the above equation and yields:

$\vec{R}=2\vec{w}+\vec{v}+\vec{w}+\vec{v}=3\vec{w}+2\vec{v}$

I end up with these two vectors. Now to obtain the resultant. I must know the angle between both. It turns out that is $150^{\circ}$. Then it can be used the cosines law:

$R=\left\|3\vec{w}+2\vec{v}\right\|^{2}=60^{2}+(20\sqrt{3})^{2}-2(60)(20 \sqrt{3})\cos 30^{\circ}$

Finally this ends up as:

$\left\|3\vec{w}+2\vec{v}\right\|^{2}=1200$

$\left\|3\vec{w}+2\vec{v}\right\|^{2}=20\sqrt{3}$

To which corresponds to the answer. @skipjack proposed to use the approach of using Pythagoras and the median. I have not tried this. But I'd like to know how would to do that. Maybe it is easier than this.