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October 13th, 2019, 04:10 AM   #1
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Question How to obtain the resultant and its modulus from a group of vectors in a circle?

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 skipjack; October 13th, 2019 at 06:00 AM. Reason: added information.
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October 13th, 2019, 06:49 AM   #2
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Quote:
Originally Posted by Chemist116 View Post
$\vec{r}=10+10+20=40$
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.
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October 15th, 2019, 10:07 PM   #3
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Question Uh I'm still confused

Quote:
Originally Posted by skipjack View Post
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.
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October 16th, 2019, 08:15 AM   #4
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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
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October 21st, 2019, 09:18 PM   #5
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Quote:
Originally Posted by skipjack View Post
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.
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.
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