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 October 13th, 2019, 04:10 AM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus 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.
October 13th, 2019, 06:49 AM   #2
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Quote:
 Originally Posted by Chemist116 $\vec{r}=10+10+20=40$
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.

October 15th, 2019, 10:07 PM   #3
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Uh I'm still confused

Quote:
 Originally Posted by skipjack 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.

 October 16th, 2019, 08:15 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,111 Thanks: 2326 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 Thanks from Chemist116
October 21st, 2019, 09:18 PM   #5
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Quote:
 Originally Posted by skipjack 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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