Physics Help Forum Adding Vectors-Calculating Magnitude and Location

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 Sep 3rd 2017, 02:30 PM #1 Junior Member   Join Date: Sep 2017 Posts: 5 Adding Vectors-Calculating Magnitude and Location Hey if someone could help me on some homework I have i'd really appreciate it. I'm stuck on this problem. Thanks!
 Sep 3rd 2017, 02:45 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 585 I assume you have heard of a vector polygon and know that the lengths of the vector arrows are proportional to teh magnitudes of the vectors to the same scale? You move the individual vectos (keeping them parallel to the original) to place in them head to tail order. So the tail of the second arrow is placed on the ehad of the first one and so on, until all have been used. This will create an open chain of vector 'links'. The sum is (also called the resultant ) is the closing link vector that makes the open chain into a closed polygon. The direction of the resultant follws the that of the rest of the chain so its tail starts from the last head in the chain and ends on the first tail. The calculation then becomes a question of geometry/trigonometry. Hint the order you take the vectors does not matter so you can rearrnage this to make it easier. Attached Thumbnails
 Sep 3rd 2017, 03:18 PM #3 Junior Member   Join Date: Sep 2017 Posts: 5 I still don't understand. Do we disregard the length and height given? (First vector 3,5) Am I trying to solve for the length of the resultant vector or am I just adding the forces? I had thought this was a two part question when it asked for magnitude and location. Thanks again for your help.
 Sep 4th 2017, 03:05 AM #4 Senior Member     Join Date: Jun 2016 Location: England Posts: 246 The 3:5 and 5:12 are indicators of the gradient of the vector. so (for example) the first vector has total length 10 If you draw a right angle triangle with the vector as the hypotenuse and with one side aligned with the X axis and the other aligned with the Y axis then the ratio of the length of the X aligned side to the length of the Y aligned side will be 3 to 5. __________________ ~\o/~
Sep 4th 2017, 04:03 AM   #5
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 I still don't understand.
It is difficult to know how answer this since you didn't answer my question.
I was not trying to trip you up, just find out what you have been taught/know about vectors and their addition.

No you don't discard the information in the triangles. As Woody says they give you the directions of the individual vectors.

The way I suggested is the way a draughtsman would solve this - by drawing to scale.
He would draw the 5,3 and 5,12 right angled triangles (What do you know about right angles triangles and these numbers?)and extend their hypotenuses to scale to represent the first two vectors.
Then he would actually draw the polygon parallel to these, head to tail, just as I said.
Finally he could then measure the magnitude and direction of the resultant.

An alternative would be to use the triangles to calculate the horizontal and vertical components of each vector.
The add up the horizontal and vertical components separately to form horizontal and vertical resultants.
Then combine the H and V resultants in a vecotr triangle to form the overall resultant.
Then calculate the length and direction of this overall resultant from the vector triangle.

Perhaps you were taught this latter method?

 Sep 4th 2017, 06:08 AM #6 Junior Member   Join Date: Sep 2017 Posts: 5 Well we studied using the Pythagorean theorem. So originally I had tried to calculate the y axis of the first vector. I was confused by the different units of measurements. So what you're saying is that because the ratio is 5 and 3 so then because the force is on the hypotenuse with 5 then the value for the x axis would be 6? The method we learned was to get the directional value of the vectors and to add them up. So taking a vector and creating a triangle to calculate its location. I am confused about the addition of the 10lbs, 4lbs, 16lbs. We learnt to add vectors like (1,2)+(3,4)+(5,6)=(9,11)
Sep 4th 2017, 07:15 AM   #7
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 What do you know about right angles triangles and these numbers?)
They have given you a couple of simple triangles with whole number sides.

There are not that many of these so it is good to know them.

The first one is a 3,4, 5 triangle.

So to make the hypotenuse = 10 we double all the lengths

That is 6,8,10.

Now we can put the first vector into form that you are familiar with.

10 is the vector itself, the horizontal bit is 6 and the vertical bit is 8 so the first vector is

Can you now use the fact that the second triangle is a 5,12,13 to put it into your format?
Note this will mean multiplying by a fraction to get the 4lbs.

Finally the last vector has only horizontal components so it is (16,0)

Now you should be able to add the vectors, and form the triangle to find the direction.

Note this is in essence exactly the same as the second method I suggested - using components.

Sep 4th 2017, 07:35 AM   #8
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Oops I missed that

 The first one is a 3,4, 5 triangle.
Ahhh,
I had misread the diagram as showing the triangle on the first vector
with X component 3 and Y component 5

It makes things much easier with having X component 3 and the hypotenuse of the triangle as 5 (leaving the Y component as 4).

But why they couldn't have just kept a standard description showing the X and Y components of gradient I don't know.
It just invites misinterpretation unnecessarily, without (in my view) significantly adding to the training effectiveness of the question.
__________________
~\o/~

 Sep 4th 2017, 11:08 AM #9 Junior Member   Join Date: Sep 2017 Posts: 5 Ok that makes things clearer. So I used the Pythagorean theorem to calculate the y component of the first vector which is 4 (I also read up on the triples). so now I have a 2:1 ratio. The x,y coordinates of the first is 6,8. The 2nd is another triple and the hypotenuse is 13. The ratio would be 4/13:1? multiply that ratio to the x and y coordinates and x becomes 1.54 and y is 3.69? The final vector is (0,16). Then I would add the x and y coordinates to solve for the final vector. (7.54, 27.69). I'm not sure if the 2nd part is done right since they aren't whole numbers.
Sep 4th 2017, 04:01 PM   #10
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 The final vector is (0,16).
Where did that come from?

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