vectors SREIOUS HELP!

[crazziekid82]

okay im doing vectors and have no idea WHAT im doing.

 

for example:

add

5.0 m north to 3.0m west..

answer

5.8m n 31 degrees west.

 

i dont understand how they got this answer. i know it has to do with resultant. can someone show me please

[Royston]

I'll show you a similar (more complicated) example, then you can figure out how to get the result you're after.

 

Suppose we want to find the resultant displacement of 200m North, then a turn of 100 degrees in the clockwise direction, with a displacement of 50m, and then a displacement of 100m heading West.

 

First we let [math]i[/math] be 1m East and [math]j[/math] be 1m North. We then label the displacement vectors [math]a[/math] [math]b[/math] [math]c[/math].

 

so [math]a[/math] will be heading North

[math]b[/math] will be the turn and displacement of 50m

[math]c[/math] will be the vector heading to the West

 

Remember North is 90 degrees, look up quadrants if you're not sure about this, so we have...

 

[math]a[/math] = 200 cos(90)[math]i[/math] + 200 sin(90)[math]j[/math] = 200[math]j[/math]

 

90 degrees - 100 degrees = -10 so...

 

[math]b[/math] = 50 cos(-10)[math]i[/math] + 50 sin(-10)[math]j[/math] = 49.24[math]i[/math] - 8.68[math]j[/math]

 

and finally we have c, which as it's heading directly West will be 180 degrees...

 

[math]c[/math] = 100 cos(180)[math]i[/math] + 100 sin(180)[math]j[/math] = -100[math]i[/math]

 

Now we just add [math]i[/math] and [math]j[/math] together, to get the resultant [math]d[/math].

 

[math]d[/math] = [math]a[/math] + [math]b[/math] + [math]c[/math] = (49.24 - 100)[math]i[/math] + (200 - 8.68)[math]j[/math] ~ -50.76[math]i[/math] + 191.32[math]j[/math] so the magnitude of vector [math]d[/math] is...

 

[math]|d|[/math] ~ [math]\sqrt{(-50.76)^2 + (191.32)^2}[/math] ~ 197.93

 

The components of [math]d[/math] are therefore d1 = -50.76 and d2 = 191.32

So [math]\phi[/math] ~ arctan (|191.32 / (-50.76)|) = arctan (191.32 / 50.76) ~ 75.14 degrees

 

Now as the direction of [math]d[/math] lies in the second quadrant, 180 - 75.14 = 104.86 or 105 degrees, which would be a bearing of North 15 degrees West, with a resultant displacement relative to the starting point of 197.93m

 

EDIT: Just to add, with any vector problem...always, always draw a diagram.

[crazziekid82]

thank you for trying to help me , but that was too complicated for me, I've only just started physics and i have limited background in it. LOL oh and never fear i always draw diagrams ..it's the only thing i don't have trouble with

but thank you again

[Royston]

Well have a look at what I've done, I take it you've studied unit circles so x = r cos [math]\theta[/math] and y = r sin [math]\theta[/math]. So the first vector for your problem will be...

 

[math]a[/math] = 5 cos(90)[math]i[/math] + 5 sin(90)[math]j[/math] = 5[math]j[/math]

 

Now work out the vector heading in the West direction, and I'll run through how you add these two together.

 

Remember treat [math]j[/math] as the y axis, and [math]i[/math] as the x axis, as this corresponds with North and East.

[5614]

[edit] this is long, but it's all words and very little maths, spend a little while reading it and I'm sure you'll understand it well!

 

 

This is the diagram of the question:

except that I have accidentally drawn 5 north + 3 east, not west, but if you understand how to do this then you will understand how to do all similar vector additions.

 

What the question is asking is: if I go 5m north, and then 5m west, where do I end up, and what's the starting to end point distance and direction?

 

As can clearly be seen in the diagram you start at the bottom and end up in the top right.

 

Hopefully you have done Pythagoras' theorem (or Pythagorean theorem), and you can therefore work out the length of the blue line, which is what you get when you add the two black lines together (the details of the blue line is what the question is asking for).

 

Pythag tells you that: length of blue line squared = length of the black line squared + the lenght of the other black line squared. Or in maths:

(length of blue line)² = 5² + 3²

length of blue line = √(5²+3²) = √(25+9) = √(34) = 5.8

 

As for the angle. Well you now have all 3 lenghts, although it is best to use the two you know exactly (the 5 and 3). Then you have to use trigonometry. Do you know SOH CAH TOA? That's what most people in the UK seem to use.

 

In your question and in my diagram we want the bottom angle, and therefore the 5 and 3 lengths are the adjacent and opposite. As we have adj and opp, we want to use TOA, or tan.

 

If you call the angle at the bottom of the triangle θ, then you get:

tanθ = opp / adj = 3 / 5

therefore: θ = atan(3/5) = 31°.

NB: atan = arctan = tan^-1 (they're all different names for the same thing).

 

Note you can click on two of the equal signs (square root and atan ones), this is where you would need a calculator.

 

I hope this answers your question. Always draw a diagram!

[NeonBlack]

5614, I made that mistake all the time in highschool. I never lost any marks as long as I drew a compass redefining the directions with east and west swapped.

[Rune175]

Snail: One question to your equation (probably a dumb question) But why this:

 

90 degrees - 100 degrees = -10 so...

 

= 50 cos(-10) + 50 sin(-10) = 49.24 - 8.68

 

Why minus 90 from 100 degrees, is that because north was 90 degrees?

[ydoaPs]

okay im doing vectors and have no idea WHAT im doing.

 

for example:

add

5.0 m north to 3.0m west..

answer

5.8m n 31 degrees west.

 

i dont understand how they got this answer. i know it has to do with resultant. can someone show me please

 

Here are a few tips that will help you get it right(and some points even if you get the answer wrong ):

1)Draw a picture. This will not only help you understand the problem better, but it will also let the grader see what you are doing and how you understand the problem.

2)Make a table. Have an x column and a y column. Under the the x colunm, right the x-component of each vector(even if it is zero). Do the same thing with the y-components in the y colunm. Draw a line under your last row and make a total row. Have a x total and a y total. Making a table not only helps you keep track of your components and see mistakes easier, it makes your work more organized and the grader will be able to find partial credit easier.

 

With your problem, the table should be pretty simple since your vectors have either only an x component or only a y component.

[thedarkshade]

*I will have committed suicide while you are reading this*

[ydoaPs]

Jeez, darkshade, can't you ever add anything to the thread that hasn't already been said?

[Royston]

Snail: One question to your equation (probably a dumb question) But why this:

 

90 degrees - 100 degrees = -10 so...

 

= 50 cos(-10) + 50 sin(-10) = 49.24 - 8.68

 

Because, like you said, 90 degrees is North, so East will be 0 degrees. Turning in a clockwise direction from the North axis would equal -10 if it was 100 degrees.