# Homework Help: Mass, magnitude, direction of vectors

1. Sep 29, 2011

### jdroidxw

1. The problem statement, all variables and given/known data
I'm given a table of vectors A, B, and C of mass, magnitude, direction, x component, and the y component. The weights of 3 masses will apply tensions to 3 cables on a force table, the object is in translational equilibrium.
Given:
Vector A: m=0.2, magnitude=?, direction 30 degrees, x comp=?, y comp=?
Vector B: m=0.2, magnitude=?, direction 120 degrees, x comp=?, y comp=?
Vector C: m=?, magnitude=?, direction 30 degrees, x comp=?, y comp=?

or here http://chemphys.calumet.purdue.edu/~napora/classes/Phys152/labs/Lab02--ForceTable.pdf [Broken] pg 3

2. Relevant equations

$\Sigma$F=0
Magnitude = sqrt(a^2+b^2)

3. The attempt at a solution
For the magnitude of vectos A and B I just multiplied by g (9.8), for x component: i did cos(theta)*magnitude, and y comp: sin(theta)*magnitude.
For vector C, for x comp I did x-comp of vector A + Vector B + vector C = 0...and same for y component. for direction i did arctan(y comp/ x comp). and the magnitude I did sqrt[(magnitude of vector A)^2 + (magnitude of vector B)^2].

So I got
Vector A: m=0.2, magnitude=1.96, direction 30 degrees, x comp=1.7, y comp=.98
Vector B: m=0.2, magnitude=1.96, direction 120 degrees, x comp=-.98, y comp=1.7
Vector C: m=.28, magnitude=2.77, direction 255 degrees, x comp=-.72, y comp=-2.68

My numbers did not match up with the lab I did, are my calculations wrong? if not maybe it was the lab.

Last edited by a moderator: May 5, 2017
2. Sep 29, 2011

### Staff: Mentor

Where do they disagree?

Also, check your angle calculation for vector C.

[EDIT]: Never mind about the angle. It's fine; I calculated a value of -105° for it, which is of course the same as +255°. Silly of me not to spot that!

Last edited: Sep 29, 2011
3. Sep 29, 2011

### jdroidxw

What is wrong with my angle calculation for vector C? I get the same angle. and the magnitudes doesn't match up on vector C

4. Sep 29, 2011

### Staff: Mentor

Can you show your calculation for the angle for vector C?

To me it looks like the magnitude that you calculated is fine for the given information. So maybe something went awry in the lab...

5. Sep 29, 2011

### BobG

Your numbers look correct to me.

(note for the heck of it - if both x & y are negative, the angle has to be in the 3rd quadrant, not the first)

6. Oct 31, 2013

### Newtype09

I know this thread is quite old but I had this same question for one of my homework assignments and would like to know how to calculate the direction for vector C.

I noticed that jdroidxw said he used arctan(y comp/ x comp) to calculate the direction.

I tried (arctan)tan^-1(-2.67 / -0.71) and got 75.1.

I'm not quite familiar with the method however I noticed 75 x 3 is equal to 255.

Any clarifications?

Last edited: Oct 31, 2013
7. Oct 31, 2013

### Staff: Mentor

The arctan function cannot distinguish between an argument that is (-2.67 / -0.71) and one that is (2.67 / 0.71). This is because the signs of the components cancel out before your calculator ever sees them. What this means is, it's up to you to place the angle in the appropriate quadrant.

If both the x and y components are negative then the angle must lie in the 3rd quadrant, and you have to adjust the result accordingly by adding or subtracting 180°.

Similarly, the arctan function can't tell if a negative argument means the sign was associated with the x or the y component. Again you need to know ahead of time which quadrant the result should fall in and adjust the result if required.