Magnitude and direction of the net force on a particle. Please help!


by hzee
Tags: direction, force, magnitude, particle
hzee
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#1
Aug18-11, 06:23 PM
P: 7
1. The problem statement, all variables and given/known data

Three point charges are arranged as shown in the figure below. (Take q1 = 5.70 nC, q2 = 4.60 nC, and q3 = -2.79 nC.)

(a) Find the magnitude of the electric force on the particle at the origin.


(b) Find the direction of the electric force on the particle at the origin.
2 (counter-clockwise from the +x-axis)


2. Relevant equations

Fab= k q1q2/r2



3. The attempt at a solution

I am attaching a photo of my last attempt to do this problem but I really think that I don't know what I am doing. I think I am using the wrong length and so I can't get the magnitude right. I can't even get to the direction part but I can't do that either because I don't know if I'm supposed to use sin or cos.
Attached Thumbnails
physics.jpg   physics2.jpg  
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PeterO
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#2
Aug18-11, 06:53 PM
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P: 2,316
Quote Quote by hzee View Post
1. The problem statement, all variables and given/known data

Three point charges are arranged as shown in the figure below. (Take q1 = 5.70 nC, q2 = 4.60 nC, and q3 = -2.79 nC.)

(a) Find the magnitude of the electric force on the particle at the origin.


(b) Find the direction of the electric force on the particle at the origin.
2 (counter-clockwise from the +x-axis)


2. Relevant equations

Fab= k q1q2/r2



3. The attempt at a solution

I am attaching a photo of my last attempt to do this problem but I really think that I don't know what I am doing. I think I am using the wrong length and so I can't get the magnitude right. I can't even get to the direction part but I can't do that either because I don't know if I'm supposed to use sin or cos.
You have said the charges are 3 m & 1 m apart when they are only 0.300 & 0.100

You won't use pythagorus on the triangle of distances, you will use it on the triangle of force, force and net force.

whether you use sin, cos or tan will depend which two forces you chose to work with: are the opposite, adjacent or hypotenuse of the triangle.
fleem
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#3
Aug18-11, 07:02 PM
P: 461
I didn't study your work in detail because the first thing I noticed was that in your calculations you were somehow using the distance between q1 and q3. That distance does not directly affect the force on q2 and is not needed for a simple solution. Also, the distances on your work paper appear to be ten times greater than those shown on the problem.

The force on a particle is the sum of the forces contributed by all other particles. So to solve this, separately calculate the force between q1 and q2 (which, conveniently, has only an X component), and the force between q3 and q2 (which, conveniently, has only a Y component) and add those two vectors together to get the force vector on q2.

hzee
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#4
Aug18-11, 07:41 PM
P: 7

Magnitude and direction of the net force on a particle. Please help!


Thank you both so much for the help. I feel like such a dummy about the distance. I think I have just been looking at these problems for too long.

Ok so I made the corrections you indicated and I'm still 10% wrong but idk where the problem is. I'm attaching my new work.
Attached Thumbnails
physics3.jpg   physics4.jpg  
gneill
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#5
Aug18-11, 08:07 PM
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In what directions are each of the forces directed? (Pay attention to both the relative positions of the charges and their signs!). How do you add forces directed in different directions to find the net force and its magnitude?
PeterO
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#6
Aug18-11, 08:08 PM
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Quote Quote by hzee View Post
Thank you both so much for the help. I feel like such a dummy about the distance. I think I have just been looking at these problems for too long.

Ok so I made the corrections you indicated and I'm still 10% wrong but idk where the problem is. I'm attaching my new work.
You can't just add the two magnitudes at the end, they are perpendicular!!!

You need to draw triangle of forces, and use pythagorus to find the resultant force.

eg 2N west plus 1N South = 2.236N "south-west-ish". Trig can find the real direction.
fleem
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#7
Aug18-11, 08:20 PM
P: 461
I didn't do the arithmetic so I'll assume its right, but you must still add those two vectors to get the sum of the forces on q2. Then convert to polar coords (magnitude and direction) because the question wants that. It looks like you entered in the 'magnitude' box only the magnitude of one of those vectors.

EDIT: Yep, what Peter said.
hzee
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#8
Aug18-11, 08:21 PM
P: 7
Quote Quote by gneill View Post
In what directions are each of the forces directed? (Pay attention to both the relative positions of the charges and their signs!). How do you add forces directed in different directions to find the net force and its magnitude?
Well I thought that when the signs are the same the force is opposing so the sign should be neg and when the signs are different the force is attractive so the sign should be pos but when I tried that it didn't work.


Quote Quote by PeterO View Post
You can't just add the two magnitudes at the end, they are perpendicular!!!

You need to draw triangle of forces, and use pythagorus to find the resultant force.

eg 2N west plus 1N South = 2.236N "south-west-ish". Trig can find the real direction.
I understand what you mean but I don't know how to calculate it. Take the sq rt of the 2 forces squared? I don't know what I am doing and I'm kinda freaking out.

Would that be sq rt of (9.15X10^-7)^2 + (1.38X10^-6)^2

I don't want to enter the wrong answer again cuz it only gives me so many chances.
PeterO
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#9
Aug18-11, 08:38 PM
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Quote Quote by hzee View Post
Well I thought that when the signs are the same the force is opposing so the sign should be neg and when the signs are different the force is attractive so the sign should be pos but when I tried that it didn't work.




I understand what you mean but I don't know how to calculate it. Take the sq rt of the 2 forces squared? I don't know what I am doing and I'm kinda freaking out.

Would that be sq rt of (9.15X10^-7)^2 + (1.38X10^-6)^2

I don't want to enter the wrong answer again cuz it only gives me so many chances.
Where on earth did you get those numbers? I don't see them written anywhere on the work sheet you showed us???
hzee
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#10
Aug18-11, 08:42 PM
P: 7
Quote Quote by PeterO View Post
Where on earth did you get those numbers? I don't see them written anywhere on the work sheet you showed us???
Sorry from a different problem (that I did get right)

It should have been 2.622X10^-6 and 1.155X10^-5

I was just looking at the wrong page when I typed that.
PeterO
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#11
Aug18-11, 08:51 PM
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Quote Quote by hzee View Post
Sorry from a different problem (that I did get right)

It should have been 2.622X10^-6 and 1.155X10^-5

I was just looking at the wrong page when I typed that.
I certainly think it is worth a shot if you use these values.
hzee
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#12
Aug18-11, 09:17 PM
P: 7
YAY!!! Thank you so much! Now I just have to figure out how to do the second half and I will be done. I just have no idea where to even start. It has been so long since I did any of this and I don't remember much. Will you please just help me get started?
gneill
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#13
Aug18-11, 09:47 PM
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Quote Quote by hzee View Post
Well I thought that when the signs are the same the force is opposing so the sign should be neg and when the signs are different the force is attractive so the sign should be pos but when I tried that it didn't work.
It works just fine, but you must also take into account the geometry of the charge arrangement! The forces individually act between pairs of charges, but those charges are located in some arrangement with respect to the coordinate system that you choose for the problem. So a force can end up with positive or negative components in the directions of the coordinate system axes whether or not it's created by an attraction or a repulsion.

Try sketching the force vectors on your diagram; place arrowed line segments with their "feet" attached to the object on which the force is acting and the pointy ends pointing in the direction that the force acts. Between charge pairs this will inevitably lie along a line joining the centers of the charge pair. After you've done this, then it's time to look at the vectors' orientation in the coordinate system and decide whether the x and y components of the vectors are positive or negative.
hzee
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#14
Aug18-11, 09:58 PM
P: 7
Quote Quote by gneill View Post
It works just fine, but you must also take into account the geometry of the charge arrangement! The forces individually act between pairs of charges, but those charges are located in some arrangement with respect to the coordinate system that you choose for the problem. So a force can end up with positive or negative components in the directions of the coordinate system axes whether or not it's created by an attraction or a repulsion.

Try sketching the force vectors on your diagram; place arrowed line segments with their "feet" attached to the object on which the force is acting and the pointy ends pointing in the direction that the force acts. Between charge pairs this will inevitably lie along a line joining the centers of the charge pair. After you've done this, then it's time to look at the vectors' orientation in the coordinate system and decide whether the x and y components of the vectors are positive or negative.
I got it with all of your help, the (a) part anyway. I just don't know what to do about the (b) part. I think I need to find a Physics for Dummies book. I don't know how I'm going to get through this class. Its going to be a long semester...
gneill
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#15
Aug18-11, 10:00 PM
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Quote Quote by hzee View Post
YAY!!! Thank you so much! Now I just have to figure out how to do the second half and I will be done. I just have no idea where to even start. It has been so long since I did any of this and I don't remember much. Will you please just help me get started?
You should review your vector math. This question is all about vector addition and converting from Cartesian to polar form.

You should be able to identify the x and y components of your resultant force vector (after all, you calculated the magnitude of the resultant from them!).
hzee
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#16
Aug18-11, 10:26 PM
P: 7
Thanks again for your help everyone!


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