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Impossible vector problem. Help! 
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#1
Jul2405, 08:38 AM

P: 70

Most of the time, one uses vectors to find an overall magnitude acting on something, but I need to go in reverse.
Say I know there are a slew of charges and the center one feels a force of some magnitude and there is one undefined vector (magnitude and angle unknown) I would think if you know the overall vector, one could calculate the missing vector in terms of magnitude and angle. I want to solve for this unknown vector but I have no idea how. It seems impossible. 


#2
Jul2405, 09:16 AM

Math
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Thanks
PF Gold
P: 39,345

Essentially, you are saying that you have the equation
[tex]x_1+ x_2+ ...+ x_n= F[/tex] where [tex]x_1,... x_{n1}[/tex] are the known vector forces, [tex]x_n[/tex] is the one unknown vector and F is the known resultant. Solve that exactly the way you would any equation: [tex]x_n= F x_1 x_2 ... x_{n1}[/tex]. Essentially that "subtraction" on the right is just like addition of vectors except that you reverse the direction of [tex]x_1,... x_{n1}[/tex]. You might find the calculation easier as [tex]x_n= x_1+ x_2+ ...+ x_{n1} F[/tex]. That is, you just reverse the direction of F, add the vectors and then reverse the direction of the result to find [tex]x_n[/tex]. 


#3
Jul2405, 09:21 AM

Mentor
P: 41,318

Consider the vector equation:
[tex]\vec{F}_{net} = \vec{F}_{known} + \vec{F}_{unknown}[/tex] Where F(net) is the net force at the center and F(known) is the sum of the known forces from each charge. To solve for the unknown vector, F(unknown), just subtract. (Looks like Halls beat me to it.) 


#4
Jul2405, 09:58 AM

P: 70

Impossible vector problem. Help!
Overall Magnitude = sqrt( (mag1*cos(A) + mag2*cos(B) + mag3*cos(C))^2 + (mag1*sin(A) + mag2*sin(B) + mag3*sin(C))^2)
I dont think it's that easy since the angle and magnitude go hand in hand. I need to solve for both mag1 and angle A. I'm sorry if I forgot to really point that out. I thought it would be something more involved along the lines of langrange multipliers. 


#5
Jul2405, 10:13 AM

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P: 41,318




#6
Jul2405, 11:06 AM

P: 70

Yes, using the horizontal and vertical components would be really easy but I only know the overall magnitude :(



#7
Jul2405, 02:43 PM

Mentor
P: 41,318

If all you know is the magnitude, then you can't solve the problem. I suspect that you know both the magnitude and the angle. Use those to find the components.



#8
Jul2405, 03:33 PM

P: 4,006

Here You Go regards marlon 


#9
Jul2605, 07:37 AM

P: 70

Thanks for all the help. I guess not having an overall angle made the problem, well, impossible. Including it brought me quickly to the answer.



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