PhysicsQ/ Adding vectors-equilateral triangle

[Ba.Ala]

Please help me with this question, I am a complete physics newbie who is lost…
Alright.
The Question states:

The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed at each corner, as shown. The 400 mC charge experiences a net force due to the charge q A and qB. This net force points vertically downward in the drawing and has a magnitude of 405 N. Determine the magnitude and algebraic sighs of the charges qA and qB.

This is what I did:

(1) I know that centimeters are supposed to be converted to meters
2.00*10-6
(2) Two vectors A & B are added together to produce a third vector C = A + B .
(3) When the triangle formed is not a right triangle:
the magnitude of the vector sum can be found by using the Law of Cosines or the Law of Sines

meaning:


4.00*10-6
-----------------------
1

so now I am lost
HELP!

 

[LowlyPion]

(1) I know that centimeters are supposed to be converted to meters
2.00*10-6
(2) Two vectors A & B are added together to produce a third vector C = A + B .
(3) When the triangle formed is not a right triangle:
the magnitude of the vector sum can be found by using the Law of Cosines or the Law of Sines

meaning:

4.00*10-6
-----------------------
1

so now I am lost
HELP!

Welcome to PF.

First off: 1 cm = .01 m

I can't see your picture. But you are right the basic idea for field values is vector addition.

The rest of it ... I can see no drawing. I'm not exactly sure what you are doing math wise.

 

[Hootenanny]

Please help me with this question, I am a complete physics newbie who is lost…
Alright.
The Question states:

The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed at each corner, as shown. The 400 mC charge experiences a net force due to the charge q A and qB. This net force points vertically downward in the drawing and has a magnitude of 405 N. Determine the magnitude and algebraic sighs of the charges qA and qB.

This is what I did:

(1) I know that centimeters are supposed to be converted to meters
2.00*10-6
(2) Two vectors A & B are added together to produce a third vector C = A + B .
(3) When the triangle formed is not a right triangle:
the magnitude of the vector sum can be found by using the Law of Cosines or the Law of Sines

meaning:


4.00*10-6
-----------------------
1

so now I am lost
HELP!

Welcome to PF,

Firstly, $2cm = 2\times10^{-2}m$ and not $2\times10^{-6}m$. Secondly, have you tried splitting the vectors into horizontal and vertical components? Since the net force acts downwards, you know that the horizontal components must sum to zero.

 

[Ba.Ala]

thank you
alright this is where my problem is
if: A B C
----= ----- = ---
sine(θA) Sine (θB) Sine(θC)

this is an equliateral triangle meaning:
2*10^-2 2*10^-2 2*10^-2
---------= ----------- -----------
1 1 1

there is no answere
and that is why I am stuck,.

 

[Hootenanny]

thank you
alright this is where my problem is
if: A B C
----= ----- = ---
sine(θA) Sine (θB) Sine(θC)

this is an equliateral triangle meaning:
2*10^-2 2*10^-2 2*10^-2
---------= ----------- -----------
1 1 1

there is no answere
and that is why I am stuck,.

There is no need to resort to the laws of sines or cosines. As I said previously, try splitting each force vector into two components, one vertical and one horizontal.

 

[Ba.Ala]

how would I do that?
could you please elaborate..