PhysicsQ/ Adding vectors-equilateral triangle

  • Thread starter Ba.Ala
  • Start date
  • #1
3
0
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!!
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,097
5
(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.
 
  • #3
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7
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, [itex]2cm = 2\times10^{-2}m[/itex] and not [itex]2\times10^{-6}m[/itex]. 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.
 
  • #4
3
0
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,.
 
  • #5
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7
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.
 
  • #6
3
0
how would I do that?
could you please elaborate..
 

Related Threads on PhysicsQ/ Adding vectors-equilateral triangle

  • Last Post
Replies
2
Views
14K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
8K
  • Last Post
Replies
5
Views
6K
Replies
2
Views
8K
Replies
10
Views
8K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
7
Views
3K
Top