I have... F 1 = F 3 = .018 F 2 = .012 F x = 0 (.018 - .018) F y = 2 F 1 cos30 -

Question

I have...

F1 = F3 = .018

F2 = .012

Fx = 0 (.018 - .018)

Fy = 2 F1 cos30 - F2

I know that direction will be 90 degrees up because the x's cancle out. Since there is no Fx value, the magnitude should be whatever the Fy value is, but where does the cos30 value come from? shouldn't it be cos60 if it's an equilateral triangle?

1. Three point charges q1 = 2 muC, q2 = 1 muC, q3 = 2 muC are located at the corners of an equilateral triangle (a = 1.0 m), as shown below. Another charge q4 = - 1 muC is placed at the base of the triangle. Calculate the net electric force (magnitude and direction) on q2. I have... F1 = F3 = .018 F2 = .012 Sigma Fx = 0 (.018 - .018) Sigma Fy = 2 F1 cos30 - F2 I know that direction will be 90 degrees up because the x's cancle out. Since there is no sigma Fx value, the magnitude should be whatever the sigma Fy value is, but where does the cos30 value come from? shouldn't it be cos60 if it's an equilateral triangle?

Explanation / Answer

You are correct in most of the assumptions you have made, but, the y component is the opposite side of the triangle. The opposite side is found by taking the sin of the angle, which is 60o.

Then remember that sin 60o is the same thing as cos 30o

So, to solve.

Fy = kqq(sin60)/d2 + kqq(sin 60)/d2 - kqq/d'2

Fy = 2(9 X 109)(2 X 10-6)(1 X 10-6)(sin 60)/(1)2 - (9 X 109)(1 x 10-6)(1 X 10-6)/(.866)2

Fy = 1.92 X 10-2 N

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---