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3 years ago

Can anyone explain how to do this to me? It is due tomorrow at 9:30am. Thanks.

Physics

1 answer:

Georgia [21]3 years ago

3 0

The 102N acting on the ropes being pulled by eric and kim have some of that force acting horizontally, and some of it vertically. By visualizing it as a right angled triangle, with the hypotenuse the length of the diagonal force, and each side the length of the horizontal and vertical forces, you can use trigonometry to calculate the length of the vertical force. You are told that it is at an angle of 30 with the vertical rope, therefore you know the length of the hypotenuse, and the angle between it and the vertical force, so using trig: (vertical force=x)
x/102=cos(30)
x=102*cos(30)
x=88.33
Therefore the diagonal ropes give a vertical force of 88.33N, and the centre rope, as it acts vertically, gives a vertical force of all 102N. The total:
88.33*2+102=278.66N

I don't know if this is very clear, I hope its good enough to help. If you don't understand, just ask, and I can answer any questions!!! :)

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Answer:

force is decreased by a factor of 4.

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3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

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- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

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F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

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3 years ago

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Sedbober [7]

Answer:

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