Three 5.00 kg masses are at the corners of an equilateral triangle and located in space far from any other....

Question

Three 5.00 kg masses are at the corners of an equilateral triangle and located in space far from any other masses.

(a) If the sides of the triangle are 2.45 m long, find the magnitude of the net force exerted on each of the three masses.
N
(b) What would your answer to part (a) be if instead the sides of the triangle are doubled in length?
N

Answer 1

a)

F1=F1,2 + F1,3
F2=F2,1 + F2,3
F3=F3,1 + F3,2

and F1=F2=F3=Ft (in magnitusde only)
F1,2=Gm1m2/(R1,2 )^2 since
All masses distances are equal
F=Gm^2/R^2 not so easy here
F1=F1,2 + F1,3
F1= Gm^2/R^2 + [Gm^2/R^2]cos(60)
F1= Gm^2/R^2(1+cos(60))=1.5 Gm^2/R^2 Actually
Absolute value of the force is
|F|=1.5 Gm^2/R^2
Note: To compute actual magnitudes adnd direction you would have to sketch and label vaiables and then solve accordingly.

b) In case r=2R
|F'|=1.5 Gm^2/(2R)^2
|F'|=0.375 Gm^2/R^2 or


F/F'=[1.5 Gm^2/R^2] /[0.375 Gm^2/R^2]

F/F'=4 so by doubling the distance we deminish the force by factor 4.

Have fun

Answer 2

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Force on one mass due to mass at one corner = F1=G*5*5/(2.45)^2

F1=10.20408 G=6.8061*10^-10 N along one side

where G is universal gravitational constant equal to 6.67*10^-11 Nm^2/kg^2

Force on one mass due to mass at another corner = F2= G*5*5/(2.45)^2

F2=10.20408 G =6.8061*10^-10 N along another side

As the triangle is equilateral, the angle between any two sides is 60 degree

The two forces F1 and F2 are inclined at 60 degree and magnitude of F1= magnitude of F2 =6.8061*10^-10 N = F

Resultant of two equal forces of magnitude F each inclined at 60 degrees = root3*F=1.732 F=1,732*6.8061*10^-10=1.1788*10^-9 N

Each of the three masses will experience same net force equal to 1.1788*10^-9 N

(a) The magnitude of the net force exerted on each of the three masses is 1.1788*10^-9 N
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(b)if the sides of the triangle are doubled in length, the net force will be (1/4) of original.

The net force after sides are doubled will be (1/4) of 1.1788*10^-9 N
The net force after sides are doubled will be 2.9471*10^-10 N
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Answer 3

Let the triangle be ABC with BC along the x-axis.
Let each side be a m (so that in the 2nd part, you just have to substitute the values.)

Force of B on A, F(BA) = Gm^2 / r^2 = 25G / a^2
Break this force into rectangular components :
theta = Angle with x-axis is 60.
So F(x) = F(BA) cos 60 = 25G / 2a^2
F(y) = F(BA) sin 60 = 43.3G / 2a^2

Similarly find F(CA) and break it into rectangular components.

So the net force exerted on A = F(BA) + F(CA)
The x-components cancel out .
Net force = 2F(y) = 43.3G / 2a^2 = 1.44 * 10^-9 / a^2

Now put the values of a for each part and find the answer.

Hope this helps.

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