[3] Force analysis A 50kg ball is attached to a 2 meter long cable which is susp

Question

[3] Force analysis A 50kg ball is attached to a 2 meter long cable which is suspended from a hook in the ceiling. The cable has a maximum load of 700N before it snaps. a) What is the maximum radius for a horizontal circle the ball can move in without breaking the cable? b) At what speed will the ball move in this circular path? Three masses are at the vertices of an equilateral triangle. Each mass is 1.21 x 1020kg and the sides of the triangle are 4.11x1015km. a) Calculate the magnitude of the net force on each mass and describe the direction the forces are in. b) How would the masses need to move to remain in a triangular arrangement of the same size at any future moment in time?

Explanation / Answer

A) maximum radius = length of the cable = 2 m

B) m*v^2/r = 700

v^2 = 700*r/m = 700*2/50 = 28

v = 5.3 m/s

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let m1 = m2 = m3 = 1.21*10^20 kg

r1 = r2 = r3 = 4.11*10^18 m

then Force on m1 due to m2 is F2 = G*m1*m2/r1^2 = (6.67*10^-11*1.21*1.21*10^40)/(4.11*4.11*10^36)
F2 = 5.78*10^-8 N

force on m1 due to m3 is F = 5.78*10^-8 m

then angle between F1 nad F2 is 60 degrees( since equilateral triangle)

then Fnet = Sqrt(F1^2+F2^2+2F1F2cos(60)) = Sqrt(3)*F1 = 5.78*10^-8*1.732 = 10.01*10^-8 N

direction is perpedicular bi sector to the base of the triangle (if our mass of consideration is at the top vertex)

B) if we want to move closer to each other then the masses must be bigger in size or vice versa

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