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Three lead balls of mass m1 = 14 kg, m2 = 26 kg, and m3 =...

Three lead balls of mass

m1 = 14

kg,

m2 = 26

kg, and

m3 = 8.7

kg are arranged as shown in the figure below. Find the total gravitational force exerted by balls 1 and 2 on ball 3. Be sure to give the magnitude and the direction of this force.

magnitude _______ N
direction __________
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Answer #1

x direction:

Fx = (-G m3 m1/(r13)^2) (3/r13) + (-G m3 m2/(r23)^2) (sqrt(2)/2)

Fx = (-6.67e-11*8.7*14/3.0414y2) * (3/3.0414) + (-6.67e-11*8.7*26/2.1213y2) * (0.7071)

Fx = -3.237 x 10^-9 N

y direction:

Fy = (G m3 m1/(r13)^2) (0.5/r13) - (G m3 m2/(r23)^2) (sqrt(2)/2)

Fy = (6.67e-11*8.7*14/3.0414y2) * (0.5/3.0414) - (6.67e-11*8.7*26/2.1213y2) * (0.7071)

Fy = -2.226 x 10^-9 N

------------------------------------

magnitude = F = sqrt(3.237*3.237 + 2.226*2.226) = 3.9 x 10^-9 N

---------------------------------------

direction = theta = 180 + arctan(2.226/3.237) = 214.515 = 210 degrees

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