Three balls A, B and C are kept in a straight line. The separation between A and C is 1 m, and B is placed at the midpoint between them. The masses of A, B, C are 100 g, 200 g and 300 g respectively. Find the net gravitational force on (a) A, (b) B, and (c) C.
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mass of body A = 100 g = 0.1 kg
mass of body B = 200 g = 0.2 kg
mass of body C = 300 g = 0.3 kg
AC = 1 m
AB = BC = 0.5 m
Force on A = Force on A due to B + Force on A due to C
= Gm(A) m(B)/AB² + Gm(A)m(C)/AC²
= 6.67 x 10^ -11 x 0.1 x 0.2 / ( 0.5 )² + 6.67 x 10^-11 x 0.1 x 0.3/1
after solving ...
= 7.34 x 10^-10 N
This process is similar for B and C
but in the case of B the forces are opposite
F = Gm(B)m(C) /BC² - Gm(B)m(A)/AB²
after solving ...
F = 1.06 x 10^-11 N
For C the process is similar ...
F = Gm(C)m(B)/BC² + Gm(C)m(A)/AC²
after solving ...
F = 1.8 x 10^-11 N
mass of body B = 200 g = 0.2 kg
mass of body C = 300 g = 0.3 kg
AC = 1 m
AB = BC = 0.5 m
Force on A = Force on A due to B + Force on A due to C
= Gm(A) m(B)/AB² + Gm(A)m(C)/AC²
= 6.67 x 10^ -11 x 0.1 x 0.2 / ( 0.5 )² + 6.67 x 10^-11 x 0.1 x 0.3/1
after solving ...
= 7.34 x 10^-10 N
This process is similar for B and C
but in the case of B the forces are opposite
F = Gm(B)m(C) /BC² - Gm(B)m(A)/AB²
after solving ...
F = 1.06 x 10^-11 N
For C the process is similar ...
F = Gm(C)m(B)/BC² + Gm(C)m(A)/AC²
after solving ...
F = 1.8 x 10^-11 N
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