The drawing shows three particles far away from any other objects and located on

Question

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 335 kg, mB = 559 kg, and mC = 137 kg. Take the positive direction to be to the right. Find the net gravitational force, including sign, acting on (a) particle A, (b) particle B, and (c) particle C

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 335 kg, mB = 559 kg, and mC = 137 kg. Take the positive direction to be to the right. Find the net gravitational force, including sign, acting on (a) particle A, (b) particle B, and (c) particle C

Explanation / Answer

mass of particle A ,m_A =335 kg

335 kg, mB= 559 kg, and mC= 137 kg





mass of particle C, m_C =137 kg

A) for particle A ,

F =F_AB+F_AC
F_AB =G(m_Am_B/r^2)
where gravitational constant G =6.67*10^-11 N.m^2/kg^2

F_AB =(6.67*10^-11 N.m^2/kg^2)(335*559/(0.5)^2)

=5.00*10^-5 N

F_AC =(6.67*10^-11 N.m^2/kg^2)(335*137/(0.75)^2)
=0.5442*10^-5 N

F =5.00*10^-5 N+0.5442*10^-5 N
=5.4442 10^-5 N
---------------------------------------------------------------------------------------------
B)
for particle B ,
F =-F_BA+F_BC
F_BA =G(m_Bm_A/r^2)
where gravitational constant G =6.67*10^-11 N.m^2/kg^2
F_BA =(6.67*10^-11 N.m^2/kg^2)(335*559/(0.5)^2)
=5.00*10^-5 N
F_BC =(6.67*10^-11 N.m^2/kg^2)(559*137/(0.25)^2)
=8.1729*10^-5 N
F =-5.00*10^-5 N+8.1729^-5 N
=3.1729*10^-5 N -------------------------------------------------------------------------------------------------- C) for particle C ,
F =-F_CB-F_CA
F_CB =(6.67*10^-11 N.m^2/kg^2)(559*137/(0.25)^2)
=8.1729*10^-5 N
F_CA =(6.67*10^-11 N.m^2/kg^2)(335*137/(0.75)^2)
=0.5442*10^-5 N
F =-8.1729*10^-5 N-0.5442*10^-5 N
F=-8.7171*10^-5 N

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