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Objects with masses of 191 kg and 572 kg are separated by 0.406 m. A 50 kg mass is placed midway between them. 0.406 m b b 191 kg 50 kg 572 kg Find the magnitude of the net gravitational force exerted by the two larger masses on the 50 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.

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Answer:

3.06 x 10^{-5} N

Explanation:

mass M1 = 191 kg

mass M2 = 572 kg

mass M3 = 50 kg

distance apart (d) = 0.406 m

gravitational constant (G) = 6.62 x 10^{-11} m^{2}/kg^{2}

Find the magnitude of the net gravitational force exerted by the two larger masses on the 50 kg mas

the net  gravitational force acting on the 50 kg mass (M3) = gravitational force exerted by the larger mass (M2) - gravitational force exerted by the smaller mass (M1)

⇒[tex]\frac{GM2.M3}{y^{2}} -\frac{GM1.M3}{y1^{2}}[/tex]

  • y = distance between masses M2 and M3 = 0.406/2 = 0.203
  • y1 = distance between masses M1 and M3 = 0.406/2 = 0.203
  • y = y1 = 0.203

net force = [tex](GM3)(\frac{M2}{y^{2}} -\frac{M1}{y1^{2}})[/tex]

net force = [tex](6.62 x 10^{-11} x 50)(\frac{572}{0.203^{2}} -\frac{191}{0.203^{2}})[/tex]

net force =  [tex](6.62 x 10^{-11} x 50)(13880.5 - 4634.9)[/tex]

net force = 3.06 x 10^{-5} N

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