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Objects with masses of 87 kg and 772 kg are separated by 0.45 m. a 63.9 kg mass is placed midway between them. 0.45 m b b 87 kg 63.9 kg 772 kg find the magnitude of the net gravitational force exerted by the two larger masses on the 63.9 kg mass. the value of the universal gravitational constant is 6.672 × 10−11 n · m2 /kg2 . answer in units of n. 005 (part 2 of 2) 10.0 points leaving the distance between the 87 kg and the 772 kg masses fixed, at what distance from the 772 kg mass (other than infinitely remote ones) does the 63.9 kg mass experience a net force of zero? answer in units of m.

Respuesta :

The net gravitational force on the object is the vector sum of all gravitation forces applied to the object.  The net gravitational force on the 3rd object will be -97.5524457254 N.

The net gravitational force:

[tex]F = F_s+ F_b[/tex]

Where,

[tex]F[/tex] - net force,

[tex]F_s[/tex] = Gravitational force by 1st object

[tex]F_b[/tex] -  Gravitational force by 2nd onject

From Newton's Law of Universal gravitation:

[tex]F=G{\frac{m_1m_2}{r^2}}[/tex]

Where,

F_1_3 = force

[tex] m_1 [/tex] =  mass of object 1 = 87 kg

[tex]m_3 [/tex]  = mass of object 3 = 63.9 kg

r = distance between centers of the masses = 0.225 m

Put the values in the formula,

[tex]F_1_3 = 6.672 \times 10^{-11 }\times \dfrac {87 \times 63.9}{0.225^2}\\\\ F_1_3 = 12.3898726686 [/tex]

Now solve it for F_23

[tex]F_2_3 = 109.942318394[/tex]

So, the net force on the 3rd mass,

F = -97.5524457254

Therefore, the net gravitational force on the 3rd object will be -97.5524457254 N.

Learn more about Net gravitational force:

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