( 1) The net gravitational force between the 74.8 kg mass is 8.09 x 10⁻⁵ N.
( 2) The distance from the from the 494 kg mass where the middle mass experiences net zero force is 0.258 m.
What is net gravitational force on 74.8 kg mass?
The net gravitational force between the 74.8 kg mass is calculated by applying Newton's law of universal gravitation as shown below.
F = ( GmM ) / ( R² )
where;
- G is universal gravitation constant
- m is the mass of the middle mass
- M is the mass of the first mass
- R is the distance of separation between the two masses
The force between the first mass and the middle mass is calculated as;
F' = ( 6.672 x 10⁻¹¹ x 141 x 74.8 ) / ( 0.198² )
F' = 1.8 x 10⁻⁵ N
The force between the third mass and the middle mass is calculated as;
F'' = ( 6.672 x 10⁻¹¹ x 494 x 74.8 ) / ( 0.198² )
F'' = 6.29 x 10⁻⁵ N
The net gravitational force on the middle mass;
F (net) = 6.29 x 10⁻⁵ N + 1.8 x 10⁻⁵ N
F ( net ) = 8.09 x 10⁻⁵ N
Let the distance of zero net force = d
F' = ( 6.672 x 10⁻¹¹ x 141 x 74.8 ) / ( d² )
F' = 0.704 x 10⁻⁶ / d²
F'' = ( 6.672 x 10⁻¹¹ x 494 x 74.8 ) / ( 0.396 - d )²
F'' = 2.47 x 10⁻⁶ / ( 0.396 - d )²
for zero net force, the two forces must be equal
0.704 x 10⁻⁶ / d² = 2.47 x 10⁻⁶ / ( 0.396 - d )²
0.704 (0.396 - d )² = 2.47d²
(0.396 - d )² = ( 2.47d² ) / ( 0.704 )
(0.396 - d )² = 3.51d²
0.396 - d = √ ( 3.51d² )
0.396 - d = 1.87d
1.87d + d = 0.396
d (1.87 + 1) = 0.396
d (2.87) = 0.396
d = 0.396 / 2.87
d = 0.138 m
The distance from the 494 kg mass = 0.396 - 0.138 m = 0.258 m
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