Answer :
Answer: [tex]3.2 × 10 ^{-9}[/tex] N
Explanation:
from the question we are given the following:
mass of each object (m) = 7.5 kg
distance of objects from each other (r) = 1.41 m
universal gravitation constant (G) = [tex]6.674 × 10^{-11} \frac{m^{3} }{kgs^{2} }[/tex]
to get the magnitude of the net force extended on each of the masses we can apply the formula below
F = [tex]\frac{G × M1 × M2 }{r^{2} }[/tex]
where M1 and M2 are the masses
take note that since all masses are the same and of the same distance from each other they will all experience the same force
F = [tex]\frac{ [tex]6.674 × 10^{-11}[/tex]× 7.5 × 7.5 }{1.41^{2} }[/tex]
F = [tex]1.89 × 10 ^{-9}[/tex] N
take note that each of the masses is attracted by the other two masses and their force component would be in the direction of their resultant force we would multiply the force above by 2cos 30
F = [tex]1.89 × 10 ^{-9}[/tex] × 2cos 30 = [tex]3.2 × 10 ^{-9}[/tex] N
The magnitude of the net force exerted on each of the three masses is 5.66 x 10⁻⁹ N.
Force between the masses
The force between the masses is determined by applying Newton's law of universal gravitation as shown below;
F = Gm₁m₂/R²
where;
- G is universal gravitation constant
- R is the distance between the masses
Force between mass 1 and mass 2
[tex]F_{12} = \frac{6.67 \times 10^{-11} \times 7.5^2}{(1.41)^2} \\\\F_{12} = 1.887 \times 10^{-9} \ N[/tex]
Force between mass 1 and mass 3
F(13) = F(12) = 1.887 x 10⁻⁹ N
Force between mass 2 and mass 3
F(23) = F(13) = 1.887 x 10⁻⁹ N
Net force between the masses
The net force between the masses is cacuated as follows;
F(net) = 3 x 1.887 x 10⁻⁹ N = 5.66 x 10⁻⁹ N.
Learn more about force between masses here: https://brainly.com/question/11359658