Question

Three 7.50-kg masses are at the corners of an equilateral triangle and located in space far from any other masses.If the sides of the triangle are 1.41 m long, find the magnitude of the net force exerted on each of the three masses?

Answer 1

Answer: $3.2 × 10 ^{-9}$ 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) = $6.674 × 10^{-11} \frac{m^{3} }{kgs^{2} }$

to get the magnitude of the net force extended on each of the masses we can apply the formula below

F = $\frac{G × M1 × M2 }{r^{2} }$

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 = $\frac{ [tex]6.674 × 10^{-11}$× 7.5 × 7.5 }{1.41^{2} }[/tex]

F = $1.89 × 10 ^{-9}$ 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 = $1.89 × 10 ^{-9}$ × 2cos 30 =  $3.2 × 10 ^{-9}$ N