Four identical masses of 2.6 kg each are located at the corners of a square with...

Four identical masses of 2.6 kg each are located at the corners of a square with 1.2 m sides. What is the net force on any one of the masses?
? ?? N toward ...............opposite corner

Solutions

Expert Solution

Let us consider the mass is m = 2.6 kg
Now the force on the mass on corner 1 due to mass on corner 2
F₁₂ = Gm*m / r²
where r is the distance between the two mass
F₁₂ = 6.754*10^-11*(2.6*2.6) /(1.2)² = 31.706*10^-11 N
Now force on mass 1 due to mass 4 will also be same because the mass and distance is same , therefore
F₁₄ = 31.706*10^-11 N
Now the force on mass 1 due to mass 3
Here the distance between the two mass r = (1.2² + 1.2²)^1/2 = 1.697 m
F₁₃ = Gm*m /(r)² = 6.754*10^-11*(2.6*2.6) /(1.697)² = 15.854*10^-11 N
Now we will take the component of the F₁₃ in horizontal and vertical direction as shown in the figure
Now the net force on mass toward opposite centre
Considering horizontal direction
F_net = F₁₂ + F₁₃Cos45 = 31.706*10^-11 + ( 15.854*10^-11 )Cos45 = 42.917*10^-11 N
Similarly in the y direction
F_net = F₁₄ + F₁₃Sin45 = 31.706*10^-11 + ( 15.854*10^-11)Sin45 = 42.917*10^-11 N
hence the net force on any mass toward opposite corner will be 42.917*10^-11 N

genius_generous answered 1 year ago

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