The gravitational force on one sphere due to the three of square is 1042.18 ×10⁻¹¹N
We are given that ,
Mass = m = 7.5 kg
Side = d = 0.6 m
Due to the square's three corner there will be up , right and diagonal three forces acting on the system,
Fup = G(m²/d²)
Fright = G(m²/d²)
Fdia = G (m²/(√2d)²)cos45°+ G (m²/(√2d)²)sin45°
To calculate the force on the left sphere in the lower left corner. From the symmetry of the problem the net forces in the x and y direction will be the same and θ = 45°
Therefore, the force will be,
Fx =Fright +Fdia
Fx = G(m²/d²) +G (m²/(√2d)²)(1/√2)
Fx = G(m²/d²)[(1+(1/2√2)]
Thus, Fy = Fx =G(m²/d²)[(1+(1/2√2)]
By the Pythagorean combination of the two component forces , due to symmetry the the net force will be along the diagonal of the square,
F = √Fx +Fy= Fx√2
F = G(m²/d²)[(1+(1/2√2)]√2
F = (6.67 ×10⁻¹¹ Nm²/kg²)(7.5kg)²/(0.6m)²
F = 1042.18 ×10⁻¹¹N
The force at points are the center of the square.
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