Question

The masses and coordinates of three spheres are as follows: 26 kg, x = 1.50 m,...

The masses and coordinates of three spheres are as follows: 26 kg, x = 1.50 m, y = 1.00 m; 42 kg, x = -2.25 m, y = -1.25 m; 50 kg, x = 0.00 m, y= -0.75 m. What is the magnitude of the gravitational force on a 16 kg sphere located at the origin due to the other spheres?

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Answer #1

Due to 26kg sphere

r = √[(1.5)2 + (1)2] = 1.803

Let 'θ' be the angle that the line joining the masses makes with positive x-axis

cos(θ) = 1.5 / r = 0.832

sin(θ) = 1 / r = 0.555

F1 = G(26)*(16) / r2 = 8.54*10-9 N

F1 = [F1*cos(θ)] i + [F1*sin(θ)] j

=> F1 = (7.1*10-9) i + (4.74*10-9) j

Due to 42kg sphere

r = √[(2.25)2 + (1.25)2] = 2.574

Let 'θ' be the angle that the line joining the masses makes with negative x-axis

cos(θ) = 2.25 / r = 0.874

sin(θ) = 1.25 / r = 0.486

F2 = G(42)*(16) / r2 = 6.76*10-9 N

F2 = -[F2*cos(θ)] i + -[F2*sin(θ)] j

=> F2 = -(5.91*10-9) i - (3.28*10-9) j

Due to 50kg sphere

r = 0.75

F3 = G(50)*(16) / r2 = 94.86*10-9 N

F3 = -[F3] j

=> F3 = -(94.86*10-9) j

F = F1 + F2 + F3

=> F = (1.19*10-9) i - (93.4*10-9) j

Magnutude = √[(1.19*10-9)2 + (-93.4*10-9)2] = 93.41*10-9 N

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