Question

In: Physics

In the figure, a square of edge length 22.0 cm is formed by four spheres of...

In the figure, a square of edge length 22.0 cm is formed by four spheres of masses m1 = 4.70 g, m2 = 2.50 g, m3 = 1.40 g, and m4 = 4.70 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.40 g?

Solutions

Expert Solution

The gravitational force between 2 objects is given by:
F = G*m1*m2 / r²
where
F = force
G = gravitational constant = 6.67E-11 Nm²/kg²
m₁ = mass of one object
m₂ = mass of other object
r = distance between the centers of the masses

By symmetry of the given situation, you can see that the force by m₁ and the force by m₄ cancel out, so you only need to consider the forces by m₂ and m₃.

The distance from m₅ to m₂ or m₃ is:
r2 = [ (0.11 m)² + (0.11 m)² ] = 0.0242 m2

The force by m₂ on m₅ is:
F₂₅ = (6.67*10-11 Nm²/kg²) (2.5*10-3 kg) (2.4*10-3 kg) / 0.0242 m2
F₂₅ = 1.65*10-14 N
You can see that the angle is 45° NE

The force by m₃ on m₅ is:
F₃₅ = (6.67*10-11 Nm²/kg²) (1.4*10-3 kg) (2.4*10-3 kg) / 0.0242 m2
F₃₅ = 9.26*10-15 N
You can see that the angle is 45° SW

The resultant of forces F₂₅ and F₃₅ is:
F = F₂₅ - F₃₅
F = (1.65*10-14 N) - (9.26*10-15 N)
F = 7.24*10-15 N
at an angle is 45° NE

The horizontal component of F is:
Fh = (7.24*10-15 N) cos(45) = 5.12*10-15 N

The vertical component of F is:
Fh = (7.24*10-15 N) sin(45) = 5.12*10-15 N

The force F in vector notation is:
F = (5.12 i + 5.12 j) ×10-15 N


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