Four small aluminum spheres form the corners of a square whose side is 1.5 cm lo

Question

Four small aluminum spheres form the corners of a square whose side is 1.5 cm long. A fifth iron sphere is in the center of the square.

m1 = 470 g

m2 = 90 g

m3 = 250 g

m4 = 470 g

m5 = 550 g

Force of Gravity 4 m5 Four small aluminum spheres form the corners of a square whose side is 1.5 cm long. A fifth iron sphere is in the center of the square. m, 470 m2 90 g m 250 3 m4 470 3 ms 550 g N Submit 2) What is total gravitational force on the fifth sphere in the x-direction? Submit 3) What is magnitude of the total gravitational force on the fifth sphere? Submit

Explanation / Answer

Forces due to m1, m4 cancel out each other.

It means we have effective 250-90=160 g at m3 location.

F = Gm5*m/r^2

= 6.67e-11 *0.550*0.160/(0.015^2 /2)

= 5.22*10^-8 N

Gravitational force in y direction = F cos 45 degree

= 5.22e-8*cos 45 degree

=3.69*10^-8 N

Total gravitational force in x direction = - 3.69*10^-8 N

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