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Four asteroids, of mass M = 3 × 106 kg, are positioned at the corners of...

Four asteroids, of mass M = 3 × 106 kg, are positioned at the corners of an imaginary square of side length s = 2.4 × 108 m. Two other asteroids of mass m = 2 × 106 kg are positioned in the middle of the left and right sides of the square. If a spaceship is positioned at the middle of the bottom side of the square what is the net force on the ship in Newtons? weight of the spaceship is 2000 kg

Solutions

Expert Solution

See the diagram :

M = 3 x 106 kg

m = 2 x 106 kg

mo = 2 x 103 kg

s = 2.4 x 108 m

I have numbered the charges which are exerting forces on mo, and the forces are directed accordingly.

F1 = Force exerted by mass at 1 =

F2 = Force exerted by mass at 2 =

F3 = Force exerted by mass at 3 = F2

F4 = Force exerted by mass at 4 = F1

F5 = Force exerted by mass at 5 =

F6 = Force exerted by mass at 6 = F5

From the diagram, it can be seen that F1 and F4 are in opposite direction having same magnitude, so they cancels each other.

The resultant of F2 and F3 is = 2F2cos along vertical symmetry of the square.

The resultant of F5 and F6 is = 2F5cos along vertical symmetry of the square.

As these resultants are in same direction, So, net Force = 2F2cos + 2F5cos

=> Fnet = 2.3 x 10-17 N. [answer]

genius_generous answered 1 year ago

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