Question

Four masses are at the corners of a square whose side is 2m long. (500kg top left, 300kg top right, 100kg bottom left, 500kg bottom right) Find the magnitude and direction of the net gravitational force on Ma(100kg) which is in the center of the square. (G=6.67*10^-11)

Answer 1

distamce of centre from the corner d = sqrt(1^2 + 1^2) = 1.414 m

considereing centre as origin and +ve x axis along right side.

Force due to 500 kg mass,

F1 = Gm1m2 / d^2 = (6.67 x 10^-11 x 500 x 100) / (1.414^2) = 1.67 x 10^-6 N at -45 degrees

Force due to 300 kg

F2 = (6.67 x 10^-11 x 300 x 100 ) / ( 1.414^2) = 1 x 10^-6 N at an angle of 180+45 deg

Force due to 100 kg

F3 = (6.67 x 10^-11 x 100 x 100 ) / ( 1.414^2) = 3.34 x 10^-7 N at an angle of 45 deg

Force due to 500 kg

F4 = (6.67 x 10^-11 x 500 x 100 ) / ( 1.414^2) = 1.67 x 10^-6 N at an angle of 180- 45 deg

Fnet = F1 + F2 + F3 + F4

= (1.67cos(-45) i + 1.67 sin(-45) j + 1cos225 i + 1 sin225 j + 0.334cos45 i + 0.334sin45j + 1.67cos135 i + 1.67 sin135j ) x 10^-6 N

= - 0.471 i - 0.471 j ) x 10^-6

magnitude = sqrt(0.471^2 + 0.471^2)^ = 0.666 x 10^-6 N

direction = 180 + 45 = 225 degrees from considered x -axis in counterclockwise direction