Four identical masses of 800 kg each are placed at the corners of a square whose side length is 10.0 cm. What is the net gravitational force (magnitude and direction) on one of the masses, due to the other three?

Solution 40P Gm m The gravitational force between two different objects is given by, F = 12 2, where m an1 r m a2e the masses of the two objects, G is gravitational constant and ris the separation between them. Let us have a look at the following figure to understand the question better. Let the masses be A, B, C and D and they are placed as shown in the figure above. Let us calculate the force on the mass A due to each at B, C and D. Length of each side = 10.0 cm = 0.1 m 11 2 2 Gravitational constant G = 6.67 × 10 N.m /kg Force on the mass at A due to the one at B, G×800 kg×800 kg F AB = (0.1) m 3 F AB = 4.26 × 10 N Similarly, the force on the mass at A due to the mass at D, F = 4.26 × 103N AD Now, the resultant of the two forces will be along the diagonal AC of the square, with which each side of the square makes an angle of 45 . Therefore, the resultant force along AC due to eachABand F AD is 0 3 0 3 F res= cos 45 × 4.26 × 10 N + cos 45 × 4.26 × 10 N F = 6.02 × 10 3 N …..(1) res Now, the force on mass A due to the mass at C is, F = G×800 kg×800 kg AC (0.1 +0.1 ) m 3 F AC = 2.13 × 10 N …..(2) If we now add equations (1) and (2), we shall get the net force on the mass at A. 3 3 3 So, the net force Fnet= F res+ F AC = 6.02 × 10 N + 2.13 × 10 N = 8.15 × 10 N F 8.2 × 10 3N net 3 Therefore, the net gravitational force is 8.2 × N . The direction of this force will be along the diagonal of the square.