Four masses of 2 kg are arranged in a square with sides of 50 cm. Find the magni

Question

Four masses of 2 kg are arranged in a square with sides of 50 cm. Find the magnitude and direction of the total forces on the mass in the upper left corner (you are to consider only the gravitational forces between the masses.)

What is the net force (magnitude and direction) on a 1 kg mass placed in the center of the square of 2 kg masses?

Now two of the masses on the right side of the square are replaced with 10 kg masses. The left side of the square still has 2 kg masses. What is the direction of the net force of a 1 kg mass placed at the center of this arrangement?

Explanation / Answer

force due to mass at the bottom of it
F = GX 2 X 2 / (0.5)2 = 16G direction is in -ve y-axis s0 - 16G j

force due to right mass ,

F = GX 2 X 2 / (0.5)2 = 16G direction is in +ve x-axis s0 16G i

due to other corner mass,

F = G X 2 X 2 / 2 X 0.52 = 8G at angle of 45 degree below to +ve x-axis

so   = 8Gcos45 i - 8Gsin45 j = 5.66G i - 5.66G j

total force = -16G j + 16G i + 5.66Gi - 5.66Gj = 21.66G i - 21.66G j

|Ft | = (2 X 21.66G2 )   = 30.63G N = 30.63 X 6.67 X 10-11 N = 204.315 X 10-11 N

at an angle of -45 degrees with +ve x -axis .

if we keep a mass at centre so all forces are equal and opposite , so net force will be zero .

( due to symmetry)

now right side has more mass and upper masses ( 2 + 10) and bottom masses (2+10) ae equal and syymetric , so upper and lower forces will be canceled but right side has more force so net force will be parrallel to +ve x-axis and in direction of +ve x-axis (i) .

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