Question

While performing the rotational equilibrium experiment, a group of students became adventurous and decided to determine where the center of gravity would be for the following masses and coordinates.

100 gm af (0,3), 75gm at (-4,7), 225gm at (-6,3) and 250gm af (2,6). Where would the expected center of gravity for this system of masses lie?

Where would the 650gram mass need to be placed to bring the table into equilibrium?

Answer 1

sr. no	mass(kg)	cordinates
1	0.1	0,3
2	0.075	-4,7
3	0.225	-6,3
4	0.250	2,6

Let (x,y) be the center of gravity for this mass system,

then, x = (m1 * x1 + m2 *x2 + m3*x3 +m4*x4 ) / (m1 + m2 +m3 + m4)

= ( 0.1 * 0 + 0.075*(-4) + 0.225 *(-6)+0.25 * 2 ) / (0.1 + 0.075 + 0.225 + 0.25) = -1.769

y = (m1 * y1 + m2 *y2 + m3*y3 +m4*y4 ) / (m1 + m2 +m3 + m4)

= ( 0.1 * 3+ 0.075*7 + 0.225 *3+0.25 * 6 ) / (0.1 + 0.075 + 0.225 + 0.25) = 4.615

so, CG(x,y) = (-1.769 , 4.615)

let the position of mass m5 = (x5,y5) about origin given, m5 = 650 g = 0.65 kg

Apply the equalibrium of torque of all masses about the origin for x-axis,

(m1 * x1 + m2 *x2 + m3*x3 +m4*x4 + m5* x5) = 0

(0.1 * 0 + 0.075*(-4) + 0.225 *(-6)+0.25 * 2 + 0.65* x5 = 0

-1.15 + 0.65 * x5 = 0

so, x5 = 1.769

similarly

Apply the equalibrium of torque of all masses about the origin for y-axis,

(m1 * y1 + m2 *y2 + m3*y3 +m4*y4 + m5* y5) = 0

(0.1 * 3 + 0.075*7 + 0.225 *3+0.25 * 6 + 0.65* y5 = 0

3 + 0.65 * y5 = 0

so, y5 = -4.615

so, (x5,y5) = (1.769, -4.615)