Question

c9/1

A 1.50 kg particle has the xy coordinates (-1.62 m, 0.165 m), and a 3.22 kg particle has the xy coordinates (0.453 m, -0.336 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.23 kg particle such that the center of mass of the three-particle system has the coordinates (-0.454 m, -0.347 m)?

Answer 1

Let´s start with the following definition. The center of mass is the point of application of the resultant of all forces of gravity acting on different material portions of a body, so that the torque from any point of this resulting applied on the center of mass is the same as that produced by the weights of all the material masses which constitute the body.

In other words, the center of mass of a body is the point regarding which forces that gravity exerts on the different material points that constitute the body produce a null result torque.

The center of mass of a system of particles is a weighted average, according to the individual mass of the positions of all the particles which compose it, i.e, the center of mass position vector is defined as:

rc = m1 r1 + m2 r2 + ...... / m1 + m2 + ...... = mi ri / mi

Of course as a vector it will have components. In this specific case, will have two components. One on th x axis an other on the y axis.

Therefore, the previos formula can be broken down in coordinates x and y (specifically for this probem) as follows:

Xc = m1 x1 + m2 x2 + m x3 / m1 + m2 + m3

Yc = m1 y1 + m2 y2 + m3 y3 / m1 + m2 + m3 where we know variables Xc, x1, x2, m1, m2 , m3, Yc, y1, y2, and we do not know x3 and y3.

If we substitute values we will notice that we will have left two equations of the first grade which will have to be solved:

m1 = 1.50 Kg, x1 = -1.62m, y1 = 0.165m

m2= 3.22 Kg, x2= 0.453m , y2 = -0.336 m

m3 = 3.23 Kg, Xc = -0.454 m, Yc = -0.347 m   Substituting:

Xc = m1 x1 + m2 x2 + m x3 / m1 + m2 + m3

-0.454 = (1.50)(-1.62) + (3.22)(0.453) + (3.23) (x3) / (1.50 + 3.22 + 3.23)

-0.454 = -0.97134 + 3.23x3 / 7.95

(-0.454)(7.95) = -0.97134 + 3.23 x3

-3.6093+0.97134 = 3.23 x3

-2.63796 = 3.23 x3

x3 = -2.63796 / 3.23 = -0.816705882 meters Now with the yc coordinate:

Yc = m1 y1 + m2 y2 + m3 y3 / m1 + m2 + m3

-0.347 = (1.50)(0.165) + (3.22)(-0.336) + (3.23) (y3) / (1.50 + 3.22 + 3.23)

-0.347 = -0.93102 + 3.23y3 / 7.95

(-0.347)(7.95) = -0.93102 + 3.23 x3

-2.75865+0.93102 = 3.23 x3

-1.82763 = 3.23 x3

y3 = -1.82763 / 3.23 = -0.565829721 meters

So the requested coordinates are:

Xc = -0.816705882

Yc = -0.565829721