A 5.60-kg particle has the xy coordinates (-1.20 m, 0.500 m), and a 5.20-kg particle has the xy coordinates (0.600 m, -0.750 m). Both lie on a horizontal plane. At what x and y coordinates must you place a 1.60-kg particle such that the center of mass of the three-particle system has the coordinates (-0.500 , -0.700 m)?
(m1+m2+m2)(-.5)= 5.6(-1.2) + 5.2(.6) + 1.6(x) solve for x.
Do the same for y.
To find the x and y coordinates at which to place the 1.60-kg particle, we need to use the concept of center of mass. The center of mass of a system of particles can be calculated using the formula:
Center of mass x-coordinate = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)
Center of mass y-coordinate = (m1y1 + m2y2 + m3y3) / (m1 + m2 + m3)
where m1, m2, and m3 are the masses of the particles, and x1, x2, x3, y1, y2, and y3 are the corresponding xy coordinates.
In this case, we have the following values:
m1 = 5.60 kg, x1 = -1.20 m, y1 = 0.500 m
m2 = 5.20 kg, x2 = 0.600 m, y2 = -0.750 m
m3 = 1.60 kg, x3 = ?, y3 = ?
We are given that the center of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m).
To find the x-coordinate, we can use the equation:
(m1 + m2 + m3)(center of mass x-coordinate) = (m1x1 + m2x2 + m3x3)
Substituting the known values, we get:
(5.60 + 5.20 + 1.60)(-0.500 m) = (5.60 kg)(-1.20 m) + (5.20 kg)(0.600 m) + (1.60 kg)(x3)
Simplifying the equation, we have:
(12.40)(-0.500 m) = (-6.72 kg·m) + (3.12 kg·m) + (1.60 kg)(x3)
(-6.20 kg·m) = (-3.60 kg·m) + (1.60 kg)(x3)
Now we can solve for x3:
-6.20 kg·m + 3.60 kg·m = 1.60 kg·m(x3)
-2.60 kg·m = 1.60 kg·m(x3)
x3 = (-2.60 kg·m) / (1.60 kg)
x3 = -1.625 m
Therefore, the x-coordinate at which to place the 1.60-kg particle is approximately -1.625 m.
Similarly, we can find the y-coordinate by using the equation:
(m1 + m2 + m3)(center of mass y-coordinate) = (m1y1 + m2y2 + m3y3)
Substituting the known values, we get:
(5.60 + 5.20 + 1.60)(-0.700 m) = (5.60 kg)(0.500 m) + (5.20 kg)(-0.750 m) + (1.60 kg)(y3)
Simplifying the equation, we have:
(12.40)(-0.700 m) = (2.80 kg·m) + (-3.90 kg·m) + (1.60 kg)(y3)
(-8.68 kg·m) = (-1.10 kg·m) + (1.60 kg)(y3)
Now we can solve for y3:
-8.68 kg·m + 1.10 kg·m = 1.60 kg·m(y3)
-7.58 kg·m = 1.60 kg·m(y3)
y3 = (-7.58 kg·m) / (1.60 kg)
y3 = -4.738 m
Therefore, the y-coordinate at which to place the 1.60-kg particle is approximately -4.738 m.
So, to make the center of mass of the three-particle system have the coordinates (-0.500 m, -0.700 m), you would place the 1.60-kg particle at approximately (-1.625 m, -4.738 m).