Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the two-dimensional Cartesian coordinate system is assumed to be at the center of the platform. A 30-kg acrobat is located at (4 m, 9 m), and a 36-kg acrobat is located at (−2 m, −2 m). Assuming that the acrobats stand still in their positions, where must a 20-kg acrobat be located so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced?
x(c.m.) =0,
y(c.m.) =0,
x(c.m.) = m1•x1+m2•x2+m3•x3/(m1+m2+m3) =0,
y(c.m.) = m1•y1+m2•y2+m3•y3/(m1+m2+m3) =0,
30•4-36•2+20•x3 =0,
30•9+36•2+20•y3 = 0 ,
x3 = - 48/20 = - 2.4 m,
y3 = -198/20 = - 9.9 m.
Well, it seems like we have a balancing act going on! Let's see where we can place our 20-kg acrobat to make it all work.
To find the center of mass of the system, we need to consider both the masses and their respective positions. The center of mass coordinates can be calculated using the following equation:
x_cm = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)
y_cm = (m1y1 + m2y2 + m3y3) / (m1 + m2 + m3)
Here, m1, m2, and m3 refer to the masses of the acrobats, while x1, y1, x2, y2, x3, and y3 are their respective coordinates.
Let's plug in the values we have. The 30-kg acrobat is located at (4 m, 9 m), the 36-kg acrobat is located at (-2 m, -2 m), and we want to find the position of the 20-kg acrobat.
Let's call the coordinates of the 20-kg acrobat as (x, y).
For the x-coordinate of the center of mass to be zero:
(30 kg * 4 m + 36 kg * -2 m + 20 kg * x) / (30 kg + 36 kg + 20 kg) = 0
Simplifying this equation will give us the value of x.
Similarly, for the y-coordinate of the center of mass to be zero:
(30 kg * 9 m + 36 kg * -2 m + 20 kg * y) / (30 kg + 36 kg + 20 kg) = 0
Simplifying this equation will give us the value of y.
Now, doing some calculations to solve these equations might not be the funniest thing in the world, but I hope it helps you find the solution and get your acrobats perfectly balanced! Good luck!
To find the location of the 20-kg acrobat, we need to ensure that the center of mass of the system is at the origin. We can do this by using the concept of moment of inertia.
The center of mass of the system is given by the formula:
Cx = (m1 * x1 + m2 * x2 + m3 * x3) / (m1 + m2 + m3)
Cy = (m1 * y1 + m2 * y2 + m3 * y3) / (m1 + m2 + m3)
Where
Cx and Cy represent the x and y coordinates of the center of mass,
m1, m2, and m3 represent the masses of the acrobats,
and (x1, y1), (x2, y2), and (x3, y3) represent their respective positions.
In this case, we have:
m1 = 30 kg, (x1, y1) = (4 m, 9 m)
m2 = 36 kg, (x2, y2) = (-2 m, -2 m)
m3 = 20 kg, (x3, y3) = (x, y)
We want the center of mass to be at the origin, so Cx = 0 and Cy = 0.
Substituting the given values and solving for x and y:
0 = (30 kg * 4 m + 36 kg * (-2 m) + 20 kg * x) / (30 kg + 36 kg + 20 kg)
0 = (30 kg * 9 m + 36 kg * (-2 m) + 20 kg * y) / (30 kg + 36 kg + 20 kg)
Simplifying the equations:
120 + (-72) + 20x = 0
270 - 72 + 20y = 0
48 + 20x = 0
198 + 20y = 0
Solving for x and y:
20x = -48
x = -2.4 m
20y = -198
y = -9.9 m
Therefore, the 20-kg acrobat must be located at (-2.4 m, -9.9 m) so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced.
To find the location of the 20-kg acrobat, we need to ensure that the center of mass of the system consisting of all three acrobats is at the origin (0, 0).
The center of mass coordinates for a system of objects can be found using the following formula:
x_cm = (m1*x1 + m2*x2 + m3*x3) / (m1 + m2 + m3)
y_cm = (m1*y1 + m2*y2 + m3*y3) / (m1 + m2 + m3)
Where:
- m1, m2, m3 are the masses of the individual acrobats.
- x1, x2, x3 are the x-coordinates of the individual acrobats.
- y1, y2, y3 are the y-coordinates of the individual acrobats.
- x_cm, y_cm represents the x and y coordinates of the center of mass.
We know the following information:
Acrobat 1: m1 = 30 kg, x1 = 4 m, y1 = 9 m
Acrobat 2: m2 = 36 kg, x2 = -2 m, y2 = -2 m
Acrobat 3 (unknown): m3 = 20 kg, x3 = ?, y3 = ?
We can substitute these values into the center of mass formula and solve for x3 and y3:
x_cm = (30*4 + 36*(-2) + 20*x3) / (30 + 36 + 20)
y_cm = (30*9 + 36*(-2) + 20*y3) / (30 + 36 + 20)
Since the center of mass must be at the origin (0, 0), we set x_cm and y_cm to 0 and solve for x3 and y3:
0 = (30*4 + 36*(-2) + 20*x3) / (30 + 36 + 20)
0 = (30*9 + 36*(-2) + 20*y3) / (30 + 36 + 20)
Simplifying these equations, we get:
120 + (-72) + 20*x3 = 0
270 - 72 + 20*y3 = 0
48 + 20*x3 = 0
198 + 20*y3 = 0
20*x3 = -48
20*y3 = -198
x3 = -48 / 20 = -2.4 m
y3 = -198 / 20 = -9.9 m
Therefore, the 20-kg acrobat must be located at approximately (-2.4 m, -9.9 m) in order to balance the system and have the center of mass at the origin.