Question

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 30.0 -kg acrobat is located at \((3.00 \mathrm{~m}, 4.00 \mathrm{~m})\), and a 40 - \(\mathrm{kg}\) acrobat is located at \((-2.00 \mathrm{~m}\) \(-2.00 \mathrm{~m})\). Assuming that the acrobats stand still in their positions, where must a 20.0 -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?

Short answer

Answer: The 20 kg acrobat must be positioned at coordinate (-3 m, 2 m).

Step-by-step solution

Find the center of mass for the system of three#acrobats

The formula for the center of mass in two-dimensional Cartesian coordinates is given by: $$ (\overline{x},\overline{y}) = (\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}, \frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}) $$ Here, \((x_i, y_i)\) are the coordinates of the \(i\)-th acrobat and \(m_i\) is their mass.

Plug in the given values and set the system's center of mass to (0, 0)

We are given the masses and positions of the first two acrobats, and the mass of the third acrobat, so we can write the center of mass equation as: $$ (0, 0) = (\frac{30(3)+40(-2)+20x}{30+40+20}, \frac{30(4)+40(-2)+20y}{30+40+20}) $$

Solve the system of equations for x and y

We now have the following system of equations: $$ 0 = \frac{30(3)+40(-2)+20x}{30+40+20} $$ $$ 0 = \frac{30(4)+40(-2)+20y}{30+40+20} $$ To solve the system, let's multiply both sides of each equation by the denominator (90), then isolate x and y. First equation: $$ 0 = 30(3)+40(-2)+20x $$ $$ 20x = (-60) $$ $$ x = -3 $$ Second equation: $$ 0 = 30(4)+40(-2)+20y $$ $$ 20y= (120-80) $$ $$ y = 2 $$

Write the final answer

The location where the 20 kg acrobat must be positioned for the center of mass to be at the origin is at coordinate \((-3 \mathrm{~m}, 2 \mathrm{~m})\).

Key concepts

Two-dimensional Cartesian coordinates

Cartesian coordinates are a way to pinpoint any location in a grid, especially in two dimensions. Imagine a graph with two lines perpendicular to each other. These lines cross at a point called the origin, denoted by (0, 0). In this coordinate system, any location is defined by two numbers. The first number tells you how far to move horizontally (left or right), along the x-axis. The second number tells you how far to move vertically (up or down), along the y-axis.
It's like giving someone directions. If you say, 'Go 3 meters to the right and 4 meters upwards,' you just used a Cartesian coordinate: (3, 4).
Using these coordinates makes it easy to describe positions and distances in a two-dimensional space. When we apply this to our acrobats on the platform, each acrobat's position is defined in terms of these two numbers. Knowing their exact spots helps us calculate how their positions affect the overall balance of the platform.

System of equations

A system of equations is just a set of equations that you need to solve together. Often these equations are related, which means they have common variables that need to satisfy all equations at once. It’s like solving a puzzle where pieces (equations) need to fit together perfectly.
In our problem, we have two such equations. These correspond to the x and y coordinates in a two-dimensional plane. Our task is to determine the unknown coordinates (where the new acrobat should stand) by using the information given about the other acrobats.
By plugging the given data into our equations, we began calculating by setting up the expressions that help us reach the right coordinates for balancing. The beauty of using a system of equations is its power to find unknown variables efficiently by considering multiple conditions at once. It's about balancing conditions across different equations to find a coherent solution.

Balancing systems

When we talk about balancing systems, it relates to how objects are positioned in such a way that they don't tip over or move unexpectedly. Imagine a seesaw; it only balances perfectly when the weights on either side are equal.
Translating this idea to our circular platform, we see that the acrobats must stand in such a manner that their combined weight is evenly distributed around the center. Otherwise, the platform would tilt. This concept is crucial when positioning the third acrobat to achieve balance.
By calculating the right spot using the system of equations, we can find a location where the weight of each acrobat contributes equally to maintaining balance. This explains why balancing isn’t just about the weight itself, but also how that weight is positioned in relation to the pivot point, or in this case, the center of the platform.

Mass distribution

Mass distribution refers to how weight is shared among different parts of a system. In our case, it shows how the mass of the three acrobats is spread around the circular platform to keep it balanced.
To achieve perfect balance, the total masses must coordinate with their distances from the center. This ensures the center of mass remains at the origin. The expression for the center of mass in two dimensions involves considering both the x and y coordinates.
The center of mass is essentially the average location of the mass. By adjusting where the third acrobat stands, we're manipulating this distribution so that the collective mass averages out to the origin. This means all physical forces acting on the platform are balanced, keeping everything steady. Through calculated mass distribution, we intuitively ensure stability and equilibrium.