Question

A two-dimensional object with uniform mass density is in the shape of a thin square and has a mass of \(2.00 \mathrm{~kg}\). The sides of the square are each \(20.0 \mathrm{~cm}\) long. The coordinate system has its origin at the center of the square. A point mass, \(m,\) of \(2.00 \cdot 10^{2} \mathrm{~g}\) is placed at one corner of the square object, and the assembly is held in equilibrium by positioning the support at position \((x, y)\), as shown in the figure. Find the location of the support.

Short answer

#Answer# Based on the calculations, the support point coordinates (x, y) that keep the 2D object and point mass in equilibrium are approximately (-0.00909 m, 0.00909 m) or (-0.909 cm, 0.909 cm).

Step-by-step solution

Convert units of the given quantities

We need to make sure that all our quantities are in the same units, so we will convert the units to SI units (meters and kilograms). We are given: - Side length of the square = 20.0 cm = 0.2 m - Mass of the square = 2.00 kg - Mass of the point mass, m = 2.00 x 10^2 g = 0.2 kg

Find the center of mass of the square object and the point mass

Let's find the individual centers of mass: 1. Square object: The center of mass of the square object is at its geometric center. So, its coordinates can be written as \((0, 0)\). 2. Point mass: The center of mass of the point mass is directly where the mass is placed. The coordinate of the corner of the square is at \((-0.1, 0.1)\).

Calculate the center of mass of the system

We will now calculate the center of mass of the entire system, (X_CM, Y_CM). We can use the following formula: $$X_{CM} = \frac{m_1x_1 + m_2x_2}{m_1+m_2}$$ $$Y_{CM} = \frac{m_1y_1 + m_2y_2}{m_1+m_2}$$ By plugging in the given values: $$X_{CM} =\frac{(2.00)(0) + (0.2)(-0.1)}{2.00+0.2} = \frac{-0.02}{2.2} $$ $$Y_{CM} = \frac{(2.00)(0) + (0.2)(0.1)}{2.00+0.2} = \frac{0.02}{2.2} $$

Balance the moments around the support point

To find the support point, we must balance the moments created by the total gravitational force acting on the system, which acts through the center of mass. The moment of a force about a point is given by: $$ M = F \times d $$ Where M is the moment, F is the force, and d is the perpendicular distance (lever arm) from the point to the line of action of the force. Here d is the distance from the center of mass to the support point (x, y). Since the gravitational force acts vertically, the x and y components of the moment will be: $$ M_x = W_{total} \times Y_{CM} $$ $$ M_y = W_{total} \times X_{CM} $$ Where \(W_{total}\) is the total weight of the system (gravitational force), which can be calculated as: $$ W_{total} = (m_1 + m_2)g $$ We find \(M_x=(2.00+0.2)(9.81)(0.02/2.2)\) and \(M_y=-(2.00+0.2)(9.81)(0.02/2.2)\). Equating them to the gravitational force at the arbitrary point \((x, y)\), we get: $$ (2.00+0.2)(9.81)(0.02/2.2) = (2.00+0.2)g \times y $$ $$ -(2.00+0.2)(9.81)(0.02/2.2) = (2.00+0.2)g \times x $$ Solving for x and y, we get the support point coordinates (x, y).

Key concepts

Mass Density

Mass density, often simply called density, is a measure of the mass of an object divided by its volume. In the case of uniform mass density, the mass is evenly distributed throughout the object. This concept is crucial when analyzing objects like the given two-dimensional square, which has a uniform mass distribution. Uniform mass density simplifies the process of finding the center of mass, as it will always be at the geometric center of the shape, provided the shape itself has a uniform structure, such as a square or a circle.

Understanding mass density also allows us to compare the 'heaviness' of materials. For example, if two objects have the same volume but different mass densities, the one with higher mass density will weigh more. This is pertinent when considering how different parts of an assembly contribute to its overall center of mass and equilibrium.

Equilibrium

Equilibrium refers to the state in which all forces and moments acting on an object are balanced, resulting in no acceleration. There are two types of equilibrium: static and dynamic. Static equilibrium implies the object is at rest with no net forces or torques, while dynamic equilibrium means the object moves with constant velocity without rotation.

In the context of the exercise, the square and the point mass system must be in static equilibrium for the setup to be stable. To achieve this, the moments, or rotational forces, about the support point must counteract the gravitational pull on the system's combined center of mass. This equilibrium is the underlying principle used to calculate the precise location of the support that would keep the object and point mass stable.

Gravitational Force

Gravitational force is the attractive force that acts between any two masses. This force is what gives weight to physical objects and causes them to fall towards the ground when dropped. Every particle of mass exerts a gravitational pull on every other mass. The more massive an object is, the stronger its gravitational pull.

In practical problems like the one being solved here, we consider the gravitational force acting on an object to be its weight, which is calculated as the product of its mass and the acceleration due to gravity (\( g \), approximately \(9.81 m/s^2\) on Earth's surface). The gravitational force on the system acts through its center of mass, and this force's effect on an object's rotation or balance is directly linked to the concept of torque or moments.

Moments and Torque

Moments and torque are concepts that describe the rotational effect of a force. A moment (also known as torque) is the measure of the tendency of a force to rotate an object about an axis or pivot. Mathematically, it is defined as the cross product of the position vector (the distance from the pivot point to the point where the force is applied) and the force vector. The formula for torque (\(\tau\)) is given by \(\tau = r \times F\), where \(r\) is the position vector and \(F\) is the force applied.

In our exercise, the torque is created by the gravitational forces acting on the mass of the square and the point mass. For the system to be in equilibrium, the sum of all torques must be zero. By calculating the torques about the support point using the distances from the center of mass to this point, we can determine where to place the support to maintain equilibrium. This involves equating the clockwise and counterclockwise torques to ensure they balance out.