Problem 83 - College Physics

Question

Three particles, each with a mass of \(0.25 \mathrm{~kg},\) are located at \((-4.0 \mathrm{~m}, 0),(2.0 \mathrm{~m}, 0),\) and \((0,3.0 \mathrm{~m})\) and are acted on by forces \(\overrightarrow{\mathbf{F}}_{1}=(-3.0 \mathrm{~N}) \hat{\mathbf{y}}, \overrightarrow{\mathbf{F}}_{2}=(5.0 \mathrm{~N}) \hat{\mathbf{y}},\) and \(\overrightarrow{\mathbf{F}}_{3}=(4.0 \mathrm{~N}) \hat{\mathbf{x}},\) respectively. Find the acceleration \((\mathrm{mag}-\) nitude and direction) of the center of mass of the system. [Hint: Consider the components of the acceleration.]

Step-by-Step Solution

Answer

The acceleration of the center of mass is approximately 5.97 m/s² at 26.57°.

Step 1: Calculate the Total Mass

Each of the three particles has a mass of \(0.25 \text{ kg}\). Thus, the total mass \(M\) of the system can be calculated by summing the masses of the individual particles:\[M = 0.25 + 0.25 + 0.25 = 0.75 \text{ kg}\]

Step 2: Calculate the Center of Mass Coordinates

The coordinates \((x_{cm}, y_{cm})\) of the center of mass are calculated using:\[x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}\]\[y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}\]For the given points:\[x_{cm} = \frac{0.25 \times (-4) + 0.25 \times (2) + 0.25 \times (0)}{0.75} = -0.67 \text{ m}\]\[y_{cm} = \frac{0.25 \times (0) + 0.25 \times (0) + 0.25 \times (3)}{0.75} = 1 \text{ m}\]Thus, the center of mass is located at \((-0.67 \text{ m}, 1 \text{ m})\).

Step 3: Calculate the Net Force Components on the System

The net force in each direction can be found by summing the force components:\[F_{x, net} = F_{3x} = 4.0 \text{ N}\]\[F_{y, net} = F_{1y} + F_{2y} = (-3.0) + 5.0 = 2.0 \text{ N}\]

Step 4: Calculate the Acceleration of the Center of Mass

The acceleration \(a_{cm}\) of the center of mass can be calculated by dividing the net force components by the total mass \(M\):\[a_{cm,x} = \frac{F_{x, net}}{M} = \frac{4.0}{0.75} \approx 5.33 \text{ m/s}^2\]\[a_{cm,y} = \frac{F_{y, net}}{M} = \frac{2.0}{0.75} \approx 2.67 \text{ m/s}^2\]

Step 5: Calculate the Magnitude and Direction of the Acceleration

The magnitude of the acceleration \(a_{cm}\) is calculated using the Pythagorean theorem:\[a_{cm} = \sqrt{a_{cm,x}^2 + a_{cm,y}^2} = \sqrt{(5.33)^2 + (2.67)^2} \approx 5.97 \text{ m/s}^2 \]The direction \(\theta\) of the acceleration with respect to the positive x-axis is:\[\theta = \tan^{-1}\left(\frac{a_{cm,y}}{a_{cm,x}}\right) = \tan^{-1}\left(\frac{2.67}{5.33}\right) \approx 26.57^\circ\]

Key Concepts

Understanding Force Components

In physics, force components are crucial in analyzing specific forces acting in different directions. When dealing with forces, it is often convenient to break them down into components, usually along the x-axis and y-axis of a coordinate system. This helps to simplify calculations and allows us to focus on the effect of forces in specific directions.

In the original exercise, the particles experience forces that are either in the x-direction or the y-direction. For example, \( \overrightarrow{\mathbf{F}}_{1} = (-3.0 \, \text{N}) \hat{\mathbf{y}} \) and \( \overrightarrow{\mathbf{F}}_{2} = (5.0 \, \text{N}) \hat{\mathbf{y}} \) indicate forces acting solely in the y-direction. Meanwhile, \( \overrightarrow{\mathbf{F}}_{3} = (4.0 \, \text{N}) \hat{\mathbf{x}} \) refers to a force acting in the x-direction.

By analyzing these components separately, it's easier to calculate their collective impact on an object's motion. Determining the sum of these components allows for the computation of a net force, which then helps to establish how much the object's center of mass accelerates in these respective directions.

Simplifying Acceleration Calculation

Acceleration calculation often involves determining how quickly an object's velocity changes due to the forces acting upon it. For a system of particles, such as in this exercise, we first determine the net force components and then relate these to the total mass of the system.

The calculated net forces influence how much the system accelerates in each direction. According to Newton's second law, acceleration is the result of the net force divided by the particle's mass. Thus, the formula becomes:

• \( a_{cm,x} = \frac{F_{x, net}}{M} \)
• \( a_{cm,y} = \frac{F_{y, net}}{M} \)

This step-by-step division makes it clear how each directional component of the net force contributes to the overall acceleration of the entire system's center of mass. The center of mass acceleration in the x and y directions are then calculated, resulting in values that can be further analyzed to derive the magnitude and direction of total system acceleration.

Adopting a Coordinate System

Using a coordinate system is fundamental when analyzing physical problems involving positions and motions. It helps in organizing the data so that calculations become more precise and logically structured.

In two-dimensional space, positions like those given - \((-4.0 \, \text{m}, 0)\), \((2.0 \, \text{m}, 0)\), and \((0, 3.0 \, \text{m})\) - provide clear spatial references to where forces are applied to the particles.

Choosing an effective coordinate system, typically one in which forces align nicely with the axes, minimizes potential errors and maximizes the efficiency in solving the problem. Once all positions and forces are correlated within this system, it becomes simple to determine the impact on the system and its center of mass. This helps compute the center's movement based on the external forces exerted upon it.

Calculating Net Force

The concept of net force is central to understanding how objects will move as a result of multiple forces acting on them. Essentially, the net force is the vector sum of all individual forces acting on an object, directing how the object accelerates.

In our exercise, we determine this by addressing each directional component separately. For the system along the y-axis, the net force was calculated as \( (F_{1y} + F_{2y} = -3.0 + 5.0 = 2.0 \, \text{N}) \), while for the x-axis, the net force is straightforward with \( F_{x, net} = 4.0 \, \text{N} \). These individual calculations help derive the resulting motion.

Once these forces are tallied, they become inputs for computing the acceleration. Notably, this net force method highlights how the resultant force affects the overall motion of the system, dictating the subsequent direction and magnitude of the center of mass's acceleration. Understanding and computing the net force is a critical step in any mechanics problem analysis.