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Chapter 8: Problem 13

Consider the following mass distribution, where \(x\)-and \(y\)-coordinates are given in meters: \(5.0 \mathrm{~kg}\) at \((0.0,0.0) \mathrm{m}\), \(3.0 \mathrm{~kg}\) at \((0.0,4.0) \mathrm{m}\), and \(4.0 \mathrm{~kg}\) at \((3.0,0.0) \mathrm{m}\). Where should a fourth object of \(8.0 \mathrm{~kg}\) be placed so that the center of gravity of the four-object arrangement will be at \((0.0,0.0) \mathrm{m}\) ?

Short Answer

Expert verified
The fourth object of 8.0 kg should be placed at (-1.5m, -1.5m) to bring the center of gravity of the four-object arrangement to the origin (0.0,0.0) m.

Step by step solution

01

Recall the definition of center of mass

The 2D center of mass for a system of particles can be calculated using the formula: \[x_{cm} = \frac{\sum m_i x_i}{\sum m_i}\] \[y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\] where \(m_i, x_i, y_i\) are the mass and coordinates of the \(i-th\) particle.
02

Set up the equations for the center of mass at the origin

In the given problem, we want the center of mass to be at the origin, i.e., \(x_{cm} = 0\) and \(y_{cm} = 0\). This gives us two equations: \(0 = \frac{(5kg * 0m) + (3kg * 0m) + (4kg * 3m) + (8kg * x)}{5kg + 3kg + 4kg + 8kg}\) \(0 = \frac{(5kg * 0m) + (3kg * 4m) + (4kg * 0m) + (8kg * y)}{5kg + 3kg + 4kg + 8kg}\) where (x,y) are the coordinates we need to find.
03

Solve the equations to find the coordinates

Isolate the expressions for x and y to get: \(x = -\frac{(5kg * 0m) + (3kg * 0m) + (4kg * 3m)}{8kg}\) and \(y = -\frac{(5kg * 0m) + (3kg * 4m) + (4kg * 0m)}{8kg}\). Solving these expressions will give us the values of x and y.
04

Calculate the values of x and y

Solving the expressions gives us: \(x = -\frac{(5kg * 0m) + (3kg * 0m) + (4kg * 3m)}{8kg} = -\frac{12 kg*m}{8kg} = -1.5 m\) and \(y = -\frac{(5kg * 0m) + (3kg * 4m) + (4kg * 0m)}{8kg} = -\frac{12 kg*m}{8kg} = -1.5 m\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Distribution
Understanding mass distribution is key when calculating the center of mass in any system of particles. Mass distribution refers to how mass is spread out within a system or an object. In our example, we have three masses placed at different coordinates:
  • 5.0 kg at (0.0, 0.0) m
  • 3.0 kg at (0.0, 4.0) m
  • 4.0 kg at (3.0, 0.0) m
Each mass contributes differently to the overall center of mass.
The approach is to first calculate the masses' net effect on the position of the center of mass in both the x and y directions. The greater the mass, the more it influences the center.
Consider how the position of the 8.0 kg mass could bring the overall center of mass to a desired coordinate, which in this case is the origin at (0.0, 0.0) m, signifying a balanced net distribution of mass.
2D Center of Mass Formula
The 2D center of mass for a set of particles provides us with a way to determine a single point where we can assume the entire mass of the system is concentrated. This mathematical balance point considers the mass of each object and its position. The formulas needed to determine the center of mass in two dimensions are:
\[x_{cm} = \frac{\sum m_i x_i}{\sum m_i}, \,\, y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\]where:
  • \(m_i\) is the mass of the i-th object,
  • \(x_i\) is the x-coordinate of the i-th object,
  • \(y_i\) is the y-coordinate of the i-th object.
In simple terms, it's a weighted average of the x and y coordinates, with the weights being the masses of the particles.
For our exercise, we use these formulas by substituting the given values and set the center of mass \((x_{cm} , y_{cm})\) at the origin, i.e., (0.0, 0.0) m. This requires solving these equations for the unknown coordinates of the 8.0 kg mass.
Coordinate Calculations
Accurate coordinate calculations are crucial to finding the correct position of the 8.0 kg mass. To ensure the center of mass is at (0.0, 0.0) m, we must calculate the x and y coordinates where the mass should be placed.
Using substitutions and simplifying, we calculate:
  • \(x = -\frac{12 \, kg \cdot m}{8 \, kg} = -1.5 \, m\)
  • \(y = -\frac{12 \, kg \cdot m}{8 \, kg} = -1.5 \, m\)
Breaking it down: - We first sum the mass times coordinate products of known masses and coordinates.- We then divide by the mass of the unidentified particle.
These calculations help us position the new mass precisely so the overall system leans towards the designated center of mass at the origin. This balance keeps the entire system stable as intended.

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Most popular questions from this chapter

QIC I S A painter climbs a ladder leaning against a smooth wall. At a certain height, the ladder is on the verge of slipping. (a) Explain why the force exerted bythe vertical wall on the ladder is horizontal. (b) If the ladder of length \(L\) leans at an angle \(\theta\) with the horizontal, what is the lever arm for this horizontal force with the axis of rotation taken at the base of the ladder? (c) If the ladder is uniform, what is the lever arm for the force of gravity acting on the ladder? (d) Let the mass of the painter be \(80 \mathrm{~kg}, L=4.0 \mathrm{~m}\), the ladder's mass be \(30 \mathrm{~kg}, \theta=53^{\circ}\), and the coefficient of friction between ground and ladder be \(0.45\). Find the maximum distance the painter can climb up the ladder.

GP A beam resting on two pivots has a length of \(L=6.00 \mathrm{~m}\) and mass \(M=90.0 \mathrm{~kg}\). The pivot under the left end exerts a normal force \(n_{1}\) on the beam, and the second pivot placed a distance \(\ell=4.00 \mathrm{~m}\) from the left end exerts a normal force \(n_{2}\). A woman of mass \(m=55.0 \mathrm{~kg}\) steps onto the left end of the beam and begins walking to the right as in Figure P8.12. The goal is to find the woman's position when the beam begins to tip. (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman \(x\) meters to the right of the first pivot, which is the origin. (b) Where is the woman when the normal force \(n_{1}\) is the greatest? (c) What is \(n_{1}\) when the beam is about to tip? (d) Use the force equation of equilibrium to find the value of \(n_{2}\) when the beam is about to tip. (e) Using the result of part (c) and the torque equilibrium equation, with torques com- puted around the second pivot point, find the woman's puted around the second pivot point, find the woman's position when the beam is about to tip. (f) Check the answer to part (e) by computing torques around the first pivot point. Except for possible slight differences due to rounding, is the answer the same?

A uniform ladder of length \(L\) and weight \(w\) is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is \(\mu_{s}=0.500\), determine the smallest angle the ladder can make with the floor without slipping.

Each of the following objects has a radius of \(0.180 \mathrm{~m}\) and a mass of \(2.40 \mathrm{~kg}\), and each rotates about an axis through its center (as in Table 8.1) with an angular speed of \(35.0 \mathrm{rad} / \mathrm{s}\). Find the magnitude of the angular momentum of each object. (a) a hoop (b) a solid cylinder (c) a solid sphere (d) a hollow spherical shell

A car is designed to get its energy from a rotating flywheel with a radius of \(2.00 \mathrm{~m}\) and a mass of \(500 \mathrm{~kg}\). Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to \(5000 \mathrm{rev} / \mathrm{min}\). (a) Find the kinetic energy stored in the flywheel. (b) If the flywheel is to supply energy to the car as a \(10.0\)-hp motor would, find the length of time the car could run before the flywheel would have to be brought back up to speed.
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