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Chapter 8: Problem 21

A door weighing \(300.0 \mathrm{N} \quad\) measures \(2.00 \mathrm{m} \times 3.00 \mathrm{m}\) and is of uniform density; that is, the mass is uniformly distributed throughout the volume. A doorknob is attached to the door as shown. Where is the center of gravity if the doorknob weighs \(5.0 \mathrm{N}\) and is located \(0.25 \mathrm{m}\) from the edge?

Short Answer

Expert verified
The center of gravity is at (1.016 m, 1.479 m).

Step by step solution

01

Concept of Center of Gravity

The center of gravity of a system of objects can be found by computing the sum of moments due to each object about a reference point and dividing by the total weight.
02

Determine Total Weight and Moment Contributions

The total weight of the door system is the sum of the weight of the door and the weight of the doorknob. Calculate the moments about the bottom left corner of the door:- Moment due to door's center of gravity: Since the door is uniform, the center of gravity is at its geometric center, i.e., \[ x_{door} = \frac{2.00}{2} = 1.00 \text{ m}, \quad y_{door} = \frac{3.00}{2} = 1.50 \text{ m} \]- Moment due to doorknob: The position of the doorknob is \[ x_{knob} = 2.00 \text{ m}, \quad y_{knob} = 0.25 \text{ m} \]
03

Calculate Moments About X-axis and Y-axis

Compute the moments about the x-axis and y-axis using the formula:\[ M_x = W_{door} \cdot x_{door} + W_{knob} \cdot x_{knob} \]and \[ M_y = W_{door} \cdot y_{door} + W_{knob} \cdot y_{knob} \]Thus:\( M_x = 300 \times 1.00 + 5 \times 2.00 = 310 \text{ Nm} \)\( M_y = 300 \times 1.50 + 5 \times 0.25 = 451.25 \text{ Nm} \)
04

Compute the Center of Gravity's Position

Using the sums of moments divided by total weight, find the center of gravity of the system:\[ x_{cg} = \frac{M_x}{W_{total}} = \frac{310}{305} \approx 1.016 \text{ m} \]\[ y_{cg} = \frac{M_y}{W_{total}} = \frac{451.25}{305} \approx 1.479 \text{ m} \]Thus, the center of gravity is at \(1.016 \text{ m}, 1.479 \text{ m} \).

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Key Concepts

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

Uniform Density
When an object possesses uniform density, its mass is evenly spread out across its volume. This is an important concept in physics, particularly when discussing long or flat objects like doors.
  • A uniformly dense object's center of gravity lies at its geometric center.
  • For a solid object like the door in our example, its mass distribution helps in simplifying complex calculations, as you can consider it behaves like a single point mass located at its center.
  • In the given exercise, the door, being uniformly dense, has its center of gravity at the midpoint of its dimensions: 1.00 m from one side and 1.50 m from the other.
Understanding uniform density allows us to apply simple geometrical assumptions, which make solving problems related to balance and moment calculation much more straightforward.
Moment Calculation
Calculating the moment, or torque, involves understanding how forces lead to rotational motion about an axis. Moments are calculated by multiplying the force applied by the perpendicular distance from the axis of rotation.
  • For instance, in the step-by-step solution, the moments are calculated from a reference point using the distances to the center of gravity of each piece (door and doorknob) along with their weights.
  • Moments around the x-axis consider horizontal distances, while those around the y-axis examine vertical ones.
  • Thus, using the formulae \( M_x = W_{door} \cdot x_{door} + W_{knob} \cdot x_{knob} \) and \( M_y = W_{door} \cdot y_{door} + W_{knob} \cdot y_{knob} \), we determine the influence of the distributed mass and additional weight on the object's balance.
This method offers a systematic approach to assess how different forces affect a system's overall stability, and is crucial to understanding weight distribution and center of gravity.
Physics Problems
Physics problems involving concepts such as center of gravity can often appear daunting, yet they are simply applications of fundamental principles. In scenarios where different parts of an object or system exert independent weights, solving for their collective effects becomes essential.
  • In problems like this one, separating the weights and considering each component individually simplifies the solution.
  • Here, the door's and doorknob's separate moments are calculated before combining them to find a practical center of gravity.
  • This piecemeal approach can be very helpful, especially in complex systems where numerous weights and distances are involved.
By breaking down the situation into smaller, more manageable parts, we reveal the broader physics concepts at play in straightforward terms, making problem-solving more approachable.
Weight Distribution
Weight distribution refers to how mass or weight is spread over a body or system. This is crucial when considering stability, balance, and mechanical advantage.
  • The weight distribution can impact an object's rotational equilibrium and its tendency to topple or remain stable in various positions.
  • In our exercise's context, the door and its knob demonstrate this concept, as their combined weight distribution determines the overall center of gravity.
  • It's important to remember that even small additional weights, like the 5 N doorknob, can shift the center of gravity significantly, affecting the object's balance.
By studying weight distribution through exercises like this, students learn to gauge stability and understand balance in everyday objects, which is an important aspect of engineering and physics.

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Most popular questions from this chapter

A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist \(15 \mathrm{cm}\) to the right of his center of gravity. If the climber weighs \(770 \mathrm{N},\) find (a) the tension in the rope and (b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.
A person is trying to lift a ladder of mass \(15 \mathrm{kg}\) and length $8.0 \mathrm{m} .$ The person is exerting a vertical force on the ladder at a point of contact \(2.0 \mathrm{m}\) from the center of gravity. The opposite end of the ladder rests on the floor. (a) When the ladder makes an angle of \(60.0^{\circ}\) with the floor, what is this vertical force? (b) A person tries to help by lifting the ladder at the point of contact with the floor. Does this help the person trying to lift the ladder? Explain.
A 68 -kg woman stands straight with both feet flat on the floor. Her center of gravity is a horizontal distance of \(3.0 \mathrm{cm}\) in front of a line that connects her two ankle joints. The Achilles tendon attaches the calf muscle to the foot a distance of \(4.4 \mathrm{cm}\) behind the ankle joint. If the Achilles tendon is inclined at an angle of \(81^{\circ}\) with respect to the horizontal, find the force that each calf muscle needs to exert while she is standing. [Hint: Consider the equilibrium of the part of the body above the ankle joint.]

In the movie Terminator, Arnold Schwarzenegger lifts someone up by the neck and, with both arms fully extended and horizontal, holds the person off the ground. If the person being held weighs \(700 \mathrm{N},\) is \(60 \mathrm{cm}\) from the shoulder joint, and Arnold has an anatomy analogous to that in Fig. \(8.30,\) what force must each of the deltoid muscles exert to perform this task?
A turntable of mass \(5.00 \mathrm{kg}\) has a radius of \(0.100 \mathrm{m}\) and spins with a frequency of 0.550 rev/s. What is its angular momentum? Assume the turntable is a uniform disk.
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