Question

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

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

Step-by-step solution

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.

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 \[ x_{knob} = 2.00 \text{ m}, \quad y_{knob} = 0.25 \text{ m} \]

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:\( M_x = 300 \times 1.00 + 5 \times 2.00 = 310 \text{ Nm} \)\( M_y = 300 \times 1.50 + 5 \times 0.25 = 451.25 \text{ Nm} \)

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 \(1.016 \text{ m}, 1.479 \text{ m} \).

Key concepts

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.