Problem 31 - Calculus

Question

31 Put a mass \(m\) at the point \((x, 0)\). Around the origin the torque from gravity is the force \(m g\) times the distance \(x\). This equals \(g\) times the \(\quad m x\).

Step-by-Step Solution

Answer

The torque is \( mgx \).

Step 1: Understanding the Problem Statement

We need to determine the torque created by gravity on a mass placed at point \((x, 0)\) around the origin. Torque is calculated as the product of force and the perpendicular distance from point to axis.

Step 2: Identifying the Given Values

Here, force due to gravity is \(mg\), with \(m\) being the mass and \(g\) being gravitational acceleration. The distance from origin is \(x\).

Step 3: Calculating the Torque

Torque \(\tau\) is given by the formula \(\tau = \text{Force} \times \text{Distance}\). Here, it becomes \(\tau = mg \times x \).

Step 4: Conclusion

The torque around the origin from gravity for the mass \( m \) located at \((x, 0)\) is \( mgx \).

Key Concepts

Gravity

Gravity is a fundamental force we experience every day, but let's explore what it actually means. At its core, gravity is the attractive force that pulls objects with mass towards each other. On Earth, it's what keeps us grounded and what gives weight to physical objects.

The force of gravity between two objects depends on both their masses and the distance between them. Specifically, this is explained by Newton's Law of Universal Gravitation. It states that every particle attracts every other particle in the universe with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The equation for this force is:

\[ F = G \frac{m_1 m_2}{r^2} \]

Where:

• \( F \) is the gravitational force between the masses,
• \( G \) is the gravitational constant \(6.674 \, \times \, 10^{-11} \, \mathrm{N(m/kg)^2}\),
• \( m_1 \) and \( m_2 \) are the masses of the objects,
• \( r \) is the distance between the centers of the two masses.

Understanding gravity helps in grasping how it influences torque, especially since the force due to gravity, \( mg \), directly impacts the torque calculation when we look into forces acting perpendicularly at a distance.

Force

In the realm of physics, force is described as any interaction that, when unopposed, changes the motion of an object. Simply put, a force can cause an object to start moving, stop, or change direction. Force can also change the shape of an object.

One of the most common examples of force is the force due to gravity. As in our exercise, this force is defined as the product of mass \( m \) and gravitational acceleration \( g \). We express it as:

\[ F = m \cdot g \]

Here, \( g \) is the acceleration due to Earth's gravity, approximately \( 9.81 \, \mathrm{m/s^2} \). This formula helps in understanding how much force gravity is exerting on a mass at rest or in free fall.

Force plays a vital role in calculating torque, which is crucial when determining rotational effects. In our problem, knowing the force due to gravity helps in assessing the torque created around a point, considering how force combines with distance to produce rotational motion.

Distance

Distance in the context of torque and force applications is more than just a measurement. It's a vital element that influences the strength of rotational effects like torque.

When we calculate torque, distance refers to the length from the axis of rotation to the point where the force is applied. This distance needs to be perpendicular to the direction of the force to accurately determine the torque. Mathematically, torque \( \tau \) is defined as:

\[ \tau = F \times d \]

Where:

• \( \tau \) is the torque,
• \( F \) is the force applied,
• \( d \) is the perpendicular distance from the axis to the line of action of the force.

Understanding distance in this role helps clarify its impact on rotational dynamics. A greater distance increases the torque, given that the same amount of force is applied. This principle is why tools like wrenches have longer handles – to maximize torque with less applied force. In the original exercise, \( x \) served as the vital distance, helping compute the total torque caused by the weight of the object around the origin.