Torque & Equilibrium: Videos & Practice Problems

In mechanics, equilibrium involves both linear and rotational aspects. Linear equilibrium occurs when the net force is zero, resulting in no acceleration. Rotational equilibrium requires the sum of all torques to also be zero, preventing angular acceleration. For complete equilibrium, both conditions must be satisfied. Understanding these principles is crucial for analyzing rigid bodies, as various force and torque configurations can lead to different states of equilibrium, influencing stability and motion.

Torque & Equilibrium

In the study of mechanics, understanding equilibrium is crucial, particularly when distinguishing between linear and rotational equilibrium. An object is in linear equilibrium when the net force acting on it is zero, which means there is no acceleration. This principle is derived from Newton's second law, expressed as \( F = ma \). If the sum of all forces equals zero, the object remains at rest or moves with constant velocity.

However, when considering rotational motion, the situation becomes more complex. An object can experience linear equilibrium while still being in a state of rotational imbalance. For instance, if a uniform bar is supported at its center and subjected to forces that cancel each other vertically, it may still rotate due to the torques produced by these forces. Torque, defined as \( \tau = r \cdot F \cdot \sin(\theta) \), depends on the distance from the axis of rotation (r), the force applied (F), and the angle (\(\theta\)) between the force and the lever arm. If the torques do not cancel, the object will rotate, indicating a lack of rotational equilibrium.

To achieve complete equilibrium, both the sum of forces and the sum of torques must equal zero. This state is often referred to as static equilibrium, where the object is not only at rest but also has no angular motion. In practical scenarios, various configurations can lead to different outcomes regarding equilibrium. For example, if two equal forces act in opposite directions on a bar, they may create linear equilibrium but not rotational equilibrium if they produce unequal torques. Conversely, if forces are balanced but create equal and opposite torques, both types of equilibrium can be achieved.

In summary, recognizing the conditions for linear and rotational equilibrium is essential for analyzing the behavior of rigid bodies. Understanding how forces and torques interact allows for a deeper comprehension of static systems and their stability.

Balancing a bar with a force

In this example of rotational equilibrium, we analyze a uniform bar with a length of 4 meters and a mass of 10 kilograms, which is free to rotate about a fulcrum positioned 1 meter from its left end. The center of mass of the bar is located at its midpoint, 2 meters from either end, where the gravitational force (mg) acts downward. To maintain equilibrium, a downward force (F) is applied at the left edge of the bar to prevent it from tipping over.

To determine the magnitude of the force (F) required for balance, we apply the principle of rotational equilibrium, which states that the sum of all torques acting on the system must equal zero. The torque due to the gravitational force (mg) acts clockwise and is considered negative, while the torque due to the applied force (F) acts counterclockwise and is positive. The equation for rotational equilibrium can be expressed as:

$$\tau_F + (-\tau_{mg}) = 0$$

This simplifies to:

$$\tau_F = \tau_{mg}$$

Next, we expand the torques:

$$\tau_F = F \cdot r_F \cdot \sin(\theta_F)$$

$$\tau_{mg} = mg \cdot r_{mg} \cdot \sin(\theta_{mg})$$

Here, the distance (r) for both forces is 1 meter, and the angle (θ) between the force and the lever arm is 90 degrees, making the sine of 90 equal to 1. Substituting the values:

$$F \cdot 1 \cdot 1 = (10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) \cdot 1 \cdot 1$$

Solving this gives:

$$F = 98 \, \text{N}$$

This indicates that a force of 98 newtons must be applied to maintain equilibrium. The relationship between the distances and forces shows that if the distances from the fulcrum are equal, the forces must also be equal.

For the second part, we calculate the normal force (N) exerted by the fulcrum. According to the equilibrium of forces in the vertical direction, the sum of the forces must also equal zero:

$$N - (F + mg) = 0$$

Rearranging gives:

$$N = F + mg$$

Substituting the values:

$$N = 98 \, \text{N} + 98 \, \text{N} = 196 \, \text{N}$$

This means the normal force exerted by the fulcrum is 196 newtons, balancing the downward forces acting on the bar. Understanding these concepts of torque and equilibrium is crucial for analyzing systems in physics.

A composite disc is made out of two concentric cylinders, as shown. The inner cylinder has radius 30 cm. The outer cylinder has radius 50 cm. If you pull on a light rope attached to the edge of the outer cylinder (shown left) with 100 N, how hard must you pull on a light rope attached to the edge of the inner cylinder (shown right) so the disc does not spin?

• A. 300 N
• B. 667 N
• C. 60 N
• D. 167 N

Answer: (D)

Pin holding a horizontal bar

In this example, we explore the concept of torque and equilibrium by analyzing a bar supported by a pin on a wall. The bar has a mass of 20 kg and a length of 3 meters, positioned horizontally. The gravitational force acting on the bar, denoted as \( mg \), is located at the midpoint, creating a torque that tends to rotate the bar downwards around a defined axis of rotation.

To maintain equilibrium, the sum of all torques acting on the bar must equal zero, which means the torque produced by the gravitational force must be countered by the torque exerted by the pin. This relationship can be expressed mathematically as:

\[ \text{Torque}_{\text{pin}} - \text{Torque}_{mg} = 0 \]

Rearranging this equation gives us:

\[ \text{Torque}_{\text{pin}} = \text{Torque}_{mg} \]

To calculate the torque due to the gravitational force, we use the formula for torque:

\[ \text{Torque} = r \cdot F \cdot \sin(\Theta) \]

In this scenario, \( r \) is the distance from the axis of rotation to the point where the gravitational force acts, which is half the length of the bar (1.5 meters). The force \( F \) is the weight of the bar, calculated as:

\[ F = mg = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

Since the angle \( \Theta \) between the force and the lever arm is 90 degrees, we have \( \sin(90^\circ) = 1 \). Thus, the torque due to the gravitational force becomes:

\[ \text{Torque}_{mg} = 1.5 \, \text{m} \times 196 \, \text{N} \times 1 = 294 \, \text{N} \cdot \text{m} \]

This value represents the torque that the pin must counteract to keep the bar in equilibrium. Therefore, the torque on the pin required to balance the bar is also 294 N·m. This example illustrates the fundamental principles of torque and equilibrium in a straightforward manner, emphasizing the importance of balancing forces to prevent rotational motion.

Here’s what students ask on this topic:

What is the difference between linear equilibrium and rotational equilibrium?

Linear equilibrium occurs when the net force acting on an object is zero, resulting in no linear acceleration. This means the object will either remain at rest or continue moving at a constant velocity. The condition for linear equilibrium is given by the equation:

$\sum F=0$

Rotational equilibrium, on the other hand, occurs when the sum of all torques acting on an object is zero, resulting in no angular acceleration. This means the object will not start rotating or will rotate at a constant angular velocity. The condition for rotational equilibrium is given by the equation:

$\sum \tau =0$

For complete equilibrium, both linear and rotational equilibrium conditions must be satisfied simultaneously.

How do you calculate torque in a rotational system?

Torque (τ) in a rotational system is calculated using the equation:

$\tau =Frsin\left(\theta \right)$

where:

• $F$ is the force applied,
• $r$ is the distance from the axis of rotation to the point where the force is applied (lever arm),
• $\theta$ is the angle between the force vector and the lever arm.

The sine function accounts for the component of the force that is perpendicular to the lever arm, which is the only component that contributes to the torque.

What are the conditions for an object to be in static equilibrium?

For an object to be in static equilibrium, two conditions must be met:

1. The sum of all forces acting on the object must be zero, ensuring no linear acceleration:

   $\sum F=0$

2. The sum of all torques acting on the object must be zero, ensuring no angular acceleration:

   $\sum \tau =0$

When both conditions are satisfied, the object will not move or rotate, maintaining a state of complete equilibrium.

How do forces and torques affect the stability of rigid bodies?

Forces and torques play a crucial role in determining the stability of rigid bodies. When the net force on a rigid body is zero, it is in linear equilibrium, meaning it will not accelerate linearly. When the sum of all torques is zero, the body is in rotational equilibrium, meaning it will not experience angular acceleration. For a rigid body to be completely stable, both linear and rotational equilibrium must be achieved. If either condition is not met, the body may translate, rotate, or both, leading to instability. Proper analysis of forces and torques helps in designing stable structures and systems.

What is the significance of the center of mass in rotational equilibrium?

The center of mass is the point where the mass of an object is considered to be concentrated. In rotational equilibrium, the position of the center of mass is crucial because it affects the distribution of torques. If forces are applied at or through the center of mass, they do not create any torque, as the lever arm distance (r) is zero. Understanding the center of mass helps in analyzing and predicting the rotational behavior of objects, ensuring that the sum of torques is zero for achieving rotational equilibrium.