5.3 Torque

Torque is a fundamental concept in physics, describing the rotational force applied to an object. It's crucial for understanding how forces cause objects to rotate, from simple machines to complex mechanical systems.

The magnitude of torque depends on the force applied, the distance from the rotation axis, and the angle between them. Force diagrams help visualize torques, while equations like τ = rF sin θ allow for precise calculations.

Torques on rigid systems

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Perpendicular force component

Static friction is an "adjustable" force that matches the applied force up to its maximum value.

• Torque is generated solely by the force component perpendicular to the position vector extending from the axis of rotation to the force's application point
• When forces are applied at angles, only the perpendicular component contributes to rotation
• The parallel component has no rotational effect

Lever arm

The lever arm is what gives mechanical advantage to many tools and machines we use daily.

• Lever arm refers to the perpendicular distance between the axis of rotation and the line along which the force is exerted
• Increasing the lever arm amplifies the torque produced by a given force
• Torque is maximized when the force is applied at a 90° angle to the lever arm
• This explains why wrenches have long handles and why door handles are placed far from hinges

Description of torques

Force diagrams

Force diagrams serve as the primary visual tool for analyzing rotational problems.

• Force diagrams provide a visual representation of the torques acting on a rigid system
• They are analogous to free-body diagrams but focus on rotational effects
• Force diagrams illustrate the relative magnitude and direction of the forces acting on a rigid system
• They show points of application in relation to the axis of rotation
• Curved arrows typically indicate the direction of rotation that would result from each force

Magnitude of torque

The mathematical relationship between force, distance, and angle determines the torque value.

• The magnitude of the torque ($\tau$) exerted on a rigid system by a force is given by the equation: $\tau = rF\sin\theta$
• $r$ represents the distance from the axis of rotation to the point where the force is applied
• $F$ denotes the magnitude of the applied force
• $\theta$ is the angle between the force vector and the position vector from the axis of rotation to the force's application point
• When the force is perpendicular to the position vector ($\theta = 90°$), the torque reaches its maximum value: $\tau_{max} = rF$
• Conversely, when the force is parallel to the position vector ($\theta = 0°$ or $180°$), the torque is zero: $\tau = 0$

Practice Problem 1: Torque Calculation

Solution:

1. Identify the given values:

• Force (F) = 15 N
• Distance from axis of rotation (r) = 0.75 m
• Angle (θ) = 90° (perpendicular force)

1. Apply the torque equation: $\tau = rF\sin\theta$ $\tau = (0.75 \text{ m})(15 \text{ N})(\sin 90°)$ $\tau = (0.75 \text{ m})(15 \text{ N})(1)$ $\tau = 11.25 \text{ N·m}$ with $\tau = 11.25 \text{ N}\cdot\text{m}$

2. The torque applied to the door is 11.25 N·m.

Practice Problem 2: Lever Arm Analysis

Solution:

1. For part (a):

• Force (F) = 50 N
• Distance (r) = 0.3 m
• Angle (θ) = 30° $\tau = rF\sin\theta$ $\tau = (0.3 \text{ m})(50 \text{ N})(\sin 30°)$ $\tau = (0.3 \text{ m})(50 \text{ N})(0.5)$ $\tau = 7.5 \text{ N·m}$

1. For part (b) - perpendicular force (θ = 90°): $\tau_{max} = rF$ $\tau_{max} = (0.3 \text{ m})(50 \text{ N})$ $\tau_{max} = 15 \text{ N·m}$

2. The increase in torque would be: $\Delta\tau = 15 \text{ N·m} - 7.5 \text{ N·m} = 7.5 \text{ N·m}$

3. The torque would double (increase by 100%) if applied perpendicularly.

Practice Problem 3: Multiple Torques

Solution:

1. When a system is balanced, the sum of clockwise torques equals the sum of counterclockwise torques.

2. Calculate the torque from the first child:

• Mass = 30 kg
• Weight = mg = (30 kg)(9.8 m/s²) = 294 N
• Distance from pivot = 1.5 m
• $\tau_1 = (1.5 \text{ m})(294 \text{ N}) = 441 \text{ N·m}$

1. For the second child:

• Mass = m (unknown)
• Weight = mg = m(9.8 m/s²) = 9.8m N
• Distance from pivot = 2 m
• $\tau_2 = (2 \text{ m})(9.8m \text{ N}) = 19.6m \text{ N·m}$

1. Set up the balance equation: $\tau_1 = \tau_2$ $441 \text{ N·m} = 19.6m \text{ N·m}$ $m = \frac{441}{19.6} = 22.5 \text{ kg}$

2. The mass of the second child is 22.5 kg.