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Chapter 9: Problem 10

A bicycle pedal is pushed straight downwards by a foot with a 27 Newton force. The shaft of the pedal is \(20 \mathrm{~cm}\) long. If the shaft is \(\pi / 3\) radians past horizontal, what is the magnitude of the torque about the point where the shaft is attached to the bicycle? __________\(\mathrm{Nm}\)

Short Answer

Expert verified
The magnitude of the torque about the point where the shaft is attached to the bicycle is approximately \(4.6764 \mathrm{~Nm}\).

Step by step solution

01

Identify the given information

We are given the following information: - The force exerted on the pedal is 27 N - The length of the pedal shaft is 20 cm - The angle between the pedal shaft and the horizontal axis is 蟺/3 radians
02

Convert the length of the pedal shaft into meters

Since the torque we need to find is in Newton meters (Nm), we need the length of the pedal shaft in meters. To do this, we divide the length in centimeters by 100: Pedal shaft length = 20 cm 梅 100 = 0.2 m
03

Determine the perpendicular distance from the point of force application to the axis of rotation

We can determine the perpendicular distance by using the sine function with the given angle. Perpendicular distance = 0.2 * sin(蟺/3)
04

Calculate the torque

Now that we have the perpendicular distance, we can find the torque using the formula: Torque = Force 脳 Perpendicular Distance Torque = 27 N 脳 (0.2 * sin(蟺/3))
05

Evaluate the expression

Evaluating the expression, we get: Torque = 27 N 脳 (0.2 * sin(蟺/3)) Torque = 27 N 脳 (0.2 * (鈭�3/2)) Torque = 27 N 脳 (0.1 * 鈭�3) Torque 鈮� 27 N 脳 0.1732 Torque 鈮� 4.6764 Nm The magnitude of the torque about the point where the shaft is attached to the bicycle is approximately 4.6764 Nm.

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Key Concepts

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

Newton's Laws of Motion
Newton's laws form the backbone of classical mechanics, helping us understand how objects behave when forces are applied. In this exercise, we are especially interested in how forces, like the 27 Newton force on the pedal, cause the bike pedal to rotate. This leads us right to our topic: torque, which is the force making things turn.
  • First Law (Law of Inertia): An object will remain at rest or in uniform motion unless acted upon by an external force. This means the pedal won't move unless someone pushes it.
  • Second Law (Law of Acceleration): The force acting on an object is equal to the mass of that object times its acceleration (F = ma). In terms of rotation, torque is the rotational equivalent of force, essentially providing the 'push' needed for rotation.
  • Third Law (Action and Reaction): For every action, there is an equal and opposite reaction. When the foot pushes the pedal, the pedal provides an equal force back on the foot, causing it to rotate.
Understanding forces and their effects on motion gives the foundation for solving this exercise: figuring out how much torque this force produces. By breaking down the forces applied鈥攚e can compute how the foot pushing on the pedal turns the bike pedal shaft around its pivot point.
Trigonometry in Torque Calculation
Trigonometry is essential for solving physics problems that involve angles, such as this torque problem with a pedal \( \pi/3 \) radians past horizontal. Torque is affected by the angle of force application, needing trigonometry for accurate calculations.
  • Sin Function Role: In this exercise, the sine function helps determine the effective 'lever arm' or the actual distance from the pivot point at which the force acts鈥攃rucial for torque calculation.
  • Angle Determination: The pedal is moved \( \pi/3 \) radians from horizontal, necessitating the sine of the angle to find the perpendicular force component.
  • Perpendicular Distance: Calculating the exact part of the 20 cm shaft length affecting rotation involves using trigonometry: converting that shaft length to meters, then multiplying by \( \sin(\pi/3) \) to get the perpendicular distance.
These trigonometric relationships clarify how certain angles influence the effectiveness of a force when causing rotation. This integration of trigonometry simplifies solving complicated physics problems by finding perpendicular components needed to calculate torque.
Physics Problem Solving Strategies
When faced with a physics problem, using systematic strategies is key to finding the solution. Let's walk through some effective problem-solving techniques applied in this torque exercise:
  • Identify and List Known Information: Clearly determine what you already know from the problem statement (e.g., force magnitude, lengths, angles) to understand the problem better.
  • Convert Units: Ensure all measurements are in the correct units for calculation, such as converting centimeters to meters.
  • Use Formulas Appropriately: Apply relevant formulas, like torque = force 脳 perpendicular distance, aligning with the type of problem.
  • Check Calculations Step-by-Step: Evaluate each computational step, ensuring granular details such as angle functions (e.g., using the correct trigonometric function) are accurate.
  • Review Your Process: Simplifying and cross-checking ensures logical consistency in your solution, building confidence in the calculated torque value.
By applying these strategies, problem-solving becomes a structured, less daunting task, enabling queries like this torque calculation to be tackled systematically and accurately.

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