**Exercise: Calculating Torque in Leg Exercises****Context:**To exercise, a man attaches a 3.0 kg weight to the heel of his foot.**Problem Statement:**Part A: When his leg is stretched out before him, what is the torque exerted by the weight about his knee, which is 50 cm away from the weight? Use \( g = 10 \, \text{m/s}^2 \).**Options:**- 1.5 N·m- 1500 N·m- 15 N·m- 150 N·m**Explanation:**Torque (\(\tau\)) is calculated using the formula:\[\tau = r \cdot F\]where \( r \) is the distance from the pivot point to the point where the force is applied (50 cm or 0.5 m), and \( F \) is the force (weight) acting perpendicular to the lever arm.The force due to the weight is calculated as \( F = m \cdot g = 3.0 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \).Using the formula for torque:\[\tau = 0.5 \, \text{m} \times 30 \, \text{N} = 15 \, \text{N·m}\]Thus, the correct answer is **15 N·m**.

Given data: The mass, m=3.0 kg The distance, r=50 cm=0.5 m g=10 m/s2.

To exercise, a man attaches a 3.0 kg weight to the heel of his foot. ▼ Part A When his leg is stretched out before him, what is the torque exerted by the weight about his knee, 50 cm away from the weight? Use g = 10 m/s². O 1.5 N·m 1500 N. m 15 N.m 150 Nm

To exercise, a man attaches a 3.0 kg weight to the heel of his foot. ▼ Part A When his leg is stretched out before him, what is the torque exerted by the weight about his knee, 50 cm away from the weight? Use g = 10 m/s². O 1.5 N·m 1500 N. m 15 N.m 150 Nm

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

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**Exercise: Calculating Torque in Leg Exercises**

**Context:**
To exercise, a man attaches a 3.0 kg weight to the heel of his foot.

**Problem Statement:**
Part A: When his leg is stretched out before him, what is the torque exerted by the weight about his knee, which is 50 cm away from the weight? Use \( g = 10 \, \text{m/s}^2 \).

**Options:**
- 1.5 N·m
- 1500 N·m
- 15 N·m
- 150 N·m

**Explanation:**
Torque (\(\tau\)) is calculated using the formula:
\[
\tau = r \cdot F
\]
where \( r \) is the distance from the pivot point to the point where the force is applied (50 cm or 0.5 m), and \( F \) is the force (weight) acting perpendicular to the lever arm.

The force due to the weight is calculated as \( F = m \cdot g = 3.0 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \).

Using the formula for torque:
\[
\tau = 0.5 \, \text{m} \times 30 \, \text{N} = 15 \, \text{N·m}
\]

Thus, the correct answer is **15 N·m**.

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