A 50-kg wheel of diameter 2.0 m is free to rotate about its center. A 20 N force is applied on the wheel at a point on the rim so that the angle between the force and R is 30What is the torque (in Nm) produced by the force?
A 50-kg wheel of diameter 2.0 m is free to rotate about its center. A 20 N force is applied on the wheel at a point on the rim so that the angle between the force and R is 30What is the torque (in Nm) produced by the force?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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A 50-kg wheel of diameter 2.0 m is free to rotate about its center. A 20 N force is applied on the wheel at a point on the rim so that the angle between the force and R is 30What is the torque (in Nm) produced by the force?
![### Question 3
A 50-kg wheel of diameter 2.0 m is free to rotate about its center. If a force \( F \) of 100 N is applied tangentially to the wheel at its edge such that the angle between the force and \( R \) is 30°, what is the torque \( \tau \) about the center of the wheel?
**Diagram:**
- A circle representing the wheel.
- A radius \( R \) is drawn from the center of the wheel to the point where the force is applied.
- The force \( F \) is applied tangentially to the edge of the wheel.
- An angle of 30° is marked between the radius \( R \) and the direction of the force \( F \).
**Answer Choices:**
- ○ 10
- ○ 50
- ○ -10
To solve the problem, consider the torque formula:
\[ \tau = R \times F \times \sin(\theta) \]
Where:
- \( \tau \) is the torque
- \( R \) is the radius of the wheel
- \( F \) is the force applied
- \( \theta \) is the angle between \( R \) and \( F \)
Given:
- Diameter of the wheel = 2.0 m, so the radius \( R \) = 1.0 m
- Force \( F \) = 100 N
- Angle \( \theta \) = 30°
\[ \tau = 1.0 \, \text{m} \times 100 \, \text{N} \times \sin(30°) \]
\[ \tau = 100 \, \text{N} \cdot \text{m} \times 0.5 \]
\[ \tau = 50 \, \text{N} \cdot \text{m} \]
Thus, the torque \( \tau \) about the center of the wheel is **50 N·m**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc154e8f7-5981-4a34-85b0-b10a7e25c57a%2F1782b032-9ee3-4a65-9590-a1cf454acfa5%2Fw6yb1mt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 3
A 50-kg wheel of diameter 2.0 m is free to rotate about its center. If a force \( F \) of 100 N is applied tangentially to the wheel at its edge such that the angle between the force and \( R \) is 30°, what is the torque \( \tau \) about the center of the wheel?
**Diagram:**
- A circle representing the wheel.
- A radius \( R \) is drawn from the center of the wheel to the point where the force is applied.
- The force \( F \) is applied tangentially to the edge of the wheel.
- An angle of 30° is marked between the radius \( R \) and the direction of the force \( F \).
**Answer Choices:**
- ○ 10
- ○ 50
- ○ -10
To solve the problem, consider the torque formula:
\[ \tau = R \times F \times \sin(\theta) \]
Where:
- \( \tau \) is the torque
- \( R \) is the radius of the wheel
- \( F \) is the force applied
- \( \theta \) is the angle between \( R \) and \( F \)
Given:
- Diameter of the wheel = 2.0 m, so the radius \( R \) = 1.0 m
- Force \( F \) = 100 N
- Angle \( \theta \) = 30°
\[ \tau = 1.0 \, \text{m} \times 100 \, \text{N} \times \sin(30°) \]
\[ \tau = 100 \, \text{N} \cdot \text{m} \times 0.5 \]
\[ \tau = 50 \, \text{N} \cdot \text{m} \]
Thus, the torque \( \tau \) about the center of the wheel is **50 N·m**.
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