You are pulling your luggage in an air port for an international flight. You know not to pack too much else they charge you extra fees so you weighted your luggage at home to be 21.6 kg. You exert a pull of 49 N in magnitude and at 22 degrees above the horizontal to ensure you can walk at a constant speed on the carpeted floor. What is the coefficient of friction between the caster wheels and the carpet?
You are pulling your luggage in an air port for an international flight. You know not to pack too much else they charge you extra fees so you weighted your luggage at home to be 21.6 kg. You exert a pull of 49 N in magnitude and at 22 degrees above the horizontal to ensure you can walk at a constant speed on the carpeted floor. What is the coefficient of friction between the caster wheels and the carpet?
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![### Physics Problem: Calculating the Coefficient of Friction
**Scenario Description:**
You are pulling your luggage in an airport for an international flight. Understanding the additional charges for excess weight, you decide to weigh your luggage at home and find it to be 21.6 kg. To move the luggage on the carpeted floor at a constant speed, you pull it with a force of 49 N, applied at an angle of 22 degrees above the horizontal plane.
**Problem Statement:**
Calculate the coefficient of friction between the caster wheels of your luggage and the carpet.
**Given Data:**
- Mass of the luggage (m): 21.6 kg
- Magnitude of applied force (F): 49 N
- Angle of applied force from horizontal (\(\theta\)): 22 degrees
- Acceleration due to gravity (g): 9.8 m/s²
**Required:**
Coefficient of friction (μ) between the caster wheels and the carpeted floor.
**Steps to Solve:**
1. **Compute the Normal Force (N):**
The normal force in this context is less than the weight of the luggage because of the vertical component of the applied force.
\[
N = mg - F\sin(\theta)
\]
Where:
- \( m \) is the mass of the luggage.
- \( g \) is the acceleration due to gravity.
- \( F \) is the force exerted.
- \( \theta \) is the angle of the applied force.
2. **Calculate the Frictional Force (f):**
At constant speed, the frictional force equals the horizontal component of the applied force.
\[
f = F\cos(\theta)
\]
Where:
- \( \cos(\theta) \) is the cosine of the angle at which the force is applied.
3. **Determine the Coefficient of Friction (μ):**
\[
\mu = \frac{f}{N}
\]
This provides the coefficient of friction needed to maintain a constant speed.
**Visualization:**
Imagine a free-body diagram with:
- **Downwards:** Weight (\(mg\))
- **Upwards:** Normal force (N) and vertical component of pulling force (\(F\sin(\theta)\))
- **Horizontal:** Pulling force (\(F\cos(\theta)\)) and frictional force](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0f85ff8-2ea0-4d43-a0c0-84de96cc63b8%2Fa3172c70-a31c-4006-9cbf-a2616273f387%2Frpfjjrj_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Calculating the Coefficient of Friction
**Scenario Description:**
You are pulling your luggage in an airport for an international flight. Understanding the additional charges for excess weight, you decide to weigh your luggage at home and find it to be 21.6 kg. To move the luggage on the carpeted floor at a constant speed, you pull it with a force of 49 N, applied at an angle of 22 degrees above the horizontal plane.
**Problem Statement:**
Calculate the coefficient of friction between the caster wheels of your luggage and the carpet.
**Given Data:**
- Mass of the luggage (m): 21.6 kg
- Magnitude of applied force (F): 49 N
- Angle of applied force from horizontal (\(\theta\)): 22 degrees
- Acceleration due to gravity (g): 9.8 m/s²
**Required:**
Coefficient of friction (μ) between the caster wheels and the carpeted floor.
**Steps to Solve:**
1. **Compute the Normal Force (N):**
The normal force in this context is less than the weight of the luggage because of the vertical component of the applied force.
\[
N = mg - F\sin(\theta)
\]
Where:
- \( m \) is the mass of the luggage.
- \( g \) is the acceleration due to gravity.
- \( F \) is the force exerted.
- \( \theta \) is the angle of the applied force.
2. **Calculate the Frictional Force (f):**
At constant speed, the frictional force equals the horizontal component of the applied force.
\[
f = F\cos(\theta)
\]
Where:
- \( \cos(\theta) \) is the cosine of the angle at which the force is applied.
3. **Determine the Coefficient of Friction (μ):**
\[
\mu = \frac{f}{N}
\]
This provides the coefficient of friction needed to maintain a constant speed.
**Visualization:**
Imagine a free-body diagram with:
- **Downwards:** Weight (\(mg\))
- **Upwards:** Normal force (N) and vertical component of pulling force (\(F\sin(\theta)\))
- **Horizontal:** Pulling force (\(F\cos(\theta)\)) and frictional force
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