Part A Review I Constants A 5000 kg truck is parked on a 7.0° slope. How big is the frictio truck? Express your answer with the appropriate units.
Part A Review I Constants A 5000 kg truck is parked on a 7.0° slope. How big is the frictio truck? Express your answer with the appropriate units.
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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![### Physics Problem Explanation
**Problem Statement:**
A 5000 kg truck is parked on a 7.0° slope. How big is the friction force on the truck? Express your answer with the appropriate units.
**Solution Steps:**
To solve this problem, we need to determine the frictional force acting on the truck that is preventing it from sliding down the slope. This is a classic static friction problem involving forces on an inclined plane. Here are the steps you would follow:
1. **Identify the Forces Acting on the Truck:**
- The gravitational force (weight) acting vertically downward: \( F_g = mg \)
- Where \( m \) is the mass of the truck (5000 kg)
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- The normal force perpendicular to the slope.
- The frictional force parallel to the slope, preventing the truck from moving.
2. **Calculate the Gravitational Force:**
\[
F_g = m \cdot g = 5000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 49050 \, \text{N}
\]
3. **Determine the Components of the Gravitational Force:**
- Parallel to the slope: \( F_{g, \parallel} = F_g \sin(\theta) \)
- Perpendicular to the slope: \( F_{g, \perp} = F_g \cos(\theta) \)
4. **Calculate the Parallel Component:**
\[
F_{g, \parallel} = 49050 \, \text{N} \times \sin(7.0^\circ)
\]
Using \( \sin(7.0^\circ) \approx 0.1219 \)
\[
F_{g, \parallel} = 49050 \, \text{N} \times 0.1219 \approx 5977 \, \text{N}
\]
5. **Static Frictional Force:**
- For a vehicle parked on a slope, the static frictional force must be equal and opposite to the component of the gravitational force parallel to the slope to prevent movement.
So, the magnitude of the friction force \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf86c848-962e-42b8-b4eb-ec16b886dd8d%2F00d3bb78-1b53-4141-8a3a-c3be2d02dfe7%2Fse98y1r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem Explanation
**Problem Statement:**
A 5000 kg truck is parked on a 7.0° slope. How big is the friction force on the truck? Express your answer with the appropriate units.
**Solution Steps:**
To solve this problem, we need to determine the frictional force acting on the truck that is preventing it from sliding down the slope. This is a classic static friction problem involving forces on an inclined plane. Here are the steps you would follow:
1. **Identify the Forces Acting on the Truck:**
- The gravitational force (weight) acting vertically downward: \( F_g = mg \)
- Where \( m \) is the mass of the truck (5000 kg)
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- The normal force perpendicular to the slope.
- The frictional force parallel to the slope, preventing the truck from moving.
2. **Calculate the Gravitational Force:**
\[
F_g = m \cdot g = 5000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 49050 \, \text{N}
\]
3. **Determine the Components of the Gravitational Force:**
- Parallel to the slope: \( F_{g, \parallel} = F_g \sin(\theta) \)
- Perpendicular to the slope: \( F_{g, \perp} = F_g \cos(\theta) \)
4. **Calculate the Parallel Component:**
\[
F_{g, \parallel} = 49050 \, \text{N} \times \sin(7.0^\circ)
\]
Using \( \sin(7.0^\circ) \approx 0.1219 \)
\[
F_{g, \parallel} = 49050 \, \text{N} \times 0.1219 \approx 5977 \, \text{N}
\]
5. **Static Frictional Force:**
- For a vehicle parked on a slope, the static frictional force must be equal and opposite to the component of the gravitational force parallel to the slope to prevent movement.
So, the magnitude of the friction force \(
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