Question
Below, a 10 N block is sliding down a ramp with friction. The frictional force is 2 N. Find the net force of the block as it slides down the ramp. Why does only the gravitational vector need to be decomposed?
Answers
GPT-4o mini
To find the net force acting on the block as it slides down the ramp, we need to take into account both the gravitational force acting down the ramp and the frictional force opposing the motion.
1. **Gravitational Force (downward)**: The weight of the block is \( W = 10 \, \text{N} \). This force can be decomposed into two components:
- One component acting parallel to the ramp (down the ramp).
- One component acting perpendicular to the ramp (normal force).
Since you mentioned we need to decompose the gravitational force, we focus on the component parallel to the ramp. However, if the angle of the ramp is given, the component could be calculated using \( W \sin(\theta) \), where \( \theta \) is the angle of the ramp.
2. **Frictional Force (opposing the motion)**: The problem states that the frictional force is \( 2 \, \text{N} \).
### Calculation of the Net Force
The net force \( F_{\text{net}} \) acting on the block as it slides down the ramp can be calculated using the formula:
\[
F_{\text{net}} = F_{\text{gravity, parallel}} - F_{\text{friction}}
\]
Assuming \( F_{\text{gravity, parallel}} \) is greater than \( F_{\text{friction}} \), we can express this as:
- Let’s denote \( F_{\text{gravity, parallel}} \) as \( F_g \).
Assuming \( F_g \) is the force of gravity acting parallel to the ramp, the equation becomes:
\[
F_{\text{net}} = F_g - 2 \, \text{N}
\]
However, since we need the actual value of \( F_g \) to find \( F_{\text{net}} \), if we do not have the angle, let's consider the net scenario.
But since \( F_g \) is not mentioned, let's assume it is greater than \( 2 \, \text{N}\) (considering common scenarios):
If \( F_g \) (the parallel component) is, for example, \( 8 \, \text{N} \) (to satisfy \( F_g > F_{\text{friction}} \)), then:
\[
F_{\text{net}} = 8 \, \text{N} - 2 \, \text{N} = 6 \, \text{N}
\]
### Why does only the gravitational vector need to be decomposed?
Friction is always opposing the motion along the surface, thus it acts in a fixed direction that does not depend on angles - it simply acts opposite to the block's intended motion. The gravitational force, however, must be decomposed because it acts vertically downwards, and when an object is on an inclined plane, the forces acting on it need to be analyzed in terms of their components relative to the incline.
That is why we only have to decompose the gravitational force: to find out how much of that force is contributing to the block's acceleration down the ramp (the parallel component) and how much is being counteracted by friction (and the normal force on the ramp, if needed).
In summary:
- **Net Force**: \( F_{\text{net}} = F_g - 2 \, \text{N} \) (with \( F_g \) being the gravitational force component along the ramp),
- **Decompose only gravitational vector**: Because it acts vertically and must be broken down into components relevant to the ramp's angle of elevation.
1. **Gravitational Force (downward)**: The weight of the block is \( W = 10 \, \text{N} \). This force can be decomposed into two components:
- One component acting parallel to the ramp (down the ramp).
- One component acting perpendicular to the ramp (normal force).
Since you mentioned we need to decompose the gravitational force, we focus on the component parallel to the ramp. However, if the angle of the ramp is given, the component could be calculated using \( W \sin(\theta) \), where \( \theta \) is the angle of the ramp.
2. **Frictional Force (opposing the motion)**: The problem states that the frictional force is \( 2 \, \text{N} \).
### Calculation of the Net Force
The net force \( F_{\text{net}} \) acting on the block as it slides down the ramp can be calculated using the formula:
\[
F_{\text{net}} = F_{\text{gravity, parallel}} - F_{\text{friction}}
\]
Assuming \( F_{\text{gravity, parallel}} \) is greater than \( F_{\text{friction}} \), we can express this as:
- Let’s denote \( F_{\text{gravity, parallel}} \) as \( F_g \).
Assuming \( F_g \) is the force of gravity acting parallel to the ramp, the equation becomes:
\[
F_{\text{net}} = F_g - 2 \, \text{N}
\]
However, since we need the actual value of \( F_g \) to find \( F_{\text{net}} \), if we do not have the angle, let's consider the net scenario.
But since \( F_g \) is not mentioned, let's assume it is greater than \( 2 \, \text{N}\) (considering common scenarios):
If \( F_g \) (the parallel component) is, for example, \( 8 \, \text{N} \) (to satisfy \( F_g > F_{\text{friction}} \)), then:
\[
F_{\text{net}} = 8 \, \text{N} - 2 \, \text{N} = 6 \, \text{N}
\]
### Why does only the gravitational vector need to be decomposed?
Friction is always opposing the motion along the surface, thus it acts in a fixed direction that does not depend on angles - it simply acts opposite to the block's intended motion. The gravitational force, however, must be decomposed because it acts vertically downwards, and when an object is on an inclined plane, the forces acting on it need to be analyzed in terms of their components relative to the incline.
That is why we only have to decompose the gravitational force: to find out how much of that force is contributing to the block's acceleration down the ramp (the parallel component) and how much is being counteracted by friction (and the normal force on the ramp, if needed).
In summary:
- **Net Force**: \( F_{\text{net}} = F_g - 2 \, \text{N} \) (with \( F_g \) being the gravitational force component along the ramp),
- **Decompose only gravitational vector**: Because it acts vertically and must be broken down into components relevant to the ramp's angle of elevation.