Problem 16 - Applied Mathematics

Question

A ring is free to slide down a rough straight wire. Find the acceleration of the ring. (a) The coefficient of friction between the wire and the ring is \(\mu\). (b) The wire is inclined at an angle \(\theta\) to the horizontal. (c) The mass of the ring is \(m\).

Step-by-Step Solution

Answer

The acceleration of the ring is \ a = g ( \sin(\theta) - \mu \cos(\theta) ) \.

Step 1: Identify forces acting on the ring

List all the forces acting on the ring. These include the gravitational force, the normal force from the wire, and the frictional force opposing the motion.

Step 2: Resolve the gravitational force

Resolve the gravitational force into two components: one parallel to the wire \( mg \sin(\theta) \) and one perpendicular to the wire \( mg \cos(\theta) \).

Step 3: Determine the frictional force

Calculate the frictional force using the normal force and the coefficient of friction. The frictional force is \( f = \mu \times N \), where \( N \) is the normal force. Here, \( N = mg \cos(\theta) \), so \( f = \mu mg \cos(\theta) \).

Step 4: Apply Newton's second law

Use Newton's second law in the direction of the ring's motion along the wire. The net force acting on the ring down the wire is \( mg \sin(\theta) \) minus the frictional force \( \mu mg \cos(\theta) \). So, the net force is \[ F_{net} = mg \sin(\theta) - \mu mg \cos(\theta) \. \]

Step 5: Calculate the acceleration

To find the acceleration of the ring, use the relation \( F_{net} = ma \. \) Therefore, \[ ma = mg \sin(\theta) - \mu mg \cos(\theta) \. \] Simplify to find \ a \: \[ a = g ( \sin(\theta) - \mu \cos(\theta) ) \]

Key Concepts

Newton's second law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed as \( F = ma \). In the given problem, the ring sliding down the wire is influenced by several forces, which we need to consider to determine its acceleration.
These forces include:
• The gravitational force acting vertically downward
• The normal force from the wire acting perpendicular to the surface
• The frictional force opposing the motion along the wire.
By resolving these forces along the direction of the ring's motion, we can use Newton's Second Law to find the ring's acceleration.

friction

Friction is the force that opposes the relative motion of two surfaces in contact. It plays a crucial role in this problem as it resists the ring’s motion down the inclined wire. The coefficient of friction, denoted by \( \mu \), quantifies the amount of friction between the wire and the ring.
To calculate the frictional force, we use the formula:
\[ f = \mu \times N \]
where \( N \) is the normal force. For our inclined plane, the normal force \( N \) is given by \( mg \cos(\theta) \). Once we have this, we can find the frictional force as \( f = \mu mg \cos(\theta) \). This force will oppose the gravitational component driving the ring down the wire.

gravitational force

Gravitational force is the weight of the ring acting downward due to the Earth's gravity. Its magnitude is given by \( mg \), where \( m \) is the mass of the ring and \( g \) is the acceleration due to gravity.
In this problem, when the wire is inclined at an angle \( \theta \) to the horizontal, the gravitational force can be decomposed into two components:
• Parallel to the wire: \( mg \sin(\theta) \)
• Perpendicular to the wire: \( mg \cos(\theta) \)
The parallel component \( mg \sin(\theta) \) tries to slide the ring down the wire, while the perpendicular component \( mg \cos(\theta) \) contributes to the normal force, which in turn affects the friction.

acceleration

Acceleration is the rate of change of velocity of an object. According to Newton's Second Law, it can be calculated when we know the net force acting on an object and its mass. In our problem, the net force along the direction of the ring's motion is the difference between the gravitational component parallel to the wire and the frictional force.
Using the formula:
\[ F_{net} = mg \sin(\theta) - \mu mg \cos(\theta) \]
and applying Newton’s Second Law \( F_{net} = ma \), we get:
\[ a = g ( \sin(\theta) - \mu \cos(\theta) ) \]
This equation gives the acceleration of the ring as it slides down the inclined wire, incorporating the effects of both gravity and friction.