Problem 18 - Applied Mathematics

Question

A uniform ladder of weight \(W\) rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is perpendicular to the wall and the ladder is inclined at an angle \(\alpha\) to the vertical. Prove that, if the ladder is on the point of slipping and \(\mu\) is the coefficient of friction between it and the ground, then \(\tan \alpha=2 \mu\).

Step-by-Step Solution

Answer

\(\tan \alpha = 2 \mu\)

Step 1: Analyze the forces acting on the ladder

Identify the forces acting on the ladder. The ladder has its weight (W) acting downwards from its center of gravity. The ladder is in contact with the ground and the wall. Let's denote: \(R_g\) as the reaction force from the ground, \(F_g\) as the friction force at the ground, and \(R_w\) as the reaction force from the wall.

Step 2: Set up the equilibrium conditions

Since the ladder is on the point of slipping, it is in static equilibrium. Therefore, the sum of the forces in both the horizontal and vertical directions must equal zero. Horizontally: \(F_g = R_w\). Vertically: \(R_g = W\).

Step 3: Consider the torques about the base of the ladder

To find the relationship between α and μ, consider the torques about the base of the ladder. The torques must also be in equilibrium: \(R_w \cdot L \cdot \cos \alpha = W \cdot (\frac{L}{2} \cdot \sin \alpha)\). Simplify this equation: \(R_w \cdot \cos \alpha = \frac{W}{2} \cdot \sin \alpha\).

Step 4: Substitute the value of \(R_w\)

From Step 2, we have \(F_g = R_w\). The friction force at the ground is given by \(F_g = \mu R_g = \mu W\). Substituting this in the torque equilibrium equation: \(\mu W \cdot L \cdot \cos \alpha = \frac{W}{2} \cdot L \cdot \sin \alpha\).

Step 5: Simplify the equation

Cancel out \(L\) and \(W\) from both sides of the equation: \(\mu \cdot \cos \alpha = \frac{1}{2} \cdot \sin \alpha\). Rearrange the equation to find \( \tan\alpha\): \(\tan \alpha = 2 \mu\).

Key Concepts

Static Equilibrium

In the context of a ladder problem, static equilibrium means the ladder is not moving at all.
For the ladder to be in static equilibrium, all the forces acting on it must balance each other out.
This includes both the forces in the vertical direction and the horizontal direction.
To put it simply:

• The upward forces should be equal to the downward forces.
• The forces pushing to the left should be equal to the forces pushing to the right.

So, if the ladder is not slipping or tipping over, we say it is in static equilibrium.
This is key for making sure the ladder stays safe and steady.
Understanding static equilibrium helps us figure out how forces interact and balance each other, which is crucial for solving many physics problems.

Torque in Equilibrium

Torque is like a turning force that makes objects rotate around a point. In the ladder problem, torque helps us understand how the ladder stays in place without tipping over.
To achieve torque equilibrium, the torques that cause clockwise rotation must balance out the torques that cause counterclockwise rotation.
We calculate torque as the force multiplied by the distance to the pivot point. For the ladder:

• The torque due to the wall's reaction force causes a clockwise rotation.
• The torque due to the ladder's weight causes a counterclockwise rotation.

If these two torques balance each other, the ladder doesn't tip over, and we have torque in equilibrium.
This balance helps us determine important relationships, like the angle at which the ladder is about to slip.

Coefficient of Friction

The coefficient of friction (often denoted as \( \mu \)) is a measure of how much frictional force exists between two surfaces.
In our ladder problem, we deal with the friction between the ladder and the ground.
The coefficient of friction is important because it tells us how much grip the ground provides to the ladder.
If the ladder is on the verge of slipping, the frictional force is at its maximum value, which equals \( \mu \)\times the normal force.
So, the friction force at the ground (\( F_g \)) can be written as:
\[ F_g = \mu R_g = \mu W \]
Here, \( W \) is the weight of the ladder, and \( R_g \) is the reaction force from the ground.
Understanding this concept helps us calculate the angle at which the ladder starts to slip only by knowing the coefficient of friction.

Force Analysis

Force analysis is about breaking down all the forces acting on an object to understand how they interact.
In the ladder problem, we consider several forces:

• The weight (\( W \)) of the ladder acting downwards at its center of gravity.
• The reaction force from the ground (\( R_g \)) acting upwards.
• The frictional force at the ground (<\( F_g \), acting horizontally from the base of the ladder).
• The reaction force from the wall (<\( R_w \), acting horizontally and pushing the ladder away from the wall).

These forces need to be analyzed both horizontally and vertically.
Horizontally, the frictional force must balance the reaction force from the wall:
\[ F_g = R_w \]
Vertically, the reaction force from the ground must balance the weight of the ladder:
\[ R_g = W \]
By setting up these equations, we can start to solve the problem and understand why the ladder is staying in place, even if it looks like it might slip.
Force analysis gives us a clear picture of all the interactions at play.