Question

A ladder of mass \(37.7 \mathrm{~kg}\) and length \(3.07 \mathrm{~m}\) is leaning against a wall at an angle \(\theta .\) The coefficient of static friction between ladder and floor is \(0.313 ;\) assume that the friction force between ladder and wall is zero. What is the maximum value that \(\theta\) can have before the ladder starts slipping?

Short answer

Answer: The maximum angle at which the ladder can lean against the wall without slipping is approximately \(38.5^\circ\).

Step-by-step solution

Analyze the forces acting on the ladder

There are three forces acting on the ladder: gravitational force (weight of the ladder) acting at the center of mass, normal force of the floor, and normal force of the wall. In addition, there is a static friction force acting at the base of the ladder. Since the friction force between ladder and wall is zero, we don't need to consider it in our analysis.

Set up the equations for moments and forces in equilibrium

At the point of maximum angle, the ladder will start slipping, meaning that the static friction will be at its maximum. The ladder is still in equilibrium, so we can set up the following equations for the moments and forces to maintain equilibrium: - Sum of horizontal forces: \(F_{wall} = 0\) - Sum of vertical forces: \(F_{floor} - W = 0\) where W is the weight of the ladder, acting at the center of mass of the ladder - Sum of moments acting on the ladder: \(F_{friction} \times L - W \times \frac{L}{2} \times \sin \theta = 0\), where \(F_{friction}\) is the static friction force, \(L\) is the length of the ladder and \(\frac{L}{2}\) represents the distance of the weight force from the wall.

Calculate the maximum static friction force

Since we know the coefficient of static friction, \(\mu_s = 0.313\), and the normal force of the floor is equal to the weight of the ladder, we can calculate the maximum static friction force as: \(F_{friction,max} = \mu_s \times F_{floor} = \mu_s \times W\)

Solve for the maximum angle \(\theta\)

Rewrite the moment equation from Step 2 using \(F_{friction,max}\) and substituting \(W\) for \(F_{floor}\): \(\mu_s \times W \times L - W \times \frac{L}{2} \times \sin \theta = 0\) Now, we can solve for the maximum angle \(\theta\): \(\sin \theta = 2 \times \mu_s\) \(\theta_{max} = \sin^{-1} (2 \times \mu_s)\) Plug in the given value of \(\mu_s = 0.313\): \(\theta_{max} = \sin^{-1} (2 \times 0.313)\) \(\theta_{max} \approx 38.5^\circ\) The maximum angle at which the ladder can lean against the wall without slipping is approximately \(38.5^\circ\).

Key concepts

Coefficient of Static Friction

Understanding the coefficient of static friction is pivotal when studying how objects interact with surfaces without motion. It's a dimensionless number that represents the ratio between the maximum static friction force and the normal force acting between two surfaces. Imagine trying to push a heavy box across the floor; the coefficient of static friction describes how much force is needed to start the box moving.

In our ladder example, the coefficient of static friction between the ladder and the floor is critical to determining the ladder's stability. The higher the coefficient, the steeper the angle at which the ladder can remain before slipping. Since the ladder's maximum angle \( \theta \) correlates with the coefficient, we can see that a larger coefficient would allow for a steeper angle. When we solve the problem, we use the coefficient of static friction \( \mu_s = 0.313 \) to calculate the maximum frictional force that prevents the ladder from slipping.

Equilibrium of Forces

The equilibrium of forces refers to the state in which all forces acting upon an object are balanced, resulting in no net force and, consequently, no acceleration. Every object remaining at rest or moving with constant velocity is in force equilibrium. In our scenario with the ladder, there are several forces at play: the weight of the ladder acting downward, the normal force exerted by the floor, and the static friction force resisting movement.

For the ladder to be in equilibrium, the sum of vertical forces must equal zero; hence, the normal force of the floor must balance the ladder's weight. Similarly, as there's no horizontal motion, the static friction force at the base of the ladder must counteract any horizontal components of other forces. This is crucial for calculating the moment equilibrium and solving for the maximum angle \( \theta \) at which the ladder does not slip.

Torque and Moments

The concept of torque, or the moment of force, is about the rotational effect produced by a force. In simpler terms, it's how much a force acting on an object causes that object to rotate. The equation for torque is \( \tau = r \times F \times \sin(\theta) \) where \( r \) is the distance from the pivot point to the point where the force is applied, \( F \) is the force magnitude, and \( \theta \) is the angle between the force direction and the lever arm.

In the case of the leaning ladder, the moments causing rotation are due to the weight of the ladder and the static friction force. We set up a moment equilibrium equation to find the point at which these moments balance each other, indicating that the ladder will not rotate (thus not slip). The ladder's weight creates a moment about the contact point with the wall, and the static friction force resists this moment. By equating the two moments, we are able to solve for the critical angle at which the ladder is on the verge of slipping — a powerful demonstration of the principles of torque and moments in real-life applications.