In this post, we will solve a numerical problem based on the *equilibrium of a leaning ladder*.

To do this you need to first understand the conditions of equilibrium. If you require some revision then you can quickly go through our physics **tutorial on the equilibrium conditions** first, before attempting this numerical.

Anyways, let’s begin solving the numerical.

## Leaning ladder equilibrium numerical

**Problem statement:**

A uniform ladder of length l rests against a smooth, vertical wall (Fig. 1).

The mass of the ladder is *m*, and the coefficient of static friction between the ladder and the ground is μ_{s}=0.40.

Find the minimum angle θ_{min} at which the ladder does not slip.

How to solve this kind of problem:

In this post, we will understand this problem first and then take a step-by-step approach to solve the numerical.

ConceptualizeThink about any ladders you have climbed. Remember that a large friction force between the bottom of the ladder and the surface is required to keep the leaning ladder in a balanced condition.If the friction force is zero, then the ladder won’t be able to stay up. You may simulate a ladder with a ruler leaning against a vertical surface. You will find that the ruler slips at some angles and stay up at others.

CategorizeWe do not wish the ladder to slip, so we model it as a rigid object in equilibrium.

AnalyzeThefree-body diagram(FBD) showingall the external forcesacting on the ladder is illustrated inFigure 2.

The force exerted by the ground on the ladder is the vector sum of a normal force **n** and the force of static friction **f**_{s}.

The force P exerted by the wall on the ladder is horizontal because the wall is frictionless.

**Apply the first condition for equilibrium to the ladder:**

Σ*F*_{x}= *f**s **– **P **=* 0 ……… (1)

Σ*F*_{y}= *n **– **mg **=* 0 ….. (2)

Solve Equation (1) for *P*:*P **=* *f*_{s}* ………**(3)*_{}

*Solve Equation (2) for n*:*n = mg ……..(4)*

When the ladder is on the verge of slipping, the force of static friction must have its maximum value, which is given by *f*_{s,max} = *μ*_{s }*n.*

Combine this equation with Equations (3) and (4):

*P =* *f*_{s}*= μ*_{s }*n = μ*_{s }*mg* …………. (5)

*Apply the second condition for equilibrium to the ladder, *taking torques about an axis through *O*:

Σ τ = P l sin θ_{min} – mg (l/2) cos θ_{min} = 0

Solve for tan θ_{min} :

sin θ_{min} / cos θ_{min} =_{ }tan θ_{min} = mg/ (2P) = mg/(2 *μ*_{s }_{}*mg) = 1/(**2 **μ*_{s}*) = 1/(2 x 0.4) = 1.25*

*hence, **θ*_{min}*= tan*^{-1}*(1.25) = 51 **°*

* Observation *Notice that the angle depends only on the coefficient of friction, not on the mass or length of the ladder.