*Notes to a video lecture on http://www.unizor.com*

__Problems on Equilibrium__

*Problem 1*

Two point-objects of mass

**and**

*M***are hanging at two opposite ends of a weightless rod of length**

*m***.**

*L*At what point on a rod should we fix a thread, so the system of a rod with two weights is hanging on this thread in equilibrium?

Does this state of equilibrium imply that a rod is horizontal?

*Answer*

At distance

**from the end with object of mass**

*m·L /(M+m)***.**

*M**Problem 2*

An object of weight

**is positioned on a flat surface. The coefficient of**

*W**static*friction is

**.**

*μ*What is the minimum pulling force

**to be applied to a**

*P*rope, attached to this object, to start pulling it forward, if the angle

between a rope and a horizontal flat surface is

**?**

*φ**Answer*

**[**

*P = μ·W /***]**

*cos(φ)+μ·sin(φ)**Problem 3*

Consider the following illustration to this problem.

A person of weight

**stands on a platform of weight**

*W***. Pulleys and ropes are arranged as on this illustration.**

*P*What is the force

**a person should apply to a rope to keep the whole system in equilibrium?**

*F**Solution*

Let

**be a reaction of a platform onto a person. It pushes the person upward.**

*N*If

**is a force of a person pulling a rope down, and the**

*F*system is in equilibrium, that is all components are at rest and all

forces are balanced, the rope pulls a person up with the same force

**, according to the Third Newton's Law.**

*F*Since a person is in equilibrium, forces directed upward (

**) should be equal in magnitude to forces directed downwards (weight of the person**

*N+F***):**

*W*

*N+F = W*The platform is also in equilibrium. The forces that push it down are its weight

**and reaction**

*P***of a person standing on it. The forces pulling the platform up are tensions of two ropes**

*N***on the left and**

*T***on the right:**

*F*

*P+N = T+F*Finally, a small pulley is in equilibrium. It's pulled down by two tensions

**, from the left and from the right. The tension**

*F***pulls it up:**

*T*

*T = 2F*We have three equations with three unknowns:

**,**

*T***and**

*N***, which we have to solve for an unknown force**

*F***, with which a person pulls a rope to balance the system:**

*F*Substituting

**from the last equation into the second one, getting a system of two equations:**

*T*

*N+F = W*

*P+N = 3F*Solving the first equation for

**and substituting into the second:**

*N*

*P+W−F = 3F*Resolving for

**:**

*F*

*F = (P+W)/4**Answer*

*F = (P+W)/4**Problem 4*

A ladder is standing at some angle to the floor. It's in equilibrium,

that is it's not slipping down to the floor. What holds it is the

*static*friction between it and the floor.

If the ladder stands almost vertically, the

*static*friction is

greater and it easily holds the ladder in the position. As we move the

bottom of a ladder away from the vertical, there will be a point when

*static*friction will no longer could hold the ladder in place and the ladder would slip down to the floor.

What is the smallest angle the ladder would hold if the coefficient of

*static*friction is

**?**

*μ**Solution*

Consider the following illustration to this problem.

While the reaction of the wall

**is smaller than the**

*N*_{1}*static*friction

**, the ladder would hold its position.**

*F*The smallest angle between the ladder and the floor will be when these two forces are equal.

To be in equilibrium, vector sum of all forces must be equal to

null-vector and a sum of all angular momentums also must be equal to

null-vector.

Along the horizontal X-axis the sum of all forces equals to zero if the horizontal reaction of the wall

**equals to the force of**

*N*_{1}*static*friction

**, which, in turn, equals to a product of the vertical reaction of the floor**

*F***by a coefficient of**

*N*_{2}*static*friction

**:**

*μ*(a)

*N*_{1}= μ·N_{2}Along the vertical Y-axis the sum of forces equals to zero if the reaction of the floor

**equals to weight of a ladder**

*N*_{2}**:**

*W*(b)

*N*_{2}= WTo analyze angular momentums, let's choose an axis of rotation. It's

easier to choose an axis perpendicular to both X- and Y-axis that goes

through a point where a ladder touches the floor. In this case angular

momentums of forces

**and**

*N*_{2}**are zero because their corresponding radiuses are zero.**

*F*Let's assume that the length of the ladder is

**and the angle between the ladder and the floor is**

*D***. The magnitude of the angular momentum of the force of weight**

*φ***will then be**

*W***. The magnitude of the angular momentum of the reaction of the wall**

*W·D·cos(φ)/2***will be**

*N*_{1}**. The condition of the sum of angular momentums to be equal to zero results in the equation:**

*N*_{1}·D·sin(φ)

*N*_{1}·D·sin(φ) = W·D·cos(φ)/2The length of the ladder can be canceled:

(c)

*N*_{1}·sin(φ) = W·cos(φ)/2Equations (a), (b) and (c) form a system of three equations with three unknowns:

(a)

*N*_{1}= μ·N_{2}(b)

*N*_{2}= W(c)

*N*_{1}·sin(φ) = W·cos(φ)/2From this

*tan(φ) = 1/(2μ)**Answer*

**[**

*φ = arctan***]**

*1/(2μ)*Interestingly, the answer does not depend on the weight or the length of the ladder.

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