# Chapter 14 - Static equilibrium, Elasticity and Fracture

Equilibrium describes a state where both the linear acceleration and angular acceleration of an object or system are zero. This means the object is either at rest or moving with constant velocity.

Here we will focus on situations where the system is a rest, so called static equilibrium, as opposed to dynamic equilibrium, where the system has constant velocity.

## Conditions for Equilibrium

For an object to have no translational acceleration

$\Sigma \vec{F}=0$

and for this to be true the sum of each component of the forces must be zero.

An object may have no net force on it, but be able to turn, which means for an object to be in equilibrium

$\Sigma \tau = 0$

Where the torques can be calculated around any axis. Some axes are more sensible than others, a good idea is to choose an axis at place where a force is acting (as the torque due to that force around that axis will be zero.)

## 2 Balls on a rod

In the above example we calculate the torques around the pivot point. This is a sensible choice, but we could have also chosen, for example, an axis that goes through the second mass $m_{2}$.

In this case

$\Sigma \tau=m_{1}g4L-({m_{1}+m_{2}})g3L=0$

$m_{1}gL=m_{2}g3L$

$\large m_{2}=\frac{m_{1}}{3}$

## Crane boom with arm at 90 degrees

There are four forces which act on the arm. Forces do not need to be directed along the arm as it is a rigid object. On the other hand, the force due to tension must be directed along the rope (otherwise it would change it's shape). To take the weight of the arm in to account we can treat the weight of the rod as if it acts through the center of mass of the object, which we will approximate as being halfway along the arm.

Sum of forces on arm

$F_{hx}=F_{Tx}$

$mg+m_{a}g=F_{Ty}+F_{hy}$

Sum of torques on arm around hinge

$F_{Ty}l=F_{T}l\sin\theta=mgl+m_{a}g\frac{l}{2}$ → $F_{T}\sin\theta=mg+\frac{1}{2}m_{a}g$

$F_{T}=\frac{mg+\frac{1}{2}m_{a}g}{\sin\theta}$

Compressive force on arm is $F_{hx}=F_{Tx}=(mg+\frac{1}{2}m_{a}g)\frac{{\cos\theta}}{\sin\theta}=\frac{mg+\frac{1}{2}m_{a}g}{\tan\theta}$

## Crane boom with rope at 90 degrees

Forces on arm

$F_{hy}=mg+m_{a}g$

$F_{hx}=F_{Tx}$

Torques on arm

$F_{T}l\sin\phi=(mgl+m_{a}g\frac{l}{2})\sin(90^{o}-\phi)=(mg+\frac{1}{2}m_{a}{g})l\cos\phi$

$F_{T}=F_{Tx}=F_{hx}=\frac{mg+\frac{1}{2}m_{a}{g}}{\tan\phi}$

Compressive force on spring

$\frac{F_{hx}}{\cos\phi}=\frac{mg+\frac{1}{2}m_{a}{g}}{\sin\phi}$

## Crane boom with arm and rope at angle

Forces on arm

$F_{hy}+F_{Ty}=F_{hy}+F_{T}\sin\theta=mg+m_{a}g$

$F_{hx}=F_{Tx}=F_{T}\cos\theta$

Torques on arm

$F_{T}l\sin(\phi+\theta)=(mgl+m_{a}g\frac{l}{2})\sin(90^{o}-\phi)=(mg+\frac{1}{2}m_{a}{g})l\cos\phi$

$F_{T}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\phi}{\sin(\phi+\theta)}$

$F_{Tx}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\phi\cos\theta}{\sin(\phi+\theta)}=F_{hx}$

Compressive force on spring

$\frac{F_{hx}}{\cos\phi}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\theta}{\sin(\phi+\theta)}$

## Hanging sign problem

After numerous unfortunate incidents involving polar bears and tourists who have seen too many CokeTM ads, Santa has decided to install a new sign at the North Pole. To do this he uses a uniform metal rod of length 0.8 m and a cable which he attaches 20 cm from the end of the metal rod. The new sign has mass 3 kg and hangs from the end of the rod, while the rod itself has mass 1 kg. The cable is attached the top of the pole and the rod is held by a hinge 30 cm down from the top of the pole.

What angle θ does the cable make with the rod?

What is the tension in the cable?

What are the horizontal and vertical components of the force exerted by the hinge on the rod? Take the positive directions to be up and to the right.

## Hanging sign solution

$\theta=\tan^{-1}\frac{30}{60}=26.57^{o}$

Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$

Horizontal forces, right is positive $-T\cos\theta+Fh_{x}=0$

Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$

$T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$

$Fh_{y}=4.981-102.24\sin26.57^{o}=-6.53\mathrm{N}$

So force points down.

$Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$

Force points to the right

## Will the ladder slip?

The forces that act on the ladder of length $l$ are the weight of the ladder that acts down at it's center of mass, $m\vec{g}$, a normal force exerted by the wall, $\vec{f}_{NW}$, an normal force exerted by the ground $\vec{f}_{NG}$, and a force due to the friction of the ground, $\vec{F}_{Fr}$. The friction between the ladder and the wall is typically neglected.

For equilibrium the sum of the forces in both the horizontal and vertical directions must be zero and also the sum of the torques around an appropriate axis must be equal to zero.

Balance of the forces in the horizontal direction implies that

$F_{NW}=F_{Fr}$

and in the vertical direction

$F_{NG}=m_{l}g$

If we calculate the torques around the point at which the ladder touches the ground then they will be

$F_{Nw}l\sin\theta-m_{l}g\frac{l}{2} \sin(90^{o}-\theta)=0$

$F_{Nw}l\sin\theta-m_{l}g\frac{l}{2}\cos\theta=0$

This can be written in terms of the frictional force by using the equation we got from the horizontal forces.

$F_{fr}l\sin\theta-m_{l}g\frac{l}{2}\cos\theta=0$

The maximum force that can be provided by friction is $\mu F_{NG}$ and so the maximum safe angle is when this frictional force is provided

$\mu F_{NG}l\sin\theta_{max}-m_{l}g\frac{l}{2}\cos\theta_{max}=0$

$\mu mgl\sin\theta_{max}-m_{l}g\frac{l}{2}\cos\theta_{max}=0$

$\tan\theta_{max}=\frac{1}{2\mu}$

Of course if a person is standing on the ladder, their weight would need to be taken in to account both in the balance of the vertical forces and the torques.

## Stable and Unstable equilibrium

An object in static equilibrium will not move from it's equilibrium position unless it is disturbed by an additional force. The question of stability concerns what happens when equilibrium is slightly disturbed. If the equilibrium is stable the system will return to it's equilibrium position, but if it instead moves away from the equilibrium then we say the equilibrium is unstable. If the small perturbation moves the system to a new equilibrium we can say the equilibrium is neutral.

We can demonstrate the different kinds of equilibrium with the two balls on a stick we looked at earlier.

## Circus Trick

This circus trick looks pretty difficult.

## Conditions for stable equilibrium

But actually it isn't!

An object whose center of gravity is below it's point of support will be in stable equilibrium so riding a unicycle on a high wire is easy if there is weight suspended from it like the one at COSI.

This is also the reason that tightrope artists will sometimes hold a heavy pole that bends down at the ends, to lower their center of gravity and increase their stability.

## Corbel arch

An object can be in stable equilibrium if a vertical line from the center of gravity lies within it's center of support.

We will demonstrate this concept by building a corbel arch.

## True Arch

The weakness of a corbel arch as compared to a true arch which should contain a catenary curve, is that not all the tensile stresses are converted in to compressive stresses. What does this mean and why it is important? We need to know a little about the properties of materials.

## Stress

When an object is exposed to a external force, it can deform. The relative size of this deformation, is called a strain $\varepsilon=\frac{\Delta l}{l_{0}}$. Often, when dealing with material properties it can be more useful to consider the force as a force per unit area, or stress $\sigma=\frac{F}{A}$

These videos of bubble rafts from DoITPoMs at Cambridge are useful for visualizing the way materials respond to the various kinds of stress.

## Elastic Deformation

Many materials exhibit elastic behavior, where they will recover from a strain to their original shape.

This can be divided in to two regimes, the linear elastic regime, below the proportional limit, where objects obey Hooke's Law

$F=k\Delta l$

and their strain is linearly proportional to the stress. As the stress is increased this is followed by a non-linear elastic regime where the stress and strain are not linearly related.

The upper limit of this non-linear elastic regime is the elastic limit. Up to this limit the object will revert to it's original shape when the stress is removed, beyond it the object will deform in a non-reversible fashion, until it eventually breaks.

The material specific plot of stress vs strain is called a stress strain curve.

## Elastic Moduli

There are a set of elastic moduli that link the deformation of an object to particular types of stress.

Young's modulus, $E$, is the constant that relates stress along a direction to strain in that direction in the elastic regime.

$\sigma=E\varepsilon$

$\frac{F}{A}=E \frac{\Delta l}{l_{0}}$

This formula can be applied to either tensile stress or compressive stress.

For shear stresses a different elastic modulus, the shear modulus, $G$ applies. $G$ is usually smaller than $E$.

$\sigma=G\varepsilon$

$\frac{F}{A}=G \frac{\Delta l}{l_{0}}$

Note: for this equation $\Delta l$ is perpendicular to $l_{0}$!

An object subject to uniform pressure change will have a volume change

$\frac{\Delta V}{V_{0}}=-\frac{1}{B}\Delta P$

$B$ is the bulk modulus.

## Strength of materials

Different materials, depending on their micro-structure have different strengths with relation to the different strains. By strength we refer to the maximum stress that can be withstood without the material breaking.

Ultimate tensile strength for ductile materials is usually determined by necking, brittle materials, like concrete typically have very low tensile strength, but have high compressive strength. Structures under compressive stress can collapse well before the material itself fails due to buckling.

Shear failure can be greatly assisted by imperfections that allow dislocation glide.

## Truss bridge

A truss bridge uses connected elements to spread the load out, some elements are under compression, others are under tension. In the diagram below we ignore the mass of the elements themseleves.

Frequently different materials are used for compression and tension elements. Long truss bridges get very heavy, a better design for these applications is a suspension bridge.

## Suspension bridge

The simplest kind of suspension bridge is one that hangs under it's own weight in the shape of the catenary, the equation for which is $y=a \cosh (\frac{x}{a})$. An inverted catenary is the ideal shape for an arch which supports only it's own weight.

A more normal suspension bridge, and also more typical arches, support weight besides their own and the ideal shape in these situations is closer to that of a parabola, in the case where the mass of the cables can be neglected the shape is entirely parabolic.

## The best bridge

The Sydney Harbour Bridge is a steel through arch bridge. As you can see it incorporates elements of both suspension and truss bridges. The deck itself is suspended from a trussed double arch.

phy131studiof15/lectures/chapter14.txt · Last modified: 2015/10/19 08:33 by mdawber