Introduction to Statics_ Equilibrium

8/3/2019 Introduction to Statics_ Equilibrium

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EN3: Introduction to Engineering and Statics

Division of Engineering

Brown University

7. Equilibrium

Engineers need to manage forces in structures and machines. There most common problems that engineers face

are:

1. To design a system that will apply a set of forces to an object. For example, many manufacturing

processes (milling, cutting, injection molding, etc) need to apply large forces to the material being

processed; and engineers have to design machines capable of applying these forces. Vehicle design is a

second example. An aircraft in flight must be designed carefully to balance gravitational and aerodynamic

forces, to ensure that the aircraft remains controllable. Similarly, a cars engine and transmission must be

designed to exert whatever forces are necessary to overcome air resistance, rolling resistance, etc while

the car travels.

2. To design a structure or machine that can support a given set of external forces for an extended period of

time without failure. Examples range from structural problems, to the design of artificial joints to design of

micromachines;

To manage forces, engineers need to be able to answer two questions: (1) What forces are required to make a

structure move or deform in some prescribed way?; and (2) What internal forces are induced in a component byexternal loading? A structure or machine fails when these internal forces become larger than the materials

strength.

Surprisingly, forces in most engineering systems can be understood using the principle ofstatic equilibrium. There

are really only three main situations where statics doesnt work, (because large accelerations occur)

1. Anything involving impact (crashing vehicles, exploding bombs, etc)

2. Anything that travels along a curved path (an airplane in a turn, cornering vehicles)

3. Systems that contain rapidly rotating parts;

4. Forces induced by vibrations.

A static analysis always solves the same problem: given a set of known forces acting on a structure or machine,

calculate a set of unknown forces. The unknown forces could be internal forces, or some subset of unknown

external forces.

The unknown forces are always calculated by solving equilibrium equations for the system.

7.1 Definition


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Four and a half simple concepts are required to understand the concept of static equilibrium

1. If a structure is stationary it is in static equilibrium

2. If a machine moves at constant speed along a straight line without rotation it is in static

equilibrium

3. The resultant force acting on a structure or machine that is in static equilibrium is zero

4. The resultant moment acting on a structure or machine that is in static equilibrium is zero.

4.5 It doesnt matter what point you choose to take moments about in a statics problem its zero

about any point. But beware! When you start solving dynamics problems, you will need to take

moments about special points to apply equations of motion (usually the center of mass).

Thats it! Theres nothing more to statics. We can all go home now

But wait! How about a few examples and applications before you go? We have to dosomethingto earn yourtuition dollars

7.2 Examples of the use of static equilibrium to calculate unknown forces

When solving statics problems, well always follow the steps below:

1. Draw a clear picture showing the forces and moments acting on the object(s) of interest. Its important to

show thepositions of the forces correctly;

2. Introduce an appropriate basis to be used for all vector calculations

3. Write down the forces acting on the system (introduce variables to describe unknown forces)4. Write down the position of the forces

5. Find the moment of the all forces about any convenient point (you must use the same point for each force).

6. Write down any pure moments or torques acting on the system

7. Find the resultant force F

8. Find the resultant moment M

9. Set F=0 and M=0.

10. If you can, then solve the equations for unknown quantities of interest (often forces, but the unknowns

could be other things too, as we shall see).

11. If you have too many unknown forces and not enough equations at this point, you need to look for more

equations. These may come from (a) Force balance for other components; (b) Force laws (eg spring law,

buoyancy force law, gravity law, etc); (c) geometry.

12. When you have enough equations, solve them.

So lets try some problems and see how this works.

Example 1: A beam with weight Wis suspended by two spring-scales, as shown in the picture below.

(i) Find an expression for the force reading on each spring scale in terms ofW,L and d.


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(ii) If the spring scales both have stiffness kandun-stretched length a, find an expression for the difference in

length b of the scales, in terms ofk, W,L and d.

This is a standard equilibrium problem. We know the weight force acting on the beam. We dont know the forces

exerted on the beam by the springs. However, we do know that the beam isnt moving and therefore must be in

static equilibrium. We can try to calculate the unknown spring forces by writing down equilibrium equations for the

beam.

To solve the problem, we follow the steps in our recipe.

(1, 2) Heres the picture showing the forces; as well as a basis for our vectors. We know the weight acts at the

center of mass, and can look up the position of the center of mass in the table provided in section 2.1. We also

know the spring scales will pull on the beam along the line of the scale, and show the forces accordingly.

Now we proceed to steps 3-9. Its convenient to assemble this information in aforce and moment balance table

as shown below.

Force and moment balance table for beam ABC. Origin at A.


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Force/Moment Position Force Moment about originWeight

Force at A 0 0

Force at B

Sum (=0)

Note that you can use any of our short-cuts to compute the moments, if you wish. For a statics problem, it doesnt

matter what point you choose for the origin. But beware when you move on to dynamics, you will need to writedown dynamical equations of motion by taking moments about special points

Finally, step 10 - we collect and (try to) solve the governing equations. We get the equations by setting the i,j and

kcomponents of resultant force and moment to zero individually. For the problem at hand, we have only two

equations

We have two equations and two unknown forces, so we are ready to solve them. If, at this point, we had more

unknowns than equations, wed have to look for other principles to apply in order to provide additional equations

for the unknowns.

The equations are easily solved for the unknown forces and

This answers part (i) of our problem.

To answer part (ii), we have to look for additional principles to apply we cant get any more information from

statics. Elementary geometry tells us that b is the difference in length of the two spring scales. Next, we recall that

the length of each spring scale can be computed from the force using the force-extension relation for a spring as

Therefore, the difference in length is given by

giving the solution to part (ii).

If youve taken courses in physics in high school, this procedure may look cumbersome and unnecessary to you.

Bear with me. Its true that simple problems like this one can be solved in your head. But the objective of this

discussion is to develop a systematic approach that can be used to solve 3D problems involving complex

engineering systems, containing tens or even hundreds of unknown forces. You can get completely lost if you try to

solve these using an ad-hoc procedure.


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The force balance table for the wheel is shown below.

Force and moment balance table for wheel. Origin at force transducer.

Force/Moment Position Force Moment about origin

Contact force

Force transducer 0

Sum (=0)

This gives us 6 equations for the 6 unknown contact force and moment components

We could solve these by hand But were lazy. Instead, well use MAPLE

So, the answer we are looking for is


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Example 3:

The figure below shows a proposed design for a torque-wrench. The system is intended to measure the moment M

that is exerted by the wrench on a bolt. This will be accomplished by mounting three small force transducers at the

locations shown. The user will place the wrench over a bolt, and apply a force to the handle. As a result, the bolt

will exert forces , and a moment on the wrench at D. The transducers will record the forces acting on

member (2) at points A, B, and C . The transducers are designed to behave like springs, exerting only one

component of force parallel to the transducer. The force readings must be used to calculate .

The force transducer readings will be converted to a torque reading using a microprocessor built into the wrench.

We have been assigned the chore of working out the formulas that the microprocessor must use to deduce the

torque.

Once again, this is a standard equilibrium problem. For a known set of forces and moment acting at D, we want to

calculate the force transducer readings. We can relate the force transducer readings to the forces and moments at

D using the condition that component ABCD is in static equilibrium.

The picture below shows the forces acting on component ABCD of the wrench. The forces acting at D are the

forces and moment exerted by the head of the bolt on the wrench they are therefore equal and opposite to the

forces and moment exerted by the wrench on the bolt.


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The transducer forces can be related to the forces acting at D using force and moment equilibrium. The force and

moment balance table is shown below

Force and moment balance table for wrench component. Origin at D.

Force/Moment Position Force Moment about originForce and moment

exerted by bolt

0

Transducer A

Transducer B

Transducer C 0

Sum (=0)

The equilibrium equations are

We have three equations, and three unknowns ( ). We can go ahead and solve them. MAPLE provides

the formulas we are looking for.


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So the formulas are

A torque wrench usually only records the moment and does not measure the forces. Our calculations showthat we could compute the moment using only two force transducers at A and B. We could cut cost by replacing

the third force transducer with a small stiff spring.

Example 4:

The figure illustrates a proposed design for a floating oil platform. The platform has total weight W, and has center

of gravity at the location shown. The platform floats in seawater with mass density . The platform floats on three

cylindrical legs, with radius , which must be extended or retracted appropriately to level the platform. Find a

formula which predicts the length of each leg, as a function ofW, ,R, H,a and the acceleration due to gravityg.


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This time it seems like the problem weve been asked to solve has nothing to do with forces we need to calculate

the length of each leg, not a bunch of forces! But of course, the length of each leg must be chosen to balance theforces acting on the oil platform properly. Recall that the buoyancy force acting on an object is proportional to its

submerged volume. The forces acting on each leg are therefore controlled by lowering (which increases the force)

or raising the leg.

So, to solve the problem, we will first calculate the force that needs to act on each leg of the platform to balance it.

Then well use the buoyancy force law to work out how much of the leg needs to be immersed.

We will begin by reviewing how buoyancy forces work.

Recall that

1. Buoyancy forces act perpendicular to the water surface

2. The force magnitude is equal to the weight of water

displaced

3. The force acts at the center of mass of the displaced water.

The volume of water displaced by leg A is the submerged

volume of the cylinder, i.e.

.

The weight of water displaced by leg A is therefore

The force acts halfway up the submerged part of the leg, on the axis of the cylinder.

With this in mind, we can draw a free body diagram for the platform


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We can now balance forces acting on the platform

Force and moment balance table for platform. Origin at center of circular platform.Force/Moment Position Force Moment about origin

Weight

Force on leg A

Force on leg B

Force on leg C

Sum (=0)


We have three unknown forces, and three equations, so we could go ahead and solve them. However, were not

interested in the forces were being paid to calculate the length of each leg.

We can get additional equations for the length of each leg using the buoyancy law.


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We now have 6 unknowns ( and ) and six equations, so were all set. We can solve them by

hand, or if lazy use Maple. Here is the MAPLE solution (lazy? Me? No way!)

(IMPORTANT NOTE: Weve used variables t60 to denote the angles (60 degrees). If you try to put in sin(60) into MAPLE, it

will calculate sin(60 radians) !! To get numbers, wed have to subs titute values for t60 in radians)

Example 5:

This is a goofy problem with no practical application whatsoever, for which we apologize But its designed to

make an important point.


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In all the problems weve solved up to now, its been fairly obvious what part of the system we should isolate to

balance forces. But thats not always the case.

For example, suppose that two identical boxes, each of weight W, are suspended by the system of spring-balances

as shown. (I said this was a stupid problem). Calculate the forces shown by each spring balance.

Obviously, we need to calculate the force readings by balancing forces for something. But what? The top box?

The bottom one? Both together?

Whenever youre stuck like this, the best thing to do is to write down equilibrium equations for everything you can

think of, and then look at the equations you end up with, and see if you can solve them.

Well work through all three possible combinations for the example. First, well get force balance equations for

each box, then well do the two together.

Heres what we get for the top box


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Force and moment balance table for top box. Origin at CG.


Weight 0 0

Force A

Force B

Force E

Force F

Sum (=0)

Next, the bottom box

Force and moment balance table for bottom box. Origin at CG.


Weight 0 0

Force A

Force B


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Force E

Force F

Sum (=0)

Finally, the two boxes taken together as a single system

Force and moment balance table for both boxes. Origin at CG for bottom box.


Weight (top) 0

Weight (bot) 0 0

Force C

Force DForce A

Force B

Sum (=0)


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Note in passing how we introduced the variable h to give the distance between the CGs of the two boxes. The

problem didnt give us this info and well find out later that the answer is actually independent of the height h bu

if you need a dimension in a problem as an intermediate step, its perfectly OK to go ahead and define one.OK, lets check the equations we ended up with. The first set (eq1) were for the bottom box; the second (eq2)

were for the top, and the third (eq3) were for the two together

(1.1

(1.2

(1.3

So now what? We have 9 equations, and only 6 unknowns. Can we solve them? Usually, no. But we could trythrowing the whole shebang into MAPLE and see what happens. It turns out that this actually works heres

MAPLEs solution. (By the way - if youve struggled to get solutions to some of the homework problems for this

course, you are not alone... It took me over an hour to fix all the sign errors in my calculations and typos in the

MAPLE formulas before I could get this to work!)


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(IMPORTANT NOTE: Weve used variables t30 and t60 to denote the two angles (30 degrees and 60 degrees). If you try to putin sin(30) into MAPLE, it will calculate s in(30 radians) !! To get numbers, wed have to subs titute values for t30 and t60 in

radians)

Its surprising that this works. Usually, when you have more linear equations to solve than unknowns, you cant

satisfy all the equations. So how come it works here?

It works because our three sets of equations are not independent. If any two of them are satisfied, the third one is

satisfied automatically. We can easily check this with MAPLE. Heres what happens if we solve the first two and

plug the solution into the third.


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Although this is a goofy example, it illustrates a general feature of complex statics problems. Heres what we can

conclude:

1. When you need to solve a statics problem that involves many different components connected together, youcan write down equilibrium equations for each component, or you can write down equilibrium equations for

combinations of the components.

2. Generally the safest way to proceed is to get equations for each component, and solve them. For example, in

the problem we just solved, we could have written down equilibrium equations for each box by itself, and

solved (1.1) and (1.2).

3. However, its perfectly OK to get equilibrium equations for group(s) of components too. We see from our

simple example that we will get the same answer, and in many cases the equilibrium equations for a group of

components may give us the answer we are interested in much more quickly than if we were to try to analyze

each component separately. For example, if you want to calculate the forces acting on the wheels of the car,

you would just write down equilibrium equations for the whole car, you wouldnt do one for each nut, bolt,engine component, etc

When you solve a complex statics problem, its often worth getting equilibrium equations for lots of possible

combinations, and then selecting the set of equations that looks the simplest.


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Example 6Another goofy example to make another point. Not all statics problems have a solution. Consider the

following the beam shown below is supported by a single spring balance. What force does the spring balance

record?

This problem is so simple we dont need to draw a force diagram heres the force balance table

Force and moment balance table for beam ABC. Origin at CG.


Weight 0 0

Force at A

Sum (=0)

So now we need to solve the equations

but theres no value of that can satisfyboth equations.

Thats because the system cannot be in static equilibrium in the configuration shown the beam would fall down.

So, if you work through a problem and find that you cant satisfy all the equilibrium equations, it means the system

you are analyzing must move.

Example 7: This is a final goofy problem to make another point. Some statics problems may not have a uniquesolution. Heres an example. The beam shown below is supported by three spring balances. Calculate the force

reading on each.


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equations. The information was always presented in one or more force and moment balance tables.

Some simple problems can be solved using short-cuts, which you may have used extensively in introductory physics

courses. Below, we list a few of the more common short-cuts.

1If only two forces act on an object that is in static equilibrium, the forces must act along the same line,

and must be equal in magnitude and opposite in direction

You can show this using force and moment balance.

The shortcut is useful, because for systems with only 2 forces acting on them, we dont need to work through all

that force balance table crud. We dont even have to mess with vectors. We can just use the fact that the two

forces must have equal magnitudes and opposite directions.

Example: Estimate the terminal velocity of a tennis ball, and human, falling through the air.

In both cases the objects are subject to only 2 forces: weight, and air drag.

1. The weight force has magnitude ;

2. the air drag has magnitude , where is air density; Vis the velocity; is the drag coefficient and

A is the projected frontal area of the object.

The forces must be equal and opposite, so

Finally, we need to make an estimate of the numbers. This is a good chance to use our formal estimation

procedures

Data/Assumptions:

1. Tennis ball

a. Mass: 0.057g (source http://physicsweb.org/article/world/12/6/3

b. Area: diameter is 0.066m (source as above) so projected area is 0.0034

c. Drag coefficient 0.1-0.5 (smooth sphere result from section 2.7 of notes)

2. Human (assumed to be upright)a. Mass 100kg

b. Area 0.125m (rough measurement of nearest human)

c. Drag coefficient 1.0-1.3 (source: http://aerodyn.org/Drag/tables.html#man

Air density


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Calculation

1. Tennis ball

2. Human

Error estimate

1. Tennis ball

a. Mass +/- 2%

b. Area +/-2%

c. Drag coeft +/- 50%

d. Air density +/- 5%

e. Total error = 59/2 % = 30% (the factor of 1/2 is because of the square root)

2. Humana. Mass +/-20%

b. Area +/- 20%

c. Drag coeft +/- 20%

d. Air density +/- 5%

e. Total error = 65/2% = 33% (again, the factor of 1/2 is the square root)

Example 2 we actually already made use of this principle when we described the forces exerted by a spring.

Remember the picture below?

We said that the force exerted by the spring always had to act along the line of the spring. Why? Well, if we look

at the spring, there are only two forces acting on it. The forces act on the ends of the spring. Now we know that

the forces have to be parallel; equal in magnitude, and opposite in direction. This means that the forces mustact


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along the line of the spring otherwise the spring wouldnt be in moment equilibrium, and would spin around.

2.If only three forces act on an object that is in static equilibrium, then (a) the forces must be coplanar and

(b) the forces must either act through the same point, or must be parallel.If the forces are not parallel,

moment balance is satisfied automatically. If the forces are parallel, two components of force balance and

one component of moment balance are satisfied automatically..

Example. The chair on a ski-lift has weight Wand is supported by two cables, which may be idealized as springs

with stiffness kand unstretched length a. Find a formula for the stretch of each cable in terms ofW, , kand a.

To calculate the extension of each cable, we clearly need to calculate the force thats stretching it, and then use the

spring law. We can calculate the cable forces by performing a force balance for the chair.

The picture below shows the forces acting on the chair. There are only three forces the weight; and the force

exerted at O by the two cables. Consequently, the three forces must be in the same plane, and must act through

the same point. Both the forces exerted by the cables clearly act through O, because thats where the cables are

attached. For the weight force to act through O, the chair must swing so that its CG is directly below O. The

forces are not parallel, and so must be calculated using force balance moment balance is satisfied automatically (if

you dont see this, try taking moments about O!)


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We can do the force balance in a table

Force balance table for chair

Force exerted by cable OA

Force exerted by cable OB

Weight

Sum (=0)

This gives us, as expected, two equations for the two unknown forces in the cables.

These can be solved (even without MAPLE!) to see that

where weve used the trig formula to get the simple formulas.

Finally, we can calculate the extension of each cable using the spring force law

, so that


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Example: A bicycle is suspended from two bungee cords (think of them as springs), as shown below. Find the

position of the center of mass of the bicycle in terms of the length dand the two angles and , expressing your

answer as a position vector relative to the attachment point A.

Only three forces act on the bike weight, and the force exerted on the bike by the two bungee cords. The forces

must therefore act through the same point and must be coplanar. The center of gravity must be in the vertical plane

and must be located at the intersection of lines through the cords. We can use elementary geometry to calculate the

components of the position vector, using the triangle sketched below

Heres one way to calculatex andy. Geometry on the triangle shown gives

We can solve these forx andy


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where weve used the trig formula to simplify the result. The position vector

is

2.If only four non-coplanar forces act on an object that is in static equilibrium, then the forces must either

act through the same point, or must be parallel.If the forces are not parallel, moment balance is satisfied

automatically. If the forces are parallel, two components of force balance and one component of momentbalance are satisfied automatically..

This short-cut is not as useful as the others It would tell you, for example, that lines through the three cords lifting

the box shown in the picture below would all have to meet at a point above or below the CG of the box

7.4 Case Study using Forces and Equilibrium: A simple aircraft control model

Aircraft designers need to know how design parameters affect aircraft performance. The aerodynamicists need to

know how the design of wings, fuselage and powerplant influence performance. The control systems engineers

need to have models of aircraft performance to design fly-by-wire systems and autopilots.

Weve covered enough theory in this course to be able to develop a quite sophisticated model of aircraft behavior.

Well work through a 2D example, which will illustrate a number of very strange aspects of aircraft control and

performance. The analysis shows beautifully the power of physics and mathematics to explain the behavior ofexceedingly complex engineering systems. This case study is presented for interest only we dont expect you to

be able to do an analysis like this one (yet. But youll be doing much more sophisticated calculations very soon!)


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The picture shows key geometrical parameters that affect aircraft performance. Some remarks:

1. The aircraft travels parallel to the green line. If the aircraft is climbing, if the aircraft is descending

2. The horizontal `tail of the aircraft is known as a `stabilator. The pilot can rotate the stabilator, by pulling or

pushing on the control column. We use to denote the stabilator angle.

3. The `angle of attack of the airplane is the angle is the angle between its longitudinal axis and its direction of

travel (actually this is not quite right, but the definition is close enough)

The aircraft designer can set the following parameters

1. Wing area

2. The distance of wing center of lift behind the center of gravity of the aircraft

3. The stabilator area

4. The distance of stabilators center of lift behind the center of gravity of the aircraft

5. The aircraft weight W

6. The maximum engine thrust (this is a distributed load acting on the propeller well model it as a statically

equivalent forcePacting on the spinner)

7. The distance from the CG to the propeller

The following issues are of interest in aircraft design

1. What wing area is required to support a reasonable aircraft weight?

2. What engine power is required to give a reasonable cruise speed and rate of climb?

3. How is performance affected by the location of the centers of lift?

If you want to design an autopilot for the plane, or if you want to know how to fly it, you need to know

1. How do you control the planes speed?

2. How do you make the plane climb or descend?

(A lot of people think that the stabilator controls climb or descent, and engine power controls speed, but as we

shall see, the reverse turns out to be true!)

We will develop formulas that answer all these questions, by analyzing the forces that act on the aircraft during


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steady flight.

To begin, we need to review lift and drag forces acting on wings. Recall that a wing is subjected to lift and drag

forces as shown in the picture below.

The formulas for lift and drag force are

Here, is the air density, Vis the speed of the air past the wing, and is the wing area.

The lift and drag coefficients and vary with the angle of attack of the airfoil as

where , and are more or less constant for any given airfoil shape, for practical ranges of Reynolds

number.

We can now draw a force diagram for the plane, and perform a force and moment balance. We assume that the

plane travels at constant speed Vat a constant climb angle. The drag and lift forces act parallel and perpendicular

to the direction of travel.

The forces acting on the plane are shown below


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Heres the force balance table. The basis vectors are parallel and perpendicular to the direction of travel

Force and moment balance table for plane. Origin at CG.


Weight 0 0

Force on wing

Force on stabilator

Engine thrust

Sum (=0)


The trig functions in these formulas are a pain. Its convenient to simplify things by noting that for any reasonable

operating conditions and are both small. Thus, approximately,

This simplifies the equilibrium equations to


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attack directly.

Secondly, in the event of an engine failure, you want to minimize your descent rate, so as to give you more options

in selecting a site for a forced landing (or at least to give your passengers more time to get their affairs in order).

Again, descent rate is minimized by flying at the optimal pitch attitude

You find this pitch attitude by flying at the published `best glide speed for the airplane. This airspeed is usually

close to the best rate of climb speed (our simple theory predicts that it should be identical, but of course we made

lots of approximations).

4. The role of the stabilator

So far, we dont know how to actually change the angle of attack . The third of our equilibrium equations shows

the pilot how to use the controls to vary .

We can simplify the equation by eliminating aircraft speed Vfrom equation (1.6), using equation (1.7). If we

rearrange the result and again make the approximation that , we find that

This shows that the angle of attack is directly proportional to the stabilator angle . The pilot can therefore set

angle of attack by rotating the stabilator.

Notice also that during flight (otherwise the wing generates negative lift not a good idea!). This means that

the stabilator angle is negative, and moreover

This means that for the stabilator, so that the lift force on the stabilator actually acts downwards.

In retrospect, this is obvious: the aircrafts weight exerts a clockwise moment about the center of lift of the wing, so

the stabilator must exert a counter-clockwise moment. You might think you could improve airplane design by

moving the CG behind the wings center of lift: in this case the tail would contribute to lift. But dont try this at

home, kids, its been found (the hard way) that moving the CG in this way makes the aircraft dangerous (it maynot recover from a stall). A better fix is to put the stabilatoraheadof the main wing. A few aircraft designs

actually do this (including the Wright brothers plane) the small wing ahead of the main wings is called a `canard.

Design an aircraft


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Finally, we can use our equations to help design an aircraft. Well try to design a small general aviation single

engine aircraft, meeting the following specifications

1. Cruise speed 120 knots

2. Max rate of climb 1500 ft/min

Parameters that we dont have much control over are

1. Aircraft weight W(determined by structural considerations). Well assume 2500lb

2. Air density . Varies with temperature, air pressure and humidity theres a neat Java calculator that will

compute air density for any given temperature at http://rshelq.home.sprynet.com/density_altitude.htm . For a

standard atmosphere, .

3. Airfoil parameters , and . Typical numbers for these were given in Section 2. We will use

, , and (the latter number was obtained by assuming that the

aspect ratio of the wing is of order 0.1).

Our design variables are:

1. Wing area



4. The distance of stabilator center of lift behind the center of gravity of the aircraft

5. The maximum engine thrust

6. The distance from the CG to the propeller

We have several design constraints. Most importantly, we need to make sure that we keep and small (less

than 10 degrees) under normal operating conditions, partly to make sure our calculations are correct, but mostly toensure that the airplane remains well away from stall.

We can estimate the wing area using equation (1.5), which gives

Lets design to cruise at a 2 degree AOA. Then, plugging in numbers for all the variables and converting to SI

units, we get

The position of the aircrafts wings lift center behind the CG , the distance to the stabilator, and the stabilator

area are determined by a number of factors that we cant address completely with our simple calculations. The

position of the CG will move as the plane is loaded we need to ensure that the CG remains ahead of the lift center

to ensure that the aircraft remains stable. We need to keep the stabilator area as small as we can, and we need to

make sure that the stabilators angle of attack is always less than that of the wing, to make sure the wing stalls

before the stabilator (pilots are trained to deal with wing stalls, but not stabilator stalls)


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Finally, we can calculate the required engine thrust from the formula

To summarize, our design variables are

1. Wing area



4. The distance of stabilator center of lift behind the center of gravity of the aircraft

5. The maximum engine thrust

6. The distance from the CG to the propeller We forgot this one Theres nothing in our calculationsthat tell us what should be. It needs to be big enough to fit the engine in. Now that we know the engine

thrust, we could calculate the size of engine required but the courses that teach that stuff will cost you another

$60000 in tuition and fees

Documents

Introduction to Statics_ Equilibrium