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THE STATICS PAGE

"The most valuable of all talents is that of never using two words where one will do"-Thomas Jefferson (1743-1826)

I am revising and updating our earlier Statics Page  and presenting the results below for our Class EGM 2511. You can use it to gain further insight into the basic concepts of statics. Also it contains extra information  of use to you such as what topics are to be discussed in upcoming lectures. The book we are using is the 9th  edition of  ENGINEERING MECHANICS-STATICS by R.C.Hibbeler. It is easy to read and understand and free of errors. The knight on horseback points to the next lecture. If you have questions please contact me at  kurzweg@ufl.edu

Do you recognize the three photos above? They are of Galileo Galilei (1564-1642) and Sir Isaac Newton (1643-1727) the two most important early contributors to classical mechanics , and the  statue of  the Thinker by Rodin(about 1910) located in Paris.

LECTURE #1:-Introductory remarks, Newton's three laws of motion, plus the universal gravitational law, SI and US systems of units.

LECTURE#2:- Properties of vectors, Forces as vectors,Vector addition , subtraction and multiplication, Position vectors.

SOME INTERESTING CONVERSIONS AND LENGTH SCALES: A few helpful conversions are: 1 inch=2.540cm, 1 km=0.6215mile, 1 lb=4.448N, 1 slug=14.593kg, 1 ft-lb=1.3558 joules, 1 hp= 550 ft-lb/sec=745.70 watts(or joules/sec), 1 mile/hr= 1.6093 km/hr, and 1 lb/in^2=6894.76 pascal. While we're at it-
One Light Year=186,000 miles/sec x 3600 sec/hr x (24 x 365) hr/yr=5.87x10^12 miles=9.46x10^12km

The diameter of our galaxy( Milky Way) is about 100,000 light years , the sun is 500 light seconds away from the earth, the next nearest star (Alpha Centauri) is 4.3 light years away, and we can see out a distance of between 9 and 18 billion light years with the best optical and radio telescopes. Radiation coming from this last distance started its journey 18 billion years ago at the beginning of the universe( big bang theory). If you want some really small distances try 1 micron=10^-6 meter for the size of a typical microchip structure, 0.529 angstrom=0.529x10^-10 meter for the radius of the first Bohr orbit of the hydrogen atom , and 2.4x10^-15 meter for the diameter of a proton. The web has available many convenient sites for converting between SI and US Customary units.

CLIMB ANGLES AT THE GREAT PYRAMID:To demonstrate the use of vectors in carrying out certain geometrical calculations, we were asking in class what are the angles associated with climbing the great pyramid of Cheops at Giza in Egypt. One can look up on the internet that the base of this pyramid has sides of length b=756 ft . The four sides of this pyramid are in the form of equilateral triangles. Placing the origin of a cartesian coordinate system at the center of the base, it is easy to show that the coordinate of the pyramid vertex is at (0,0,b/sqrt(2)). Thus the position vector of length b running along one of its edges is R=-bi/2-bj/2+bk/sqrt(2). Taking a dot product of this vector with k yields cos(theta)=1/sqrt(2), or a climb angle of 45 deg. Note that the climb angle starting from the middle of one of the sides is much steeper and equals 54.73 deg. , again shown by dot product manipulations. If you ever get a chance to go over there sometime you will be truly amazed at this engineering marvel of the ancient world. They don't let you climb the pyramids of Giza anymore but you can go inside those of Cheops and Cephren (as I remember doing back in 1983) .

LECTURE#3:-More on Forces, Equlibrium of a particle in 2D , Concept of unit vectors along a given direction.

LECTURE#4:-3D Equilibrium Problems. Working out in detail several equilibrium problems from the book.

LECTURE#5:-Concept of a Couple. Working out several more problems in static equilibrium.

WHAT ARE MOMENTS AND COUPLES?:A moment M is defined as the cross product between a position vector from the point about which the moment is taken and a force F, that is M=rxF. The simplest way to evaluate this product is via a three by three determinant in which the first row elements are the base vectors i,j and k. The second row consists of the position vector components and the third row are the force components. Thus r=2i+5j-3k and F=20i-40j+30k yields M=30i-120j-180k. The dimensions of M are length-force and so given in ft-lb in the US system of units and in met-newton in SI. A couple C consists of two equal but opposite forces F and -F separated by a position vector r. The couples value is simply rxF and will retain this same value no matter about which point the moments for the two forces are taken. The couple produced by F=10j acting at (3,0,0) and -F=-10j acting at (-3,0,0)is C=60k.

LECTURE#6:-Completion of the combined couple, moment and force problem acting on a 2D slab.
Several additional problems from the book worked out in detail.

BASIC LAWS OF STATICS: We have now reached the point in the course where we can state the basic Laws of Static Equilibrium and the rest of the course will be their application to all sorts of problems, which, as you will find, is not always that easy. The Laws are-
(1)-THE SUM OF ALL FORCES ACTING ON A BODY IN STATIC EQUILIBRIUM MUST BE ZERO.
(2)-THE SUM OF ALL MOMENTS(INCLUDING COUPLES) ACTING ABOUT ANY POINT ON THE BODY MUST BE ZERO.

LECTURE#7:-Component of a moment about a given line. Examples worked out in detail.

MOMENT ABOUT A SPECIFIED LINE IN SPACE:-In solving some statics equlibrium problems one can often determine an unknown force(such as a tension in a rope attached to the body) by taking moments about a line passing through two points on the body at which there exist other unknown forces . By so doing you avoid having to solve for these other forces. To carry out this procedure requires knowledge of the moment about such a line. Let me demonstrate how this is done. Consider a cube extending over the range 0<x<1,0<y<1,0<z<1 . A force F1=10i+20j-30k acts at corner(x=1,y=0,z=1)and a second force F2=-10i+40k acts at corner (1,1,1). The moment produced by these two forces about the origin at (0,0,0) is M=(i+k)x(10i+20j-30k)+(i+j+k)x(-10i+40k)=20i-10j+30k. Now lets ask what is the projection of this moment on a line connecting the origin (0,0,0) and (1,1,0). We have this line defined by the unit vector u=(i+j)/sqrt(2) so that dotting it with M yields 10/sqrt(2). Thus the moment about the line becomes 5i+5j. Run through this calculation yourself using a sketch and also see if you can get this result another way. Note that we took the original moment about a point on the line chosen. Would you be able to obtain a moment projection along a given line if the original moment is taken about a point not on the line?

LECTURE#8:-Reduction of Force and Couple Systems. Concurrent, Coplanar, and Parallel Force  Systems. The Wrench. Location of Resultant Forces.

EQUIVALENT FORCE SYSTEMS: In many statics problem involving the effect of several forces acting on a body, it is often convenient to replace these with a single resultant force plus a couple, before proceding further in the calculations. To show you how one can establish such an equivalent system, look at the simple example HERE.

LECTURE#9:-Free Body Diagrams . Forces and Moments acting at different Supports . Pins, Rollers, Rockers, Hinges, plus Ball and Socket Joints.

HOW TO HANDLE DISTRIBUTED FORCES: When dealing with continous force distributions on extended bodies it is convenient to replace these with a single resultant force R acting at a given point P. To see how one determines the value of the equivalent force and its location, we consider the case of a weighless horizontal beam of length x=L which is loaded by a distributed downward linear force distribution w(x)=a+bx  (expressed in force per length) and supported at end points  at A and B. In this case the moment about the z axis produced by this distributed force about the left pivot point at A has magnitude M=-Int[xw(x)dx]=-[(1/2)aL^2+(1/3)bL^3]. In static equilibrium this plus LBy, where By is the vertical force on the beam at the right support, must be zero. Hence the resultant downward force R=Int[w(x)dx] acts at a distance x*=LBy/R. This turns out to be at x*=L[(a/2+bL/3)/(a+bL/2)] for this linear force distribution and thus at( /3)L when a=0. Click HEREto see a picture of the configuration.

LECTURE#10:-Working out more Statics problems in 2D and 3D for all sorts of different supports. Don't forget about your First Statics Exam coming up next week.

LECTURE#11:-Working out more statics problems in 2D and 3D from Chapter 5.
Some of you have asked me about posing an old #1 exam.  Here is one from 1997 which you can use to practice and get up to speed .

STACKING OF BRICKS TO FORM A STATICALLY STABLE ARCH: Every once in a while one discovers certain facinating problems even in topics as elementary as statics. One of these I ran across this weekend is an extention of solving the problem of the furthest distance three bricks  stacked on top of each other can have their centers of gravity moved to the right without causing the pile to tumble. The three brick problem is easy to solve . It is clear from a moment balance that the second brick can be moved a half brick length to the right of the top brick without falling. Next the two bricks in the top two rows have their center of gravity at 1/4 of a brick length so that the bottom brick can be placed at 3/4 of a brick length to the right of the top brick. Now comes the interesting extention. Consider n rows containing one brick each. Again starting with the top row, the next to the top row is placed at 1/2 to the right for a unit length brick. The one after that is at 3/4. Continuing on, we find the brick in the n+1 row from the top is at x[n+1]=x[n]+1/(2n) with x[1]=0, x[2]=1/2, x[3]=3/4, x[4]=11/12, x[5]=25/24, etc. Using the canned program MAPLE to evaluate this for a 20 row stack, we find the result shown HERE. This result looks very much like half of a parabolic arch and becomes a full arch when we also allow the same type of stacking on the other side . Note that you wouldn't need any binder to build such an arch. Undoubtably such stacking by early man is how he first came up with the idea of the arch. It is interesting to note that the ancient Greeks did not  use a proper  arch in their architecture* and that it was not until Roman times that the circular arch was invented and used in bridge and building construction. The arch reached its high point during the midle ages with the discovery of the gothic arch. This last structure, when used in conjunction with a flying buttress, is extremely stable since all compression loads lie strictly along the arch and accounts for the fact that many of the European cathederals have remained intact for nearly a thousand years.
*-Several years ago I visited an ancient tomb in Mycenae in Greece built some 3000 years ago. The tomb has an entrance arch formed by stacking up stone plates, but making the overlap the same for all rows. In view of the problem solved above, you can see why this was not the best solution and not really a true arch.

REVIEW FOR THE FIRST EXAM. The exam will be closed book and cover chapters 1 through 5. You can answer any three of the four questions.

Lecture #12:-Structural Analysis(beginning of  Chapter 6). Simple Trusses. Concept of Compression and Tension in Members. Method of Joints.

ANALYSIS OF A TRUSS: We have shown in class that the concatenation of triangular shapes formed by three members connected at their ends produces a stable truss which is the basic building block for most structures be they roof trusses, railroad bridges , furniture, etc. In the analysis of such structures one first solves for all the external forces acting on the truss and then looks at the internal forces by either the method of joints or method of sections. We give HERE a very simple example of such an analysis. The important thing to remember is that the internal forces produce a compression in a member when the calculated force points toward the pin at the joint. When your calulations show the force going away from the pin then the member is in tension.(just the opposite of what a cursory glance would suggest)

Lecture#13:-More on Trusses. The Crane Problem. Method of Sections.

Lecture #14:-Completion of discussion on trusses in 2D by application of both the Method of Joints and the Method of Sections.

APPLICATION OF THE METHOD OF SECTIONS: When one is interested in just one of the internal forces of a member of a truss, especially when the truss has many members, it is convenient to use the method of sections. This involves putting a cut through the truss including the member in whose internal force we are interested. Next treating the unknown forces in the cut members as external forces, one can rapidly find the value of the force in the member and generally can do so much faster than when using the method of joints. We demonstrate this solution method for one particular truss HERE. Have you ever wondered how many members M and joints J a truss composed of N triangles has? You can arrive at a relation between these three quantities by slowly working up from N=1. Thus when N=2 you have M=5 and J=4. Next for N=3 you find M=7 and J=5. Continuing on, you soon realize that J=N+2 and M=2N+1. Thus a truss with 80 triangle components will have 161 members and 82 joints.

Lecture #15: Trusses in 3D(Space Trusses). Introduction to Frames.

THE "A" FRAME: A prime example of a frame stuctrure is the A frame used in a variety of structures including tables and vacation homes. it consists of two two side legs plus one cross bar anchored together by pins at the three joints. You analyze a problem of this type by looking at the three members individually after determining the external reactions on the frame. If you go HERE you will see the analysis when the members are weightless and a weight W is hung from the center of the cross-bar. Note how one always reverses the forces at a joint in going from one member to the other. If you should assume the wrong direction of the force at a joint, this will come clearly at the end of the calculations by giving you a minus sign.

Lecture #16: More on Frames and Machines. The Crank-Piston Problem. Analysis of multiple pulley systems.

ANALYSIS OF A PULLEY: We all learned back in high school physics that there are six simple machines, namely, (1)the lever, (2)the screw, (3)the wheel and axle, (4)the wedge, (5)the inclined plane, and (6)the pulley. Of these , the pulley is usually the one hardest to understand. In view of our discussion on machines it is actually quite easy to grasp by analyzing each wheel sub-section individually as shown HERE. At wheel A we see, via a vertical force balance,(assuming negligible weight of the wheel) that the force in the rope going about the wheel is equal to F while the rope holding wheel A has a tension of 2F. going on to wheel B we see that the rope hung from the wheel axis must be 2F by a force balance. Continnuing this analysis on through wheel D, one sees that it can carry a weight of 8F. By using n wheels in such a pulley the mechanical advantage gained will be 2^(n-1). That is, one could use such a pulley to lift a 640lb car engine with a force of just F=40lb, when the pulley has five wheels.

Lecture #17: Completion of discussion on Frames and Machines, plus Introduction to Internal Forces(Chapter 7).

Drawing Shear and Bending Moment Diagrams: The basic formula for determining shear in a 1D beam is dV/dx=-w(x), where w(x) is the x dependent downward loading and V is the shear. The shear is thus a constant where the loading is zero and will vary linearly with x for constant loading. The bending moment formula is dM/dx=V(x), with M vanishing at the end supports(except at those which can sustain a moment such as a cantilever). Note that if V is a polynomial of order n then the bending moment M will be one of order n+1 . The jump in V at a point load is just equal to the value of this load. By clicking HERE you can see a typical example of a shear(orange) and bending moment(blue) diagram for a cantilever beam subjected to a linearly varying load.

Lecture #18-Working out Shear and Bending Moment Diagrams for a variety of different loadings.

No classes next week due to spring break. Enjoy the vacation!

Lecture #19-More on Shear and Bending Moment Diagrams. Also Shear and Normal Force in Curved Beams.

N,V, AND M FOR A CURVED BEAM:-So far we have looked at normal (N) and  shear(V) forces plus the bending moments(M) in only 1D beams for which the simple formulas dV/dx=-w(x) and dM/dx=V(x) apply. Let us next extend the discussion to curved beams. Take the case of a weightless semicircular beam of radius r=a pushed against a smooth floor by a single force Fo applied at its top as shown HERE. The analysis proceeds as usual by first determining the external forces. These are simply Ay=By=Fo/2. To now find N, V, and M at some point C along the curved beam, we apply to the beam segment B-C the conditions that the summation of forces in the x and y direction are zero and that the sum of the moments taken at  C must also  be zero . A simple calculation then yields V=-(Fo/2)sin(q), N=(Fo/2)cos(q) and M=(aFo/2)[1-cos(q)], in the range 0<q<p/2. For p/2<q<p we have a change in sign for V and the bending moment becomes M=(aFo/2)[1+cos(q)]. Note the maximum bending moment occurs at the top of the semicircle where q=p/2 and vanishes where the semicircle touches the floor.

Lecture #20: Completion of stuff on Shear and Bending Moments. How to handle a hook input. Also a discussion on Cables and the Shape of a Suspension Bridge Cable.

Shape of a Suspension Bridge Cable: In discussing cables during the previous lecture we asked the question what would be the shape of a cable of if it is subjected to a uniform  x dependent loading w(x)=Const. This is the well known suspension bridge problem which predicts, as shown HERE, that the shape of the cable is that of a parabola y=const x2. Note that a cable hanging under its own weight has a different shape , namely, that of a catenary curve y=cosh(a*x).

Lecture #21: Introduction to Coulomb Friction. Angle of Repose and Problem of  Ladder Equilibrium when resting on a rough floor.

Lecture #22: More on Dry Friction including discussion of the Wedge Problem.

Lecture #23: Belt Friction, Tipping and Sliding.

THE WEDGE PROBLEM: One of the problems which can be well explained by using Coulomb's friction law f=mN is that of the wedge. Consider a wedge with wedge angle a used to hold up a weight W as shown HERE. The external forces acting are indicated and we are  particularly interested in determining the ratio of F/W as a function of m and a , where the coefficient of friction at the wall is zero but has the same value of m on both wedge surfaces. Looking at the system as two bodies with friction forces f1 and f2 parallel to the two wedge surfaces, one finds a total of six equations governing equilibrium which can be solved to find F, A, B, f1, f2, and alpha for a given weight W and friction coefficient m. We have neglected the wedge weight w in our calulations since typically W>>w.

Lecture #24: Completion  of discussion on dry friction by working some problems from the book.

BELT FRICTION: When you wrap a rope about the circumference of a cylinder, one typically finds that there is a difference in the rope tension at the two points where the rope leaves the cylinder. This tension difference is due to friction effects and can become quite large when wrapping the rope  multiple times around the cyliner. The equation for the tension change is T=Toexp(mq), where m is the coefficient of friction between the rope and the cylinder surface and q the total angle the rope is wrapped around the cylinder. Since the exponential term is greater than unity, it is clear that T>To. This means that the friction  force f will always be opposite to the direction of impending motion. Go  HERE to see the derivation of the belt friction equation. Such belt friction is employeed by capstans used to tie up large boats at harbour docks and in devices for slowly lowering heavy weights.

REVIEW FOR SECOND HOUR EXAM: Same format as exam #1

Lecture #25: Beginning of Chapter 9 on Center of Gravity and Centroids. Basic definitions and use of calculus to find CGs for various 2D and 3D bodies.

Lecture #26: Centroids and CGs for Composite Bodies.

CENTROID OF COMPOSITE BODIES WITH HOLES: We have shown in class how the centroid of a 2D composite body is calculated by looking at the quotient of the sum of the products of xbar and the sub-area divided by the total area. When a body contains a hole the procedure stays the same except that the sub-area representing the hole is preceded by a negative sign. To demonstrate, take the circular disc x^2+y^2=4 into which is cut a square hole 0<x<1, -1/2<y<1/2. Here by symmetry ybar is zero but xbar becomes [0*4*Pi-0.5*1]/[4*Pi-1]= -0.043228. So xbar is negative as expected. Click HERE to see a schematic of this calculation.

CENTROID FOR A COMPOSITE BODY: To get some practice on finding the centroid of some 2D systems, consider the centroid of the block letter combination UF ( University of Florida ).What is x bar and y bar for this set up assuming both letters U and F are constructed of unit width rectangles and the two letters are separated by a unit distance? To make the calculation we break the U up into three rectangles. The first of these has corners at (-4,5), (-4,0),(-3,0) and(-3,5). The second has corners at (-3,0), (-1,0), (-1,1) and (-3,1) and the third corners at (-1,0), (0,0), (0,5) and (-1,5). Applying our composite body formula to these three rectangles yields a centroid for the U of xbarU=[xbarI*AI+xbarII*AII+xbarIII*AIII]/[AI+AII+AIII]=-2.00 and ybarU=[ybarI*AI+ybarII*AII+ybarIII*AIII]/[AI+AII+AIII]=13/6. Thus the centroid of U is at (-2.00,2.1667). Next look at F. Breaking it also up into three rectangles with the largest having corners at (1,0),(2,0),(2,5)and(1,5). The other rectangle has corners at (2,4), (5,4), (5,5) and (2,5) with the remaining square bounded by corners (2,2), (3,2), (3,3) and (2,3). One finds for the F the centroid is at (xbarF, ybarF)=(2.2778,3.16667). Finally combining the centroids for U and F yields xbar=[-2*12+(41/18)*9]/[12+9]=-0.1667 and ybar=[2.2778*12+3.16667*9]/21=2.5952. Thus the UF combination will have zero net moment about the point (-0.16667, 2.5952). I gave this problem several years ago as one of the problems on a statics exam and, if I remember correctly , about half got it right.

Lecture #27: More calculations for the CG and Centroid of Bodies plus the Theorems of Pappus and Guldinus.

THEOREMS OF PAPPUS AND GULDINUS: There are two theorems associated with the the names of Pappus of Alexandria( 290-350AD) and Guldinus( 1577-1643AD). The first of these states that "the product of the length of a curve and the distance moved by its centroid when the curve is rotated about an axis equals the surface area generated". As an example consider the half circle x^2+y^2=R^2, y>0. Its length is R*Pi and its centroid is at 2*R/Pi. Thus by the first theorem the surface area will be R*Pi*(2*Pi)*(2*R/Pi)=4*Pi*R^2, a well known result from calculus. Use this theorem to calculate the surface area of a toroid. The second theorem states that "The product of an area and the distance travelled by its centroid when this area is rotated about an axis equals the volume of the solid generated". As a demonstration of this consider the rectangle -a<x<a, 0<y<b rotated about the x axis. Its area is 2ab and its ybar is b/2. The volume generated is thus V=2a*b^2*Pi. This agrees, as expected, with the volume of a cylinder of height 2a and radius b. Go HERE for a derivation of the Pappus Theorems.

Lecture #28: Resultants for Distributed  Loadings. Fluid Pressure, Pascal's Law and Center of Pressure.

FORCE ON SUBMERGED SURFACES: The centroid concept can be applied to loading of surfaces due to hydrodynamic forces. Click HERE to see the general development for the resultant force on a submerged window. Also consider the case of the resultant force R on a vertical dam of height h and width w. By Pascal's law we know the fluid pressure at depth z will be p=rgz so the force on the dam will be R=Int[rgz dz, z=0..h]=rgwh2/2. That is, the force equals the dam area w*h times the average hydrostatic pressure  rgh/2. Note, R is not dependent on how much water is stored behind the dam. On the other hand, R becomes very large for dams such as the Hoover dam in Nevada where h=725ft. Note the resultant force R acts at the center of pressure which lies at the centroid of the pressure triangle and thus at h/3 from the bottom for a filled , vertical walled, dam.

Lecture #29: Introduction to Moments of Inertia(Chapter 10). Basic definition and parallel axis theorem. Using calculus to find Ixx, Iyy, and Izz.  Go HERE to see how calculus is used to determine these for various bodies.

AREA MOMENTS OF INERTIA FOR A RECTANGULAR PLATE: A good demonstration of area moments of inertia is given by looking at the rectangular plate 0<x<a, 0<y<b. In this case the values Ixx=Int[y^2dA]=a*b3/3 and Iyy=Int[x^2dA]=b*a3/3. Also Izz=Jo=Int[r2dA]=Ixx+Iyy=[(a*b)/3]*[b2+a2]. This last relation between the three area moments of inertia holds only for lamina and is often referred to as the perpendicular axis theorem. We also have from the parallel axis theorem that the Izz will have its minimum value at the plate centroid at x=a/2, y=b/2. The explicit value is Izz(at lower left corner)=Izz(at  plate centroid)+(plate area)*(square of the distance between the two points). Here this yields Izz(at plate centroid)=(a*b/12)*(a2+b2) which is seen to be four times less than that about any of the plate's corners. Note the area moments of inertia have the dimension of length to the 4th power, while mass moments of inertia have dimensions of mass times length squared.

Lecture #30: Moments of Inertia of Composite Bodies.

AREA MOMENT OF INERTIA FOR A COMPOSITE LAMINA: Moments of inertia are additive so that one can readily calculate the I values of a composite body about any chosen axis by knowing the values for the sub-bodies and then applying the parallel axis theorem. HERE is an example of such a calculation.

Lecture #31: Product of Inertia Ixy, Moment of Inertia about any Axis, Principal Moments of Inertia, Mohr Circle.

MOMENT OF INERTIA ABOUT AN ARBITRARY AXIS IN THE PLANE OF A LAMINA: Consider a lamina with a superimposed x and y axis. Introduce a second pair of orthogonal axis u and v where the u axis makes an angle of A with respect to the x axis. From the geometry it is easy to express u and v as functions of x and y , as we have done in class, and then work out the area moments of inertia Iuu and Ivv, plus the product of inertia Iuv. Go HEREto see the details. Note also the Mohr Circle follows from this analysis.

Lecture #32: Mass Moments of Inertia defined and evaluated using Calculus. Radius of Gyration.

CALCULATING THE MASS MOMENT OF INERTIA FOR A UNIFORM SPHERE: We have defined the mass moment of inertia about the z axis  of a body as Izz=Int[(x^2+y^2)*dm], where the integral extends over the entire body with the mass increment dm=rho*dxdydz. In many cases one does not need to carry out the integration of such triple integrals directly if one knows the value of dIzz of the sub-bodies making up the body of interest. We show you HEREthe calculation for finding Izz of a uniform sphere of radius R=a using the known dIzz for the discs making up this sphere.

Lecture #33:-Concept of Virtual Work(Chapter 11)and some Elementary  Applications.

EXAMPLE OF A VIRTUAL WORK PROBLEM: We have shown in class that an alternative method for solving static equilibrium problems is that of Virtual Work. The technique is based on the fact that the product of the forces acting on a body times their virtual displacements must add up to zero under static equlibrium. The virtual work will be W=F.dr  and applies only to those points where the force F can cause virtual displacements dr. We show you HEREa simple application.

Lecture #34:- More on Virtual work. Conservative Forces, and Potential Energy. Also filling out of teacher and course evaluation forms for Dr. Kurzweg's section 1576.

SMALL SIDE DIVERSION ON PI

REVIEW FOR THIRD EXAM : This exam will emphasize Chapters  9, 10 and 11 on Centroids, Center of Gravity, Moments of Inertia, and Virtual Work. One of the problems will also contain material (such as possibly frames and friction)covered in the earlier chapters.

November 27, 2010