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Static Equilibrium Force and Moment
2.1 Concept of Force
Equilibrium of a Particle You are standing in an elevator, ascending at a constant velocity, what is the resultant force acting on you as a particle? The correct response is zero: For a particle at rest, or moving with constant velocity relative to an inertial frame, the resultant force acting on the isolated particle must be zero, must vanish. We usually attribute this to the unquestion- able authority of Newton. The essential phrases in the question are constant velocity, resultant force and particle. Other words like “standing”, “elevator”, “ascending”, and “you” seem less important, even distracting, but they are there for a reason: The world that you as an engineer will analyze, re-design, and systematize is filled with people and elevators, not isolated particles, velocity vectors, or resultant forces — or at least, not at first sight. The latter concepts are abstractions which you must learn to identify in the world around you in order to work effectively as an engineer, e.g., in order to design an elevator. The problems that appear in engineering text books are a kind of middle ground between abstract theory and everyday reality. We want you to learn to read and see through the superficial appearances, these descriptions which mask certain scientific concepts and principles, in order to grasp and appropriate the underlying forms that provide the basis for engineering analysis and design. The key phrase in Newton’s requirement is isolated particle : It is absolutely essential that you learn to abstract out of the problem statement and all of its rele- vant and irrelevant words and phrases, a vision of a particle as a point free in space. It’s best to render this vision, this abstraction “hard”by drawing it on a clean sheet of paper. Here is how it would look.
You, in an elevator.
An Isolated Particle:
An Isolated Particle
R
W
All Forces Acting
10 Chapter 2
This is a non-trivial step, akin to a one month old’s apprehension that there are other egos in the world. You are to take the dot drawn as the representation of a thing, all things, that can be thought of as an isolated particle. Now show all the forces acting on the particle. We have the force due to grav- ity, W = Mg , acting vertically down, toward the center of the earth.... (Who said the elevator was oriented vertically? Who said it was on the surface of the earth? This information is not given; indeed, you could press the point, arguing that the ques- tion is not well posed. But is this information essential? We return to this point at the end of this chapter). We have the reaction force of the elevator floor acting vertically upward on you, on you as an abstraction, as an isolated particle. This is how our particle looks with all forces acting upon it. The resultant force is the vector sum of all the forces acting on the isolated particle. For static equilibrium of the isolated particle, the resultant of the two forces – W acting downward and R acting upward – must be zero.
This leads to the not very earth shaking conclusion that the magnitude of the reaction force, acting up, must equal the weight.
This apparently trivial result and simplicity of the problem, if indeed it can be called a problem, ought not to be allowed to deceive us: The introduction of the reaction of the floor on you, the passenger in the elevator, is characteristic of the most difficult step in applying the requirement of static equilibrium to an isolated particle. You will find it takes courage, as well as facility with the language of engineering mechanics, to venture forth and construct reaction forces out of thin air. They are there, hidden at the interface of your particle with the rest of the world. Some, like gravity, act at a distance, across all boundaries you may draw.
Exercise 2. Estimate the lift force acting on the wings of a Boeing 747 traveling from New York to Los Angeles during rush hour. We can use the same isolation, or free-body diagram , of the figures above where now the point represents the Boeing 747, rather than you in an elevator, and the reaction force represents the lift force acting on the airplane, rather than the force acting on you at your interface with the elevator floor. From the requirement of static equilibrium, (we implicitly acknowledge that the 747 is moving with con- stant velocity), we conclude that lift force is equal to the weight, so to estimate the lift force we estimate the weight. Constructing this argument is half the problem. Now the other half: To estimate the weight we can guess... 100 tons? (More than 10 tons; we have heard of a 10 ton truck). But perhaps we can do better by slic- ing up the total weight, and try to estimate its ingredients. Passengers: How many passengers? I estimate 500. (More than 100, less than 1000). Then
R – W = 0
R = W
weight ⁄ passenger ≈ 250 ( pounds )
12 Chapter 2
I throw in another 50,000 pounds for the motors and the tail (plus stabilizers) and so estimate the total structural weight to be
Fuel: How much does the fuel weigh? The wings hold the fuel. I estimate the total volume enclosed by the wings to be the wing area times 1 foot. I take the density of fuel to be the same as water, 62.4 pounds per cubic foot. The total weight of the fuel is then estimated to be
This looks too big. I can rationalize a smaller number citing the taper of the wing in its thickness as I move from the fuselage out to wing tip. I cut this esti- mate in half, not knowing anything more than that the tip volume must be near zero. So my fuel weight estimate is now
All together then I estimate the lift force on a Boeing 747 at rush hour (fully loaded) to be
Is this estimate correct? Is it the right answer? Do I get an A? That depends upon the criteria used to differentiate right from wrong. Certainly we must allow for more than one numerical answer since there is no one numerical answer. If we admit a range, say of 20% either up or down, I may or may not pass. If we accept anything within a factor of 2, I am more confident, even willing to place a bet at 2 to 1 odds, that I am in the right. But go check it out: Jane’s All the World’s Air- craft will serve as a resource. Is the method correct? The criteria here are more certain: In the first place, it is essential that I identify the lift force as the weight. Without this conceptual leap, without an abstraction of the plane as a particle, I am blocked at the start. This is a nontrivial and potentially argumentative step. More about that later. Second, my method is more than a guess. It has a rationale, based upon a dis- section of the question into pieces – passenger weight, structural weight, fuel weight – each of which in turn I might guess. But, again, I can do better: I dissect the passenger weight into a sum of individual weights. Here now I am on firmer ground, able to construct an estimate more easily and with confidence because an individual’s weight is close at hand. So too with the structural and fuel weight; I reduce the question to simpler, more familiar terms and quantities. Fuel is like water in weight. The fuselage is a big aluminum can of football field dimensions. Here I have made a significant mistake in taking the specific gravity of aluminum as 8, which is that for steel. I
3 ⋅ ( pounds )
3 × × × × , ≈ ⋅ ( pounds )
3 ⋅ ( pounds )
970 ⋅ 10 3 pounds or approximately 500 tons.
Static Equilibrium Force and Moment 13
ought to have halved that factor, better yet, taken it as 3. My total estimate changes but not by a factor of 2. The method remains correct. Is this the only method, the only route to a rational estimate? No. A freshman thought of the weight of a school bus, fully loaded with forty passengers, and scaled up this piece. A graduate student estimated the lift force directly by consid- ering the change in momentum of the airstream (free stream velocity equal to the cruising speed of the 747) as it went over the wing. There are alternative routes to follow in constructing an estimate; there is no unique single right method as there is no unique, single right number. This does not mean that there are no wrong methods and estimates or that some methods are not better than others. Often you will not be able to develop a feel for the ingredients of an estimate or the behavior of a system, often because of a disjunction between the scale of things in your experience and the scale of the problem at hand. If that be the case, try to breakdown the system into pieces of a more familiar scale, building an asso- ciation with things you do have some feel for. More seriously, the dictates of the fundamental principles of static equilibrium might run counter to your expecta- tions. If this is the case, stick with it. In time what at first seems counter-intuitive will become familiar.
Exercise 2. What do you need to know to determine the force in cables AB and BC?
- You need to know that the cables support only tension. A cable is not able to support compression nor does it offer any resistance to bending. Web- ster’s New Collegiate Dictionary notes the “great tensile strength” of a cable but says nothing about bending or compression. The cable’s inability to support other than tension is critical to our understanding, our vocabu- lary. Bending itself will require definition... in time. - You need to know the weight of the block. - You need to know the angles AB and BC each make with the horizontal. (I will call them ΘA and ΘB). That’s what you need to know.
W
A C
B
Static Equilibrium Force and Moment 15
learning one’s way in Engineering Mechanics. The convention brings to the fore the necessity of isolating a particle before applying the equilibrium requirement. We see that what we need to know to determine the force in cable AB and in cable BC are the angles θA and θC and the weight of the block, W. These are the givens ; the magnitudes of the two forces, F A and F C are our two scalar unknowns. We read the above then as two scalar equations in two scalar unknowns. We have reduced the problem ... show that ... to a task in elementary algebra. To proceed requires a certain versatility in this more rarefied language. There are various ways to proceed at this point. I can multiply the first equa- tion by sinθA, the second by cosθA and add the two to obtain
Making use of an appropriate trigonometric identity, we can write: 1
Similarly, we find:
And thus we have shown what was asked to be shown. We have an answer, a unique answer in the sense that it is the only acceptable answer to the problem as stated, an answer that would merit full credit. But it is also an answer that has some depth, richness, a thick answer in that we can go beyond show that to show and tell and tease out of our result several interesting features.
- First, note that the derived equations are dimensionally correct. Both sides have the same units, that of force (or ML / T^2 ). In fact, we could easily obtain non-dimensional expressions for the cable tensions by dividing by the weight W. This linear relationship between the unknown forces in the cables and the applied load W will characterize most all of our discourse. It is a critical feature of our work in that it simplifies our task: If your boss asks you what will happen if some idiot accidentally doubles the weight hanging from the cables, you simply respond that the tensions in the cables will double.As always, there are caveats: We must assure ourselves first that the cables do not deform to any significant degree under double the weight. If they do then the angle θA and θC might change so a factor of 2.0 might not be quite exact. Of course if the cables break all bets are off.
- This is an example of textbook rhetoric cryptically indicating a skipped step in the analysis. The author’s pre- sumption is that you, the reader, can easily recognize what’s been left out. The problem is that it takes time, sometimes a long time, to figure out the missing step, certainly more time than it takes to read the sentence. If you are befuddled, an appropriate response then is to take some time out to verify the step.
F (^) C ⋅ ( sin θ A ⋅ cos θ C + sin θ C ⋅ cos θ A ) = W cos θ A
F (^) C = W ⋅ cos θ A ⁄ sin ( θ A +θ C )
F (^) A = W ⋅ cos θ C ⁄ sin ( θ A +θ C )
16 Chapter 2
- Second, note that the more vertically oriented of the two cables, the cable with its θ closer to a right angle, experiences the greater of the two ten- sions; we say it carries the greatest load. - Third, note that the tension in the cables can be greater than the weight of the suspended body. The denominator sin(θA + θC) can become very small, approaching zero as the sum of the two angles approaches zero. The numerators, on the other hand, remain finite; cosθ approaches 1.0 as θ approaches 0. Indeed, the maximum tension can become a factor of 10 or 100 or 1000... whatever you like... times the weight. - Fourth, note the symmetry of the system when θA is set equal to θC. In this case the tensions in the two cables are equal, a result you might have guessed, or should have been able to claim, from looking at the figure with the angles set equal. 2 - Fifth, if both angles approach a right angle, i.e., θA → π/2 and θC→ π/2, we have the opportunity to use “L’Hospital’s rule”. In this case we have
so each cable picks up half the weight. Other observations are possible: What if one of the angles is negative? What if a bird sits on a telephone wire? Or we might consider the graphical representation of the three vectors in equilibrium, as in the following: The figure below shows how you can proceed from knowing magnitude and direction of the weight vector and the directions of the lines of action of the forces in the two cables to full knowledge of the cable force vectors, i.e., their magni- tudes as well as directions. The figure in the middle shows the directions of the lines of action of the two cable forces but the line of action of the force in cable AB, inclined at an angle θA to the horizontal, has been displaced downward.
- Buridan, a medieval scholar in Mechanics would have cited the Principle of Sufficient Reason in explaining how the forces must be equal for the symmetric configuration. There is no reason why one or the other cable tensions should be greater or less than the other. Buridan’s ass, confronting two symmetrically placed bales of hay in front of his nose, starved to death. There was no sufficient reason to go left or right, so the story goes.
F (^) C F (^) A ( W cos θ⁄ sin 2 θ) θ →π ⁄ 2
lim (– Wsin θ⁄ 2 cos 2 θ) θ →π ⁄ 2
= = = lim = W ⁄ 2
F A
F C
W W
θA
θA
θC
θC
W
18 Chapter 2
We can also speak of a single force vector as being the resultant of its compo- nents, usually its three mutually perpen- dicular (or orthogonal ) rectangular, cartesian components. In the figure, the vector F has components F x , F y and F z.
The latter three mutually perpendicular vectors are usefully written as the product of a scalar magnitude and a unit vector indicating the direction of the vector com- ponent. The three unit vectors are often indicated by i, j and k and that is the con- vention we will follow in this text. The vector resultant or sum can be written out as
Note the convention for designating a vector quantity using bold face. This is the convention we follow in the text. In lecture it is difficult to write with chalk in bold face. It is also difficult for you to do so on your homework and exams. In these instances we will use the convention of placing a bar (or twiddle) over or under the letter to indicate it is a vector quantity.
Exercise 2. For each of the three force systems shown below, estimate the magnitude of the resultant, of F 1 and F 2. What is the direction of the resultant in each case?
For the force system at the left, the vertical component of the resultant force must be zero since the vertical components of the two forces are equal but one is positive (upward) and the other negative (downward). The symmetry with respect to the horizontal also leads to the result that the horizontal component of the resultant will be the sum of the two (equal) horizontal components of the two
k
F x
F y
F z.
F
y x
z
i
j
F = F x + F y + F z
or, equivalently
F = F x i + F y j + F z k
F 1 F^2^ F^1
F 2
θ θ
θ
F 1
F 2
θ
θ
10N
10N
10N
10N
10N 10N
θ
Static Equilibrium Force and Moment 19
forces. I estimate this sum to be 8 Newtons since the angle θ looks to be something
less than 30 o. The resultant is directed horizontally to the right. For the system in the middle, the resultant is zero since the two forces are equal but oppositely directed. If we were concerned with the static equilibrium of these three systems, only this system would be in equilibrium.
For the system at the far right, with the two forces symmetric with respect to the vertical, we now have that the horizontal component of the resultant force must be zero while the vertical component of the resultant will be the sum of the two (equal) vertical components of the two forces. I estimate this sum to be 16 ~ 18 Newtons. The resultant is directed vertically upward as shown above.
Friction Force Often the greatest challenges in applying the requirements of static equilibrium to useful purpose is isolating the particle (or later the body) and showing on your isolation, your free-body diagram , all the forces (and later all the torques as well) that act. These include reaction forces as well as forces applied like the weight due to gravity. Imagining and drawing these forces takes a certain facility in the creative use of the language of Engineering Mechanics, in particular, a facility with the characteristics of different kinds of forces. One of the kinds you will be responsible for reading out of a problem statement and writing into your free-body diagram is the force due to friction. Friction is tricky because sometimes it can be anything it needs to be; it’s direction as well as magnitude have a chameleon quality, taking on the colors that best meet the requirements of static equilibrium. But it can only be so big; it’s magnitude is limited. And when things begin to move and slide, it’s something else again. Friction is even more complicated in that its magnitude depends upon the surface materials which are in contact at the interface you have constructed in your free-body diagram but not upon the area in contact. This means that you have to go to a table in a reference book, ask a classmate, or call up a supplier, to obtain an appropriate value for the coefficient of friction.
F (^2) θ F 1
F 1
F 2
θ
θ
~ (2 *4N)
F = F 1 + F 2
~ (2 *8 or9N)
F = F 1 + F 2
θ
Static Equilibrium Force and Moment 21
remains one more unknown over the number of independent equations available. Now it’s not that we can’t find a solution; indeed we can find any number of sets of the three unknowns that serve: Just pick a value for any one of the three – FA ,
F (^) f , or N as some fraction of W, say W/2, and use the equilibrium equations to
solve for the remaining two. The problem is we can not find a unique solution. We say that system is under determined, or indeterminate. This is where impending motion comes to our rescue. We add the condition that we are at the point of impending motion. At impending motion the frictional force is related to the normal force by
where μ is called the coefficient of static friction.
Note that μ is a dimensionless quantity since both the normal force N and its associated friction force F (^) f have the same dimensions, that of force. The particular
value of the coefficient of static friction depends upon the materials in contact at the interface where the friction force acts. Another coefficient of friction is defined to cover the case of sliding at constant velocity. It is labeled the coefficient of sliding friction. It too depends upon the charac- ter of the materials in contact; it’s value is nominally less than μ. This is the third equation that allows us to estimate the force that will throw my back out. In fact, solving the three equations we find:
From this expression I can estimate the weight of the block I might be able to
move by pulling on the cable AB. For example, if θ = 60 o^ and I take μ = 0.25, as an estimate for sliding blocks along the ground, then setting FA = 50 lb. — an esti- mate of the maximum force I can exert without disastrous results — and solving for W, I find from the above
The friction force at impending motion is in this case, from the first equation of equilibrium,
If I pull with a force less than fifty pounds, say twenty pounds, still at an angle
of 60 o , on a block weighing 143.3 pounds or more, the block will not budge. The friction force F (^) f is just what it needs to be to satisfy equilibrium, namely F (^) f= 20
cos 60 o^ = 10 lb. This is what was meant by the statement “ ... can be anything it has to be.”The block in this case does not move, nor is it just about to move. As I increase the force with which I pull, say from twenty to fifty pounds, the frictional force increases proportionally from ten to twenty-five pounds, at which point the block begins to move and we leave the land of Static Equilibrium. That’s okay; I know now what weight block I can expect to be able to drag along the ground without injury. I need go no further.
F (^) f =μ ⋅ N
F (^) A =μ ⋅ W ⁄( cos θ+μ ⋅ sin θ)
W =143.3 lb.
F (^) f F (^) A cos θ 50 60 o = = ⋅ cos =25 lb.
22 Chapter 2
But wait! What, you say, if I push instead of pull the block? Won’t pushing be easier on my back? You have a point: I will now analyze the situation given that the AB is no longer a cable which I pull but signifies my arms pushing. In this we
keep θ equal to 60 o. At first glance, you might be tempted as I was, when I was a youth with a good back but little facility in speaking Engi- neering Mechanics, to simply change the sign of FA in the equilibrium equations and let it go at that: Solving would simply change the sign in our final expression for FA in terms of W. But that will not do. Friction is trickier: Friction always acts in a direction resisting the impend- ing motion. Here is another way it changes its colors to suit the context. No, I can’t get away so simply; I must redraw my free-body diagram carefully showing the new directions of the forces FA and F (^) f. In this I will label the force I apply, F (^) B
Equilibrium now gives, taking horizontal components to be positive when directed to the right and vertical components positive when directed upward:
These two, again supplemented by the relationship between the friction force and the normal force, namely
now yield
This is the force I must push with in order to just start the block sliding to the left – a state of impending motion. If I push with a force less than this, the block will not budge, the friction force is whatever the first equilibrium equation says it has to be. What force must I push with in order to move a block of weight W =
143.3 lb? Again, with θ = 60 o , the above expression gives
Note well the result! I must push with more than twice the force I must pull with in order to move the block! There is no mystery here. The reasons for this are all contained in the equations of equilibrium and the rules we have laid out which govern the magnitude and direction of the force due to friction. It is the latter that adds so much spice to our story. Note, I might go on and construct a story about how pushing down at an angle adds to the normal force of reaction which in turn implies that the frictional force resisting motion at the point of impending motion
θ
F B
F f
W
N
- F (^) B cos θ+ F (^) f = 0 and -F B sin θ+ N – W = 0
F (^) f =μ ⋅ N
F (^) B =μ ⋅ W ⁄( cos θ–μ ⋅ sin θ)
F (^) B =126.3 lb.
24 Chapter 2
In this, we choose the x - y axes shown as a reference frame. We make this choice to simplify our analysis. Only the weight vector W has both x and y compo- nents. Make a mental note of this way of crafting in setting up a problem. It is a bit of nuance of language use that can help you express your thoughts more effi- ciently than otherwise and yield a rich return, for example, on a quiz when time is precious. Equilibrium in the x direction, positive down the plane, then requires
while equilibrium in the y direction yields
I manipulate these to obtain
Now we know from the previous friction problem we analyzed that the friction force can only get so large relative to the normal force before motion will ensue. For the problem at hand, once the ratio of friction force to normal force reaches a value equal to the coefficient of static friction, μ, appropriate for the chair’s leg tips interfacing with the material of the table’s top, motion of the “particle”down the plane will follow. We can state this condition as an inequality. The student and chair will not slide down the plane as long as
This immediately yields the conclusion that as long as the angle φ is less than a certain value, namely if
the particle will not move. Note that, on the basis of this one-to-one relationship, we could define the condition of impending motion between two materials in terms of the value of the angle φ as easily as in terms of μ. For this reason, φ is sometimes called the friction angle. For example, if μ = 0.25 then the angle at which the chair and student will begin to slide is φ = 14 o^ Note, too, that our result is independent of the weights of the student and the chair. All students should begin to slide down the plane at the same angle. This was to be a central point in Professor X ’s demonstration: He planned to have a variety of students take a slide down the table top. Unfortunately the tallest person in class volunteered to go first.
N – W ⋅ cos φ= 0
F (^) f = N tan φ where N = W ⋅ cos φ
F (^) f = N ⋅ tan ϕ<μ ⋅ N
tan φ<μ
Static Equilibrium Force and Moment 25
Why did the demonstration fail? It failed because Professor X saw a particle where he should have envisioned an extended body. The figure at the below is an adaptation of a sketch drawn by a student in the front row just at the instant before the student and chair tipped for- ward. Note that the line of action of the weight vector of the chair-student combination, which I have added to the student’s sketch, passes through the point of contact of the front legs of the chair with the table top, point B. Note, too, that the angle φ is less than the friction angle, less than the value at which the chair would begin to slide. In the next instant, as the students charged with lifting the left end of the table did as they were told and raised their end up an inch, the line of action of W fell forward of point B and the accident ensued.
When is a particle a particle? The question perhaps is better phrased as “When is a body a particle?”The last exercise brings forcibly home how you can go wrong if you mistakenly read a par- ticle where you ought to imagine something of more substance. We have here a failure in modeling. Modeling is a process that requires you to represent “reality”in the language of Engineering Mechanics, to see in the world (or in the text in front of you) the conceptual ingredients of force, now of torque or moment and how the laws of static equilibrium and subsidiary relations like those that describe the action of a force due to friction, are to apply. It was Professor X’s failure to see the tipping moment about point B that led to his, rather the student’s, downfall. Modeling failures are common, like the cold. And there is no easy fix nor med- icine to prescribe that will ensure 100% success in modeling. One thing is essen- tial, at least here at the start: You must draw an isolation, a free-body diagram, as a first, critical step in your response to a problem. That until now, has meant, not just drawing a point on a clean sheet of paper – anyone can do that – but that you imagine all the force vectors acting on the particle and draw them too on your sheet of paper. This requires some thought. You must imagine; you must take risks; you must conjecture and test out your conjecture. In this you have available the beginnings of a vocabulary and some grammar to help you construct an appropriate isolation of, at least, a particle:
- gravity acts downward; - friction force acts to resist motion; - the normal force acts perpendicular to the plane of contact; - a cable can only sustain a tensile force;
W
φ=10ο •B
Static Equilibrium Force and Moment 27
neers, in their work, must deal with situations in which there is an excess of information while, at other times, situations in which there is insufficient informa- tion and conjecture and estimation is necessary. It’s best you learn straight off a bit more about the real world than the traditional text allows.
2.2 Concept of Moment
Force is not enough. You know from your studies in physics of the dynamics of bodies other than particles, that you must speak about their rotation as well as translation through space; about how they twist and turn.
Equilibrium of a Body We turn, then, to consider what we can say about forces, applied and reactive, when confronted with a body that cannot be seen as a particle but must be taken as having finite dimensions, as an extended body. Crucial to our progress will be the concept of moment or torque which can be interpreted as the turning effect of a force. We start again with a block on the ground. Instead of pushing or pulling, we explore what we can do with a lever. In particular we pose, as did Galileo (who also had a bad back),
Exercise 2. Estimate the magnitude of the force I must exert with my foot pressing down at B to just lift the end of the block at A up off the ground?
We isolate the system, this time as an extended body, showing all the applied and reaction forces acting on the system. The applied forces are the weight acting downward along a vertical line of action passing through the c.g. , the center of
28 Chapter 2
gravity of the block, and the force of my foot acting downward along a line of action through the point B at the right end of the lever AB.
The reaction forces are two: (1) the force of the ground acting up on the left end of the block at E and (2), the force of the ground acting up through the pivot at C upon the lever AB. Our quest is to determine the magnitude of the (vertical) force we must apply at B in order to just lift the end A off the ground. We start by applying our known requirement for static equilibrium – for a body at rest, or moving with constant velocity, the resultant force acting on the isolated body must be zero, must vanish. We have, taking up as positive,
We read this as one (scalar) equation with three unknowns, the applied force F (^) B and the two reactions F (^) E and F (^) C. Clearly we need to say something more. That “more”is contained in the following equilibrium requirement for an extended body – for a body at rest, or moving with constant velocity, the resultant moment of all forces acting on the isolated body must be zero, must vanish. I will find the resultant moment or torque of all the forces about the left-most point E. I will take as positive, a torque which tends to rotate the extended body of block and lever – all that lies within the dotted envelope – clockwise. For exam- ple, the moment about point E of the reaction force F (^) C is negative since it tends to rotate the system counter-clockwise about the reference point E. Its value is given by (x (^) ED+x (^) DA+x (^) AC ) F (^) C , the product of the force F (^) C and the perpendicular distance from the point E to the line of action of the force F (^) C. The resultant moment of all the forces acting on the isolated system is
We may read this equation as a second scalar equation in terms of the three unknown force quantities if we take the x ’ s , the distance measures, as known. We might, at this point, estimate the distances: the block length looks to be about one
meter. 4 Then, from the figure, estimate the other lengths by measuring their mag- nitudes relative to the length of the block. I will not do this. Instead, for reasons that will become evident, I will not state the block length but simply label it L then figure the x’s in terms of L.
- A better estimate might be obtained if the reader could identify the shrub at the left of the block. But that’s beyond the scope of the course.
x ED x DA x AC x CB F B
F C
W
F E
E (^) C
B D A
F E – W + F C – F B = 0
x (^) ED ⋅ W – ( x (^) ED + x (^) DA + x (^) AC ) ⋅ F (^) C + ( x (^) ED + x (^) DA + x (^) AC + x (^) CB ) ⋅ F (^) B = 0
