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We establish the direction and sense of M by applying the right-hand rule to the sequence r F. If the filgers of the right hand are curled in the direction of rotation from the positive sense of r to the positive sense of F, then the thumb points in the positive sense of M.

We must maintain the sequence r F, because the sequence F r would produce a vector with a sense opposite to that of the correct moment. As was the case with the scalar approach, the moment M may be thought of as the moment about point A or as the moment about the line O-O which passes through point A and is perpendicular to the plane containing the vectors r and F.

When we evaluate the moment of a force about a given point, the choice between using the vector cross product or the scalar expression depends on how the geometry of the problem is specified. If we know or can easily determine the perpendicular distance between the line of action of the force and the moment center, then the scalar approach is generally simpler.

If, however, F and r are not perpendicular and are easily expressible in vector notation, then the cross-product expression is often preferable. In Section B of this chapter, we will see how the vector formulation of the moment of a force is especially useful for determining the moment of a force about a point in three-dimensional situations.

Varignon's TheoremOne of the most useful principles of mechanics is Varignon's theorem, which states that the moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.

To prove this theorem, consider the force R acting in the plane of the body shown in Fig. Thus we could have used any number of concurrent components of R in the foregoing proof. The moment of R about point O is Rd. However, if d is more difficult to determine than p and q, we can resolve R into the components P and Q, and compute the moment as where we take the clockwise moment sense to be positive.

III By the principle of transmissibility, move the N force along its line of action to point B, which eliminates the moment of the component F 2.

The moment arm of F 1 becomes and the moment is Ans. IV Moving the force to point C eliminates the moment of the component F 1. The moment arm of F 2 becomes and the moment is Ans.

V By the vector expression for a moment, and by using the coordinate system indicated on the figure together with the procedures for evaluating cross products, we haveThe minus sign indicates that the vector is in the negative z-direction.

The magnitude of the vector expression is Ans. The tension everywhere in the cable is T, and this tension applied at A causes a moment M O about the hinge at O. What is the physical significance of this ratio?

We begin by constructing a figure which shows the tension force T acting directly on the door, which is shown in an arbitrary angular position. It should be clear that the direction of T will vary as varies.

In order to deal with this variation, we write a unit vector n AB which "aims" T:Using the x-y coordinates of our figure, we can write which is plotted in the accompanying graph. It has a maximum value of 0. The expression is valid even if T varies. This sample problem treats moments in two-dimensional force systems, and it also points out the advantages of carrying out a solution for an arbitrary position, so that behavior over a range of positions can be examined.

Here, r OB is more convenient than r OA. The moment produced by two equal, opposite, and noncollinear forces is called a couple. Couples have certain unique properties and have important applications in mechanics. These two forces cannot be combined into a single force because their sum in every direction is zero.

Their only effect is to produce a tendency of rotation. The combined moment of the two forces about an axis normal to their plane and passing through any point such as O in their plane is the couple M.

This couple has a magnitude or Its direction is counterclockwise when viewed from above for the case illustrated.

Note especially that the magnitude of the couple is independent of the distance a which locates the forces with respect to the moment center O. It follows that the moment of a couple has the same value for all moment centers. Vector Algebra MethodWe may also express the moment of a couple by using vector algebra.

With the cross-product notation of Eq. Thus, we may represent M by a free vector, as show in Fig. Because the couple vector M is always perpendicular to the plane of the forces which constitute the couple, in two-dimensional analysis we can represent the sense of a couple vector as clockwise or counterclockwise by one of the conventions shown in Fig.

Later, when we deal with couple vectors in three-dimensional problems, we will make full use of vector notation to represent them, and the mathematics will automatically account for their sense. Equivalent CouplesChanging the values of F and d does not change a given couple as long as the product Fd remains the same.

Likewise, a couple is not affected if the forces act in a different but parallel plane. Figure 2 In each of the four cases, the couples are equivalent and are described by the same free vector which represents the identical tendencies to rotate the bodies.

We can represent this dual effect more easily by replacing the given force by an equal parallel force and a couple to compensate for the change in the moment of the force.

The replacement of a force by a force and a couple is illustrated in Fig. Thus, we have replaced the original force at A by the same force acting at a different point B and a couple, without altering the external effects of the original force on the body. The combination of the force and couple in the right-hand part of Fig. By reversing this process, we can combine a given couple and a force which lies in the plane of the couple normal to the couple vector to produce a single, equivalent force.

Replacement of a force by an equivalent force-couple system, and the reverse procedure, have many applications in mechanics and should be mastered.

Determine the proper angle. We apply two equal and opposite N forces at O and identify the counterclockwise couple Ans. Thus, the original force is equivalent to the N force at O and the couple as shown in the third of the three equivalent figures. Proceeding in reverse is the same as replacing the couple by two forces, one of which is equal and opposite to the N force at O.

To show the effect of the N force on the weld, Wooden Boat House Signs 2019 replace the force by its equivalent of a force and couple M at O. Express M in vector notation. What force F must be exerted by the ground on each of the main braked wheels at A and B to counteract the turning effect of the two propeller thrusts? The weld at O can support a maximum of N of force along each of the n-and t-directions and a maximum of of moment. Determine the allowable range for the direction of the N force applied at A.

The part is subjected to the 4-N force exerted at A and a restoring moment exerted by a hidden torsional spring. Determine the y-intercept of the line of action of the single equivalent force. Now we are ready to describe the resultant action of a group or system of forces. Most problems in mechanics deal with a system of forces, and it is usually necessary to reduce the system to its simplest form to describe its action.

The resultant of a system of forces is the simplest force combination which can replace the original forces without altering the external effect on the rigid body to which the forces are applied.

Equilibrium of a body is the condition in which the resultant of all forces acting on the body is zero. This condition is studied in statics. When the resultant of all forces on a body is not zero, the acceleration of the body is obtained by equating the force resultant to the product of the mass and acceleration of the body.

This condition is studied in dynamics. Thus, the determination of resultants is basic to both statics and dynamics. The most common type of force system occurs when the forces all act in a single plane, say, the x-y plane, as illustrated by the system of three forces F 1 , F 2 , and F 3 in Fig.

We obtain the magnitude and direction of the resultant force R by forming the force polygon shown in part b of the figure, where the forces are added head-to-tail in any sequence. We see this in part a of the figure for the case of three forces where the sum R 1 of F 2 and F 3 is added to F 1 to obtain R.

The principle of transmissibility has been used in this process. The last equation specifies the distance d from point O to the line of action of R, and states that the moment of the resultant force about any point O equals the sum of the moments of the original forces of the system about the same point. This extends Varignon's theorem to the case of nonconcurrent force systems; we call this extension the principle of moments.

For a parallel force system, select a coordinate axis in the direction of the forces. If the resultant force R for a given force system is zero, the resultant of the system need not be zepo because the resultant may be a couple. The three forces in Fig. Point O is selected as a convenient reference point for the force-couple system which is to represent the given system.

The force-couple system consisting of R and M O is shown in Fig. We now determine the final line of action of R such that R alone represents the original system. Hence, the resultant R may be applied at any point on the line which makes a The resultant R may also be located by determining its intercept distance b to point C on the x-axis, Fig.

With R x and R y acting through point C, only R y exerts a moment about O so that Alternatively, the y-intercept could have been obtained by noting that the moment about O would be due to R x only.

Determine the y-intercept of the resultant of this force system. The force F acting at point O in Fig. In solving three-dimensional problems, one must usually find the x, y, and z scalar components of a force.

In most cases, the direction of a force is described a by two points on the line of action of the force or b by two angles which orient the line of action. If the coordinates of points A and B of Fig. Then resolve the horizontal component F xy into x-and y-components. The choice of orientation of the coordinate system is arbitrary, with convenience being the primary consideration.

However, we must use a right-handed set of axes in our three-dimensional work to be consistent with the right-hand-rule definition of the cross product. This is equivalent to the right-hand rule. Dot ProductWe can express the rectangular components of a force F or any other vector with the aid of the vector operation known Wooden Boat House Signs Jersey as the dot or scalar product see item 6 in Art.

The dot product of two vectors P and Q, Fig. In either case the dot product of the two vectors is a scalar quantity. In more general terms, if n is a unit vector in a specified direction, the projection of F in the n-direction, Fig.

If the direction cosines of F with respect to reference axes x-y-z are l, m, and n, then the projection of F in the n-direction becomes because and The latter two sets of equations are true because i, j, and k have unit length and are mutually perpendicular. The dot-product relationship applies to nonintersecting vectors as well as to intersecting vectors.

Thus, the dot product of the nonintersecting vectors P and Q in Fig. The line of action of F passes through a point A whose coordinates are 3 m, 4 m, and 5 m. Part a. We begin by writing the force vector F as its magnitude F times a unit vector n OA.

The desired scalar components are thus Ans. Part b. Part c. Write the force which the cable exerts on the boom BC as a vector T. Determine the angles x , y , and z which the line of action of T forms with the positive x-, y-, and z-axes. Ans Ans. In three dimensions, however, the determination of the perpendicular distance between a point or line and the line of action of the force can be a tedious computation.

A vector approach with cross-product multiplication then becomes advantageous. Moments in Three DimensionsConsider a force F with a given line of action acting on a body, Fig The vector M O is normal to the plane and is directed along the axis through O.

We can describe both the magnitude and the direction of M O by the vector cross-product relation introduced in Art. Refer to item 7 in Art. The vector r runs from O to any point on the line of action of F.

As described in Art. The correct direction and sense of the moment are established by the right-hand rule, described previously in Arts. Thus, with r and F treated as free vectors emanating from O, Fig. Note the symmetry and order of the terms, and note that a right-handed coordinate system must be used. This figure shows the three components of a force F acting at a point A located relative to O by the vector r.

The scalar magnitudes of the moments of these forces about the positive x-, y-, and z-axes through O can be obtained from the moment-arm rule, and are which agree with the respective terms in the determinant expansion for the cross product r F. If n is a unit vector in the -direction, then we can use the dot-product expression for the component of a vector as described in Art.

This scalar is the magnitude of the moment M of F about. The expression is known as a triple scalar product see item 8 in Art.

It need not be written because a cross product cannot be formed by a vector and a scalar. Thus, the association would have no meaning. The theorem is easily extended to three dimensions. The sum of the moments about O of these forces is This equation states that the sum of the moments of a system of concurrent forces about a given point equals the moment of their sum about the same point.

As mentioned in Art. Thus, the moment of a couple is the same about all points. The moment of a couple is a free vector, whereas the moment of a force about a point which is also the moment about a defined axis through the point is a sliding vector whose direction is along the axis through the point. As in the case of two dimensions, a couple tends to produce a pure rotation of the body about an axis normal to the plane of the forces which constitute the couple.

Couple vectors obey all of the rules which govern vector quantities. Thus, in Fig. In Art. You should also be able to carry out this replacement in three dimensions. The procedure is represented in Fig. Thus, we see that the couple vector is simply the moment of the original force about the point to which the force is being moved. We emphasize that r is a vector which runs from B to any point on the line of action of the original force passing through A.

It should be clear that the moment arm from the x-axis to the line of action of F is c and that the moment of F about the x-axis is negative. Similarly, the moment arm from the z-axis to the line of action of F is a and the moment of F about the z-axis is positive. So we have Ans. Determine the moment about point O of the cable force acting on point A and the magnitude of this moment. We begin by writing the described force as a vector.

The moment of this force about point O isAns. This vector has a magnitudeAns. The minus sign indicates that the vector M z is in the negative z-direction. Solution b. The force of magnitude T is resolved into components T z and T xyin the x-y plane. Since T z is parallel to the z-axis, it can exert no moment about this axis. The cosine of the angle between T and T xy is , and therefore,The moment arm d equals multiplied by the sine of the angle between T xy and OA, or Hence, the moment of T about the z-axis has the magnitude Ans.

Solution c. The component T xy is further resolved into its components T x and T y. It is clear that T y exerts no moment about the z-axis since it passes through it, so that the required moment is due to T x alone.

The N forces act parallel to the y-z plane. The direction of M 1 is normal to the plane defined by the two forces, and the sense, shown in the figure, is established by the right-hand convention. The two couple vectors combine to give the components Thus,Ans. In determining the effect of the force on the shaft at a cross section such as that at O, we may replace the force by an equivalent force at O and a couple. Describe this couple as a vector M.

The couple vector is perpendicular to the plane in which the force is shifted, and its sense is that of the moment of the given force about O. The direction of M in the y-z plane is given by Ans. Force P and the normal n to the face of the door lie in a vertical plane. Compute the moment of P about the z-axis. We found the magnitude and direction of the resultant force for the two-dimensional force system by a vector summation of forces, Eq.

These same principles can be extended to three dimensions. In the previous article we showed that a force could be moved to a parallel position by adding a corresponding couple. Thus, for the system of forces F 1 , F 2 , F 3. When all forces are shifted to O in this manner, we have a system of concurrent forces at O and a system of couple vectors, as represented in part b of the figure.

The concurrent forces may then be added vectorially to produce a resultant force R, and the couples may also be added to produce a resultant couple M, Fig. The point O selected as the point of concurrency for the forces is arbitrary, and the magnitude and direction of M depend on the particular point O selected. The magnitude and direction of R, however, are the same no matter which point is selected. In general, any system of forces may be replaced by its resultant force R and the resultant couple M.

In dynamics we usually select the mass center as the reference point. The change in the linear motion of the body is determined by the resultant force, and the change in the angular motion of the body is determined by the resultant couple.

In statics, the body is in complete equilibrium when the resultant force R is zero and the resultant couple M is also zero. Thus, the determination of resultants is essential in both statics and dynamics. We now examine the resultants for several special force systems. Coplanar Forces. Concurrent Wrench Resultant. When the resultant couple vector M is parallel to the resultant force R, as shown in Fig.

By definition a wrench is positive if the couple and force vectors point in the same direction and negative if they point in opposite directions. A common example of a positive wrench is found with the application of a screwdriver, to drive a right-handed screw. Any general force system may be represented by a wrench applied along a unique line of action.

This reduction is illustrated in Fig The wrench is the simplest form in which the resultant of a general force system may be expressed.

This form of the resultant, however, has limited application, because it is usually more convenient to use as the reference point some point O such as the mass center of the body or another convenient origin of coordinates not on the wrench axis. We choose point O as a convenient reference point for the initial step of reducing the given forces to a force-couple system. The resultant force isThe sum of the moments about O is Hence, the resultant consists of a couple, which of course may be applied at any point on the body or the body extended.

Solve with a vector approach. The value of y may, of course, be any value, as permitted by the principle of transmissibility. Thus, as expected, the variable y drops out of the above vector analysis. The resultant force has the components Thus, andAns.

The couple to be added as a result of moving the N force is where r is the vector from A to B. The term-by-term, or determinant, expansion givesThe moment of the N force about A is written by inspection of its x-and zcomponents, which givesThe moment of the N force about A is easily obtained from the moments of the x-and z-components of the force. Calculate the coordinates of the point P in the x-y plane through which the resultant force of the wrench acts.

Also find the magnitude of the couple M of the wrench. We see that M turned out to be negative, which means that the couple vector is pointing in the direction opposite to R, which makes the wrench negative. In addition, with his left hand he exerts a N force as shown in order to secure the socket onto the bolt head.

Determine the equivalent force-couple system at O. Then find the point in the x-y plane through which the line of action of the resultant force of the wrench passes. Replace the two forces acting on the frame by a wrench. Write the moment associated with the wrench as a vector and specify the coordinates of the point P in the y-z plane through which the line of action of the wrench passes.

Note that the force of magnitude F is parallel to the x-axis. Replace this loading system by a force-couple system at point A. Show that R is not perpendicular to M A. Then replace this force-couple system by a wrench. Determine the magnitude M of the moment of the wrench, the magnitude of the force R of the wrench, and the coordinates of the point in the x-z plane through which the line of action of the wrench passes.

Mastery of this material is essential for our study of equilibrium in the chapters which follow. Failure to correctly use the procedures of Chapter 2 is a common cause of errors in applying the principles of equilibrium. When difficulties arise, you should refer to this chapter to be sure that the forces, moments, and couples are correctly represented. ForcesThere is frequent need to represent forces as vectors, to resolve a single force into components along desired directions, and to combine two or more concurrent forces into an equivalent resultant force.

Specifically, you should be able to Resolve a given force vector into its components along given directions, and express the vector in terms of the unit vectors along a given set of axes. Express a force as a vector when given its magnitude and information about its line of action. This information may be in the form of two points along the line of action or angles which orient the line of action.

Use the dot product to compute the projection of a vector onto a specified line and the angle between two vectors. Compute the resultant of two or more forces concurrent at a point.

MomentsThe tendency of a force to rotate a body about an axis is described by a moment or torque , which is a vector quantity. We have seen that finding the moment of a force is often facilitated by combining the moments of the components of the force.

When working with moment vectors you should be able to Determine a moment by using the moment-arm rule. Use the vector cross product to compute a moment vector in terms of a force vector and a position vector locating the line of action of the force. Utilize Varignon's theorem to simplify the calculation of moments, in both scalar and vector forms.

Use the triple scalar product to compute the moment of a force vector about a given axis through a given point. CouplesA couple is the combined moment of two equal, opposite, and noncollinear forces. The unique effect of a couple is to produce a pure twist or rotation regardless of where the forces are located.

To solve problems involving couples you should be able to Compute the moment of a couple, given the couple forces and either their separation distance or any position vectors locating their lines of action. Replace a given force by an equivalent force-couple system, and vice versa. ResultantsWe can reduce an arbitrary system of forces and couples to a single resultant force applied at an arbitrary point, and a corresponding resultant couple.

We can further combine this resultant force and couple into a wrench to give a single resultant force along a unique line of action, together with a parallel couple vector.

To solve problems involving resultants you should be able to Compute the magnitude, direction, and line of action of the resultant of a system of coplanar forces if that resultant is a force; otherwise, compute the moment of the resultant couple. Apply the principle of moments to simplify the calculation of the moment of a system of coplanar forces about a given point. Replace a given general force system by a wrench along a specific line of action.

EquilibriumYou will use the preceding concepts and methods when you study equilibrium in the following chapters. Let us summarize the concept of equilibrium This means that its center of mass is either at rest or moving in a straight line with constant velocity. When both resultants are zero, the body is in complete equilibrium.

If N forces are applied as shown, determine the magnitude F of the equal forces exerted on the 6-mm rod by each of the four cutting surfaces so that their external effect on the rod is equivalent to that of the two N forces. Write each tension force acting on point A as a vector if the mass m is 60 kg. The quantities r, h, , and F are known.

Using the x-, y-, and z-coordinates shown, express F as a vector. Consider the projection of F onto line EF as a function of s. Determine and plot as a function of the tension T required to make the net moment about O zero. The effects of the radius of the pulley at B are negligible.

Determine and plot as a function of the moment at O due to the combined effects of the weight of part P, the N weight of member OA mass center at G 1 , and the N weight of member AB mass center at G 2.

The end grip is included as a part of member AB. The lengths L 1 and L 2 are mm and mm, respectively. What is the maximum value of M O and at what value of does this maximum occur? In many applications of mechanics, the sum of the forces acting on a body is zero, and a state of equilibrium exists. This apparatus is designed to hold a car body in equilibrium for a wide range of orientations during vehicle production. This chapter on equilibrium, therefore, constitutes the most important part of statics, and the procedures developed here form the basis for solving problems in both statics and dynamics.

We will make continual use of the concepts developed in Chapter 2 involving forces, moments, couples, and resultants as we apply the principles of equilibrium.

When a body is in equilibrium, the resultant of all forces acting on it is zero. All physical bodies are three-dimensional, but we can treat many of them as two-dimensional when the forces to which they are subjected act in a single plane or can be projected onto a single plane.

When this simplification is not possible, the problem must be treated as threedimensional. Omission of a force which acts on the body in question, or inclusion of a force which does not act on the body, will give erroneous results. A mechanical system is defined as a body or group of bodies which can be conceptually isolated from all other bodies.

A system may be a single body or a combination of connected bodies. The bodies may be rigid or nonrigid. The system may also be an identifiable fluid mass, either liquid or gas, or a combination of fluids and solids. In statics we study primarily forces which act on rigid bodies at rest, although we also study forces acting on fluids in equilibrium.

Once we decide which body or combination of bodies to analyze, we then treat this body or combination as a single body isolated from all surrounding bodies. This isolation is accomplished by means of the free-body diagram, which is a diagrammatic representation of the isolated system treated as a single body.

The diagram shows all forces applied to the system by mechanical contact with other bodies, which are imagined to be removed. If appreciable body forces are present, such as gravitational or magnetic attraction, then these forces must also be shown on the free-body diagram of the isolated system.

Only after such a diagram has been carefully drawn should the equilibrium equations be written. Because of its critical importance, we emphasize here that Before attempting to draw a free-body diagram, we must recall the basic characteristics of force. These characteristics were described in Art. Forces can be applied either by direct physical contact or by remote action.

Forces can be either internal or external to the system under consideration. Application of force is accompanied by reactive force, and both applied and reactive forces may be either concentrated or distributed. The principle of transmissibility permits the treatment of force as a sliding vector as far as its external effects on a rigid body are concerned.

We will now use these force characteristics to develop conceptual models of isolated mechanical systems. These models enable us to the free-body diagram is the most important single step in the solution of problems in mechanics. Each example shows the force exerted on the body to be isolated, by the body to be removed. Newton's third law, which notes the existence of an equal and opposite reaction to every action, must be carefully observed.

The force exerted on the body in question by a contacting or supporting member is always in the sense to oppose the movement of the isolated body which would occur if the contacting or supporting body were removed.

Because of its flexibility, a rope or cable is unable to offer any resistance to bending, shear, or compression and therefore exerts only a tension force in a direction tangent to the cable at its point of attachment. The force exerted by the cable on the body to which it is attached is always away from the body. When the tension T is large compared with the weight of the cable, we may assume that the cable forms a straight line.

When the cable weight is not negligible compared with its tension, the sag of the cable becomes important, and the tension in the cable changes direction and magnitude along its length. Type of Contact and Force Origin 6. Pin connection 7. Built-in or fixed support 8. Gravitational attraction 9. Spring action Action on Body to Be Isolated A freely hinged pin connection is capable of supporting a force in any direction in the plane normal to the pin axis.

We may either show two components R x and R y or a magnitude R and direction. A pin not free to turn also supports a couple M. A built-in or fixed support is capable of supporting an axial force F, a transverse force V shear force , and a couple M bending moment to prevent rotation. Spring force is tensile if spring is stretched and compressive if compressed.

For a linearly elastic spring the stiffness k is the force required to deform the spring a unit distance. Pin free to turnPin not free to turn fectly smooth, we can assume this to be so for practical purposes in many instances. Example 4 illustrates a number of forms of mechanical support which effectively eliminate tangential friction forces.

In these cases the net reaction is normal to the supporting surface. Example 5 shows the action of a smooth guide on the body it supports. There cannot be any resistance parallel to the guide. Example 6 illustrates the action of a pin connection.

Such a connection can support force in any direction normal to the axis of the pin. We usually represent this action in terms of two rectangular components.

The correct sense of these components in a specific problem depends on how the member is loaded. When not otherwise initially known, the sense is arbitrarily assigned and the equilibrium equations are then Wooden Boat House Signs Journal written. If the solution of these equations yields a positive algebraic sign for the force component, the assigned Wooden Boat House Signs sense is correct.

A negative sign indicates the sense is opposite to that initially assigned. If the joint is free to turn about the pin, the connection can support only the force R. If the joint is not free to turn, the connection can also support a resisting couple M. The sense of M is arbitrarily shown here, but the true sense depends on how the member is loaded.

Example 7 shows the resultants of the rather complex distribution of force over the cross section of a slender bar or beam at a build-in or fixed support.

The sense of the reactions F and V and the bending couple M in a given problem depends, of course, on how the member is loaded. One of the most common forces is that due to gravitational attraction, Example 8. This force affects all elements of mass in a body and is, therefore, distributed throughout it. The location of G is frequently obvious from the geometry of the body, particularly where there is symmetry. When the location is not readily apparent, it must be determined by experiment or calculations.

Similar remarks apply to the remote action of magnetic and electric forces. These forces of remote action have the same overall effect on a rigid body as forces of equal magnitude and direction applied by direct external contact.

Example 9 illustrates the action of a linear elastic spring and of a nonlinear spring with either hardening or softening characteristics. The representations in Fig. Study these nine conditions and identify them in the problem work so that you can draw the correct free-body diagrams. Construction of Free-Body DiagramsThe full procedure for drawing a free-body diagram which isolates a body or system consists of the following steps.

Step 1. Decide which system to isolate. The system chosen should usually involve one or more of the desired unknown quantities. Step 2. Next isolate the chosen system by drawing a diagram which represents its complete external boundary. This boundary defines the isolation of the system from all other attracting or contacting bodies, which are considered removed. This step is often the most crucial of all.

Make certain that you have completely isolated the system before proceeding with the next step. Step 3. Identify all forces which act on the isolated system as applied by the removed contacting and attracting bodies, and represent them in their proper positions on the diagram of the isolated system.

Make a systematic traverse of the entire boundary to identify all contact forces. Include body forces such as weights, where appreciable. Represent all known forces by vector arrows, each with its proper magnitude, direction, and sense indicated.

Each unknown force should be represented by a vector arrow with the unknown magnitude or direction indicated by symbol. If the sense of the vector is also unknown, you must arbitrarily assign a sense. The subsequent calculations with the equilibrium equations will yield a positive quantity if the correct sense was assumed and a negative quantity if the incorrect sense was assumed.

It is necessary to be consistent with the assigned characteristics of unknown forces throughout all of the calculations. If you are consistent, the solution of the equilibrium equations will reveal the correct senses. Step 4. Show the choice of coordinate axes directly on the diagram.

Pertinent dimensions may also be represented for convenience. Note, however, that the free-body diagram serves the purpose of focusing attention on the action of the external forces, and therefore the diagram should not be cluttered with excessive extraneous information. Clearly distinguish force arrows from arrows representing quantities other than forces. For this purpose a colored pencil may be used. Completion of the foregoing four steps will produce a correct freebody diagram to use in applying the governing equations, both in statics and in dynamics.

Be careful not to omit from the free-body diagram certain forces which may not appear at first glance to be needed in the calculations. It is only through complete isolation and a systematic representation of all external forces that a reliable accounting of the effects of all applied and reactive forces can be made. Very often a force which at first glance may not appear to influence a desired result does indeed have an influence.

Thus, the only safe procedure is to include on the free-body diagram all forces whose magnitudes are not obviously negligible. The free-body method is extremely important in mechanics because it ensures an accurate definition of a mechanical system and focuses attention on the exact meaning and application of the force laws of statics and dynamics. Review the foregoing four steps for constructing a freebody diagram while studying the sample free-body diagrams shown in Fig.

Dimensions and magnitudes are omitted for clarity. Thus, we may remove the entire truss from its supporting foundation and treat it as a single rigid body. In addition to the applied external load P, the free-body diagram must include the reactions on the truss at A and B.

The rocker at B can support a vertical force only, and this force is transmitted to the structure at B Example 4 of Fig. The pin connection at A Example 6 of Fig. If the total weight of the truss members is appreciable compared with P and the forces at A and B, then the weights of the members must be included on the free-body diagram as external forces.

In this relatively simple example it is clear that the vertical component A y must be directed down to prevent the truss from rotating clockwise about B. Also, the horizontal component A x will be to the left to keep the truss from moving to the right under the influence of the horizontal component of P. Thus, in constructing the free-body diagram for this simple truss, we can easily perceive the correct sense of each of the components of force exerted on the truss by the foundation at A and can, therefore, represent its correct physical sense on the diagram.

When the correct physical sense of a force or its component is not easily recognized by direct observation, it must be assigned arbitrarily, and the correctness of or error in the assignment is determined by the algebraic sign of its calculated value. In Example 2 the cantilever beam is secured to the wall and subjected to three applied loads. When we isolate that part of the beam to the right of the section at A, we must include the reactive forces applied to the beam by the wall.

The resultants of these reactive forces are shown acting on the section of the beam Example 7 of Fig. A vertical force V to counteract the excess of downward applied force is shown, and a tension F to balance the excess of applied force to the right must also be included. Then, to prevent the beam from rotating about A, a counterclockwise couple M is also required. The weight mg of the beam must be represented through the mass center Example 8 of Fig.

In the free-body diagram of Example 2, we have represented the somewhat complex system of forces which actually act on the cut section of the beam by the equivalent force-couple system in which the force is broken down into its vertical component V shear force and its horizontal component F tensile force. The couple M is the bending moment in the beam. The free-body diagram is now complete and shows the beam in equilibrium under the action of six forces and one couple.

The force exerted by the corner A on the beam is normal to the smooth surface of the beam Example 2 of Fig. To perceive this action more clearly, visualize an enlargement of the contact point A, which would appear somewhat rounded, and consider the force exerted by this rounded corner on the straight surface of the beam, which is as-sumed to be smooth. If the contacting surfaces at the corner were not smooth, a tangential frictional component of force could exist.

In addition to the applied force P and couple M, there is the pin connection at B, which exerts both an x-and a y-component of force on the beam. The positive senses of these components are assigned arbitrarily. In Example 4 the free-body diagram of the entire isolated mechanism contains three unknown forces if the loads mg and P are known.

Any one of many internal configurations for securing the cable leading from the mass m would be possible without affecting the external response of the mechanism as a whole, and this fact is brought out by the free-body diagram.

This hypothetical example is used to show that the forces internal to a rigid assembly of members do not influence the values of the external reactions. We use the free-body diagram in writing the equilibrium equations, which are discussed in the next article. When these equations are solved, some of the calculated force magnitudes may be zero. This would indicate that the assumed force does not exist. In Example 1 of Fig. A zero reaction force is often difficult to identify by inspection, but can be determined by solving the equilibrium equations.

Similar comments apply to calculated force magnitudes which are negative. Such a result indicates that the actual sense is the opposite of the assumed sense. The assumed positive senses of B x and B y in Example 3 and B y in Example 4 are shown on the free-body diagrams. The correctness of these assumptions is proved or disproved according to whether the algebraic signs of the computed forces are plus or minus when the calculations are carried out in an actual problem.

The isolation of the mechanical system under consideration is a crucial step in the formulation of the mathematical model.

The most important aspect to the correct construction of the all-important free-body diagram is the clear-cut and unambiguous decision as to what is included and what is excluded.

This decision becomes unambiguous only when the boundary of the free-body diagram represents a complete traverse of the body or system of bodies to be isolated, starting at some arbitrary point on the boundary and returning to that same point. The system within this closed boundary is the isolated free body, and all contact forces and all body forces transmitted to the system across the boundary must be accounted for. The following exercises provide practice with drawing free-body diagrams.

This practice is helpful before using such diagrams in the application of the principles of force equilibrium in the next article. The weights of the bodies are negligible unless otherwise indicated. Dimensions and numerical values are omitted for simplicity. The weights of the bodies are significant only if the mass is stated.

Stated in another way, a body is in equilibrium if all forces and moments applied to it are in balance. These requirements are contained in the vector equations of equilibrium, Eqs. They are necessary conditions because, if they are not satisfied, there can be no force or moment balance. They are sufficient because once they are satisfied, there can be no imbalance, and equilibrium is assured. The equations relating force and acceleration for rigid-body motion are developed in Vol.

Consequently, if a body moves with constant velocity zero acceleration , the resultant force on it must be zero, and the body may be treated as in a state of translational equilibrium. For complete equilibrium in two dimensions, all three of Eqs. However, these conditions are independent requirements, and one may hold without another. Take, for example, a body which slides along a horizontal surface with increasing velocity under the action of applied forces.

The force-equilibrium equations will be satisfied in the vertical direction where the acceleration is zero, but not in the horizontal direction. Also, a body, such as a flywheel, which rotates about its fixed mass center with increasing angular speed is not in rotational equilibrium, but the two force-equilibrium equations will be satisfied.

Categories of EquilibriumApplications of Eqs. The categories of force systems acting on bodies in two-dimensional equilibrium are summarized in Fig. Category 1, equilibrium of collinear forces, clearly requires only the one force equation in the direction of the forces x-direction , since all other equations are automatically satisfied.

Category 2, equilibrium of forces which lie in a plane x-y plane and are concurrent at a point O, requires the two force equations only, since the moment sum about O, that is, about a z-axis through O, is necessarily zero. Included in this category is the case of the equilibrium of a particle. Category 3, equilibrium of parallel forces in a plane, requires the one force equation in the direction of the forces x-direction and one moment equation about an axis z-axis normal to the plane of the forces.

Two-and Three-Force MembersYou should be alert to two frequently occurring equilibrium situations. The first situation is the equilibrium of a body under the action of two forces only. Two examples are shown in Fig. The shape of the member does not affect this simple requirement. In the illustrations cited, we consider the weights of the members to be negligible compared with the applied forces. The second situation is a three-force member, which is a body under the action of three forces, Fig.

We see that equilibrium requires the lines of action of the three forces to be concurrent. If they were not concurrent, then one of the forces would exert a resultant moment about the point of intersection of the other two, which would violate the requirement of zero moment about every point. The only exception occurs when the three forces are parallel. In this case we may consider the point of concurrency to be at infinity. The principle of the concurrency of three forces in equilibrium is of considerable use in carrying out a graphical solution of the force equations.

In this case the polygon of forces is drawn and made to close, as shown in Fig. Frequently, a body in equilibrium under the action of more than three forces may be reduced to a three-force member by a combination of two or more of the known forces.

The first way is illustrated in Fig. For the body shown in Fig. Therefore, an alternative set of equilibrium equations is where the two points A and B must not lie on a line perpendicular to the x-direction.

A third formulation of the equilibrium conditions may be made for a coplanar force system. This is illustrated in Fig. Thus, we may write the equations of equilibrium aswhere A, B, and C are any three points not on the same straight line. For example, for a general problem in two dimensions with three unknowns, three moment equations written about three points which lie on the same straight line are not independent.

Constraints and Statical DeterminacyThe equilibrium equations developed in this article are both necessary and sufficient conditions to establish the equilibrium of a body.

However, they do not necessarily provide all the information required to calculate all the unknown forces which may act on a body in equilibrium. Whether the equations are adequate to determine all the unknowns depends on the characteristics of the constraints against possible movement of the body provided by its supports.

By constraint we mean the restriction of movement. In Example 4 of Fig. Thus, a tangential force cannot be supported. For the collar and slider of Example 5, constraint exists only normal to the guide.

In Example 6 the fixed-pin connection provides constraint in both directions, but offers no resistance to rotation about the pin unless the pin is not free to turn.

The fixed support of Example 7, however, offers constraint against rotation as well as lateral movement. If the rocker which supports the truss of Example 1 in Fig. The three scalar conditions of equilibrium, Eqs. These two components of force would be dependent on the deformation of the members of the truss as influenced by their corresponding stiffness properties. The horizontal reactions A x and B x would also depend on any initial deformation required to fit the dimensions of the structure to those of the foundation between A and B.

Thus, we cannot determine A x and B x by a rigid-body analysis. Again referring to Fig. Therefore, there would be four unknown supporting reactions acting on the beam, namely, the force at A, the two components of force at B, and the couple at B.

A rigid body, or rigid combination of elements treated as a single body, which possesses more external supports or constraints than are necessary to maintain an equilibrium position is called statically indeterminate. Supports which can be removed without destroying the equilibrium condition of the body are said to be redundant. The number of redundant supporting elements present corresponds to the degree of statical indeterminacy and equals the total number of unknown external forces, minus the number of available independent equations of equilibrium.

On the other hand, bodies which are supported by the minimum number of constraints necessary to ensure an equilibrium configuration are called statically determinate, and for such bodies the equilibrium equations are sufficient to determine the unknown external forces. The problems on equilibrium in this article and throughout Vol. We must be aware of the nature of the constraints before we attempt to solve an equilibrium problem.

A body can be recognized as statically indeterminate when there are more unknown external reactions than there are available independent equilibrium equations for the force system involved. It is always well to count the number of unknown variables on a given body and to be certain that an equal number of independent equations can be written; otherwise, effort might be wasted in attempting an impossible solution with the aid of the equilibrium equations only.

The unknown variables may be forces, couples, distances, or angles. Adequacy of ConstraintsIn discussing the relationship between constraints and equilibrium, we should look further at the question of the adequacy of constraints. The existence of three constraints for a two-dimensional problem does not always guarantee a stable equilibrium configuration. In part a of the figure, point A of the rigid body is fixed by the two links and cannot move, and the third link prevents any rotation about A.

Thus, this body is completely fixed with three adequate proper constraints. In part b of the figure, the third link is positioned so that the force transmitted by it passes through point A where the other two constraint forces act.

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