Statics 2

. Compute the resultant of two or more forces concurrent at a point.

. 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.

. Use the triple scalar product to compute the moment of a force vector about a given axis through a given point.

1. Compute the moment of a couple, given the couple forces and either their separation distance or any position vectors locating their lines of action.

1. Determine a moment by using the moment-arm rule.

arignon's Theorem in Three Dimensions In Art. 2/4 we introduced Varignon's theorem in two dimensions. The theorem is easily extended to three dimensions. Figure 2/24 shows a system of concurrent forces F1, F2, F3, . . . . The sum of the moments about O of these forces is r ΣF r F1 r F2 r F3 r (F1 F2 F3 ) M M rx Fx ry Fy rz Fz r (F n) (r F)n r F n M (r F n)n MO n Mx ry Fz rz Fy My rz Fx rx Fz Mz rx Fy ry Fx Article 2/8 Moment and Couple 75 Figure 2/22 Fy Fz Fx F O A x y z rx ry Mx My Mz rz r F2 F1 F3 A O r MO λ r O Mλ n F Figure 2/24 Figure 2/23 where we have used the distributive law for cross products. Using the symbol MO to represent the sum of the moments on the left side of the above equation, we have (2/18) 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. 2/4, this principle has many applications in mechanics.

2. Replace a given force by an equivalent force-couple system, and vice versa

2. 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.

3. Utilize Varignon's theorem to simplify the calculation of moments, in both scalar and vector forms

According to Newton's third law, the action of a force is always accompanied by an equal and opposite reaction. It is essential to distinguish between the action and the reaction in a pair of forces. To do so, we first isolate the body in question and then identify the force exerted on that body (not the force exerted by the body). It is very easy to mistakenly use the wrong force of the pair unless we distinguish carefully between action and reaction.

Actual problems do not come with reference axes, so their assignment is a matter of arbitrary convenience, and the choice is frequently up to the student. The logical choice is usually indicated by the way in which the geometry of the problem is specified. When the principal dimensions of a body are given in the horizontal and vertical directions, for example, you would typically assign reference axes in these directions.

As a familiar example of the concept of moment, consider the pipe wrench of Fig. 2/8a. One effect of the force applied perpendicular to the handle of the wrench is the tendency to rotate the pipe about its vertical axis. The magnitude of this tendency depends on both the magnitude F of the force and the effective length d of the wrench handle. Common experience shows that a pull which is not perpendicular to the wrench handle is less effective than the right-angle pull shown.

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. 2/10d. 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.

Before dealing with a group or system of forces, it is necessary to examine the properties of a single force in some detail. A force has been defined in Chapter 1 as an action of one body on another. In dynamics we will see that a force is defined as an action which tends to cause acceleration of a body. A force is a vector quantity, because its effect depends on the direction as well as on the magnitude of the action. Thus, forces may be combined according to the parallelogram law of vector addition.

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.

Changing 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 2/11

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. 2. Apply the principle of moments to simplify the calculation of the moment of a system of coplanar forces about a given point. 3. Replace a given general force system by a wrench along a specific line of action.

Concurrent Forces. When forces are concurrent at a point, only the first of Eqs. 2/20 needs to be used because there are no moments about the point of concurrency.

Consider a force F with a given line of action acting on a body, Fig. 2/21a, and any point O not on this line. Point O and the line of F establish a plane A. The moment MO of F about an axis through O normal to the plane has the magnitude MO Fd, where d is the perpendicular distance from O to the line of F. This moment is also referred to as the moment of F about the point O.

Consider the action of two equal and opposite forces F and F a distance d apart, as shown in Fig. 2/10a. 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

Coplanar Forces. Article 2/6 was devoted to this force system.

Couple vectors obey all of the rules which govern vector quantities. Thus, in Fig. 2/26 the couple vector M1 due to F1 and F1 may be added M r F M rA F rB (F) (rA rB) F MO Σ(r F) r R 76 Chapter 2 Force Systems Figure 2/25 M1 M M2 F1 -F1 ≡ -F F M1 M2 F2 -F2 Figure 2/26 rA rB -F A d O B r F M as shown to the couple vector M2 due to F2 and F2 to produce the couple M, which, in turn, can be produced by F and F. In Art. 2/5 we learned how to replace a force by its equivalent force-couple system. You should also be able to carry out this replacement in three dimensions. The procedure is represented in Fig. 2/27, where the force F acting on a rigid body at point A is replaced by an equal force at point B and the couple M r F. By adding the equal and opposite forces F and F at B, we obtain the couple composed of F and the original F. 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.

Dimensions are not always given in horizontal and vertical directions, angles need not be measured counterclockwise from the x-axis, and the origin of coordinates need not be on the line of action of a force. Therefore, it is essential that we be able to determine the correct components of a force no matter how the axes are oriented or how the angles are measured. Figure 2/6 suggests a few typical examples of vector resolution in two dimensions

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

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.

Figure 2/8b shows a two-dimensional body acted on by a force F in its plane. The magnitude of the moment or tendency of the force to rotate the body about the axis O-O perpendicular to the plane of the body is proportional both to the magnitude of the force and to the moment arm d, which is the perpendicular distance from the axis to the line of action of the force. Therefore, the magnitude of the moment is defined as

For a concurrent system of forces where the lines of action of all forces pass through a common point O, the moment sum ΣMO about that point is zero. Thus, the line of action of the resultant R ΣF, determined by the first of Eqs. 2/10, passes through point O. 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 zero because the resultant may be a couple. The three forces in Fig. 2/15, for instance, have a zero resultant force but have a resultant clockwise couple M F3d

Forces are classified as either contact or body forces. A contact force is produced by direct physical contact; an example is the force exerted on a body by a supporting surface. On the other hand, a body force is generated by virtue of the position of a body within a force field such as a gravitational, electric, or magnetic field. An example of a body force is your weight.

Forces may be further classified as either concentrated or distributed. Every contact force is actually applied over a finite area and is therefore really a distributed force. However, when the dimensions of the area are very small compared with the other dimensions of the 24 Chapter 2 Force Systems Figure 2/2 P A θ θ (a) (b) Cable tension P Figure 2/1 C B A O P P The forces associated with this lifting rig must be carefully identified, classified, and analyzed in order to provide a safe and effective working environment. © mrfotos/iStockphoto body, we may consider the force to be concentrated at a point with negligible loss of accuracy. Force can be distributed over an area, as in the case of mechanical contact, over a volume when a body force such as weight is acting, or over a line, as in the case of the weight of a suspended cable.

Graphically, the correct line of action of R may be obtained by preserving the correct lines of action of the forces and adding them by the parallelogram law. We see this in part a of the figure for the case of three forces where the sum R1 of F2 and F3 is added to F1 to obtain R. The principle of transmissibility has been used in this process.

If the angle between the force F and the direction specified by the unit vector n is , then from the dot-product definition we have Fn cos F cos , where n n 1. Thus, the angle between F and n is given by (2/13) In general, the angle between any two vectors P and Q is (2/13a) If a force F is perpendicular to a line whose direction is specified by the unit vector n, then cos 0, and 0. Note that this relationship does not mean that either F or n is zero, as would be the case with scalar multiplication where (A)(B) 0 requires that either A or B (or both) be zero. 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. 2/20 is Q times the projection of P on Q, or PQ cos PQ cos because P and P are the same when treated as free vectors.

In Art. 2/6 we defined the resultant as the simplest force combination which can replace a given system of forces without altering the external effect on the rigid body on which the forces act. We found the magnitude and direction of the resultant force for the two-dimensional force system by a vector summation of forces, Eq. 2/9, and we located the line of action of the resultant force by applying the principle of moments, Eq. 2/10. These same principles can be extended to three dimensions

In Chapter 2 we have established the properties of forces, moments, and couples, and the correct procedures for representing their effects. 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.

In addition to combining forces to obtain their resultant, we often need to replace a force by its vector components in directions which are convenient for a given application. The vector sum of the components must equal the original vector. Thus, the force R in Fig. 2/3a may be replaced by, or resolved into, two vector components F1 and F2 with the specified directions by completing the parallelogram as shown to obtain the magnitudes of F1 and F2.

In addition to the tendency to move a body in the direction of its application, a force can also tend to rotate a body about an axis. The axis may be any line which neither intersects nor is parallel to the line of action of the force. This rotational tendency is known as the moment M of the force. Moment is also referred to as torque.

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.

In more general terms, if n is a unit vector in a specified direction, the projection of F in the n-direction, Fig. 2/19b, has the magnitude Fn . If we want to express the projection in the n-direction as a vector quantity, then we multiply its scalar component, expressed by , by the unit vector n to give Fn . We may write this as Fn without ambiguity because the term nn is not defined, and so the complete expression cannot be misinterpreted as . If the direction cosines of n are , , and , then we may write n in vector component form like any other vector as where in this case its magnitude is unity. 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.

In some two-dimensional and many of the three-dimensional problems to follow, it is convenient to use a vector approach for moment calculations. The moment of F about point A of Fig. 2/8b may be represented by the cross-product expression

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 F1, F2, F3 . . . acting on a rigid body in Fig. 2/28a, we may move each of them in turn to the arbitrary point O, provided we also introduce a couple for each force transferred. Thus, for example, we may move force F1 to O, provided we introduce the couple M1 r1 F1, where r1 is a vector from O to any point on the line of action of F1. 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. 2/28c. The general force system, then, is reduced to

In this and the following chapters, we study the effects of forces which act on engineering structures and mechanisms. The experience gained here will help you in the study of mechanics and in other subjects such as stress analysis, design of structures and machines, and fluid flow. This chapter lays the foundation for a basic understanding not only of statics but also of the entire subject of mechanics, and you should master this material thoroughly

In two-dimensional analyses it is often convenient to determine a moment magnitude by scalar multiplication using the moment-arm rule. 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.

It is usually helpful to master the analysis of force systems in two dimensions before undertaking three-dimensional analysis. Thus the remainder of Chapter 2 is subdivided into these two categories.

Many problems in mechanics require analysis in three dimensions, and for such problems it is often necessary to resolve a force into its three mutually perpendicular components. The force F acting at point O in Fig. 2/16 has the rectangular components Fx, Fy, Fz, where (2/11) The unit vectors i, j, and k are in the x-, y-, and z-directions, respectively. Using the direction cosines of F, which are l cos x, m cos y, and n cos z, where l 2 m2 n2 1, we may write the force as (2/12) We may regard the right-side expression of Eq. 2/12 as the force magnitude F times a unit vector nF which characterizes the direction of F, or (2/12a) It is clear from Eqs. 2/12 and 2/12a that nF li mj nk, which shows that the scalar components of the unit vector nF are the direction cosines of the line of action of F. 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.

Memorization of Eqs. 2/3 is not a substitute for understanding the parallelogram law and for correctly projecting a vector onto a reference axis. A neatly drawn sketch always helps to clarify the geometry and avoid error.

Moment directions may be accounted for by using a stated sign convention, such as a plus sign () for counterclockwise moments and a minus sign () for clockwise moments, or vice versa. Sign consistency within a given problem is essential. For the sign convention of Fig. 2/8d, the moment of F about point A (or about the z-axis passing through point A) is positive. The curved arrow of the figure is a convenient way to represent moments in two-dimensional analysis.

Moments The 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:

One 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.

Rectangular components are convenient for finding the sum or resultant R of two forces which are concurrent. Consider two forces F1 and F2 which are originally concurrent at a point O. Figure 2/7 shows the line of action of F2 shifted from O to the tip of F1 according to the triangle rule of Fig. 2/3. In adding the force vectors F1 and F2, we may write or from which we conclude that (2/4) The term ΣFx means "the algebraic sum of the x scalar components". For the example shown in Fig. 2/7, note that the scalar component would be negative.

Suppose the two concurrent forces lie in the same plane but are applied at two different points as in Fig. 2/3b. By the principle of transmissibility, we may move them along their lines of action and complete their vector sum R at the point of concurrency A, as shown in Fig. 2/3b. We can replace F1 and F2 with the resultant R without altering the external effects on the body upon which they act.

The action of the cable tension on the bracket in Fig. 2/1a is represented in the side view, Fig. 2/1b, by the force vector P of magnitude P. The effect of this action on the bracket depends on P, the angle , and the location of the point of application A. Changing any one of these three specifications will alter the effect on the bracket, such as the force in one of the bolts which secure the bracket to the base, or the internal force and deformation in the material of the bracket at any point. Thus, the complete specification of the action of a force must include its magnitude, direction, and point of application, and therefore we must treat it as a fixed vector.

The concept of the couple was introduced in Art. 2/5 and is easily extended to three dimensions. Figure 2/25 shows two equal and opposite forces F and F acting on a body. The vector r runs from any point B on the line of action of F to any point A on the line of action of F. Points A and B are located by position vectors rA and rB from any point O. The combined moment of the two forces about O is However, rA rB r, so that all reference to the moment center O disappears, and the moment of the couple becomes

The correct direction and sense of the moment are established by the right-hand rule, described previously in Arts. 2/4 and 2/5. Thus, with r and F treated as free vectors emanating from O, Fig. 2/21b, the thumb points in the direction of MO if the fingers of the right hand curl in the direction of rotation from r to F through the angle . Therefore, we may write the moment of F about the axis through O as (2/14) The order r F of the vectors must be maintained because F r would produce a vector with a sense opposite to that of MO; that is, F r MO.

The effect of a force acting on a body is the tendency to push or pull the body in the direction of the force, and to rotate the body about any fixed axis which does not intersect the line of the force. 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 first two of Eqs. 2/10 reduce a given system of forces to a force- couple system at an arbitrarily chosen but convenient point O. 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.

The moment M obeys all the rules of vector combination and may be considered a sliding vector with a line of action coinciding with the moment axis. The basic units of moment in SI units are newton-meters and in the U.S. customary system are pound-feet (lb-ft)

The moment is a vector M perpendicular to the plane of the body. The sense of M depends on the direction in which F tends to rotate the body. The right-hand rule, Fig. 2/8c, is used to identify this sense. We represent the moment of F about O-O as a vector pointing in the direction of the thumb, with the fingers curled in the direction of the rotational tendency

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

The most common two-dimensional resolution of a force vector is into rectangular components. It follows from the parallelogram rule that the vector F of Fig. 2/5 may be written as (2/1) where Fx and Fy are vector components of F in the x- and y-directions. Each of the two vector components may be written as a scalar times the F Fx Fy *Perpendicular projections are also called orthogonal projections. F2 F1 R1 R2 R1 R2 R F - F Figure 2/5 y x Fy F j F i x θ appropriate unit vector. In terms of the unit vectors i and j of Fig. 2/5, Fx Fxi and Fy Fy j, and thus we may write (2/2) where the scalars Fx and Fy are the x and y scalar components of the vector F. The scalar components can be positive or negative, depending on the quadrant into which F points. For the force vector of Fig. 2/5, the x and y scalar components are both positive and are related to the magnitude and direction of F by

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 F1, F2, and F3 in Fig. 2/13a. 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. Thus, for any system of coplanar forces we may write

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.

The properties of force, moment, and couple were developed in the previous four articles. 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

The quantities Fx, Fy, and Fz are the desired scalar components of F. 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. When we rotate from the x- to the y-axis through the 90 angle, the positive direction for the z-axis in a right-handed system is that of the advancement of a right-handed screw rotated in the same sense. This is equivalent to the right-hand rule

The replacement of a force by a force and a couple is illustrated in Fig. 2/12, where the given force F acting at point A is replaced by an equal force F at some point B and the counterclockwise couple M Fd. The transfer is seen in the middle figure, where the equal and opposite forces F and F are added at point B without introducing any net external effects on the body. We now see that the original force at A and the equal and opposite one at B constitute the couple M Fd, which is counterclockwise for the sample chosen, as shown in the right-hand part of the figure. 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. 2/12 is referred to as a force-couple system.

The vector MO is normal to the plane and is directed along the axis through O. We can describe both the magnitude and the direction of MO by the vector cross-product relation introduced in Art. 2/4. (Refer to item 7 in Art. C/7 of Appendix C.) The vector r runs from O to any point on the line of action of F. As described in Art. 2/4, the cross product of r and F is written r F and has the magnitude (r sin )F, which is the same as Fd, the magnitude of MO

The weight of a body is the force of gravitational attraction distributed over its volume and may be taken as a concentrated force acting through the center of gravity. The position of the center of gravity is frequently obvious if the body is symmetric. If the position is not obvious, then a separate calculation, explained in Chapter 5, will be necessary to locate the center of gravity.

There 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:

This conclusion is summarized by the principle of transmissibility, which states that a force may be applied at any point on its given line of action without altering the resultant effects of the force external to the rigid body on which it acts. Thus, whenever we are interested in only the resultant external effects of a force, the force may be treated as a sliding vector, and we need specify only the magnitude, direction, and line of action of the force, and not its point of application. Because this book deals essentially with the mechanics of rigid bodies, we will treat almost all forces as sliding vectors for the rigid body on which they act.

Thus, the moment of a couple is the same about all points. The magnitude of M is M Fd, where d is the perpendicular distance between the lines of action of the two forces, as described in Art. 2/5. 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

To gain more confidence in the cross-product relationship, examine the three components of the moment of a force about a point as obtained from Fig. 2/22. 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.

To obtain the resultant when the two forces F1 and F2 are parallel as in Fig. 2/4, we use a special case of addition. The two vectors are combined by first adding two equal, opposite, and collinear forces F and F of convenient magnitude, which taken together produce no external effect on the body. Adding F1 and F to produce R1, and combining with the sum R2 of F2 and F yield the resultant R, which is correct in magnitude, direction, and line of action. This procedure is also useful for graphically combining two forces which have a remote and inconvenient point of concurrency because they are almost parallel.

To prove this theorem, consider the force R acting in the plane of the body shown in Fig. 2/9a. The forces P and Q represent any two nonrectangular components of R. The moment of R about point O is Because R P Q, we may write Using the distributive law for cross products, we have (2/8) which says that the moment of R about O equals the sum of the moments about O of its components P and Q. This proves the theorem. Varignon's theorem need not be restricted to the case of two components, but it applies equally well to three or more. Thus we could have used any number of concurrent components of R in the foregoing proof.*

Two or more forces are said to be concurrent at a point if their lines of action intersect at that point. The forces F1 and F2 shown in Fig. 2/3a have a common point of application and are concurrent at the point A. Thus, they can be added using the parallelogram law in their common plane to obtain their sum or resultant R, as shown in Fig. 2/3a. The resultant lies in the same plane as F1 and F2.

Use the dot product to compute the projection of a vector onto a specified line and the angle between two vectors.

We can also use the triangle law to obtain R, but we need to move the line of action of one of the forces, as shown in Fig. 2/3c. If we add the same two forces as shown in Fig. 2/3d, we correctly preserve the magnitude and direction of R, but we lose the correct line of action, because R obtained in this way does not pass through A. Therefore this type of combination should be avoided.

We can express the rectangular components of a force F (or any other vector) with the aid of the vector operation known as the dot or scalar product (see item 6 in Art. C/7 of Appendix C). The dot product of two vectors P and Q, Fig. 2/19a, is defined as the product of their magnitudes times the cosine of the angle between them. It is written as

We can measure a force either by comparison with other known forces, using a mechanical balance, or by the calibrated movement of an elastic element. All such comparisons or calibrations have as their basis a primary standard. The standard unit of force in SI units is the newton (N) and in the U.S. customary system is the pound (lb), as defined in Art. 1/5.

We can now obtain an expression for the moment M of F about any axis through O, as shown in Fig. 2/23. 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. 2/7 to obtain , the component of MO in the direction of . This scalar is the magnitude of the moment M of F about . To obtain the vector expression for the moment M of F about , multiply the magnitude by the directional unit vector n to obtain (2/16) where r F replaces MO. The expression is known as a triple scalar product (see item 8 in Art. C/7, Appendix C). 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 triple scalar product may be represented by the determinant (2/17) where , , are the direction cosines of the unit vector n.

We 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:

We can separate the action of a force on a body into two effects, external and internal. For the bracket of Fig. 2/1 the effects of P external to the bracket are the reactive forces (not shown) exerted on the bracket by the foundation and bolts because of the action of P. Forces external to a body can be either applied forces or reactive forces. The effects of P internal to the bracket are the resulting internal forces and deformations distributed throughout the material of the bracket. The relation between internal forces and internal deformations depends on the material properties of the body and is studied in strength of materials, elasticity, and plasticity

We can use algebra to obtain the resultant force and its line of action as follows: 1. Choose a convenient reference point and move all forces to that point. This process is depicted for a three-force system in Figs. 2/14a and b, where M1, M2, and M3 are the couples resulting from the transfer of forces F1, F2, and F3 from their respective original lines of action to lines of action through point O. 2. Add all forces at O to form the resultant force R, and add all couples to form the resultant couple MO. We now have the single force- couple system, as shown in Fig. 2/14c. 3. In Fig. 2/14d, find the line of action of R by requiring R to have a moment of MO about point O. Note that the force systems of Figs. 2/14a and 2/14d are equivalent, and that Σ(Fd) in Fig. 2/14a is equal to Rd in Fig. 2/14d.

We can view this product either as the orthogonal projection P cos of P in the direction of Q multiplied by Q, or as the orthogonal projection Q cos of Q in the direction of P multiplied by P. In either case the dot product of the two vectors is a scalar quantity. Thus, for instance, we can express the scalar component Fx F cos x of the force F in Fig. 2/16 as , where F i is the unit vector in the x-direction.

We express the magnitude of a vector with lightface italic type in print; that is, F is indicated by F, a quantity which is always nonnegative. However, the scalar components, also denoted by lightface italic type, will include sign information. See Sample Problems 2/1 and 2/3 for numerical examples which involve both positive and negative scalar components.

We may also express the moment of a couple by using vector algebra. With the cross-product notation of Eq. 2/6, the combined moment about point O of the forces forming the couple of Fig. 2/10b is where rA and rB are position vectors which run from point O to arbitrary points A and B on the lines of action of F and F, respectively. Because rA rB r, we can express M as Here again, the moment expression contains no reference to the moment center O and, therefore, is the same for all moment centers. Thus, we may represent M by a free vector, as shown in Fig. 2/10c, where the direction of M is normal to the plane of the couple and the sense of M is established by the right-hand rule

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

We see from Fig. 2/30 that the axis of the wrench resultant lies in a plane through O normal to the plane defined by R and M. 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.

When both a force and its vector components appear in a diagram, it is desirable to show the vector components of the force with dashed lines, as in Fig. 2/5, and show the force with a solid line, or vice versa. With either of these conventions it will always be clear that a force and its components are being represented, and not three separate forces, as would be implied by three solid-line vectors

When dealing with the mechanics of a rigid body, we ignore deformations in the body and concern ourselves with only the net external effects of external forces. In such cases, experience shows us that it is not necessary to restrict the action of an applied force to a given point. For example, the force P acting on the rigid plate in Fig. 2/2 may be applied at A or at B or at any other point on its line of action, and the net external effects of P on the bracket will not change. The external effects are the force exerted on the plate by the bearing support at O and the force exerted on the plate by the roller support at C.

Wrench Resultant. When the resultant couple vector M is parallel to the resultant force R, as shown in Fig. 2/29, the resultant is called a wrench. 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. 2/30, where part a of the figure represents, for the general force system, the resultant force R acting at some point O and the corresponding resultant couple M. Although M is a free vector, for convenience we represent it as acting through O. In part b of the figure, M is resolved into components M1 along the direction of R and M2 normal to R. In part c of the figure, the couple M2 is replaced by its equivalent of two forces R and R separated by a distance Article 2/9 Resultants 89 M R R Positive wrench Negative wrench M Figure 2/29 d M2/R with R applied at O to cancel the original R. This step leaves the resultant R, which acts along a new and unique line of action, and the parallel couple M1, which is a free vector, as shown in part d of the figure. Thus, the resultants of the original general force system have been transformed into a wrench (positive in this illustration) with its unique axis defined by the new position of R

You will use the preceding concepts and methods when you study equilibrium in the following chapters. Let us summarize the concept of equilibrium: 1. When the resultant force on a body is zero (ΣF 0), the body is in translational equilibrium. This means that its center of mass is either at rest or moving in a straight line with constant velocity. 2. In addition, if the resultant couple is zero (ΣM 0), the body is in rotational equilibrium, either having no rotational motion or rotating with a constant angular velocity. 3. When both resultants are zero, the body is in complete equilibrium.

e point of concurrency. Parallel Forces. For a system of parallel forces not all in the same plane, the magnitude of the parallel resultant force R is simply the magnitude of the algebraic sum of the given forces. The position of its line of action is obtained from the principle of moments by requiring that r R MO. Here r is a position vector extending from the force-couple reference point O to the final line of action of R, and MO is the sum of the moments of the individual forces about O. See Sample Problem 2/17 for an example of parallel-force systems.

iven axes must not be confused with the relationship between a force and its perpendicular* projections onto the same axes. Figure 2/3e shows the perpendicular projections Fa and Fb of the given force R onto axes a and b, which are parallel to the vector components F1 and F2 of Fig. 2/3a. Figure 2/3e shows that the components of a vector are not necessarily equal to the projections of the vector onto the same axes. Furthermore, the vector sum of the projections Fa and Fb is not the vector R, because the parallelogram law of vector addition must be used to form the sum. The components and projections of R are equal only when the axes a and b are perpendicular.

ly speak of the moment about a point. By this we mean the moment with respect to an axis normal to the plane and passing through the point. Thus, the moment of force F about point A in Fig. 2/8d has the magnitude M Fd and is counterclockwise

oof.* Figure 2/9b illustrates the usefulness of Varignon's theorem. 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. Sample Problem 2/5 shows how Varignon's theorem can help us to calculate moments

ouples A 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. The couple is useful in replacing a force acting at a point by a force-couple system at Article 2/10 Chapter Review 99 a different point. To solve problems involving couples you should be able to:

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