Whichever team can generate the larger force, wins. As you can see, a competitive tug-of-war can be quite intense. Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law. Vectors are represented by arrows in diagrams and are distinguished from scalar quantities in this text through the use of boldface type P.
In this text, we use italic type to denote the magnitude of a vector. Thus, the magnitude of the vector P is denoted by P. A vector used to represent a force acting on a given particle has a well-defined point of application—namely, the particle itself.
Such a vector is said to be a fixed, or bound, vector and cannot be moved without modifying the conditions of the problem. Other physical quantities, however, such as couples see Chap. Still other physical quantities, such as forces acting on a rigid body see Chap. Two vectors that have the same magnitude and the same direction are said to be equal, regardless if they have the same point of application Fig.
The negative vector of a given vector P is defined as a vector having the same magnitude as P and a direction. Clearly, we have. Because the parallelogram constructed on the vectors P and Q does not depend on the order in which P and Q are selected, we conclude that the addition of two vectors is commutative, and we write. From the parallelogram law, we can derive an alternative method for determining the sum of two vectors, known as the triangle rule.
Consider Fig. Because the side of the parallelogram opposite Q is equal to Q in magnitude and Page 21 direction, we could draw only half of the parallelogram Fig.
The sum of the two vectors thus can be found by arranging P and Q in tip-to-tail fashion and then connecting the tail of P with the tip of Q. If we draw the other half of the parallelogram, as in Fig.
We define subtraction of a vector as the addition of the corresponding negative vector. Thus, we determine the. We write. Here again we should observe that, although we use the same sign to denote both vector and scalar subtraction, we avoid confusion by taking care to distinguish between vector and scalar quantities.
We now consider the sum of three or more vectors. The sum of three vectors P, Q, and S is, by definition,. Similarly, we obtain the sum of four vectors by adding the fourth vector to the sum of the first three. It follows that we can obtain the sum of any number of vectors by applying the parallelogram law repeatedly to successive pairs of vectors until all of the given vectors are replaced by a single vector. If the given vectors are coplanar, i. For this case, repeated application of the triangle rule is simpler than applying the parallelogram law.
In Fig. The triangle rule is first applied to obtain. This is known as the polygon rule for the addition of vectors. The result would be unchanged if, as shown in Fig. Recalling that vector addition also has been shown to be commutative in the case of two vectors, we can write. This expression, as well as others we can obtain in the same way, shows that the order in which several vectors are added together is immaterial Fig.
Product of a Scalar and a Vector. Therefore, we define the. Extending this definition to include all scalars and recalling the definition of a negative vector given earlier, we define the product kP of a scalar k and a vector P as a vector having the same direction as P if k is positive or a direction opposite to that of P if k is negative and a magnitude equal to the product of P and the absolute value of k Fig.
Because the forces all pass through A, they are also said to be concurrent. We can add the vectors representing the forces acting on A by the polygon rule Fig. Because the use of the polygon rule is equivalent to the repeated application of the parallelogram law, the vector R obtained in this way represents the resultant of the given concurrent forces.
That is, the single force R has the same effect on the particle A as the given forces. As before, the order in which we add the vectors P, Q, and S representing the given forces is immaterial. Conversely, a single force F acting on a particle may be replaced by two or more forces that, together, have the same effect on the particle. These forces are called components of the original force F, and the process of substituting them for F is called resolving the force F into components.
Clearly, each force F can be resolved into an infinite number of possible sets of components. Sets of two components P and Q are the most important as far as practical applications are concerned.
However, even then, the number of ways in which a given force F may be resolved into two components is unlimited Fig. In many practical problems, we start with a given vector F and want to determine a useful set of components. Two cases are of particular interest:. We obtain the second component, Q, by applying the triangle rule and joining the tip of P to the tip of F Fig.
We can determine the magnitude and direction of Q graphically or by trigonometry. Once we have determined Q, both components P and Q should be applied at A. We obtain the magnitude and sense of the components by applying the parallelogram law and drawing lines through the tip of F that are parallel to the given lines of action Fig.
This process leads to two well-defined components, P and Q, which can be determined graphically or computed trigonometrically by applying the law of sines.
You will encounter many similar cases; for example, you might know the direction of one component while the magnitude of the other component is to be as small as possible see Sample Prob. In all cases, you need to draw the appropriate triangle or parallelogram that satisfies the given conditions. Sample Problem 2. Determine their resultant. You can solve the problem graphically or by trigonometry. For a trigonometric solution, you can use the law of cosines and law of sines or use a right-triangle approach.
Draw to scale a parallelogram with sides equal to P and Q Fig. Measure the magnitude and direction of the resultant. They are. You can also use the triangle rule. Draw forces P and Q in tip-to-tail fashion Fig.
Again measure the magnitude and direction of the resultant. The answers should be the same. Trigonometric Solution. Using the triangle rule again, you know two sides and the included angle Fig. Apply the law of cosines. Now apply the law of sines: Page Alternative Trigonometric Solution. Construct the right triangle BCD Fig.
However, it is helpful to use a graphical solution as a check. Two tugboats are pulling a barge. If the resultant of the forces exerted by the tugboats is a lb force. You can solve the first part either graphically or analytically. In the second part, a graphical approach readily shows the necessary direction for rope 2, and you can use an analytical approach to complete the solution. For part b, use a variation of the triangle rule. Tension for. Graphical Solution.
Use the parallelogram law. The resultant the diagonal of the parallelogram is equal to lb and is directed to the right. Draw the sides parallel to the ropes Fig. If the drawing is done to scale, you should measure. Use the triangle rule. Note that the triangle in Fig. Using the law of sines,. With a calculator, compute and store the value of the last quotient. Multiply this value successively by.
The minimum value of occurs when and are. The key to part b is recognizing that the minimum value of occurs when and are. Knowing that both members are in tension.
Knowing that , determine by. Knowing that. Page 28 2. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of N. Knowing that the tension is lb in AB and 40 lb in AD, determine by trigonometry the magnitude and direction of the resultant of the forces exerted by the stays at A. Here we discuss how to add forces by using their components, especially rectangular components.
This method is often the most convenient way to add forces and, in practice, is the most common approach. Note that we can readily extend the properties of vectors established in this section to the rectangular components of any vector quantity, such as velocity or momentum.
Figure 2. The x and y axes are usually chosen to be horizontal and vertical, respectively, as in Fig. In determining the rectangular components of a force, you should think of the construction lines shown in Figs. This practice will help avoid mistakes in determining oblique components, as in Sec.
Force in Terms of Unit Vectors. To simplify working with rectangular components, we introduce two vectors of unit magnitude, directed respectively along the positive x and y axes. These vectors are called unit vectors and are denoted by i and j, respectively Fig. Recalling the definition of the product of a Page We have. However, when there exists no possibility of confusion, we may refer to the vector. A similar conclusion holds for the sign of the scalar. Scalar Components. Concept Application 2.
Determine the horizontal and vertical components of the force. What are the horizontal and vertical components of the force exerted by the rope at point A? Direction of a Force. The last calculation is easier if you store the value of when originally. From this law, we derived two other methods that are more readily applicable to the graphical solution of problems: the triangle rule for the addition of two forces and the polygon rule for the addition of three or more forces.
We also explained that the force triangle used to define the resultant of two forces could be used to obtain a trigonometric solution. However, when we need to add three or more forces, we cannot obtain any practical trigonometric solution from the force polygon that defines the resultant of the forces. In this case, the best approach is to obtain an analytic solution of the problem by resolving each force into two rectangular components.
Their resultant R is defined by the relation. From this equation, we can see that Page We thus conclude that when several forces are acting on a particle, we obtain the scalar components!
Clearly, this result also applies to the addition of other vector quantities, such as velocities, accelerations, or momenta. In practice, determining the resultant R is carried out in three steps, as illustrated in Fig. Resolve the given forces Fig. The procedure just described is most efficiently carried out if you arrange the computations in a table see Sample Prob. Although this is the only practical analytic method for adding three or more forces, it is also often preferred to the trigonometric solution in the case of adding two forces.
Determine the resultant of the forces on the bolt. In the table here, we entered the x and y components of each force as determined by trigonometry Fig. According to the convention adopted in this section, the scalar number representing a force component is positive if the force component has the same sense as the corresponding coordinate axis. Thus, x components acting to the right and y components acting upward are represented by positive numbers.
You can now determine the magnitude and direction of the resultant. From the triangle shown in Fig. Problems 2. Knowing that P must have a lb horizontal component, determine a the magnitude of the force P, b its vertical component. Knowing that P must have a N horizontal component, determine a the magnitude of the force P, b its vertical component. Knowing that P must have a lb vertical component, determine a the magnitude of the force P, b its horizontal component. Knowing that P must have a N component perpendicular to the pole AC, determine a the magnitude of the force P, b its component along line AC.
Knowing that Page The connection between equilibrium and the sum of forces is very direct: a particle can be in equilibrium only when the sum of the forces acting on it is zero. Treating the carabiner as a particle, it is in equilibrium because the resultant of all forces acting on it is zero.
Although it has not occurred in any of the problems considered so far, it is quite possible for the resultant to be zero. In such a case, the net effect of the given forces is zero, and the particle is said to be in equilibrium. We thus have the definition: When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium. A particle acted upon by two forces is in equilibrium if the two forces have the same magnitude and the same line of action but opposite sense.
The resultant of the two forces is then zero, as shown in Fig. Another case of equilibrium of a particle is represented in Fig. Starting from point. Thus, the resultant R of the given system of forces is zero, and the particle is in equilibrium. The closed polygon drawn in Fig.
To express algebraically the conditions for the equilibrium of a particle, we write. Resolving each force F into rectangular components, we have Page We conclude that the necessary and sufficient conditions for the equilibrium of a particle are Equilibrium of a particle scalar equations.
Returning to the particle shown in Fig. The first of these laws can be stated as: If the resultant force acting on a particle is zero, the particle will remain at rest if originally at rest or will move with constant speed in a straight line if originally in motion. From this law and from the definition of equilibrium just presented, we can see that a particle in equilibrium is either at rest or moving in a straight line with constant speed.
If a particle does not behave in either of these ways, it is not in equilibrium, and the resultant force on it is not zero. In the following section, we consider various problems concerning the equilibrium of a particle. In practice, this means designing a bridge or a building that remains stable and does not fall over. It also means understanding the forces that might act to disturb equilibrium, such as a strong wind or a flood of water. The basic idea is pretty simple, but the applications can be quite complicated.
A sketch showing the physical conditions of the problem is known as a space diagram. The methods of analysis discussed in the preceding sections apply to a system of forces acting on a particle. A large number of problems involving actual structures, however, can be reduced to problems concerning the equilibrium of a particle.
The method is to choose a significant particle and draw a separate diagram showing this particle and all the forces acting on it.
Such a diagram is called a free-body diagram. As an example, consider the kg crate shown in the space diagram of Fig. This crate was lying between two buildings, and is now being lifted onto a truck, which will remove it. The crate is supported by a vertical cable that is joined at A to two ropes, which pass over pulleys attached to the buildings at B and C.
We want to determine the tension in each of the ropes AB and AC. Note that the forces form a closed triangle because the particle is in equilibrium and the resultant force is zero. To solve this problem, we first draw a free-body diagram showing a particle in equilibrium. Because we are interested in the rope tensions, the free-body diagram should include at least one of these tensions or, if possible, both tensions.
You can see that point A is a good free body for this problem. The free-body diagram of point A is shown in Fig. It shows point A and the forces exerted on A by the vertical cable and the two ropes. The force exerted by the cable is directed downward, and its magnitude is equal to the weight W of the crate. The forces exerted by the two ropes are not known. Because they. No other detail is included in the free-body diagram.
Because point A is in equilibrium, the three forces acting on it must form a closed triangle when drawn in tip-. We have drawn this force triangle in Fig. If we choose trigonometry, we use the law of sines:. When a particle is in equilibrium under three forces, you can solve the problem by drawing a force triangle. When a particle is in equilibrium under more than three forces, you can solve the problem graphically by drawing a force polygon. If you need an analytic solution, you should solve the equations of equilibrium given in Sec.
These equations can be solved for no more than two unknowns. Similarly, the force triangle used in the case of equilibrium under three forces can be solved for only two unknowns. The most common types of problems are those in which the two unknowns represent 1 the two components or the magnitude and direction of a single force or 2 the magnitudes of two forces, each of known direction.
Problems involving the determination of the maximum or minimum value of the magnitude of a force are also encountered see Probs. A worker ties a rope to the cable at A and pulls on it to center the automobile over its intended position on the dock.
At the moment illustrated,. You can treat point A as a particle and solve the problem using a force triangle. Choose point A as the particle and draw the complete free-body diagram Fig. Equilibrium Condition. Because only three forces act on point A, draw a force triangle to express that it is in equilibrium Fig. Multiplying this value successively by. This basic type of problem will occur often as part of more complicated situations in this text. Note that the force exerted by the rollers on the package is perpendicular to the incline.
The new wrinkle is to determine a minimum force. You can approach this part of the solution in a way similar to Sample Prob. Choose the package as a free body, assuming that it can be treated as a particle. Then draw the corresponding free-body diagram Fig. Because only three forces act on the free body, draw a force triangle to express that it is in equilibrium Fig.
Line ' represents the known direction of P. To obtain the minimum value of the force F, choose the direction of F to be perpendicular to that of P. From the geometry of this triangle,. To do so, she places a model of the proposed hull in a test channel and uses three cables to keep its bow on the centerline of the channel. Dynamometer readings indicate that for a given speed, the tension is 40 lb in cable AB and 60 lb in cable AE.
Determine the drag force exerted on the hull and the tension in cable AC. Because four forces act at A tensions in three cables and the drag force , you should use the equilibrium conditions and sum forces by components to solve the unknown forces. Free-Body Diagram. Choosing point A as a free body, draw the free-body diagram Fig. It includes the forces exerted by the three cables on the hull, as well as the drag force exerted by the flow.
Because point A is in equilibrium, the resultant of all forces is zero: 1. Because more than three forces are involved, resolve the forces into x and y components Fig. This equation is satisfied if, and only if, the coefficients of i and j are each equal to zero. You obtain the following two equilibrium equations, which express, respectively, that the sum of the x components and the sum of the y components of the given forces must be zero.
Substituting this value into Eq. A positive sign in the answer indicates that the assumed sense is correct. You can draw the complete force polygon Fig. Case Study 2. As shown in CS Fig. A dance competition was held in the atrium on July 17, , with many guests congregating on the main floor as well as the three suspended walkways.
Suddenly, the fourth-floor walkway connections failed, causing this walkway to fall onto the second-floor walkway, with both then crashing onto the main floor see CS Photo 2. Tragically, people lost their lives and another were injured; in terms of human casualties, this was the worst structural failure in U. CS Photo 2. Note that the third-floor walkway remained intact. The support system of each walkway consisted of transverse beams, which were then attached to the hanger rods depicted in CS Fig.
Also shown is the critical fastener that was involved in the connection failure. The initial connection design for the fourth-floor walkway is illustrated in CS Fig. This would require turning the fastener on the threaded hanger rod all the way Page 45 from the second-floor end to the fourth-floor level.
During construction, it was realized that this would be impractical, and a new connection detail was implemented in the field, as shown in CS Fig. Then, treating the fastener and a small portion of the hanger as a particle, draw a free-body diagram and perform an equilibrium analysis. The loads involved are shown in CS Fig.
The hanger system supports a portion of the second-floor and fourth-floor walkways, and the resulting. Treating the critical fastener and a small portion of the hanger as a particle, the free-body diagrams in each case are shown in CS Fig. Original Connection Design.
From CS Fig. In the same manner, one should avoid shortcuts in analyzing engineering mechanics problems, and instead employ the complete SMART methodology, even for very simple situations like the one considered here. It should also be noted that there were other important factors that contributed to this tragedy besides the circumstance examined in this Case Study.
Along with the report cited earlier, these factors have been discussed in a number of other publications as well. The reader is strongly encouraged to study further. Determine the tension a in cable AC, b in cable BC. Knowing that the.
Knowing that the connection is. In each case, determine the tension in each cable. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable. Page 50 Determine the magnitude of the force P required to maintain the equilibrium of the collar when a. Knowing that the mass of the crate is kg, determine the tension in the cable for each of the arrangements shown.
Determine Page 51 for each arrangement the tension in the rope. Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chap.
Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. See the hint for Prob. The pulley is held in the position shown by a second cable CAD, which passes over pulley A and supports a load P.
Determine a the tension in cable ACB, b the magnitude of load P. In the last part of this chapter, we discuss problems involving the three dimensions of space. This plane passes through the vertical y. We can resolve the force F into a vertical component '" and a. This operation, shown in Fig. The force F is then represented by the main. These triangles occupy positions in the box comparable with that of triangle OAB.
Introducing the unit vectors i, j, and k, which are directed respectively along the x, y, and z axes Fig. As in the case of two-dimensional problems, a plus sign indicates that the component has the same sense as the corresponding axis, and a minus sign indicates that it has the opposite sense. The angle a force F forms with an axis should be measured from the positive side of the axis and is always. An angle 1! In Concept Application 2.
Substituting into Eq. This equation shows that the force F can be expressed as the product of the scalar F and the vector. It follows from Eq. Also shown are the components of F and its unit vector. Recalling that the sum of the. From the direction cosines, we can find the angles 1! These computations can be carried out easily with a calculator. It follows that the scalar components of F are, respectively, Scalar components of force F.
The relations in Eq. Graphical or trigonometric methods are generally not practical in the case of forces in space. The method followed here is similar to that used in Sec. Page 58 Sample Problem 2. The tension in the wire is N. Determine a the. From that, we can find the components of the tension and the angles defining its direction.
Components of the Force. The line of action of the force acting on the bolt passes through. The components of the vector , which has the same. Direction of the Force. Using Eqs. Calculating each quotient and its arc cosine, you obtain Page Note: You could have obtained this same result by using the components and magnitude of the vector. The methods in this section allow you to translate back and forth between forces and geometry.
If the tension is lb in cable AB and lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. The simplest approach is to first resolve the forces into components and to then sum the components and find the resultant.
First resolve the force exerted by each cable on stake A into x, y, and z components. To do this, determine the components and magnitude of the vectors and , measuring.
Denoting the unit vectors along the coordinate axes by i, j, k, these vectors are. The outcome of was not as. Knowing that the tension in wire AC is lb, determine a the components of the. Knowing that the tension in wire AD is 85 lb, determine a the components of the force.
Knowing that the tension in cable BG is N, determine the components of the force exerted by cable BG on the frame at B. Knowing that the tension in cable BH is N, determine the components of the force exerted by cable BH on the frame at B.
Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable. Knowing that the tension in cable AC is 1. Knowing that the tension in the cable is N, determine the magnitude and direction of the resultant of the forces exerted by the cable at B.
Knowing that the tension in cable AB is lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. The components , , and of the resultant of forces in space are given by Eqs. We can use them to solve problems dealing with the equilibrium of a particle involving no more than three unknowns. The first step in solving three-dimensional equilibrium problems is to draw a free-body diagram showing the particle in equilibrium and all of the forces acting on it.
You can then write the equations of equilibrium 2. In the more common types of problems, these unknowns will represent 1 the three components of a single force or 2 the magnitude of three forces, each of known direction. Page 65 Sample Problem 2. A horizontal force P perpendicular to the wall holds the cylinder in the position shown. Determine the magnitude of P and the tension in each cable. You can use the given geometry to express the force components of the cables and then apply equilibrium conditions to calculate the tensions.
Choose point A as a free body; this point is subjected to four forces, three of which are of unknown magnitude. Introducing the unit vectors i, j, and k, resolve each force into rectangular components Fig. Page 66 For and , it is first necessary to determine the components and magnitudes of the vectors. Setting the coefficients of i, j, and k equal to zero, you can write three scalar equations, which express that the sums of the x, y, and z components of the forces are respectively equal to zero.
Conversely, if one of the cable force results had been negative, thereby reflecting compression instead of tension, you should recognize that the solution is in error. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. Determine the weight W of the container, knowing that the tension in cable AD is 4.
Knowing that the balloon exerts an N vertical force at A, determine the tension in each cable. Determine the weight W of the crate, knowing that the tension in cable AB is lb. Determine the tension in each wire when a lb cylinder is suspended from point D, as shown.
Determine the tension in each wire,. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. Knowing that the tension in cable AD is N, determine the weight of the plate. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.
Knowing Page Cable BAC passes through the ring and is. Hint: The tension is the same in both portions of. Knowing that a lb vertical load P is applied to ring A, determine the tension in each of the three cables. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is. Knowing that , determine the magnitude of.
Knowing that at the instant shown the counterweight is kept from moving and that the. Hint: Because there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute. If a lb force Q is applied to collar B, as shown, determine a the tension in the wire when. Determine the distances x and z for which the equilibrium of the system is maintained when. Page 73 2. Resultant of Two Forces Forces are vector quantities; they are characterized by a point of application, a magnitude, and a direction, and they add according to the parallelogram law Fig.
We can determine the magnitude and direction of the resultant R of two forces P and Q either graphically or by trigonometry using the law of cosines and the law of sines [Sample Prob. Components of a Force Any given force acting on a particle can be resolved into two or more components, i. A force F can be resolved into two components P and Q by drawing a parallelogram with F for its diagonal; the components P and Q are then represented by the two adjacent sides of the parallelogram Fig. Again, we can determine the components either graphically or by trigonometry [Sec.
Rectangular Components; Unit Vectors A force F is resolved into two rectangular components if its components and are perpendicular to each.
Introducing the unit vectors i and j along the x and y axes, respectively, we can write the components and the vector as [Sec. These components, which can be positive or negative, are. We can obtain the magnitude F of the force by solving one of the Eqs. Resultant of Several Coplanar Forces When three or more coplanar forces act on a particle, we can obtain the rectangular components of their resultant R by adding the corresponding components of the given forces algebraically [Sec.
The magnitude and direction of R then can be determined from relations similar to Eqs. Forces in Space A force F in three-dimensional space can be resolved into rectangular components , , and [Sec. Denoting by , , and , respectively, the angles that F forms with the x, y, and z axes Fig. Direction Cosines The cosines of , , and are known as the direction cosines of the force F. Introducing the unit vectors i,. This last equation shows Fig. When we are given the rectangular components , , and of a force F, we can find the magnitude F.
When a force F is defined in three-dimensional space by its magnitude F and two points M and N on its. From this equation it follows [Sample Probs. Resultant of Forces in Space When two or more forces act on a particle in three-dimensional space, we can obtain the rectangular components of their resultant R by adding the corresponding components of the given forces algebraically [Sec.
We can then determine the magnitude and direction of R from relations similar to Eqs. Equilibrium of a Particle A particle is said to be in equilibrium when the resultant of all the forces acting on it is zero [Sec.
The particle remains at rest if originally at rest or moves with constant speed in a straight line if originally in motion [Sec. Free-Body Diagram To solve a problem involving a particle in equilibrium, first draw a free-body diagram of the particle showing all of the forces acting on it [Sec.
If only three coplanar forces act on the particle, you can draw a force triangle to express that the particle is in equilibrium. Using graphical methods of trigonometry, you can solve this triangle for no more than two unknowns [Sample Prob.
If more than three coplanar forces are involved, you should use the equations of equilibrium:. Equilibrium in Space When a particle is in equilibrium in three-dimensional space [Sec. These equations can be solved for no more than three unknowns [Sample Prob. Review Problems 2. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.
This force develops due to the moment M applied at A, as shown; the concept of moments will be introduced in Chap. Knowing that P must have a N component perpendicular to member AB, determine a the magnitude of the force P, b its component along line AB. Page 78 2. The directions of the two N forces may vary,.
Determine a the x, y, and z. Determine the vertical force P exerted Page Determine the vertical force P exerted by the balloon at A, knowing that the tension in cable AC is N.
If the tension in wire AB is lb, determine the vertical force P exerted by the tower on the pin at A. If the tension in wire AC is lb, determine the vertical force P exerted by the tower on the pin at A. It will be shown in this chapter that the forces exerted on the ship by the tugboats could be replaced by an equivalent force exerted by a single, more powerful, tugboat.
Objectives Discuss the principle of transmissibility that enables a force to be treated as a sliding vector. Define the moment of a force about a point. Examine vector and scalar products, useful in analysis involving moments. Define the mixed triple product and use it to determine the moment of a force about an axis.
Define the moment of a couple, and consider the particular properties of couples. Resolve a given force into an equivalent force-couple system at another point. Reduce a system of forces into an equivalent force- couple system. Examine circumstances where a system of forces can be reduced to a single force.
Introduction 3. Introduction In Chap. Such a view, however, is not always possible. In general, a body should be treated as a combination of a large number of particles. In this case, we need to consider the size of the body as well as the fact that forces act on different parts of the body and thus have different points of application.
Most of the bodies considered in elementary mechanics are assumed to be rigid. We define a rigid body as one that does not deform. Actual structures and machines are never absolutely rigid and deform under the loads to which they are subjected. However, these deformations are usually small and do not appreciably affect the conditions of equilibrium or the motion of the structure under consideration. They are important, though, as far as the resistance of the structure to failure is concerned and are considered in the study of mechanics of materials.
In this chapter, you will study the effect of forces exerted on a rigid body, and you will learn how to replace a given system of forces by a simpler equivalent system. It follows that forces acting on a rigid body can be represented by sliding vectors, as indicated in Sec. Page 82 Two important concepts associated with the effect of a force on a rigid body are the moment of a force about a point Sec.
The determination of these quantities involves computing vector products and scalar products of two vectors; so in this chapter, we introduce the fundamentals of vector algebra and apply them to the solution of problems involving forces acting on rigid bodies. Another concept introduced in this chapter is that of a couple, i. As you will see, we can replace any system of forces acting on a rigid body by an equivalent system consisting of one force acting at a given point and one couple.
This basic combination is called a force-couple system. In the case of concurrent, coplanar, or parallel forces, we can further reduce the equivalent force-couple system to a single force, called the resultant of the system, or to a single couple, called the resultant couple of the system.
However, the effects of the force can be very different, depending on factors such as the point of application or line of action of that force. As a result, calculations involving forces acting on a rigid body are generally more complicated than situations involving forces acting on a point.
We begin by examining some general classifications of forces acting on rigid bodies. External forces are exerted by other bodies on the rigid body under consideration. They are entirely responsible for the external behavior of the rigid body, either causing it to move or ensuring that it remains at rest. We shall be concerned only with external forces in this chapter and in Chaps. Internal forces hold together the particles forming the rigid body.
If the rigid body is structurally composed of several parts, the forces holding the component parts together are also defined as internal forces. We will consider internal forces in Chaps. As an example of external forces, consider the forces acting on a disabled truck that three people are pulling forward by means of a rope attached to the front bumper Fig. The external forces acting on the truck are shown in a free-body diagram Fig. Note that this free-body diagram shows the entire object, not just a particle representing the object.
Let us first consider the weight of the truck. The point of application of this force—that is, the point at which the force acts—is Page 83 defined as the center of gravity of the truck. The weight W tends to make the truck move vertically downward. In fact, it would actually cause the truck to move downward, i. The ground opposes the downward motion of the. The people pulling on the rope exert the force F. The point of application of F is on the front bumper.
The force F tends to make the truck move forward in a straight line; the force actually makes it move, because no external force opposes this motion. We are ignoring rolling resistance here for simplicity.
This forward motion of the truck, during which each straight line keeps its original orientation the floor of the truck remains horizontal, and the walls remain vertical , is known as a translation. Other forces might cause the truck to move differently. For example, the force exerted by a jack placed under the front axle would cause the truck to pivot about its rear axle.
Such a motion is a rotation. We conclude, therefore, that each external force acting on a rigid body can, if unopposed, impart to the rigid body a motion of translation or rotation, or both. This principle, which states that the action of a force may be transmitted along its line of action, is based on experimental evidence.
It cannot be derived from the properties established so far in this text and therefore must be accepted as an experimental law. We indicated in Chap. These vectors had a well-defined point of application—namely, the particle itself—and were therefore fixed, or bound, vectors. In the case of forces acting on a rigid body, however, the point of application of the force does not matter, as long as the line of action remains unchanged. Thus, forces acting on a rigid body must be represented by a different kind of vector, known as a sliding vector, because forces are allowed to slide along their lines of action.
Note that all of the properties we derive in the following sections for the forces acting on a rigid body are valid more generally for any system of sliding vectors. To keep our presentation more intuitive, however, we will carry it out in terms of physical forces rather than in terms of mathematical sliding vectors. Returning to the example of the truck, we first observe that the line of action of the force F is a horizontal line passing through both the front and rear bumpers of the truck Page 84 Fig.
In other words, the conditions of motion are. Thus, in terms of the external behavior of the bar, the original system of forces shown in Fig. Both sets produce the same external effect equilibrium in this case but different internal forces and deformations.
However, the internal forces and deformations produced by the two systems are clearly different. The bar of Fig. Thus, although we can use the principle of transmissibility to determine the conditions of motion or equilibrium of rigid bodies and to compute the external forces Page 85 acting on these bodies, it should be avoided, or at least used with care, in determining internal forces and deformations. However, this concept is more clearly understood and is applied more effectively if we first add to the mathematical tools at our disposal the vector product of two vectors.
The vector product of two vectors P and Q is defined as the vector V that satisfies the following conditions. The line of action of V is perpendicular to the plane containing P and Q Fig. We thus have Magnitude of a vector product.
The direction of V is obtained from the right-hand rule. Your thumb then indicates the direction of the vector V Fig.
Note that if P and Q do not have a common point of application, you should first redraw them from the same point.
The three vectors P, Q, and V—taken in that order —are said to form a right-handed triad. As stated previously, the vector V satisfying these three conditions which define it uniquely is referred to as the vector product of P and Q. It is represented by the mathematical expression.
The vector product 1 g 2 is therefore unchanged if. We saw that the commutative property does not apply to vector products. However, it can be demonstrated that the distributive property. A third property, the associative property, does not apply to vector products; we have in general. To do this, we use the unit vectors i, j, and k that were defined in Chap.
This unit vector must be k, because the vectors i, j, and k are mutually perpendicular and form a right-handed triad. Similarly, it follows from the right-hand rule given in Page Thus, we can list the vector products of all the various possible pairs of unit vectors:.
We can determine the sign of the vector product of two unit vectors simply by arranging them in a circle and reading them in the order of the multiplication Fig. The product is positive if they follow each other in counterclockwise order and is negative if they follow each other in clockwise order. You can use the order of letters for the three unit vectors in a vector product to determine its sign.
We can now easily express the vector product V of two given vectors P and Q in terms of the rectangular components of these vectors. Resolving P and Q into components, we first write. Returning to Eq. As we know, the force F is represented by a vector that defines its magnitude and direction. However, the effect of the force on the rigid body depends also upon its point of application A. The position of A can be conveniently defined by the vector r that joins the fixed reference point O with A; this vector is known as the position vector of A.
The position vector r and the force F define the plane shown in Fig. Experimentally, the tendency of a force F to make a rigid body rotate about a fixed axis perpendicular to the force depends upon the distance of F from that axis, as well as upon the magnitude of F.
If you want to apply a small moment to turn a nut on a pipe without breaking it, you might use a small pipe wrench that gives you a small moment arm Page 90 Fig. But if you need a larger moment, you could use a large wrench with a long moment arm Fig. The magnitude of measures the tendency of the force F to make the rigid body. Conversely, the moment of a. Indeed, the line of action of F must lie in a plane through O perpendicular to the moment ; its distance d from O must be equal to the.
Recall from Sec. We can now restate this principle:. We should observe that if the relations of Eqs. Two-Dimensional Problems. Many applications in statics deal with two-dimensional structures.
Such structures have length and breadth but only negligible depth. Often, they are subjected to forces contained in the plane of the structure. We can easily represent two- dimensional structures and the forces acting on them on a sheet of paper or on a blackboard. Their analysis is therefore considerably simpler than that of three-dimensional structures and forces.
Consider, for example, a rigid slab acted upon by a force F in the plane of the slab Fig. The moment of F about a point O, which is chosen in the plane of the figure, is represented by a vector. In the case of. As we look at the figure, we observe in the first case that F tends to rotate the slab counterclockwise and in the second case that it tends to rotate the slab clockwise.
Page 91 Therefore, it is natural to refer to the sense of the moment of F about O in Fig. Because the moment of a force F acting in the plane of the figure must be perpendicular to that plane, we need only specify the magnitude and the sense of the moment of F about O. Indian Polity Notes — Click. Economics Notes — Click. General Science Notes — Click.
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