Concurrent Force Systems In this section we will examine various aspects of concurrent force systems. The topics examined here are principally aimed at the discussion of statics of particles. Resultant of Coplanar Forces: When we are examining a system involving two or more forces, we are usually interested in finding the resultant force in terms of its magnitude as well as direction. The graphical, trigonometric, and vector approaches discussed earlier can be applied to problems involving coplanar (two-dimensional) forces. We will expand on this discussion with the help of the following example problems. Example 1: Two tugboats are towing a cargo ship as shown below. Tugboat A exerts a force of 15,000 N at a 30º angle while tugboat B exerts a force of 20,000 N at a 50º angle. Determine the magnitude and direction of the resultant force acting on the cargo ship. Solution: We begin the analysis by drawing the known force vectors. We then construct the force triangle by a head-to-tail connection of the two force components. Graphical Approach: If the two known sides of the force triangle are drawn to scale, then we can simply measure the length of the resultant vector and multiply it times the scale factor, used for the other two sides, to find its magnitude. To find its direction, we can use a compass to measure its angle from the same reference line. The accuracy of graphical approach depends on the accuracy in drawing the force triangle and the accuracy in measuring the length and angle of the resultant. Hence, it could be subject to a considerable error. Trigonometric Approach: An alternative approach is to use the laws of sines and cosines to solve for the resultant. To do this, we need to first determine the angle b in the force triangle. With the help of the force parallelogram shown below, we determine the value of b knowing that the opposite corners of a parallelogram have equal angles. With b known, we can use the law of cosines given as to solve for the magnitude of the resultant force We then use the law of sines to solve for angle a Therefore, the direction of the resultant force is 16.8º below the horizontal reference line. Scalar Approach: In this approach, we resolve the force exerted by each tugboat into its x and y components as shown below We then add the force components in the x direction together, and those in the y direction together to obtain the x and y components of the force resultant, respectively. In doing this, we must pay close attention to the sign convention on individual force components. With its components known, we can now solve for the magnitude of the force resultant as The direction of the force resultant, q, is found as Since q is measured positive in the counter clockwise direction from x axis, the force resultant is, therefore, directed below the x axis as shown below. We see that the solution found by this approach matches that found by the trigonometric approach. We next examine the vector approach. Vector Approach: In this approach, each force is represented by its components in the rectangular Cartesian coordinates as We can then solve for the force resultant by adding the two force vectors together. The magnitude and direction of the force resultant are then found in the same manner as that described in the scalar approach. Having found the resultant force vector on the cargo ship, we know its direction of motion. Example 2: This example is a variant of the problem considered in Example 1. In this case, tugboat A is exerting a force of 15,000 N at 30º angle. We are interested in knowing the magnitude and direction of forces exerted by tugboat B on the cargo ship such that the resultant towing force is 30,000 N in the horizontal direction, as shown in the figure. Solution: Having discussed, in previous example, various approaches to analyze a co-planar concurrent force system, we focus here on the scalar approach. The schematic drawing of the forces is shown below. The magnitude and direction of the force exerted by Tugboat A and those of the force resultant are known. The unknown quantities are the magnitude and direction of the force exerted by tugboat B. The force equations in x and y directions are used to find the components of . With the two components of force known, we can solve for its magnitude and direction as Therefore, tugboat B should apply a force of 18,590 N at an angle of 24º below the x axis. Resultant of Non-Coplanar Forces: The discussion in this section mainly applies to systems involving more than two concurrent forces. A simple experiment of holding two pencils (as a model of two vectors) end to end and rotating them around at various angles will show that two concurrent vectors are always coplanar. Therefore, we could use the graphical approach in finding the resultant. The complexity arises, however, when the two forces are located in a plane other than xy, xz, or yz plane, or when the system involves three or more non-coplanar forces. In that case, it would be easier to use trigonometric or vector approach to find the resultant force. Here, we make use of direction cosines and/or unit vector to help define the exact direction of a force vector. The analysis of such a system is demonstrated in the following example. Example: An antenna tower is held in the vertical position with the help of three cables. The tensile force in cable AB is 2,000 lb, in AC is 2,500 lb, and in AD is 2,200 lb. Based on the geometry shown below, determine a. the magnitude and direction of the resultant force acting on the antenna at A, b. the angle between cables AB and AC. Solution: We begin the analysis by drawing the known force vectors acting through point A for clarity. Part a: We are given the magnitude of tension in each cable, but not its direction. So we begin by first calculating the unit vector associated with each force. To do this, we make use of the position vector measured from point A to each of the three base support points in xy plane. Knowing the coordinates A(0,0,100), B(-8,-30,0), C(25,40,0), and D(-35,10,0), we get Next, we divide each position vector by its magnitude to find the corresponding unit vector. With each unit vector known, we can express the tension in each cable as Now, a simple vector addition will give us the resultant force vector The magnitude of the resultant force is found as The direction of the resultant force can be defined in terms of the unit vector in the same direction. The unit vector indicates that the resultant force is very close to being completely in the -z direction. To check the validity of this result, we can go back and take a look at the individual forces found previously. When we look at the three forces, we find that not only all three have a negative z component, but that the z component in each case is much larger that the other two. Part b: To find the angle between cables AB and AC, we use the dot product between any vector along AB with any vector along AC. Recall that the dot product between two vectors can be expressed as where represents the angle between vectors and . Using the position vectors along AB and AC we find the desired angle as Modification of Antenna Tower Design: Having done the force analysis, we can determine whether the tower will be held in the vertical position as was originally intended. And if not, what modifications can be made to it to keep it vertical. A tower of this type is usually supported at its base by a ball and socket so that there is no moment at the base, and the cables can be adjusted to position the tower in the desired orientation. Therefore, to keep the tower vertical, the direction of the resultant force has to be completely vertical (in the ­z direction). To achieve this objective, we must adjust the support cables. Since a force vector consists of both magnitude and direction, we can either adjust the magnitude of forces in the cables, or move the base end of one or more cables to change the direction of the resultant force. Alternative 1: Let¹s consider the case of adjusting the tension in the cables with the help of turnbuckles. Also let us assume that a resultant force of 6,300 lb is acceptable in this case. Therefore, we can proceed as follows. The intent is to have the resultant force vector be defined as Since we are keeping the direction of each cable force the same, we can write the cable force vectors as Knowing that we can write This system of linear equations can be expressed in matrix form as The solution of this matrix equation gives As can be seen, a considerable tension increase in cable AB is needed to meet this objective. If we find this change unacceptable, as the tension in cable AB may be greater than what it can handle, then we must consider moving the base point of one or more cables as discussed next. Alternative 2: In this case, we will keep the cable forces as originally stated, and consider changing the location of point B to a point in xy plane such that the direction of the resultant force will be in the ­z axis. We can start from the scalar equations We know the magnitude of force in each cable. We also know the unit vectors in direction of AC, AD and R. Making the proper substitutions gives In addition to the above equations we know that From the solution of these four equations we get Having found the components of the force vector in cable AB, we can express the unit vector along that cable as Recognizing the equality of corresponding terms in the right sides of the two equations and the fact that point B is in xy plane, we can write Recognizing that we find Comparing the new location of point B to that specified in the original problem, we see that it needs to be moved 1.56 ft more in the ­x direction but 37.12 ft more in the ­y direction. It is important that we always be aware of such design issues as discussed in this example, and make rational decisions in modifying the design based on proper engineering analysis. Free-Body Diagram and Equilibrium of a Particle The principle of equilibrium was discussed previously. Let¹s recall that the equilibrium of a particle is defined by the following equations (1) Before we can apply these equations, we must have a good understanding of the forces acting on a particle. We do this by drawing its free-body diagram and showing all the forces, be it known or unknown. Since force equilibrium is the only condition that applies in this case, we cannot have more than three unknown non-coplanar forces acting on a particle. In this section, we will examine the approach that one must take in examining the equilibrium of a particle and the development and use of free-body diagrams. Free-Body Diagrams: As was stated earlier, a particle is an object whose physical dimensions have no influence on the analysis of forces acting on it. Thus, in representing the object as a particle, all the forces must form a concurrent system. It is also possible to analyze a section of a structure by cutting it away from the rest of the structure. When we draw a free-body diagram, we must be careful not to omit any existing force, and just as important, not to include a force that does not exist. When we are interested in analyzing forces in a particular section of the structure that can be modeled as a particle, we draw the free-body diagram of it by cutting (isolating) it from the rest of the structure. We must be careful with showing the forces in members that have been cut. These forces typically have unknown magnitudes and some times even unknown directions. If we know the slope of a force, but not its magnitude and sense. What we need to do is to make an initial guess on the sense of the force in the free-body diagram, then use the equilibrium equations to solve for its magnitude. If the magnitude is found to be positive, then the guessed sense is correct. Otherwise, the correct sense is opposite to that assumed. If both magnitude and direction of a force are unknown, then we have to show the force in terms of its two rectangular components, usually one horizontal and one vertical. Similar to the previous case, we make an initial assumption on the sense of each component. With the help of equilibrium equations we solve for the magnitude of each force component. If found positive, the assumed sense is correct. When two bodies are in contact, and we want to draw the free-body diagram of one of them, then we have to account for the contact force or forces. If the surface of contact is frictionless, then there is only a normal force at the point of contact. In the presence of friction, then there is also a friction force that is tangential to the surface of contact. Example 1: A 200-kg sign is supported by the cable system shown below. Determine the tension developed in each cable to hold the sign in equilibrium. Solution Step 1: A close examination of the cable system indicates that the two connection rings can be represented as two particles with the free-body diagrams shown below. Each free-body diagram represents a coplanar force system. Step 2: We continue by analyzing the free-body diagram of the lower ring next. This is because there are only two unknown forces in this case, and we can solve for them using the two equations of equilibrium. Knowing the slope of cable É is 5/2, we find the hypotenuse of that triangle as . Now, we can easily express the horizontal and vertical components of F3 in terms of the slope and magnitude of F3. => => With F3 known, the free-body diagram of the upper ring now has only two unknown forces, which are calculated similar to the previous case as follows Solving the two equations simultaneously gives Step 3, Analysis of Results: The results indicate that the tension in cable 3 is the highest among the four cables. With cable 4 being horizontal, the tension in cable 3 is 7.7% greater than the weight of the sign. This leads to the following question. What happens to the forces in cables 1, 2, and 3 if we were to eliminate cable 4? This scenario is examined next. Design Modification and Analysis: By eliminating cable 4, the lower ring and the sign will move to the left and cable 3 becomes vertical. This change, however, has no impact on the position of the upper ring, and angles of cables 1 and 2 will stay the same as before. Solution: The two free-body diagrams are shown below. Since cable 3 is now vertical, it is easy to conclude that its magnitude, based on the equilibrium of forces in the vertical direction, is equal to the weight of the sign => Using the Free-body diagram of the upper ring, we obtain Solving the two equilibrium equations simultaneously gives Therefore, by eliminating cable 4, we reduced the tensions in cables 3 and 1 by 7.2% and 22%, respectively, but increased the tension in cable 2 by 67.6%. Example 2: In this example we will analyze a three-dimensional or non-coplanar force system. We want to determine the tension in each cable to hold the 250-lb crate in equilibrium. Note that points B, C, and D are all in x-y plane. Solution: We begin the analysis by drawing the free-body diagram of the ring at point A. The forces acting on the ring include the three cable forces as well as the weight of the crate. Since the forces are in three dimensions, we use the corresponding direction cosines in formulating the equations of equilibrium in x,y,z directions. Knowing the coordinates of points A, B, C, and D, we calculate the length of each cable as Having found the length of each cable, we can now write the equilibrium equations in terms of the individual force components. We have to be careful with the direction of each component. We can either use the back substitution to solve for cable force, or simply write the three equations in matrix form and solve them simultaneously as shown below