Objectives
The most important part of analyzing any structure begins with the application of Newton's first law. Newton's first law states: An object at rest or moving at constant velocity continues its current state unless acted upon by a net force. The guiding questions that students need to determine are: What is a force? Are there different types of forces? And how do we determine the force in a structure?
Any force has three characteristics; they are magnitude, a point and line of action and a direction along its line of action. The magnitude is the amount of force say, 100 Newtons or pounds of force. The point of application is where on an object the force is applied, such as at the corner or in the center of mass of the object. The line of action is in the direction the force is acting such at 30 degrees above the horizontal plane through the point. There are two types of forces, external forces and internal forces. The effect of an external force on an object is independent of its point of application on the line of action. The effect of an internal force on an object is very much dependent on the point of application. Forces will be categorized as either contact or long range forces. Contact forces are all forces acting on the object that result from the contact between the object and its surroundings at the object's boundaries. These forces include forces of static and kinetic friction, tension and compressive forces and the normal force. Long range forces result from the object's interaction with a force field, such as magnetic, electric or gravitational fields the most common in introductory physics is the force due to gravity or the weight of the object.
For students to apply Newton's laws to solve force problems they first must be able to represent the forces that act on the object. This representation is called a free-body-diagram (FBD). Analysis of free body diagrams are one of the most difficult skills for most introductory physics students to grasp. This curriculum unit was developed to help students become proficient at drawing and analyzing static force problems and apply these skills to design, analyze and build a bridge. The purpose of developing force diagram is that the diagram simplifies the problem for analysis. Consider, for example, a weight of 100 Newtons hanging from a wire. We begin the analysis by drawing the free body diagram of just the weight, free of the connection to the wire.
The arrow indicates the direction of the force and the 100 Newtons is the magnitude of the force. The point of application is at the center of mass of the object or the centroid. Labels are used to represent the forces; I will use F g for the force due to gravity, T for a force in tension, C for a force in compression, F f for a frictional force, F N for the normal force and a F for a general push or pull force. The weight is in static equilibrium which means that the force due to gravity is just balanced by the tension in the wire. Therefore the tension in the wire is equal to 100 N.
In this simple example determining the tensile force in the wire was straight forward and almost common sense but, the physics principles that was applied was Newton's laws of motions and specifically the condition for static equilibrium. The condition of static equilibrium can be expressed mathematically, as the sum of the forces in that act on the object is equal to zero. To do this we arbitrary assign upward forces as positive and downward forces as negative and the equation is simply:
∑ F y = 0.
The subscript y represents the vertical direction or the y-axis on the coordinate system. This elementary application of Newton's first law, that the sum of the forces in the vertical direction equals zero is the basics of design of every structure, from the simplest to the most complex. It is one of three fundamental conditions of static equilibrium that apply to all objects in a two dimensional system. The other two equations are;
∑ F x = 0
∑Τ o = 0
The first equation is the sum of the forces in the horizontal direction or x-axis are equal to zero and is a logical extension of the sum of the forces in the y-axis. The second equation is the sum of the moments or torque around any point of rotation, o, are equal to zero. The torque is the rotation force on an object caused by forces where their lines of action do not cross a common point. For my introductory physics all lines of action will cross a common point of application therefore the torque will be zero. Torque or moment will be addressed when the students analyze the forces in a section of the truss bridge they design.
Referring back to the free-body diagram of the weight held by a wire, some students will observe that there is a 100 Newtons pulling down and 100 Newtons pulling up and conclude that the tensile force in the wire is 200 Newtons; this is not true. If we make an imaginary cut through the wire at any point and draw a free-body diagram of only that portion of the wire that lies above or below the cut, the force inside the wire is always 100 Newtons. 1
Once the students have an understanding of the simple system of one wire holding a weight free-body diagram, we will explore what would change if two vertical wires support the weight, then three and so forth. The next step in teaching analysis of static structures using free-body diagrams would be to introduce a horizontal force in addition to vertical forces. For example, instead of supporting the weight with one vertical wire I will have the student analyze the same weight supported by two wires pulling at the same angle as shown the Figure 2a below.
The completed free-body diagram of this situation is show in Figure 2b. The student must draw an arrow to represent each of the wires at the point they attach to the 100 Newton weight. They need to include an x-y coordinate axis and the 100 Newton gravitation force.
The previous example dealt with collinear force, forces in only one direction, which allow us to easily add and subtract the vectors. The forces in the above example are not collinear even though they are concurrent, which means that their lines of action all pass through a common point. To find the sum of concurrent forces that are not collinear, we must use vector addition.
Vector Addition
There are five vector quantities that are discussed in introductory physics. They are displacement, velocity, acceleration, force and momentum. A vector is a physical quantity that has both magnitude and direction, such as force. There are two methods of vector addition, graphical and mathematical.
Graphical addition of vectors
Graphical addition of vectors requires students to draw each vector to a set scale in a head to tail method using a ruler and protractor. The resultant (sum) is the vector drawn from the starting point to the head of the last vector; using the scale the student determines the magnitude and direction for the force using a ruler and protractor. Figure 3 below demonstrates how the three vectors in the free-body diagram above are added using a graphical method.
Notice how the vector addition results in a closed triangle, if a structure is static where the sum of all the forces are equal to zero then the vector addition will always have a sum of zero. We can use this fact to graphically solve for unknown forces. When graphically adding vectors the order in which you draw the scaled arrow does not matter, you will always have the same resultant independent of order. Figure 4 below shows the graphical addition of a five force system solving for one unknown force added in two different orders.
Both orders of adding the vectors concluded that the fifth vector resultant force was in a downward right direction and of the same length which represents the magnitude of the force. Both dashed line in figure 4 are identical.
Mathematical addition of vectors
To add vectors mathematically, the student must first resolve any force not in a vertical or horizontal direction into their horizontal, x, and vertical, y, components using trigonometry. For example in Figure 5 the tension in the wire is composed of a horizontal and vertical component, the horizontal component is equal to T x = TcosΘ and the vertical component is equal to T y = TsinΘ . To solve for the unknown horizontal pull of the second wire, T 2 , the horizontal components are then summed to zero resulting in the following mathematical formula which can be solved for the unknown horizontal quantities.
∑F x = 0, 0 = T 2 - Tcos60 o, or T y = Tcos60 o = 58 N
The vertical components are also summed to zero resulting in a mathematical formula which can be solved for any unknown vertical, y, quantities. If there is more than one unknown quantity in each mathematical formula a series of systematic equations is required to solve for the two unknown quantities.
As you can see, these problems can become very complex and students will struggle with the mathematics initially, but the more they practice with resolving vectors into their horizontal and vertical component and applying Newton's first law, the more proficient they will become at solving static force problems.
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