Engineering Statics

Subsection 3.3.1 A simple case

Consider the weight suspended by a rope shown in Figure 3.3.1. Diagrams of this type are called space diagrams; they show the objects as they exists in space.

In mechanics we are interested in studying the forces acting on objects and in this course, the objects will be in equilibrium. The best way to do this is to draw a diagram which focuses on the forces acting on the object, not the mechanisms that hold it in place. We call this type of diagram a free-body diagram because it shows the object disconnected or freed from its supporting mechanisms. You can see the free-body diagram for this situation by moving the slider in the interactive to position two. This shows that there are two forces acting on the object; the force of the rope holding it up, and the weight of the object which is trying to pull it to earth and which we treat as acting at its center of gravity.

Drawing free-body diagrams can be surprisingly tricky. The reason for this is that you must identify all the forces acting on the object and correctly represent them on the free-body diagam. If you fail to account for all the forces, include additional ones, or represent them incorrectly, your analysis will surely be wrong.

So what kind of analysis can we do here? Admittedly not much. We can find the tension in the rope caused by a particular weight and use it to select an appropriately strong rope, or we can determine the maximum weight a particular rope can safely support.

The actual analysis is so trivial that you've probably already done it in your head, nevertheless, there are several ways to approach it which will be shown next.

In the vector approach we will use the equation of equilibrium.

We can also formulate this example in terms of unit vectors. Recall that \(\jhat\) is the unit vector which points up. It has a magnitude of one with no units associated. So in terms of unit vector \(\jhat\text{,}\) \(\vec{T} = T \jhat\) and \(\vec{W} = -W \jhat\text{.}\)

In the previous example, the unit vector \(\jhat\) completely dropped out of the equation leaving only the coefficients of \(\jhat\text{.}\) This will be the case whenever you add vectors which all act along the same line of action.

The coefficients of \(\ihat\text{,}\) \(\jhat\text{,}\) and \(\khat\) are known as the scalar components. A scalar component times the associated unit vector is a force vector.

When you use scalar components, the forces are represented by scalar values and the equilibrium equations are solved using normal algebraic addition rather than vector addition. This leads to a slight simplification of the solution as shown in the next example.

Subsection 3.3.2 Scalar Components

The scalar component of a vector is a signed number which indicates the vector's magnitude and sense, and is usually identified by a symbol with a subscript that indicates the line of action of the vector.

So for example, \(F_x = \N{10}\) is a scalar component. We can tell it's not a vector because it \(F_x\) is not bold. \(\N{10}\) is the magnitude of the associated vector; the subscript \(x\) indicates that the force acts "in the \(x\) direction," in other words it acts on a line of action which is parallel to the \(x\) axis; and the (implied) positive sign means that the vector points towards the positive end of the \(x\)- axis — towards positive infinity. So a scalar component, while not a vector, contains all the information necessary to draw the corresponding vector. Be careful not to confuse scalar components with vector magnitudes. A force with a magnitude of \(\N{10}\) can point in any direction, but can never have a negative magnitude.

Scalar components can be added together algebraically, but only if they act “in the same direction.” It makes no sense to add \(F_x\) to \(F_y\text{.}\) If that's what you want to do, first you must convert the scalar components to vectors, then add them according to the rules of vector addition.

Example 3.3.4. 1-D Scalar Addition.

If \(A_x = \lb{10}\) and \(B_x\) = \(\lb{-15}\text{,}\) find the magnitude and direction of their resultant \(\vec{R}\text{.}\)

Answer

\begin{gather*} \vec{R} = \lb{5} \leftarrow \end{gather*}

Solution

Start by sketching the two forces. The subscripts indicate the line of action of the force, and the sign indicates the direction along the line of action. A negative \(B_x\) points towards the negative end of the \(x\) axis.

\begin{align*} R \amp = A_x + B_x\\ \amp = \lb{10} + \lb{-15} \\ \amp= \lb{-5} \end{align*}

\(R\) is the scalar component of the resultant \(\vec{R}\text{.}\)

The negative sign on the result indicates that the resultant force acts to the left.

Example 3.3.5. 2-D Scalar Addition.

If \(F_x\) = \(\N{-40}\) and \(F_y\) = \(\N{30}\text{,}\) find the magnitude and direction of their resultant \(\vec{F}\text{.}\)

Answer

\begin{align*} \vec{F} \amp = \N{50} \text{ at } 143.1^\circ \measuredangle \end{align*}

Solution

In this example the scalar components have different subscripts indicating that they act along different lines of action, and this must be accounted for when they are added together.

Make a sketch of the two vectors and add them using the parallelogram rule to get

\begin{align*} F \amp = \sqrt{F_x^2 + F_y^2} \amp \theta \amp = \tan^{-1} \left|\frac{B_y}{B_x}\right|\\ \amp = \sqrt{(-40)^2 + (30)^2} \amp \amp = \tan^{-1}\left |\frac{30}{-40}\right |\\ \amp = \N{50} \amp \amp = 36.9^\circ \end{align*}

These are the magnitude and direction of vector \(\vec{F}\text{.}\)

Subsection 3.3.3 Two-force Bodies

As you might expect, a two-force body is a body with two forces acting on it, like the weight just discussed. As we just saw, in order for a two-force body to be in equilibrium the two forces must add to zero. There are only three possible ways that this can happen:

The two forces must either

• share the same line of action, have the same magnitude, and point away from each other, or
• share the same line of action, have the same magnitude, and point towards each other, or
• both forces must have zero magnitude.

When two forces have the same magnitude but act in diametrically opposite directions, we say that they are equal and opposite. When equal and opposite forces act on an object and they point towards each other we say that the object is in compression, when they point away from each other the object is in tension. Tension and compression describe the internal state of the object.

Two force bodies appear frequently in multipart structures and machines which will be covered in Chapter 6. Some examples of two force bodies are struts and linkages, ropes, cables and guy wires, and springs.

Example 3.3.6. Tug of War.

Marines and Airmen at Goodfellow Air Force Base are competing in a tug of war and have reached a stalemate. The Marines are pulling with a force of \(\lb{1500}\text{.}\) How hard are the Airmen pulling? What is the tension in the rope?

This is a simple question, but students often get it wrong at first.

Answer

The tension in the rope is \\(lb{1500}\text{.}\) Both teams are pulling with the same force.

Solution

Assumptions.

A free-body diagram of the rope is shown.

We'll solve this with scalar components because there's no need for the additional complexity of the vector approaches in this simple situation.

We'll align the \(x\) axis with the rope with positive to the right as usual to establish a coordinate system.

Assume that the pull of each team can be represented by a single force. Let force \(M\) be supplied by the Marines and force \(A\) by the Airmen; call the tension in the rope \(T\text{.}\)

Assume that the weight of the rope is negligible; then the rope can be considered a particle because both forces lie along same line of action.

Givens.

\(M = \lb{1500}\text{.}\)

Procedure.

Since they're stalemated we know that the rope is in equilibrium.

Applying the equation of equilibrium gives:

\begin{align*} \Sigma F_x \amp= 0\\ -M + A \amp= 0\\ A \amp= M\\ \amp = \lb{1500} \end{align*}

We find out that both teams pull with the same force. This was probably obvious without drawing the free-body diagram or solving the equilibrium equation.

It may seem equally obvious that if both teams are pulling with \(\lb{1500}\) in opposite directions that the tension in the rope must be\(\lb{3000}\text{.}\) This is wrong however.

The tension in the rope \(T\) is an example of an internal force and in order to learn its magnitude we need a free-body diagram which includes force \(T\text{.}\) To expose the internal force we take an imaginary cut through the rope and draw (or imagine) a free-body diagram of either half of the rope.

The correct answer is easily seen to be \(T = A = M = \lb{1500}\text{.}\)

Example 3.3.7. Hanging Weight.

The wire spool being lifted into the truck consists of \(\m{750}\) of three strand medium voltage (5 kV) 1/0 AWG electrical power cable with a 195 amp capacity at 90°C, weighing 927 kg/km, on a \(\kg{350}\) steel reel.

How much weight is supported by the hook and high tension polymer lifting sling?

Answer

\begin{equation*} W = \N{10300} \end{equation*}

Solution

The entire weight of the wire and the spool is supported by the hook and sling.

Remember that weight is not mass and mass is not force. The total weight is found by multiplying the total mass by the gravitational constant \(g\text{.}\)

\begin{align*} W \amp = m g\\ \amp = (m_w + m_s )\,g\\ \amp=\left((\km{0.75}) (\kgperkm{927}\right) + \kg{350})\,g \\ \amp = (\kg{1045})(\aSI{9.81})\\ \amp = \N{10300} \end{align*}