Statics: 3D Particle Equilibrium

Section 3.5 3D Particle Equilibrium

The world we live in has three dimensions. One and two-dimensional textbook problems have been useful for learning the principles of engineering mechanics, but to model real-world problems we will have to consider all three.

Fortunately, all the principles you have learned so far still apply, but many students have difficulty visualizing three-dimensional problems drawn on two-dimensional paper and the mathematics becomes a bit harder. It is especially important to have good diagrams and keep your work neat and organized to avoid errors.

Subsection 3.5.1 Three-Dimensional Coordinate Frame

We need a coordinate frame for three dimensions, just as we did in two dimensions, so we add a third orthogonal axis \(z\) to our existing two-dimensional frame.

For equilibrium of a particle, usually the origin of the coordinate frame is at the particle, the \(x\) axis is horizontal, and the \(y\) axis is vertical just as in a two-dimensional situation. The orientation of the \(z\) axis is determined by the right-hand rule. Using your right hand, put your palm at the origin and point your fingers along the positive \(x\) axis. Then curl your fingers towards the positive \(y\) axis. Your thumb will point in the direction of the positive \(z\) axis. For example, in the plane of the page with the positive \(x\) axis horizontal and to the right and the positive \(y\) axis vertical and upwards, the positive \(z\) axis will point towards you out of the page. Remember that the three axes are mutually perpendicular, i.e. each axis is perpendicular to both of the others. The right-hand rule is important in many aspects of engineering, so make sure that you understand how it works. Mistakes will lead to sign errors.

Figure 3.5.1. Point-and-curl right-hand rule technique.

Subsection 3.5.2 Free-Body Diagrams

As we did before, we begin our analysis by drawing a free-body diagram that shows all forces and moments acting on the object of interest. Drawing a FBD in three dimensions can be difficult. It is sometimes hard to see things in three dimensions when they are drawn on a two-dimensional sheet. Consequently, it is important to carefully label vectors and angles, but not to clutter up the diagram with too much and/or unnecessary information. When working in two dimensions, you only need one angle to determine the direction of the vector, but when working in three dimensions you need two or three angles.

Subsection 3.5.3 Angles

As stated above, when working in three dimensions you need three angles to determine the direction of the vector, namely, the angle with respect to the \(x\) axis, the angle with respect to the \(y\) axis and the angle with respect to the \(z\) axis. The three angles mentioned above are not necessarily located in any of the coordinate planes. Think of it this way — three points determine a plane, and in this case, the three points are: the origin, the tip of the vector, and a point on an axis. The plane made by those three points is not necessarily the \(xy\text{,}\) \(yz\text{,}\) or \(xz\) plane. It is most likely a “tilted” plane.

As you learned in Subsection 2.4.2, one way to quantify the direction of a vector is with direction cosine angles. These direction cosine angles are measured from the positive x, y, and z axes and are often labeled \(θ_x\text{,}\) \(θ_y\text{,}\) and \(θ_z\text{,}\) respectively.

As with two dimensions, angles can be determined from geometry — a distance vector going in the same direction as the force vector. This is the three-dimensional equivalent of similar triangles that you used in the two-dimensional problems.

Instructions.

Rotate the diagram to see the force and angles in space.

Figure 3.5.2. Direction Cosine Angles

If you know that the line of action of a force vector goes between two points, then you can use the distance vector that goes from one point to the other to determine the angles.

Let’s suppose that the line of action goes through two points \(A\) and \(B\text{,}\) and the direction of the force is from \(A\) towards \(B\text{.}\) The first step in determining the three angles is to write the distance vector from point \(A\) towards point \(B\text{.}\) Let’s call this vector \(\vec{r}_{AB}\text{.}\) Starting at point \(A\text{,}\) you need to determine how to get to point \(B\) by moving in each of the three directions. Ask yourself: to get from point \(A\) to point \(B\) do I have to move in the \(x\) direction? If so, how far do I have to travel? This becomes the \(x\) component of the vector \(\vec{r}_{AB}\) namely \(r_{AB_x}\text{.}\) Next, to get from point \(A\) to point \(B\) how far do I move in \(y\) direction? This distance is \(r_{AB_y}\text{.}\) Finally, to get from point A to point B how far do I move in the z-direction? This distance is \(r_{AB_z}\text{.}\)

When writing these scalar components pay attention to which way you move along the axes. If you travel toward the positive end of an axis, the corresponding scalar component gets a positive sign. Travel towards the negative end results in a negative sign. The sign is important.

Once you have determined the components of the distance vector \(r_{AB}\text{,}\) you can determine the total distance from point \(A\) to \(B\) using the three-dimensional Pythagorean Theorem

\begin{equation} {r_{AB}}= \sqrt{(r_{AB_x})^2 + (r_{AB_y})^2 +(r_{AB_z})^2}\tag{3.5.1} \end{equation}

Lastly, the angles are determined by the direction cosines, namely

\begin{align*} \cos \theta_x \amp = \dfrac{r_{AB_x}}{r_{AB}} \amp \cos \theta_y \amp = \dfrac{r_{AB_y}}{r_{AB}} \amp \cos \theta_z \amp = \dfrac{r_{AB_z}}{r_{AB}} \end{align*}

Since the force vector has the same line of action as the distance vector, by the three-dimensional version of similar triangles,

\begin{align*} \dfrac{r_{AB_x}}{r_{AB}} \amp = \dfrac{F_x}{F} \amp \dfrac{r_{AB_y}}{r_{AB}} \amp = \dfrac{F_y}{F} \amp \dfrac{r_{AB_z}}{r_{AB}} \amp = \dfrac{F_z}{F} \end{align*}

. So,

\begin{align*} F_x \amp = \left(\dfrac{r_{AB_x}}{r_{AB}}\right) F \amp F_y \amp = \left(\dfrac{r_{AB_y}}{r_{AB}}\right) F \amp F_z \amp = \left(\dfrac{r_{AB_z}}{r_{AB}}\right) F \end{align*}

Now, that is a bit of math there, but the important things to remember are:

You can use three angles to determine the direction of a force in three dimensions.
You can use the geometry to get them from a distance vector that lies along the line of action of the force.

The three direction cosine angles are not mutually independent. From (3.5.1) you can easily show that

\begin{equation} \cos \theta_x^2 + \cos \theta_y^2 + \cos \theta_z^2 = 1\text{,}\tag{3.5.2} \end{equation}

so if you know two direction cosine angles you can find the third from this relationship.

Subsection 3.5.4 General Procedure

The general procedure for solving three-dimensional particle equilibrium is essentially the same as for two-dimensional particle equilibrium using the components method. The major differences are that you must carefully find each vector component using the techniques from Section 2.4. The process follows the same five-step method for creating a free-body diagram, followed by steps to solve your equilibrium equations.

Draw a Free-body Diagram:

Select and isolate the particle. The “free-body” in free-body diagram means that a concurrent force particle or connection must be isolated from the supports that are physically holding it in place. This means creating a separate free-body diagram from your problem sketch.
Establish a coordinate system. Draw a right-handed coordinate system to use as a reference for your equilibrium equations. Look ahead and select a coordinate system that minimizes the number of force components. This will simplify your vector algebra. The choice is technically arbitrary, but a good choice will simplify your calculations and reduce your effort.
Identify all loads. Add force vectors to your free-body diagram representing each applied load pushing or pulling the body, in addition to the body’s weight, if it is non-negligible. If a force vector has a known direction, draw it. If its direction is unknown, assume one, and your later algebra will check your assumption. Every vector should have a descriptive variable name and a clear arrowhead indicating its direction.
Identify all reactions. Reactions represent the resistance of the physical supports you cut away by isolating the body in step 1. All particle supports are some type of two-force members with tension or compression reaction forces. These reactions will all be concurrent with the body loads from Step 2. Label each reaction with a descriptive variable name and a clear arrowhead. Again, if a vector’s direction is unknown, just assume one.
Label the diagram. Verify that every dimension, angle, force, and moment is labeled with either a value or a symbolic name if the value is unknown. In our eyes, dimensioning is optional. Having the information needed for your calculations is helpful, but don’t clutter the diagram up with unneeded details. Your final free-body diagram should be a stand-alone presentation and is the basis of your equilibrium equations.

Create and Solve Equilibrium Equations

Break vectors into components. Compute each force’s \(x\text{,}\) \(y\text{,}\) and \(z\) components using the tools outlined in Section 2.4. While the components in two-dimensional problems can often be found with right triangle trigonometry, three-dimensional problems often use unit vectors.
Write equilibrium equations. Now represent your free-body diagram as equilibrium equations. For a three-dimensional particle equilibrium problem, you can have up to three force equilibrium equations corresponding to a force balance in the three independent \(x\text{,}\) \(y\text{,}\) and \(z\) directions. Each equation should start with the governing equation, like \(\Sigma F_x=0\text{.}\)
Count knowns and unknowns. At this point, you should have at most three unknowns remaining. If you have over three, reread the problem and look for overlooked information.
Solve for unknowns. Use algebra to simplify the equilibrium equations and solve for unknowns. With multiple unknowns scattered across multiple equations, linear algebra may be more efficient than substitution. Assume that all answers have units - unless you prove that they don’t. Finally, underline or box your answers.
Check your work. If you add the components of the forces, do they add to zero? Do the results seem reasonable given the situation? Have you included appropriate units?

Now let’s see how that process looks on an example problem.

Example 3.5.3. Balloon.

A hot air balloon \(\ft{30}\) above the ground is tethered by three cables as shown in the diagram.

If the balloon is pulling upwards with a force of \(\lb{900}\text{,}\) what is the tension in each of the three cables?

The grid lines on the ground plane are spaced \(\ft{10}\) apart.

A hot air balloon is 30 ft above the ground (coordinates (0, 30, 0)) and tethered by three cables. Assume the origin is directly below the balloon on the ground. The three cables are attached from the basket of the balloon to the following points (ft): A (-20, 0, 0), B (30, 0, -20), C (0, 0, -20).

Answer.

\begin{align*} A \amp = \lb{464} \amp B \amp = \lb{402} \amp C \amp = \lb{309} \end{align*}

Solution.

Strategy.

The three tensions are the unknowns which we can find by applying the three equilibrium equations.

We’ll establish a coordinate system with the origin directly below the balloon and the \(y\) axis vertical, then draw and label a free-body diagram.

Next we’ll use the given information to find two points on each line of action to find the components of each force in terms of the unknowns.

When the \(x\text{,}\) \(y\) and \(z\) components of all forces can be expressed in terms of known values, the equilibrium equations can be solved.
Geometry.
From the diagram, the coordinates of the points are

\begin{align*} \text{A} \amp= (-20,0,0)\amp \text{B}\amp= (30,0,20) \amp \text{C}\amp= (0,0,-20) \amp \text{D} \amp = (0,30,0) \end{align*}

Use the point coordinates to find the \(x\text{,}\) \(y\) and \(z\) components of the forces.

\begin{align*} A_x \amp = \frac{-20}{L_A} A \amp A_y \amp = \frac{-30}{L_A} A \amp A_z \amp = \frac{0}{L_A} A \\ B_x \amp = \frac{30}{L_B} B \amp B_y \amp = \frac{-30}{L_B} B \amp B_z \amp = \frac{20}{L_B} B \\ C_x \amp = \frac{0}{L_C} C\amp C_y \amp = \frac{-30}{L_C} C\amp C_z \amp = \frac{-20}{L_C} C \end{align*}

Where \(L_A\text{,}\) \(L_B\) and \(L_C\) are the lengths of the three cables found with the distance formula.

\begin{align*} L_A \amp= \sqrt{(-20)^2 + (-30)^2+ 0^2} \amp= \ft{36.1}\\ L_B \amp= \sqrt{30^2 + (-30)^2 +20^2} \amp= \ft{46.9}\\ L_C \amp= \sqrt{0^2+(-30)^2 + (-20)^2} \amp= \ft{36.1} \end{align*}
Equilibrium Equations.
Applying the three equations of equilibrium yields three equations in terms of the three unknown tensions.

\begin{align*} \Sigma F_x \amp = 0\\ \amp A_x + B_x + C_x = 0\\ \amp - \frac{20}{36.1} A + \frac{30}{46.9} B + 0\, C = 0\\ A \amp= 1.153\, B \amp (1)\\\\ \Sigma F_z \amp= 0\\ \amp A_z + B_z + C_z = 0\\ \amp 0\, A + \frac{20}{46.9}B -\frac{20}{36.1} C = 0\\ C \amp= 0.769 \,B \amp (2)\\\\ \Sigma F_y \amp= 0\\ \amp A_y + B_y + C_y + D = 0\\ \amp -\frac{30}{36.1} A -\frac{30}{46.9} B - \frac{30}{36.1} C + 900 = 0\\ \amp 0.832\, A + 0.640\, B + 0.832\, C = \lb{900} \amp (3) \end{align*}

Solving these equations simultaneously yields the answers we are seeking. One way to do this is to substitute equations (1) and (2) into (3) to eliminate \(A\) and \(C\) and solve the resulting equation for \(B\text{.}\)

\begin{align*} 0.832\, (1.153\, B) + 0.640\, B + 0.832\, (0.769 \,B) \amp= \lb{900}\\ 2.24 B \amp = \lb{900}\\ B \amp = \lb{402} \end{align*}

With \(B\) known, substitute it into equations (1) and (2) to find \(A\) and \(C\text{.}\)

\begin{align*} A \amp = 1.153\, B \amp C \amp =0.769\, B \\ \amp = \lb{464} \amp \amp = \lb{309} \end{align*}

Example 3.5.4. Skycam.

The skycam at Stanford University Stadium has a mass of \(\kg{20}\) and is supported by three cables as shown. Assuming that it is currently in equilibrium, find the tension in each of the three supporting cables.

Skycam has mass of 20 kg and is in equilibrium. Cable A is in the x-y plane, acting 35 degrees from the positive x-axis. Cable B acts -30 degrees from the negative x-axis in the x-z plane, and 15 degrees above the x-z plane. Cable C acts 20 degrees in the y-z plane, -20 degrees from the positive z-axis.

Instructions.

Rotating the interactive diagram may help you visualize the situation.

Answer.

\begin{align*} A \amp = \N{196.4} \amp B \amp = \N{192.2} \amp C \amp = \N{98.5} \end{align*}

Solution.

In this situation, the directions of all four forces are specified by the angles in the free-body diagram, and the magnitude of the weight is known. The three unknowns are the magnitudes of forces \(\vec{A}\text{,}\) \(\vec{B}\text{,}\) and \(\vec{C}\text{.}\)

\begin{equation*} W = m g = \kg{20}\ \aSI{9.81} = \N{196.2} \end{equation*}

We will first find unit vectors in the directions of the four forces by inspection of the free-body diagram. This step requires visualizing the component’s unit vectors and determining the angles each makes with the coordinate axis.

\begin{align*} \hat{\vec{W}} \amp = \langle 0, -1, 0 \rangle \\ \hat{\vec{A}} \amp = \langle \cos \ang{35}, \cos{55}, 0 \rangle\\ \hat{\vec{B}} \amp = \langle -\cos \ang{15} \cos \ang{30}, \cos \ang{75}, -\cos \ang{15} \cos \ang{60} \rangle\\ \hat{\vec{C}} \amp = \langle 0, \cos{70}, \cos \ang{20} \rangle \end{align*}

Particle equilibrium requires that \(\sum \vec{F} = 0\text{.}\)

\begin{equation*} A\ \hat{\vec{A}} + B\ \hat{\vec{B}} + C\ \hat{\vec{C}} = - W\ \hat{\vec{W}} \end{equation*}

This is a 3 \(\times\) 3 system of three simultaneous equations, one for each coordinate direction, which needs to be solved for \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

\begin{align*} A \cos \ang{35} - B \cos \ang{15} \cos \ang{30} + 0 \amp = 0 \amp \amp \amp (\Sigma F_x \amp= 0)\\ A \cos \ang{55} + B \cos \ang{75} + C \cos \ang{70} \amp = \N{196.2} \amp \amp \amp (\Sigma F_y \amp= 0)\\ 0 - \cos \ang{15} \cos \ang{60} + C \cos \ang{20} \amp = 0 \amp \amp \amp (\Sigma F_z \amp= 0) \end{align*}

These can be solved by any method you choose. Here we will use Sage. Evaluating the coefficients and expressing the equations in matrix form gives

\begin{equation*} \begin{bmatrix} 0.819 \amp -0.837 \amp 0 \\ 0.574 \amp 0.259 \amp 0.342 \\ 0 \amp -0.482 \amp 0.940 \end{bmatrix} \begin{bmatrix} A \\ B \\ C\\ \end{bmatrix} = \begin{bmatrix} 0 \\ \N{196.2} \\ 0 \end{bmatrix}\text{.} \end{equation*}

This is an equation in the form

\begin{equation*} [A][x] =[B]\text{.} \end{equation*}

Entering the coefficient matrices into Sage.

After evaluating, we learn that

\begin{align*} A \amp = \N{196.4} \amp B \amp = \N{192.2} \amp C \amp = \N{98.5} \end{align*}

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