Skip to main content
Engineering LibreTexts

2.5: Equilibrium Analysis for Concurrent Force Systems

  • Page ID
    50576
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    If a body is in static equilibrium, then by definition that body is not accelerating. If we know that the body is not accelerating then we know that the sum of the forces acting on that body must be equal to zero. This is the basis for equilibrium analysis for a particle.

    In order to solve for any unknowns in our sum of forces equation, we actually need to turn the one vector equation into a set of scalar equations. For two dimensional problems, we will split our one vector equation down into two scalar equations. We do this by summing up all the \(x\) components of the force vectors and setting them equal to zero in our first equation, and summing up all the \(y\) components of the force vectors and setting them equal to zero in our second equation.

    \[ \sum \vec{F} \, = \, 0 \]

    \[ \sum F_x \, = \, 0 \, ; \,\,\, \sum F_y \, = \, 0 \]

    We do something similar in three dimensional problems except we will break all our force vectors down into \(x\), \(y\), and \(z\) components, setting the sum of \(x\) components equal to zero for our first equation, the sum of all the \(y\) components equal to zero for our second equation, and the sum of all our \(z\) components equal to zero for our third equation.

    \[ \sum \vec{F} \, = \, 0 \]

    \[ \sum F_x \, = \, 0 \, ; \,\,\, \sum F_y \, = \, 0 \, ; \,\,\, \sum F_z \, = \, 0 \]

    Once we have written out the equilibrium equations, we can solve the equations for any unknown forces.

    Finding the Equilibrium Equations:

    The first step in finding the equilibrium equations is to draw a free body diagram of the body being analyzed. This diagram should show all the known and unknown force vectors acting on the body. In the free body diagram, provide values for any of the know magnitudes or directions for the force vectors and provide variable names for any unknowns (either magnitudes or directions).

    Example of reducing the picture accompanying with a force analysis problem into a free body diagram.
    Figure \(\PageIndex{1}\): The first step in equilibrium analysis is drawing a free body diagram. This is done by removing everything but the body and drawing in all forces acting on the body. It is also useful to label all forces, key dimensions, and angles.

    Next you will need to chose the \(x\), \(y\), and \(z\) axes. These axes do need to be perpendicular to one another, but they do not necessarily have to be horizontal or vertical. If you choose coordinate axes that line up with some of your force vectors you will simplify later analysis.

    Once you have chosen axes, you need to break down all of the force vectors into components along the \(x\), \(y\) and \(z\) directions (see the vectors page in Appendix 1 if you need more guidance on this). Your first equation will be the sum of the magnitudes of the components in the \(x\) direction being equal to zero, the second equation will be the sum of the magnitudes of the components in the \(y\) direction being equal to zero, and the third (if you have a 3D problem) will be the sum of the magnitudes in the \(z\) direction being equal to zero. Collectively these are known as the equilibrium equations.

    Once you have your equilibrium equations, you can solve them for unknowns using algebra. The number of unknowns that you will be able to solve for will be the number of equilibrium equations that you have. In instances where you have more unknowns than equations, the problem is known as a statically indeterminate problem and you will need additional information to solve for the given unknowns.

    Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/Dbd9SvdfoN8.

    Example \(\PageIndex{1}\)

    The diagram below shows a 3-lb box (Box A) sitting on top of a 5-lb box (box B). Determine the magnitude and direction of all the forces acting on box B.

    A 3-pound box (A) rests on top of a 5-pound box (B), which rests on top of a flat surface.Figure \(\PageIndex{2}\): problem diagram for Example \(\PageIndex{1}\); two stacked boxes sitting on a flat surface.
    Solution
    Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/J54OZSitzzM.

    Example \(\PageIndex{2}\)

    A 600-lb barrel rests in a trough as shown below. The barrel is supported by two normal forces (\(F_2\) and \(F_3\)). Determine the magnitude of both of these normal forces.

    A barrel is wedged in a gap shaped like a point-down triangle, with the left side of that triangle making a 45-degree angle with the horizontal and the right side of the triangle making a 30-degree angle to the vertical.
    Figure \(\PageIndex{3}\): problem diagram for Example \(\PageIndex{2}\); a barrel resting in a trough with straight, angled sides.
    Solution
    Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/qKhZvf55Bc0.

    Example \(\PageIndex{3}\)

    A 6-kg traffic light is supported by two cables as shown below. Find the tension in each of the cables supporting the traffic light.

    A traffic light held in midair by two cables: one horizontal cable on the left and one on the right raised at 15 degrees above the horizontal.
    Figure \(\PageIndex{4}\): problem diagram for Example \(\PageIndex{3}\); a traffic light is held in midair by two cables, one horizontal and one angled..
    Solution
    Video \(\PageIndex{4}\): Worked solution to example problem \(\PageIndex{3}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/Oi2yDg1SmrI.

    Example \(\PageIndex{4}\)

    A 400-kg wrecking ball rests against a surface as shown below. Assuming the wrecking ball is currently in equilibrium, determine the tension force in the cable supporting the wrecking ball and the normal force that exists between the wrecking ball and the surface.

    A wrecking ball rests on a surface slanted at 45 degrees above the whole, with its cable stretched taut and held at a 15 degree angle away from the vertical, slanting towards the supporting surface.
    Figure \(\PageIndex{5}\): problem diagram for Example \(\PageIndex{4}\); a wrecking ball on a cable is resting against an angled surface.
    Solution
    Video \(\PageIndex{5}\): Worked solution to example problem \(\PageIndex{4}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/gETMTfy5Sew.

    Example \(\PageIndex{5}\)

    Barrels A and B are supported in a foot truck as seen below. Assuming the barrels are in equilibrium, determine all forces acting on barrel B.

    A 150-lb barrel (A) is stacked on top of a 200-lb barrel (B), with both placed on a handcart. The cart is tilted so the bottom is 30 degrees above the horizontal.
    Figure \(\PageIndex{6}\): problem diagram for Example \(\PageIndex{5}\); two barrels stacked on their sides are on a handcart, whose bottom is tilted upwards.
    Solution
    Video \(\PageIndex{6}\): Worked solution to example problem \(\PageIndex{5}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/8DgrClhT4AM.

    Example \(\PageIndex{6}\)

    Three soda cans, each weighing 0.75 lbs and having a diameter of 4 inches, are stacked in a formation as shown below. Assuming no friction forces, determine the normal forces acting on can B.

    Two soda cans of equal radius, B and C, are lying on their sides next to each other on a flat surface 9 inches wide. The flat surface is bounded on the left and right by a vertical wall. A third soda can of the same size, A, is stacked on its side on top of B and C.
    Figure \(\PageIndex{7}\): problem diagram for Example \(\PageIndex{6}\); three soda cans are stacked lying on their sides, in a flat area bounded on two sides by walls.
    Solution
    Video \(\PageIndex{7}\): Worked solution to example problem \(\PageIndex{6}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/lAUahV7Mml4.

    Example \(\PageIndex{7}\)

    The skycam shown below is supported by three cables. Assuming the skycam has a mass of 20 kg and that it is currently in a state of equilibrium, find the tension in each of the three cables supporting the skycam.

    A skycam, represented as a rectangular box, is drawn on a 3-dimensional coordinate plane and shown being held up by 3 cables. One makes a 10-degree angle above the x-axis in the negative direction, one makes a 15-degree angle above the x-axis in the positive direction, and one makes a 15-degree angle above the xz plane with its projection onto said plane making a 60-degree angle with the negative z-axis.
    Figure \(\PageIndex{8}\): problem diagram for Example \(\PageIndex{7}\); a skycam is held in midair by 3 cables, whose angles in relation to a three-dimensional coordinate plane are shown. Image by Jrienstra CC-BY-SA 3.0.
    Solution
    Video \(\PageIndex{8}\): Worked solution to example problem \(\PageIndex{7}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/FD3yKyfXkGU.

    Example \(\PageIndex{8}\)

    A hot air balloon is tethered to the ground with three cables as shown below. If the balloon is pulling upwards with a force of 900 lbs, what is the tension in each of the three cables?

    A hovering hot-air balloon, attached to the ground by 3 cables, is shown on a 3-dimensional coordinate plane. The balloon is 30 ft above the origin, cable A is attached to the ground 20 ft to the left of the origin, cable B is attached to the ground 30 feet to the right of and 20 feet in front of the origin, and cable C is attached to the ground 20 ft behind the origin.
    Figure \(\PageIndex{9}\): problem diagram for Example \(\PageIndex{8}\); a hot-air balloon is tethered to the ground by 3 cables, whose points of contact with the ground are given in relation to a three-dimensional coordinate plane. Adapted from image by L. Aragon CC-BY-SA 3.0.
    Solution
    Video \(\PageIndex{9}\): Worked solution to example problem \(\PageIndex{8}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/HQqNGJR3ybQ.

    This page titled 2.5: Equilibrium Analysis for Concurrent Force Systems is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jacob Moore & Contributors (Mechanics Map) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.