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3.6: Equilibrium Analysis for a Rigid Body

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    51709
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    For an rigid body in static equilibrium—that is, a non-deformable body where forces are not concurrent—the sum of both the forces and the moments acting on the body must be equal to zero. The addition of moments (as opposed to particles, where we only looked at the forces) adds another set of possible equilibrium equations, allowing us to solve for more unknowns as compared to particle problems.

    Moments, like forces, are vectors. This means that our vector equation needs to be broken down into scalar components before we can solve the equilibrium equations. In a two-dimensional problem, the body can only have clockwise or counterclockwise rotation (corresponding to rotations about the \(z\) axis). This means that a rigid body in a two-dimensional problem has three possible equilibrium equations; that is, the sum of force components in the \(x\) and \(y\) directions, and the moments about the \(z\) axis. The sum of each of these will be equal to zero.

    For a two-dimensional problem, we break our one vector force equation into two scalar component equations. \[ \sum \vec{F} \, = \, 0 \]

    \[ \sum F_x \, = \, 0\, ; \,\, \sum F_y \, = \, 0 \] The one moment vector equation becomes a single moment scalar equation. \[ \sum \vec{M} \, = \, 0 \]

    \[ \sum M_z \, = \, 0 \]

    If we look at a three-dimensional problem we will increase the number of possible equilibrium equations to six. There are three equilibrium equations for force, where the sum of the components in the \(x\), \(y\), and \(z\) directions must be equal to zero. The body may also have moments about each of the three axes. The second set of three equilibrium equations states that the sum of the moment components about the \(x\), \(y\), and \(z\) axes must also be equal to zero.

    We break the forces into three component equations. \[ \sum \vec{F} \, = \, 0 \]

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

    Then we also break the moments into three component equations. \[ \sum \vec{M} \, = \, 0 \]

    \[ \sum M_x \, = \, 0 \, ; \,\, \sum M_y \, = \, 0 \, ; \,\, \sum M_z \, = \, 0 \]

    Finding the Equilibrium Equations

    As with particles, 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 force vectors acting on the body. In the free body diagram, provide values for any of the known magnitudes, directions, and points of application for the force vectors and provide variable names for any unknowns (either magnitudes, directions, or distances).

    Next you will need to choose 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 page for more details on this process). 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.

    Next you will need to come up with the the moment equations. To do this you will need to choose a point to take the moments about. Any point should work, but it is usually advantageous to choose a point that will decrease the number of unknowns in the equation. Remember that any force vector that travels through a given point will exert no moment about that point. To write out the moment equations, simply sum the moments exerted by each force (adding in pure moments shown in the diagram) about the given point and the given axis, and set that sum equal to zero. All moments will be about the \(z\) axis for two-dimensional problems, though moments can be about the \(x\), \(y\) and \(z\) axes for three-dimensional problems.

    Once you have your equilibrium equations, you can solve these formulas for unknowns. The number of unknowns that you will be able to solve for will again be the number of equations that you have.

    Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/OiJ2xbMIixY.
    Example \(\PageIndex{1}\)

    The car below has a weight of 1500 lbs with the center of mass 4 ft behind the front wheels of the car. What are the normal forces on the front and the back wheels of the car?

    Side view of a car facing left, with the center of mass marked as being 4 feet to the right of point A, the front wheel, and 3 feet to the left of point B, the rear wheel.
    Figure \(\PageIndex{1}\): problem diagram for Example \(\PageIndex{1}\). Adapted from the public domain image by Ebaychatter0.
    Solution
    Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/1LD5QW-70PA.
    Example \(\PageIndex{2}\)

    A 5-meter-long beam has a fixed connection to a wall at point A and a force acting as shown at point B. What are the reaction forces acting on the beam at point A?

    A horizontal 5-meter-long beam has its left end, point A, attached to a wall. At its free end, point B, a force of magnitude 6 kN is applied downward and to the left, making a 20 degree angle with the vertical.
    Figure \(\PageIndex{2}\): problem diagram for Example \(\PageIndex{2}\). A horizontal beam attached to the wall at one end experiences a force applied at its free end.
    Solution
    Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/JrVV7k1aQEk.
    Example \(\PageIndex{3}\)

    A ladder with negligible mass is supporting a 120-lb person as shown below. If the contact point at A is frictionless, and the contact point at B is a rough connection, determine the forces acting at contact points A and B.

    A ladder is propped against a wall, with its base 5 feet away from the wall and its top 20 feet above the ground. A person weighing 120 lbs stands on the ladder at a spot 15 feet above the ground.
    Figure \(\PageIndex{3}\): problem diagram for Example \(\PageIndex{3}\). A ladder with its base on a rough floor leans against a frictionless wall, with a person standing on the ladder partway up.
    Solution
    Video \(\PageIndex{4}\): Worked solution to example problem \(\PageIndex{3}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/WzkAnPdhao4.
    Example \(\PageIndex{4}\)

    Member ABC is 6 meters long, with point B being at its midpoint. Determine all forces acting on member ABC.

    A structural member consists of point A, the intersection between a horizontal beam attached to a wall and a 6-meter-long diagonal beam at a 45-degree angle above the horizontal; point B, the midpoint of the diagonal beam and the point of attachment for a cable fixed to the wall making a 45-degree angle below the horizontal; and point C, the free end of the diagonal beam with a 300-kg load hanging from it.
    Figure \(\PageIndex{4}\): problem diagram for Example \(\PageIndex{4}\). A diagonal structural member is attached to the wall at one end, connected to the wall via cable at its midpoint, and holding up a 300-kg load at its free end.
    Solution
    Video \(\PageIndex{5}\): Worked solution to example problem \(\PageIndex{4}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/sMQrjwUMpSQ.
    Example \(\PageIndex{5}\)

    While sitting in a chair, a person exerts the forces in the diagram below. Determine all forces acting on the chair at points A and B. (Assume A is frictionless and B is a rough surface).

    Side view of a chair facing left, with a seat 18 inches above the ground, the top edge of the chair back 18 inches above the seat, and 18 inches between the left and right legs. Points A and B are the points where the left and right chair legs touch the ground respectively. A downward force of 180 lbs is exerted on the chair seat, 12 inches behind the seat's front edge, and a rightwards force of 15 lbs is exerted against the top edge of the seat back.
    Figure \(\PageIndex{5}\): problem diagram for Example \(\PageIndex{5}\). A chair is placed on a flat surface, with that surface assumed to be frictionless where it contacts the chair's front leg (point A) and exerting friction where it contacts the back leg (point B).
    Solution
    Video \(\PageIndex{6}\): Worked solution to example problem \(\PageIndex{5}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/nSOxK1ZMggA.
    Example \(\PageIndex{6}\)

    The trailer shown below consists of a deck with a weight of 250 lbs on an axle with wheels with a weight of 350 lbs. Assume the weight forces act in the center of each component. If we wish the tongue weight (\(F_T\)) of the unloaded trailer to be 50 lbs, what is the distance \(d\) from the front where we must place the axle?

    A trailer consisting of a flat rectangular deck weighing 250 lbs, supported on 2 wheels weighing 350 lbs connected by an axle.
    Figure \(\PageIndex{6}\): problem diagram for Example \(\PageIndex{6}\). A trailer consists of a flat rectangular deck on top of two wheels on an axle.
    Solution
    Video \(\PageIndex{7}\): Worked solution to example problem \(\PageIndex{6}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/wpEBuitLD5s.
    Example \(\PageIndex{7}\)

    A 12-inch-by-24-inch flat steel sign is supported by two cables, each 6 inches from the edge of the sign. The sign has a weight of 10 lbs, and the wind is causing the sign to sit at an angle of 10 degrees from vertical (the \(y\) axis). If we treat the wind as a point force acting in the negative \(z\) direction on the center of the sign, how strong must the wind force be to cause this ten-degree angle?

    A 10-lb metal rectangular sign, 12 by 24 inches, is hanging with its longer side horizontal. It is supported from the top by 2 cables, each attached 6 inches from an edge of the sign. A wind pushes the sign into the screen, so it makes a 10-degree angle with the plane of the screen.
    Figure \(\PageIndex{7}\): problem diagram for Example \(\PageIndex{7}\). A hanging sign experiences a wind whose direction points into the screen, causing the sign to make a 10° angle with the plane of the screen.
    Solution
    Video \(\PageIndex{8}\): Worked solution to example problem \(\PageIndex{7}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/pR-0xbj8wF0.
    Example \(\PageIndex{8}\)

    A sixty-kilogram acoustic panel is suspended by three cables as shown below. Assuming the panel has a uniformly distributed weight, what is the tension in each of the cables?

    A 60-kg rectangular panel, of dimensions 10 by 4 meters with one shorter side facing the viewer, lies in the xy plane (in the plane of the floor). It is suspended from the ceiling by 3 cables: cable 1 at the front right corner, cable 2 7 meters behind the location of cable 1, and cable 3 at the midpoint of the left side of the panel.
    Figure \(\PageIndex{8}\): problem diagram for Example \(\PageIndex{8}\). A uniform rectangular panel is suspended from above by 3 cables located at different points along its edges.
    Solution
    Video \(\PageIndex{9}\): Worked solution to example problem \(\PageIndex{8}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/Kbsc1m0f9pQ.

    This page titled 3.6: Equilibrium Analysis for a Rigid Body 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.