Statics and Dynamics Quiz: 3d Rigid Body Equilibrium

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3d Rigid Body EquilibriumQuestion 1 of 12

A rigid L-shaped bracket lies in 3D space. The short arm extends along the x-axis from the origin O to point A at $(0.2, 0, 0)$ m, and the long arm extends from A along the z-axis to point B at $(0.2, 0, 0.6)$ m. A force $\mathbf{F} = 100\hat{j}$ N is applied at B. The bracket is fixed at O by a built-in (cantilever) support.

What is the resultant moment vector (in N·m) that the fixed support at O must exert on the bracket to maintain equilibrium?

\mathbf{M}_O = -60\hat{i} + 0\hat{j} + 20\hat{k}

N·m, found by computing

\mathbf{r}_{OB} \times \mathbf{F}

directly and treating the result as the required support moment, without applying the equilibrium sign reversal.

\mathbf{M}_O = 0\hat{i} + 20\hat{j} - 60\hat{k}

N·m, found by crossing the position vector with the force but assigning components according to the arm lengths projected onto each axis independently rather than using the determinant form of the cross product.

\mathbf{M}_O = -60\hat{i} + 0\hat{j} - 20\hat{k}

N·m, found by computing

\mathbf{r}_{OB} \times \mathbf{F}

but using only the z-arm length (0.6 m) to determine the i-component and only the x-arm length (0.2 m) for the k-component, then negating both.

\mathbf{M}_O = 60\hat{i} + 0\hat{j} - 20\hat{k}

N·m, found by computing the moment of the applied force about O as

\mathbf{r}_{OB} \times \mathbf{F}

, then negating it to obtain the equal-and-opposite reaction moment required for equilibrium.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: 3d Rigid Body Equilibrium

Practice 3d Rigid Body Equilibrium in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 3d Rigid Body Equilibrium, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the resultant moment vector (in N·m) that the fixed support at O must exert on the bracket to maintain equilibrium?

$\mathbf{M}_O = -60\hat{i} + 0\hat{j} + 20\hat{k}$ N·m, found by computing $\mathbf{r}_{OB} \times \mathbf{F}$ directly and treating the result as the required support moment, without applying the equilibrium sign reversal.
$\mathbf{M}_O = 0\hat{i} + 20\hat{j} - 60\hat{k}$ N·m, found by crossing the position vector with the force but assigning components according to the arm lengths projected onto each axis independently rather than using the determinant form of the cross product.
$\mathbf{M}_O = -60\hat{i} + 0\hat{j} - 20\hat{k}$ N·m, found by computing $\mathbf{r}_{OB} \times \mathbf{F}$ but using only the z-arm length (0.6 m) to determine the i-component and only the x-arm length (0.2 m) for the k-component, then negating both.
$\mathbf{M}_O = 60\hat{i} + 0\hat{j} - 20\hat{k}$ N·m, found by computing the moment of the applied force about O as $\mathbf{r}_{OB} \times \mathbf{F}$ , then negating it to obtain the equal-and-opposite reaction moment required for equilibrium. (correct answer)

Explanation: When a fixed support holds a structure in equilibrium, it must supply both a reaction force and a reaction moment. The reaction moment exactly cancels the moment that the applied loads create about the support point — meaning you first compute the moment due to the applied force, then negate it. Start by finding

\mathbf{r}_{OB}

, the position vector from O to the point of force application B:

\mathbf{r}_{OB} = 0.2\hat{i} + 0\hat{j} + 0.6\hat{k}

m. Now cross it with

\mathbf{F} = 100\hat{j}

N using the determinant form:

\mathbf{r}_{OB} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.2 & 0 & 0.6 \\ 0 & 100 & 0 \end{vmatrix} = (0\cdot0 - 0.6\cdot100)\hat{i} - (0.2\cdot0 - 0.6\cdot0)\hat{j} + (0.2\cdot100 - 0\cdot0)\hat{k}

= -60\hat{i} + 0\hat{j} + 20\hat{k} \text{ N·m}

For moment equilibrium, the support moment must be the negative of this:

\mathbf{M}_O = 60\hat{i} + 0\hat{j} - 20\hat{k}

N·m. That's answer D. Choice A makes the cross product correctly but skips the negation — it reports the moment of the force rather than the reaction moment the support must provide. Choice B scrambles the component assignments, essentially guessing which arm length belongs to which axis instead of trusting the determinant. Choice C gets the magnitudes right but incorrectly negates both components of the cross product before the equilibrium step, producing the wrong signs. Your go-to strategy: always separate the two steps — compute

\mathbf{r} \times \mathbf{F}

first, then negate for the support reaction. Conflating these steps is the most common trap in 3D moment problems.

Question 2

In a 3D rigid-body equilibrium problem, a student takes moments about a carefully chosen axis (not just a point) to eliminate four of the six unknown reactions simultaneously, leaving a single equation with one unknown. Under what conditions is this strategy valid and sufficient to determine that one unknown directly?

The strategy is valid whenever the chosen axis passes through the point of application of the unknown force; the moment of any force about an axis through its point of application is always zero, so all other forces with lines of action intersecting that axis are also eliminated simultaneously.
The strategy is valid only when all six reactions are force reactions (no moment reactions), because a moment reaction about any axis is a free vector that contributes to moment equations about all axes equally, making it impossible to eliminate moment reactions by choice of moment axis.
The strategy is valid only if the chosen axis is one of the three coordinate axes, because the scalar moment equation about an arbitrary axis requires projection onto a unit vector along that axis, introducing additional geometric unknowns that prevent direct solution.
The strategy is valid when the chosen axis intersects the lines of action of four reactions (making their moments about that axis zero) and the remaining unknown's line of action does not intersect the axis; the single resulting equation is sufficient to solve for that unknown without solving the full 6×6 system. (correct answer)

Explanation: When tackling 3D rigid-body equilibrium, you have six scalar equations available. The power of choosing a moment axis (rather than just a moment point) is that any force whose line of action intersects that axis contributes zero moment about it — because the perpendicular distance from the axis to that line of action is zero. This lets you strategically silence multiple unknowns at once. Answer D captures exactly why this works: if four reactions have lines of action that intersect your chosen axis, their moment contributions vanish, leaving a single equation in one unknown — provided that unknown's line of action does not intersect the axis (giving it a nonzero moment arm). That single equation is then directly solvable without touching the rest of the system. Answer A contains a seductive half-truth. Yes, a force's moment about an axis through its own point of application can be zero, but only if the force's line of action actually intersects (or is parallel to) that axis. A force applied at a point on the axis but directed away from it still produces a moment. The moment-elimination condition depends on the line of action, not just the point of application. Answer B is wrong because moment reactions (couples) are indeed free vectors — but this makes them harder to eliminate, not universally impossible. The issue is real but doesn't invalidate the entire strategy; it just means you must account for couple-moment reactions carefully. Answer C is wrong because moment equations about an arbitrary axis are perfectly valid using the scalar triple product

M_\lambda = \hat{\lambda} \cdot (\mathbf{r} \times \mathbf{F})

. No coordinate-axis restriction exists. Study tip: Always ask two questions when choosing a moment axis: "Does this axis intersect four unknown lines of action?" and "Does my target unknown have a nonzero moment arm about it?" Both must be true for the strategy to work.

Question 3

Taking moments about a ball-and-socket support at O, how many of its support reactions appear?

All three components
None, since they act at O (correct answer)
Two, as couple moments
Only one component acts

Explanation: A ball-and-socket support exerts reaction force components at point O. When you take moments about O, the moment arm for each of those forces is zero, so none of the support reactions appear in the moment equation. The tempting error is counting all three force components, but forces acting at the moment center produce no moment.

Question 4

A rigid body in 3D is acted on by exactly two forces. For equilibrium, the forces must be

equal, opposite, and collinear (correct answer)
equal in magnitude, opposite
opposite, coplanar, concurrent
equal, parallel, and coplanar

Explanation: For a rigid body, two forces must first cancel each other, so they must be equal in magnitude and opposite in direction. They must also share the same line of action: if they're parallel but offset, they create a couple and the body rotates. The tempting 'equal in magnitude and opposite' answer misses that collinearity requirement.

Question 5

In 3D equilibrium, a force parallel to the x-axis not on it has what x-axis moment?

$F$ times the distance to axis
Zero, regardless of location (correct answer)
$F z$ , the y-axis moment
$F y$ , the z-axis moment

Explanation: The moment is r cross F. With F along x, F_y and F_z are zero, so the x-component of the moment is y F_z minus z F_y, which is zero no matter where the force is located. It creates only y- and z-axis moment components. The tempting mistake is F times distance to the axis, but that confuses the magnitude of the moment about a point or other axes with the moment about the x-axis.

Question 6

For a 3D rigid body, if force resultant is zero and moment about A is zero, the moment about B is

nonzero in general
$r_{AB}\times F$
zero only if B is A
zero at every point B (correct answer)

Explanation: Since the resultant force is zero, moving the reference point adds the term r_BA x 0, which is zero. With the moment already zero at A, the moment stays zero no matter which point B you choose. The tempting r_AB x F formula cannot change this because F itself is zero, so that term vanishes too.

Question 7

A 3D rigid body is held by independent supports with 8 unknown reactions. It is

Statically determinate system
Statically indeterminate by 1
Statically indeterminate by 2 (correct answer)
Statically unstable always

Explanation: A free 3D rigid body has exactly 6 equilibrium equations: 3 force sums and 3 moment sums. With 8 unknown support reactions, the excess is 8 - 6 = 2, so the system is indeterminate by 2. The tempting wrong answer, indeterminate by 1, comes from using only 7 equations or forgetting the three independent moment conditions.

Question 8

A rigid body in 3D is supported by support system S1 consisting of: a ball-and-socket at A (3 unknowns), a smooth journal bearing at B with shaft along the z-axis (2 unknowns: $F_x, F_y$ ), and a single cable at C (1 unknown). A second student proposes replacing S1 with support system S2: two smooth journal bearings, one at A and one at B, both with shafts along the z-axis (2 unknowns each: $F_x, F_y$ ), plus one cable at C.

Comparing S1 and S2, which statement correctly characterizes the two systems?

S1 has 6 unknowns and is statically determinate; S2 has 5 unknowns and is a mechanism (under-constrained), because replacing the ball-and-socket at A with a journal bearing removes the axial (z-direction) force reaction, leaving the body free to translate in the z-direction unless the cable supplies that component. (correct answer)
Both S1 and S2 have 6 unknowns and are statically determinate; however, S2 has a geometric deficiency because both bearing axes are parallel (both along z), so the system cannot resist a net moment about the z-axis and is improperly constrained despite the correct unknown count.
S1 has 6 unknowns and is properly constrained; S2 also has 6 unknowns (2 from each journal bearing and 2 from the cable if it is not axis-aligned), and is therefore statically determinate provided the cable direction is specified.
S2 has 6 unknowns and is statically determinate; it improves upon S1 because distributing reactions over two journal bearing locations rather than a single ball-and-socket always reduces the maximum reaction magnitude and eliminates the risk of under-constraint.

Explanation: When analyzing 3D support systems, always count unknowns first, then check for geometric deficiency — a system can have exactly 6 unknowns and still be improperly constrained if the supports cannot resist every possible load direction or moment. S1 combines a ball-and-socket at A (

F_x, F_y, F_z

— 3 unknowns), a journal bearing at B with shaft along z (

F_x, F_y

— 2 unknowns), and one cable (1 unknown), totaling 6 unknowns for 6 equilibrium equations. Critically, the ball-and-socket provides the

F_z

reaction at A. S2 replaces that ball-and-socket with a journal bearing (also shaft along z), which only provides

F_x

and

F_y

— dropping

F_z

at A. Now S2 has only 5 unknowns: 2 + 2 + 1. Unless the cable has a z-component, no support resists translation in the z-direction, making the body a mechanism. Answer A correctly identifies this: S1 has 6 unknowns and is properly constrained; S2 has 5 and is under-constrained. Answer B incorrectly claims both systems have 6 unknowns — it misses that swapping the ball-and-socket for a journal bearing loses one unknown. Answer C compounds this error by suggesting S2 can reach 6 unknowns if the cable is "not axis-aligned," but a single cable still contributes only 1 unknown regardless of orientation. Answer D fabricates an engineering advantage for S2 — distributing reactions doesn't compensate for a missing force component, and S2 is strictly worse, not better. Your study tip: always list each support's specific reaction components before summing unknowns. A journal bearing with a z-axis shaft provides no z-force — don't assume it behaves like a ball-and-socket.

Question 9

A 3D frame is in equilibrium. It is supported by a smooth journal bearing at A (which constrains displacement perpendicular to the shaft axis but allows rotation and axial displacement), a thrust bearing at B (which constrains all three translational displacements but allows rotation about the shaft axis and supplies no moment reactions), and a single cable attached at point C. The shaft axis runs along the x-direction.

How many scalar unknowns are introduced by this support configuration, and is the system statically determinate?

7 unknowns total (2 from the journal bearing, 3 from the thrust bearing, 1 from the cable, and 1 moment reaction from the thrust bearing about the shaft axis); the system is statically indeterminate because 7 unknowns exceed the 6 equilibrium equations.
6 unknowns total (2 from the journal bearing, 3 from the thrust bearing, and 1 from the cable); the system is statically determinate because exactly 6 unknowns match the 6 equilibrium equations, provided the unknown directions are geometrically independent. (correct answer)
5 unknowns total (2 from the journal bearing, 2 from the thrust bearing, and 1 from the cable); the system has fewer unknowns than equilibrium equations, indicating the body is a mechanism and cannot be in stable equilibrium under a general loading.
6 unknowns total (2 from the journal bearing, 3 from the thrust bearing, and 1 from the cable), but the system is statically indeterminate because the journal bearing and thrust bearing share the same shaft axis, making two of their reactions linearly dependent and reducing the effective rank of the equation system.

Explanation: When analyzing 3D support reactions, your first job is to carefully count the scalar unknowns each support contributes based on its physical constraints, then compare that total to the 6 equilibrium equations (

\sum F_x, \sum F_y, \sum F_z, \sum M_x, \sum M_y, \sum M_z

). Here's how the count works for answer B: A smooth journal bearing constrains displacement in the two directions perpendicular to the shaft axis (here, the x-axis), giving 2 force reactions (in y and z). It cannot resist axial force or supply any moment. A thrust bearing constrains all three translational displacements, giving 3 force reactions (x, y, and z), but supplies no moment reactions about any axis. A single cable can only pull in tension along its length, contributing 1 scalar unknown. Total:

2 + 3 + 1 = 6

unknowns, exactly matching the 6 equilibrium equations — making the system statically determinate, provided those reactions are geometrically independent (non-parallel, non-concurrent in a degenerate way). A is wrong because it invents a moment reaction from the thrust bearing about the shaft axis. By definition, thrust bearings allow free rotation about the shaft axis and supply no moment there — that's 0 moment unknowns, not 1. C is wrong because it undercounts the thrust bearing as supplying only 2 reactions. A thrust bearing resists axial translation too, so it contributes 3 force reactions, not 2. D is a tempting trap: sharing a shaft axis does not automatically make reactions linearly dependent. The journal bearing provides y- and z-forces, while the thrust bearing additionally provides an x-force at a different location, so their contributions are geometrically independent. Study tip: Always map each support to its physical action — what motions does it prevent? Each prevented translation adds one force unknown; each prevented rotation adds one moment unknown. Don't let shared geometry fool you into assuming dependence.

Question 10

A 3D rigid body is supported by exactly six scalar reaction unknowns from its supports, and the six equilibrium equations ( $\sum \mathbf{F} = 0$ and $\sum \mathbf{M}_A = 0$ about some point A) are written. A student correctly solves the system and finds that one of the six unknowns is negative. Which statement best describes the physical and mathematical interpretation of this result?

The negative value indicates a mathematical inconsistency: since the system has exactly 6 equations and 6 unknowns, a negative reaction implies the equilibrium equations were set up with at least one sign convention error, and the problem should be re-solved with corrected directions.
The negative value is physically meaningful and indicates the actual direction of that reaction is opposite to the assumed positive direction; the equilibrium solution is valid, and no error has occurred if the support can physically exert a force or moment in that reversed direction. (correct answer)
The negative value is physically meaningful only if the unknown is a moment reaction; a negative force reaction always implies the support is being pulled rather than pushed, which is impossible for a contact support such as a bearing, and the problem must be reconsidered for a different support configuration.
The negative value means the system is actually statically indeterminate because a properly determinate system always yields non-negative reactions when the loading direction is correctly identified, and a negative result signals a redundant constraint has been overlooked.

Explanation: Whenever you solve a 3D equilibrium problem, remember that assumed directions for unknown reactions are just that — assumptions. You assign a positive direction to each unknown at the start, write your six equations (

\sum F_x = 0

\sum F_y = 0

\sum F_z = 0

, and three moment equations), and solve. The sign of the result tells you whether your assumption was correct. A negative value simply means the actual reaction acts opposite to your assumed direction — nothing more. This is the core of B, the correct answer. The solution is mathematically valid and physically meaningful, provided the support is capable of exerting a force or moment in that reversed direction (e.g., a pin can push or pull, a fixed wall can reverse a moment). A statically determinate system with a unique solution is still determinate regardless of the signs of its answers. A is wrong because a negative result is not a sign of a setup error. Sign conventions exist precisely so that negative answers can reveal reversed directions — this is a feature, not a bug. Re-solving because you got a negative value would be a mistake. C is wrong because it incorrectly singles out force reactions as uniquely problematic. Both force and moment reactions can be negative, and a negative force reaction at a pin simply means the pin pulls rather than pushes — which a pin absolutely can do. Whether that's physically realizable depends on the specific support type, not on whether the unknown is a force or moment. D is wrong because the sign of a reaction has no bearing on static determinacy. A system is determinate when the number of independent equations equals the number of unknowns — negative results don't change that count. Study tip: Always state your assumed positive direction clearly at the start of any support-reaction problem. A negative answer is your solution's way of correcting your assumption — trust it.

Question 11

For a 3D rigid body in equilibrium supported by a combination of supports that together provide exactly 6 independent scalar reactions, which of the following statements about the moment-equation choices is FALSE?

Taking moments about three different non-collinear points (rather than a single point) can replace the standard $\sum \mathbf{M}_A = 0$ vector equation and still yield three independent scalar equations, provided the geometry is chosen so that the force equilibrium conditions are implicitly enforced by the moment equations about those points.
Taking moments about three mutually perpendicular axes all passing through the same point A is exactly equivalent to writing the single vector equation $\sum \mathbf{M}_A = 0$ , and both approaches yield the same three independent scalar equations with identical physical content.
Taking moments about six different axes can replace the standard six equilibrium equations entirely, provided the six axes are chosen such that each isolates a different unknown reaction; this approach is always valid regardless of the geometry of the support reactions, and no force-equilibrium equations are needed. (correct answer)
If the six scalar equilibrium equations are written using $\sum F_x=0$ , $\sum F_y=0$ , $\sum F_z=0$ , and $\sum M_x=0$ , $\sum M_y=0$ , $\sum M_z=0$ about a common point, and the six reactions are truly independent, then the solution is unique and the choice of moment point affects the algebraic form of individual equations but not the final numerical values of the unknowns.

Explanation: Whenever you see a question about 3D equilibrium and moment equations, ask yourself: does this approach guarantee both force and moment equilibrium, or could it leave some conditions unchecked? The standard six scalar equilibrium equations —

\sum F_x=0

\sum F_y=0

\sum F_z=0

\sum M_x=0

\sum M_y=0

\sum M_z=0

— are required because a rigid body can both translate and rotate. Moment equations alone, no matter how many axes you choose, cannot guarantee translational equilibrium unless the geometry of those axes is specifically arranged to implicitly enforce it. C is false (and therefore the correct answer) because it claims you can always replace all six equilibrium equations with six moment-axis equations "regardless of the geometry." This is the critical flaw. While it is sometimes possible to select axes cleverly enough that moment equations implicitly enforce force balance, this only works for specific geometric configurations — it is not a universal guarantee. The word "always" makes C definitively false. A is actually true: taking moments about three carefully chosen non-collinear points can replace the standard moment vector equation, provided the geometry ensures force equilibrium is implicitly captured — exactly the condition C incorrectly claims is always satisfied. B is true: moments about three mutually perpendicular axes through a single point are literally the three scalar components of

\sum \mathbf{M}_A = 0

, so both approaches are identical. D is true: with truly independent reactions, the system has a unique solution; changing the moment point only reorganizes the algebra. Your strategy tip: watch for absolute words like "always" or "regardless of geometry" — in statics, geometric conditions almost always matter, and such sweeping claims are usually the trap.

Question 12

A uniform rectangular sign of weight $W$ is attached to a wall by a ball-and-socket joint at corner A and two cables. Cable 1 runs from corner B (diagonally opposite A on the sign) to a point on the wall directly above A, and Cable 2 runs from the midpoint M of the top edge of the sign to a point on the ceiling. A student argues that the ball-and-socket joint at A must supply a moment reaction to prevent the sign from rotating about the axis AB. Which of the following most precisely identifies the error in the student's reasoning?

A ball-and-socket joint does supply moment reactions about all three axes; the student's error is in claiming rotation would occur about AB specifically rather than about a horizontal axis perpendicular to the wall.
A ball-and-socket joint supplies three force reactions but zero moment reactions; however, the student's conclusion is also wrong because without a moment reaction at A, the sign cannot be in rotational equilibrium about AB unless the two cables happen to be collinear with AB.
A ball-and-socket joint supplies three force reactions but zero moment reactions; the sign's rotational equilibrium about any axis through A, including AB, is maintained by the moments produced by the cable tension forces acting at their respective attachment points, with no moment reaction required at A. (correct answer)
A ball-and-socket joint supplies three force reactions but zero moment reactions; the student's specific error is that rotation about axis AB is prevented by the weight W alone acting through the center of mass, making the two cables entirely redundant for rotational equilibrium.

Explanation: Whenever you see a problem involving a ball-and-socket joint, your first instinct should be to recall exactly what reactions it provides: three force components (

F_x, F_y, F_z

) and zero moment reactions. A ball-and-socket allows free rotation in any direction, so it cannot resist moments. This is the foundation of the entire problem. The student's error is two-layered. First, they incorrectly assume the joint supplies a moment reaction. Second — and more subtly — they assume one is even needed. For any axis passing through A, rotational equilibrium is governed by the moments of all other forces about that axis. Here, the tensions in Cable 1 and Cable 2 each act at points away from A (corner B and midpoint M, respectively), so they produce nonzero moment arms about the axis AB. The weight

W

also contributes its moment about AB through the center of mass. Together, these external force moments balance each other, satisfying

\sum M_{AB} = 0

with no moment reaction required at A. That's exactly what C describes, making it correct. A is wrong because it misrepresents the ball-and-socket joint entirely — it provides no moment reactions, not three. B correctly identifies the joint's properties but then invents a false condition: cables do not need to be collinear with AB to maintain equilibrium; they just need to produce balancing moments about AB. D incorrectly claims weight alone maintains rotational equilibrium and that the cables are redundant — this is generally false and would only hold in a very special geometric case. Your study tip: always separate two questions — what does this support provide? and what does equilibrium actually require? They're not always the same thing.