Varignon's Theorem - Statics and Dynamics

1. Using Varignon's theorem, the moment about O can be written as $M_O = r_x F_y - r_y F_x$. A student sets up the calculation as $M_O = F_x(-4) + F_y(3)$, where the sign convention assigns counterclockwise as positive. Which of the following correctly evaluates $M_O$ and identifies the physical meaning of the sign?
   Answer: $M_O = (120\cos30°)(-4) + (120\sin30°)(3) \approx -235.3\,\text{N}\cdot\text{m}$; the negative sign indicates a clockwise moment about O, consistent with the geometry of the force placement relative to O. Whenever you see a question involving moments and Varignon's theorem, your job is to apply the cross-product structure carefully: $M_O = r_x F_y - r_y F_x$. The key insight is that each position coordinate multiplies the opposite-axis force component, and the signs of the coordinates matter — they encode the geometry that determines rotational direction. Here, the position is $(r_x, r_y) = (3\,\text{m}, 4\,\text{m})$ and the force components are $F_x = 120\cos30° \approx 103.9\,\text{N}$ and $F_y = 120\sin30° = 60\,\text{N}$. Plugging into the student's correctly structured setup: $M_O = F_x(-r_y) + F_y(r_x) = (103.9)(-4) + (60)(3) = -415.6 + 180 \approx -235.3\,\text{N}\cdot\text{m}$ The negative result confirms a clockwise moment about O, which is physically reasonable: the dominant horizontal component of the force acts downward-left relative to the moment arm in the y-direction, overpowering the counterclockwise contribution of $F_y$. Answer A captures this correctly. B is wrong because it takes both coordinates as positive, dropping the sign on $r_y = -4$, which destroys the directional information embedded in the cross product. C is wrong because it matches each force component to its same-axis coordinate ($F_x$ with $r_x$, $F_y$ with $r_y$) — the exact mistake Varignon's theorem is designed to avoid; the correct pairing swaps axes. D is wrong because it incorrectly negates $F_y$, based on a false rule that upward forces always produce clockwise moments in the first quadrant — sign is determined by geometry, not force direction alone. Study tip: Always write $M_O = r_x F_y - r_y F_x$ explicitly before substituting. The cross-axis pairing and coordinate signs will protect you from the traps in B, C, and D.
2. A force $\vec{F}$ is applied to a rigid body, and the moment about point O is computed as $\vec{M}_O$. A second point Q lies on the line of action of $\vec{F}$. Using Varignon's theorem and the properties of the moment, which of the following statements is necessarily true about the moment $\vec{M}_Q'$ computed using position vector $\vec{r}_{Q/O}$ (from O to Q) in place of the original position vector?
   Answer: $\vec{M}_Q' = \vec{M}_O$ because the moment of a force about a fixed point is independent of which point on the line of action of the force is used as the tip of the position vector, since the perpendicular distance from O to the line of action remains unchanged. Whenever you see a question about moments and position vectors, anchor yourself to one fundamental principle: the moment of a force about a fixed point depends only on the perpendicular distance from that point to the line of action — not on which specific point along the line of action you use to construct your position vector. Here's why B is correct. If Q lies on the line of action of $\vec{F}$, then the vector from O to Q can be written as $\vec{r}_{Q/O} = \vec{r}_{P/O} + t\vec{F}/|\vec{F}|$ for some scalar $t$, where P is the original application point. When you compute the cross product: $\vec{r}_{Q/O} \times \vec{F} = (\vec{r}_{P/O} + t\hat{F}) \times \vec{F} = \vec{r}_{P/O} \times \vec{F} + t(\hat{F} \times \vec{F}) = \vec{M}_O + \vec{0}$ The extra term vanishes because a vector crossed with a parallel vector is zero. Therefore $\vec{M}_Q' = \vec{M}_O$ always, in any dimension. A misreads Varignon's theorem. That theorem concerns decomposing a force into components — it says nothing about restricting which point on the line of action is used. C is a trap: it suggests the moment grows with position vector magnitude, ignoring that the angle between the position vector and $\vec{F}$ also changes, keeping the perpendicular distance (and thus the moment) constant. D invents a 2D-only restriction that has no mathematical basis — the cross product argument above holds fully in three dimensions. Your study tip: whenever you add a vector parallel to $\vec{F}$ to your position vector, remember that its cross product with $\vec{F}$ is zero — this "sliding" property is why forces can be transmitted along their line of action without changing the moment.
3. A student applies Varignon's theorem to compute the resultant moment about O by summing the moments of all x- and y-components separately. The student writes: $M_O = (80)(3) + (-60)(2) + (-50)(-1) + (40)(5)$. What error has the student made, and what is the correct resultant moment (CCW positive)?
   Answer: The student paired each force component with its same-axis coordinate rather than using the cross-product form $r_x F_y - r_y F_x$. The correct sum is $M_O = [(2)(-60)-(3)(80)] + [(5)(40)-(-1)(-50)] = [-120-240]+[200-50] = -360+150 = -210\,\text{N}\cdot\text{m}$ (clockwise). Whenever you see a moment calculation involving multiple force components, your first instinct should be the 2D cross-product formula: $M_O = r_x F_y - r_y F_x$. This is the heart of Varignon's theorem — the moment of each component equals its force magnitude times the perpendicular distance, which means pairing each force component with the other coordinate axis, not its own. The student's error was pairing each component with its same-axis coordinate: writing $(80)(3)$ pairs $F_x$ with the x-coordinate, and $(-60)(2)$ pairs $F_y$ with the y-coordinate — both wrong. The correct expression applies $r_x F_y - r_y F_x$ to each force separately. For $\vec{F}_1$ at $(2,3)$: $(2)(-60) - (3)(80) = -120 - 240 = -360\,\text{N}\cdot\text{m}$. For $\vec{F}_2$ at $(5,-1)$: $(5)(40) - (-1)(-50) = 200 - 50 = 150\,\text{N}\cdot\text{m}$. Summing gives $-360 + 150 = -210\,\text{N}\cdot\text{m}$ (clockwise), confirming C is correct. Choice A negates every term arbitrarily without fixing the structural pairing error — a sign patch on a formula that's fundamentally wrong. Choice B partially corrects only the $F_x$ sign terms, which still doesn't address the pairing problem and produces an incorrect intermediate expression. Choice D claims the student's original expression is structurally valid — it isn't, because the pairings violate the cross-product definition. Study tip: Memorize $M = r_x F_y - r_y F_x$ as a single unit. The minus sign and the axis-swap are inseparable — if you remember only one part, you'll fall into exactly the trap this question is testing.
4. A force $\vec{F}$ passes through point A at position $\vec{r}_A = 5\hat{i} - 2\hat{j}\,\text{m}$ from origin O, and through point B at $\vec{r}_B = 5\hat{i} + 6\hat{j}\,\text{m}$ from O. The magnitude of $\vec{F}$ is $200\,\text{N}$. An engineer uses Varignon's theorem to compute the moment of $\vec{F}$ about O by decomposing $\vec{F}$ along the line AB. Which statement most accurately characterizes the application of Varignon's theorem in this scenario?
   Answer: Varignon's theorem applies directly: the moment of $\vec{F}$ about O equals the sum of moments of its components, and since any point on the force's line of action may serve as the tip of the position vector, using either A or B yields the same moment about O. Whenever you see a question combining Varignon's theorem with a force defined by two points, focus on two core principles: (1) Varignon's theorem states that the moment of a force about a point equals the sum of moments of its components about that same point, and (2) a force's moment about a point depends only on its line of action, not which specific point along that line you choose as your position vector tip. Answer A is correct for both reasons. Varignon's theorem places no restriction on how you decompose a force — into perpendicular components, oblique components, or anything else — as long as they sum to $\vec{F}$. The moment is additive over components: $\vec{M}_O = \vec{r} \times \vec{F} = \vec{r} \times (\vec{F}_1 + \vec{F}_2) = \vec{r} \times \vec{F}_1 + \vec{r} \times \vec{F}_2$. Furthermore, since moment depends on the line of action, any position vector pointing to any point on that line — here, A or B — produces the identical moment about O. Answer B is built on a false restriction. Varignon's theorem applies to forces along a line of action, not just "single-point" forces — that concept doesn't exist in rigid-body statics. Answer C invents a nonexistent rule about using the midpoint; the moment is invariant along the line of action, so A, B, the midpoint, or any other point on AB all give the same result. Answer D incorrectly limits Varignon's theorem to axis-aligned components — the theorem is fully general and works for components in any directions that sum to $\vec{F}$. Your study tip: remember that "line of action" is the key phrase. Whenever a force's line of action is defined, moment calculations are free to use any convenient point on it.
5. A force system consists of three concurrent forces acting at point P: $\vec{F}_1 = 200\hat{i}\,\text{N}$, $\vec{F}_2 = -150\hat{j}\,\text{N}$, and $\vec{F}_3 = 100\hat{i} + 100\hat{j}\,\text{N}$. Point P is located at $\vec{r} = 3\hat{i} + 4\hat{j}\,\text{m}$ from moment center O. An engineer uses Varignon's theorem and computes the moment of the resultant by first finding $\vec{R} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3$ and then computing $\vec{r} \times \vec{R}$. A second engineer instead sums the individual moments $\vec{r} \times \vec{F}_1 + \vec{r} \times \vec{F}_2 + \vec{r} \times \vec{F}_3$. Which of the following is true about the two approaches, and what is the resultant moment (CCW positive)?
   Answer: Both approaches yield identical results by Varignon's theorem: $\vec{R} = 300\hat{i} - 50\hat{j}\,\text{N}$, and $M_O = (3)(-50) - (4)(300) = -150 - 1200 = -1350\,\text{N}\cdot\text{m}$ (clockwise), confirming that summing moments of components equals the moment of the resultant for concurrent forces. Whenever you see a question involving moments of concurrent forces, your first instinct should be to recall Varignon's Theorem: the moment of a resultant equals the sum of the moments of its components. This isn't just a shortcut — it's a fundamental consequence of the distributive property of the cross product. Here's the core reasoning for answer D. First, find the resultant: $\vec{R} = (200+100)\hat{i} + (-150+100)\hat{j} = 300\hat{i} - 50\hat{j}\,\text{N}$. Then compute the moment using $\vec{r} \times \vec{R}$, where $\vec{r} = 3\hat{i} + 4\hat{j}$. For a 2D cross product, $M_O = r_x R_y - r_y R_x = (3)(-50) - (4)(300) = -150 - 1200 = -1350\,\text{N}\cdot\text{m}$. The negative sign means clockwise. Varignon's theorem guarantees the second engineer's summation of individual moments produces the exact same result, since $\vec{r} \times (\vec{F}_1 + \vec{F}_2 + \vec{F}_3) = \vec{r} \times \vec{F}_1 + \vec{r} \times \vec{F}_2 + \vec{r} \times \vec{F}_3$ by linearity. Answer A gets the resultant right but botches the cross product formula, swapping which component multiplies which. It incorrectly writes $(3)(300) - (4)(-50)$, confusing $r_x R_x$ for a moment arm — a critical sign/formula error. Answer B falsely claims Varignon's theorem only works component-by-component, which misunderstands the theorem entirely. The theorem applies to the full vector resultant without restriction. Answer C incorrectly claims the individual-moments method double-counts $\vec{F}_3$. The cross product is linear — no force is ever double-counted. Study tip: Always memorize the 2D moment formula as $M_O = r_x F_y - r_y F_x$, and verify your sign by checking whether the force tends to rotate the body clockwise or counterclockwise.
6. Using Varignon's theorem, the student decomposes $\vec{F}$ and sets the moment about $(x_0, 0)$ equal to zero. However, the student reasons: 'Since the force is at 45° and acts symmetrically, the moment arm must be zero, so $x_0 = 3\,\text{m}$.' Which of the following correctly finds $x_0$ and identifies the flaw in the student's reasoning?
   Answer: The correct value is $x_0 = 0\,\text{m}$; the student's error is neglecting that zero moment requires the line of action of $\vec{F}$ to pass through the point, and the 45° line through $(3,3)$ with slope 1 intersects the x-axis at the origin, not at $x = 3\,\text{m}$. When a moment equals zero about a point, it means that point lies on the line of action of the force — the infinite line along which the force vector acts. This is the key geometric insight for this problem. The force $\vec{F} = 100\hat{i} + 100\hat{j}$ points at 45°, meaning its line of action is a straight line with slope 1 passing through the point of application $P = (3, 3)$. That line's equation is $y - 3 = 1(x - 3)$, or simply $y = x$. To find where this line crosses the x-axis, set $y = 0$: you get $x = 0$. So $x_0 = 0\,\text{m}$, confirming C is correct. You can also verify using Varignon's theorem: the moment about $(x_0, 0)$ is $M = (3)(100) - (3 - x_0)(100) = 0$, which gives $3 - (3 - x_0) = 0$, so $x_0 = 0$. Both approaches agree. A is wrong because $x_0 = 3$ does not place the point on the line of action $y = x$ — the moments from the two components do not cancel at $x = 3$. B sets up the moment equation incorrectly, mixing up which component contributes which moment arm, and arrives at the wrong answer of 6. D fabricates a "half perpendicular distance" rule that has no basis in statics. Study tip: Whenever you need a zero-moment point, ask yourself: where does the line of action cross that axis? Sketch the force's line of action first — it instantly reveals the answer and guards against algebraic setup errors.
7. The engineer computes the resultant force magnitude as $R = \frac{1}{2}w_0 L = 24\,\text{kN}$. To find the location $\bar{x}$ of the resultant from $x = 0$, the engineer applies Varignon's theorem: $R \cdot \bar{x} = \int_0^L x \cdot w(x)\,dx$, where $w(x) = \frac{w_0}{L}x$. Which of the following gives the correct $\bar{x}$ and explains the role of Varignon's theorem in this calculation?
   Answer: $\bar{x} = \frac{2L}{3} = 2.67\,\text{m}$; Varignon's theorem justifies replacing the distributed load with its resultant R acting at $\bar{x}$, because the theorem guarantees that the moment of R about any point equals the sum (integral) of moments of the infinitesimal load elements about that same point. Whenever you see a distributed load problem asking for the location of a resultant, your two-step framework is: (1) find the resultant magnitude from the area of the load diagram, and (2) use Varignon's theorem to locate it via moment equivalence. Varignon's theorem states that the moment of a resultant equals the sum of moments of its components — extended to continuous distributions, this becomes an integral. Here, $w(x) = \frac{w_0}{L}x$, so each infinitesimal element $dF = w(x)\,dx$ contributes a moment $x \cdot w(x)\,dx$ about the origin. Setting $R\bar{x} = \int_0^L x \cdot \frac{w_0}{L}x\,dx$ gives: $\int_0^4 \frac{12}{4}x^2\,dx = 3\cdot\frac{x^3}{3}\Big|_0^4 = \frac{4^3}{1} \cdot 1= 64\,\text{kN}\cdot\text{m}$ Then $\bar{x} = \frac{64}{24} = \frac{8}{3} = \frac{2L}{3} \approx 2.67\,\text{m}$. This is D, and it correctly identifies Varignon's theorem as the theoretical justification for the moment-integral approach. A gets the numerical answer backwards. A triangle's centroid is at $\frac{1}{3}$ of the base from the maximum-load end, which is $\frac{2L}{3}$ from the zero end — not $\frac{L}{3}$. B is doubly wrong: Varignon's theorem is not restricted to concurrent forces, and a triangular distribution is definitively not centered at $L/2$. C sets up a valid alternate integral (measuring from $x = L$) but then misreports the result — the distance $L/3$ from the peak corresponds to $2L/3$ from the origin, not $L/3$. Study tip: For any triangular load, always ask "measured from which end?" — the centroid is $L/3$ from the peak and $2L/3$ from the zero end. Confusing these two is the most common trap on resultant-location problems.
8. The engineer using Strategy II writes the position vector from B to A as $(x_A - x_B,\, y_A - y_B) = (-3,\, 4)\,\text{m}$ and computes $M_B = F_x \cdot (y_A - y_B) - F_y \cdot (x_A - x_B) = (60)(4) - (80)(-3) = 240 + 240 = 480\,\text{N}\cdot\text{m}$. Is this result correct, and which of the following best explains why or why not?
   Answer: The result is incorrect; the engineer swapped the pairing in the cross product. The correct Varignon expression is $M_B = (r_{BA,x})(F_y) - (r_{BA,y})(F_x) = (-3)(80) - (4)(60) = -240 - 240 = -480\,\text{N}\cdot\text{m}$ (clockwise). The engineer's formula computes $r_{BA,y} F_x - r_{BA,x} F_y$, which is the negative of the true moment, giving the wrong sign (direction). When computing a moment using the cross product, the order of terms matters — and this question tests whether you can catch a subtle sign error hidden inside a numerically "clean" result. The standard 2D cross-product formula for the moment of $\vec{F}$ about point B is: $M_B = r_{BA,x} F_y - r_{BA,y} F_x$ where $\vec{r}_{BA} = (x_A - x_B,\, y_A - y_B) = (-3,\, 4)\,\text{m}$. Plugging in: $M_B = (-3)(80) - (4)(60) = -240 - 240 = -480\,\text{N}\cdot\text{m}$. The negative sign indicates a clockwise moment. This confirms B is correct — the engineer swapped the pairings, computing $r_y F_x - r_x F_y$ instead of $r_x F_y - r_y F_x$, which flips the sign entirely. A is wrong in a subtle way: it correctly identifies that the engineer's formula is the negative of the standard cross product, but then claims the magnitude is still valid while only the sign is wrong. That's actually true — but A frames this as a partial error rather than recognizing that the sign is the full answer here. More importantly, A never identifies which formula is actually correct, leaving the student without the right expression. C is incorrect because no such "sign-reversal convention" exists when shifting moment centers. The position vector $\vec{r}_{BA}$ already accounts for the shift from O to B; no additional sign flip occurs. D is a fabricated rule. Varignon's theorem applies to any moment center — no "transfer correction term" is needed. You simply use the position vector from the new moment center to the point of application. Strategy tip: Always write the cross-product formula explicitly as $r_x F_y - r_y F_x$ before substituting numbers. When the answer comes out positive and "clean," that's exactly when a sign error is easiest to miss.

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