Statics Quiz: Varignons Theorem

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Varignons TheoremQuestion 1 of 17

A distributed load is replaced by its resultant force R acting at the centroid of the load distribution. When applying Varignon's theorem to find the moment about a point, which limitation must be considered?

Varignon's theorem cannot be applied to distributed loads because it only applies to concentrated forces acting at discrete points

The theorem applies only if the distributed load is uniform; non-uniform distributions require integration of individual force elements

Varignon's theorem is valid for the resultant force system, but the original distributed load must be discretized into finite elements for moment calculations

The theorem applies directly since the resultant force and its centroidal location fully represent the moment characteristics of the distributed load

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Statics Quiz

Statics Quiz: Varignons Theorem

Practice Varignons Theorem in Statics 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 Varignons Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

How to use this quiz

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

Varignon's theorem cannot be applied to distributed loads because it only applies to concentrated forces acting at discrete points
The theorem applies only if the distributed load is uniform; non-uniform distributions require integration of individual force elements
Varignon's theorem is valid for the resultant force system, but the original distributed load must be discretized into finite elements for moment calculations
The theorem applies directly since the resultant force and its centroidal location fully represent the moment characteristics of the distributed load (correct answer)

Explanation: Varignon's theorem applies to any force system, including distributed loads represented by their resultants. The fundamental principle of replacing a distributed load with its resultant force at the centroid ensures that both the net force and the net moment about any point remain unchanged. This is precisely what makes the resultant-centroid representation valid. Choice D correctly recognizes this principle. Choice A incorrectly limits Varignon's theorem to point forces. Choice B incorrectly suggests the theorem doesn't work for non-uniform distributions - the resultant-centroid method handles any distribution. Choice C unnecessarily complicates the approach - discretization isn't required when the resultant is properly located at the centroid.

Question 2

In applying Varignon's theorem to a force system, a student decomposes a 400 N force at point (3, 5) m into components $F_x = 320$ N and $F_y = 240$ N. When calculating the moment about point (1, 2) m, which statement correctly describes the application of the theorem?

The sum of moments of components equals the moment of the resultant force about the same reference point
The moment of each component is calculated using the original point of application and the reference point
The perpendicular distance from reference point to each component's line of action must be determined separately
All of the above statements are correct requirements for applying Varignon's theorem (correct answer)
Only statements A and C are correct for proper application of the theorem

Explanation: When you encounter a problem involving force components and moments, you're dealing with Varignon's theorem, one of the most powerful tools in statics. This theorem states that the moment of a resultant force about any point equals the sum of the moments of its component forces about the same point. Let's examine why answer D is correct by verifying each statement. Statement A captures the essence of Varignon's theorem:

M_{resultant} = M_{F_x} + M_{F_y}

. For this problem, the moment of the 400 N force about point (1,2) must equal the sum of moments created by the 320 N and 240 N components about the same reference point. Statement B correctly emphasizes that both components act at the original point of application (3,5) m, not at the reference point. This is crucial because changing the point of application would alter the moment calculations entirely. Statement C addresses the geometric requirement: you must find the perpendicular distance from the reference point (1,2) to each component's line of action. The horizontal component

F_x

creates a moment arm based on the vertical distance, while

F_y

uses the horizontal distance. Since A, B, and C are all essential requirements for properly applying Varignon's theorem, D is the complete answer. Each individual statement represents only part of the complete process. Study tip: Remember that Varignon's theorem is both a computational tool and a check for your work. When decomposing forces, always verify that the sum of component moments equals the original force's moment about the same reference point.

Question 3

A force of magnitude 600 N acts at point R(4, 3) m making an angle of 60° with the positive x-axis. If Varignon's theorem is used to find the moment about point S(2, 1) m, what is the moment contribution from the force component parallel to the vector SR?

$M_{S,parallel} = 0$ N⋅m (correct answer)
$M_{S,parallel} = 300$ N⋅m counterclockwise
$M_{S,parallel} = 519.6$ N⋅m clockwise
$M_{S,parallel} = 600$ N⋅m counterclockwise
$M_{S,parallel} = 200$ N⋅m clockwise

Explanation: When applying Varignon's theorem to find moments, you're breaking a force into components and calculating each component's moment contribution separately. The key insight here is understanding how the direction of a force component affects its ability to create a moment about a point. First, let's find the position vector

\vec{SR}

. From S(2,1) to R(4,3):

\vec{SR} = (2, 2)

m. The force components are:

F_x = 600\cos(60°) = 300

N and

F_y = 600\sin(60°) = 519.6

N. Here's the crucial principle: a force component that acts parallel to the line connecting the moment center to the point of application creates zero moment. This is because the perpendicular distance (moment arm) from the line of action to the moment center is zero. The parallel component of the force passes directly through point S, so it cannot create any rotational effect about S. Looking at the wrong answers: Answer B (300 N⋅m) incorrectly assumes the x-component of force contributes to the moment. Answer C (519.6 N⋅m) wrongly uses the y-component's magnitude. Answer D (600 N⋅m) mistakenly uses the entire force magnitude, ignoring that only the perpendicular component matters. The correct answer is A:

M_{S,parallel} = 0

N⋅m because forces acting along the line from the moment center to the point of application always produce zero moment. Study tip: Remember that only force components perpendicular to the position vector create moments. Parallel components always contribute zero—this is a fundamental principle that appears frequently in statics problems involving Varignon's theorem.

Question 4

A force $\vec{F} = 200\hat{i} + 150\hat{j}$ N acts at point (6, 4) m. According to Varignon's theorem, what is the magnitude of the difference between the moment about point A(2, 1) m and the moment about point B(3, 2) m?

$|\Delta M| = 50$ N⋅m (correct answer)
$|\Delta M| = 100$ N⋅m
$|\Delta M| = 150$ N⋅m
$|\Delta M| = 200$ N⋅m
$|\Delta M| = 250$ N⋅m

Explanation: When you encounter moment calculations involving multiple points, Varignon's theorem provides a powerful shortcut. Instead of calculating each moment separately, you can find the difference by recognizing that moments about different points differ by the force applied over the distance between those points. To solve this systematically, first find the position vectors from each point to where the force acts. From point A(2,1) to (6,4):

\vec{r_A} = (6-2)\hat{i} + (4-1)\hat{j} = 4\hat{i} + 3\hat{j}

m. From point B(3,2) to (6,4):

\vec{r_B} = (6-3)\hat{i} + (4-2)\hat{j} = 3\hat{i} + 2\hat{j}

m. The moments are:

M_A = \vec{r_A} \times \vec{F} = (4)(150) - (3)(200) = 0

N⋅m and

M_B = \vec{r_B} \times \vec{F} = (3)(150) - (2)(200) = 50

N⋅m. Therefore,

|\Delta M| = |0 - 50| = 50

N⋅m. Answer A (50 N⋅m) correctly captures this difference. Answer B (100 N⋅m) likely results from calculation errors in the cross products or incorrectly doubling the result. Answer C (150 N⋅m) suggests using only the j-component of the force rather than performing the full cross product. Answer D (200 N⋅m) indicates using only the i-component of the force. Remember: when comparing moments about different points, focus on accurate vector cross products. The key insight is that Varignon's theorem allows you to break complex moment problems into manageable vector calculations.

Question 5

A 400 N force acts at 45° to the horizontal at point P(8, 6) m. When applying Varignon's theorem to find the moment about point Q(3, 2) m, a student incorrectly calculates the moment as 800 N⋅m. What error did the student most likely make?

Used the wrong reference point for calculating position vector components
Failed to decompose the force into rectangular components before applying the theorem
Used the magnitude of the force instead of individual component magnitudes
Applied the cross product formula incorrectly for moment calculation
Used an incorrect moment arm distance instead of the perpendicular distance to the line of action (correct answer)

Explanation: When applying Varignon's theorem to calculate moments, you need to understand that the theorem states the moment of a resultant force equals the sum of moments of its components. This requires careful attention to both force decomposition and position vector calculation. Let's work through this correctly. The force components are:

F_x = 400\cos(45°) = 282.8 \text{ N}

and

F_y = 400\sin(45°) = 282.8 \text{ N}

. The position vector from Q(3,2) to P(8,6) is:

\vec{r} = (8-3)\hat{i} + (6-2)\hat{j} = 5\hat{i} + 4\hat{j}

meters. Using Varignon's theorem:

M = r_x F_y - r_y F_x = (5)(282.8) - (4)(282.8) = 282.8 \text{ N⋅m}

The student got 800 N⋅m, which suggests they likely used the full force magnitude (400 N) with an incorrect moment arm calculation, possibly using the distance between points (

\sqrt{5^2 + 4^2} ≈ 6.4

m) in an oversimplified approach. Answer A is incorrect because using the wrong reference point would give a completely different geometric setup. Answer B is wrong because you must decompose forces to properly apply Varignon's theorem - this is a requirement, not an error. Answer C doesn't match the 800 N⋅m result pattern. Answer D is incorrect because the cross product formula, when applied correctly, yields the right answer. Study tip: Always decompose angled forces into rectangular components when using Varignon's theorem, and double-check your position vector direction from the moment center to the point of application.

Question 6

According to Varignon's theorem, when a 500 N force acting at point (4, 3) m at 37° above the horizontal is replaced by its rectangular components, the sum of the moments of the components about any point must equal what quantity?

The moment of the original 500 N force about the same reference point
The product of the force magnitude and the distance to the reference point
The vector sum of the individual moment vectors about the reference point
The moment of the resultant force about the centroid of the force system
Both options A and C represent equivalent valid expressions for this quantity (correct answer)

Explanation: When you encounter questions about Varignon's theorem, you're dealing with one of the fundamental principles of moment analysis in statics. This theorem states that the moment of a force about any point equals the sum of the moments of its components about that same point. Let's apply this to the 500 N force. When you break this force into its rectangular components (

F_x = 500\cos(37°)

and

F_y = 500\sin(37°)

), Varignon's theorem tells us that

M_{original} = M_{F_x} + M_{F_y}

about any reference point. This means the moment of the original 500 N force about any point equals the algebraic sum of the moments created by its x and y components about that same point. Looking at the incorrect options: Choice A describes exactly what Varignon's theorem states - the sum of component moments equals the original force's moment. Choice B incorrectly suggests the answer is simply force times distance, ignoring the perpendicular distance requirement and angle considerations. Choice C mentions "vector sum of moment vectors," but moments about a point in 2D problems are scalars (positive or negative), not vectors requiring vector addition. Choice D incorrectly focuses on the centroid, which isn't relevant to Varignon's theorem - the theorem works for any reference point, not just centroids. The question asks what the sum "must equal," and based on Varignon's theorem, this is the moment of the original force about the same reference point. Remember: Varignon's theorem is your tool for simplifying moment calculations - you can work with components instead of the original force and get the same result.

Question 7

According to Varignon's theorem, if a 500 N force at point (6, 8) m is decomposed into components $F_x = 300$ N and $F_y = 400$ N, and the moment about point (3, 4) m is calculated, which statement about the moment contributions is correct?

The x-component contributes 900 N⋅m and the y-component contributes 1200 N⋅m to the total moment
The x-component contributes 1200 N⋅m and the y-component contributes 900 N⋅m to the total moment
The x-component contributes 1200 N⋅m and the y-component contributes 1200 N⋅m to the total moment
The contributions depend on the signs of the force components and their directions (correct answer)
Each component contributes equally since the force magnitude is the same

Explanation: When applying Varignon's theorem to calculate moments, you must carefully consider both the magnitude and direction of each force component's contribution. Varignon's theorem states that the moment of a resultant force equals the sum of moments of its components, but the sign and direction matter crucially. Let's examine this systematically. The moment arm from point (3, 4) to point (6, 8) creates perpendicular distances of 3 m in the x-direction and 4 m in the y-direction. The x-component force

F_x = 300

N creates a moment of

300 \times 4 = 1200

N⋅m, while the y-component force

F_y = 400

N creates a moment of

400 \times 3 = 1200

N⋅m in magnitude. However, answer D is correct because the actual contributions depend critically on the directions of the force components. If

F_x

acts in the positive x-direction, its moment about the reference point could be clockwise or counterclockwise depending on the relative positions. Similarly,

F_y

's contribution depends on its direction. Without knowing whether these components are positive or negative (their actual directions), you cannot determine the signs of their moment contributions. Answer A incorrectly calculates the magnitudes (900 and 1200). Answer B gives the correct magnitudes (1200 and 1200) but ignores the direction issue. Answer C also gives incorrect magnitudes and ignores direction. Remember: when calculating moments using Varignon's theorem, always account for the directional signs of both force components and their resulting moment contributions. The magnitudes alone don't tell the complete story.

Question 8

A student applies Varignon's theorem to find the moment of a 450 N force about point A. The force acts at point B and makes a 53° angle with the line AB. If the distance AB is 8 m, and the student incorrectly calculates the moment as 3600 N⋅m, which error was most likely made?

Used the total distance AB instead of the perpendicular distance as the moment arm (correct answer)
Failed to decompose the force into components parallel and perpendicular to AB
Calculated the moment using the component parallel to AB instead of perpendicular to AB
Used the sine of the angle instead of the cosine when finding the moment arm
Applied the force magnitude without considering the directional component

Explanation: When applying Varignon's theorem to calculate moments, you need to identify the correct moment arm, which is always the perpendicular distance from the point of rotation to the line of action of the force. Let's work through this problem correctly. The moment should be calculated as

M = F \times d_{\perp}

, where

d_{\perp}

is the perpendicular distance from point A to the force's line of action. Since the force makes a 53° angle with line AB, the perpendicular distance is

d_{\perp} = 8 \sin(53°) = 8 \times 0.8 = 6.4 \text{ m}

. The correct moment is

M = 450 \times 6.4 = 2880 \text{ N⋅m}

. However, if you incorrectly use the full distance AB as the moment arm, you get

M = 450 \times 8 = 3600 \text{ N⋅m}

, which matches the student's incorrect answer. This confirms that A is correct. B is wrong because Varignon's theorem doesn't require force decomposition—you can calculate moments directly using the perpendicular distance. C is incorrect because using the parallel component would give zero moment (parallel forces create no moment about their line). D is wrong because using sine instead of cosine for the moment arm would actually give the correct answer, not the incorrect one shown. Study tip: Always remember that moment arm means perpendicular distance. When you see a force at an angle, immediately think about finding the perpendicular distance using trigonometry—this is the most common source of moment calculation errors.

Question 9

A force $\vec{F} = 100\hat{i} + 200\hat{j}$ N acts at point A(3, 4) m. Using Varignon's theorem, if the moment about point B(1, 2) m is calculated by decomposing the force, what is the contribution of the x-component to the total moment?

$M_{B,x} = 200$ N⋅m counterclockwise (correct answer)
$M_{B,x} = 300$ N⋅m clockwise
$M_{B,x} = 400$ N⋅m counterclockwise
$M_{B,x} = 100$ N⋅m clockwise
$M_{B,x} = 0$ N⋅m

Explanation: When you encounter moment problems with force decomposition, you're applying Varignon's theorem, which states that the moment of a resultant force equals the sum of moments from its components. This is particularly useful for breaking down complex force calculations. To find the x-component's contribution to the moment about point B, you need the x-component of the force (

F_x = 100

N) and the perpendicular distance from B to the line of action of this component. Since the x-component acts horizontally at point A(3, 4), and B is at (1, 2), the perpendicular distance is the vertical separation:

4 - 2 = 2

m. The moment magnitude is

M_{B,x} = F_x \times d = 100 \text{ N} \times 2 \text{ m} = 200

N⋅m. Using the right-hand rule, a horizontal force pointing in the positive x-direction creates a counterclockwise moment about a point below it, confirming answer A is correct. Looking at the wrong answers: B incorrectly uses

F_x

times the full horizontal distance (

100 \times 3 = 300

) instead of the perpendicular distance, and gets the wrong direction. C uses the y-component of force (

F_y = 200

) times the perpendicular distance, confusing which force component contributes to this moment. D uses the correct perpendicular distance concept but applies it incorrectly, getting both magnitude and direction wrong. Study tip: Always identify the perpendicular distance carefully—it's the shortest distance from the moment center to the line of action, not just any distance between points.

Question 10

In using Varignon's theorem, a force $\vec{F} = 300\hat{i} - 400\hat{j}$ N acts at point (5, 2) m. If the moment about the origin is calculated by treating this as two separate force components, what is the ratio of the moment due to the x-component to the moment due to the y-component?

$\frac{M_x}{M_y} = \frac{3}{4}$
$\frac{M_x}{M_y} = \frac{2}{5}$
$\frac{M_x}{M_y} = \frac{4}{3}$
$\frac{M_x}{M_y} = \frac{5}{2}$
$\frac{M_x}{M_y} = \frac{3}{10}$ (correct answer)

Explanation: When you encounter Varignon's theorem problems, you're applying the principle that the moment of a resultant force equals the sum of moments from its components. This requires calculating moments from individual force components acting at their respective perpendicular distances. To find the moment about the origin from each component, you need the perpendicular distance from the origin to each force's line of action. The x-component

F_x = 300

N acts horizontally at point (5, 2), so its perpendicular distance is the y-coordinate: 2 m. The y-component

F_y = -400

N acts vertically at the same point, so its perpendicular distance is the x-coordinate: 5 m. The moment from the x-component is

M_x = F_x \times d = 300 \times 2 = 600

N⋅m. The moment from the y-component is

M_y = F_y \times d = 400 \times 5 = 2000

N⋅m. Therefore, the ratio is

\frac{M_x}{M_y} = \frac{600}{2000} = \frac{3}{10}

. Answer A (

\frac{3}{4}

) incorrectly uses the ratio of force magnitudes rather than moments. Answer B (

\frac{2}{5}

) appears to mix up coordinate values with force ratios. Answer C (

\frac{4}{3}

) inverts the force magnitude ratio. Answer D (

\frac{5}{2}

) uses the ratio of perpendicular distances rather than the actual moments. Remember: in moment calculations, always multiply the force component by its correct perpendicular distance to the reference point. Don't confuse force ratios with moment ratios – the perpendicular distances make all the difference.

Question 11

Two forces $F_1 = 40$ N and $F_2 = 60$ N act at points A(2, 3) and B(5, 1) respectively, both directed along the positive y-axis. A student attempts to apply Varignon's theorem by moving both forces to point C(3, 2) and claims the moment about the origin remains unchanged. What is the error in this approach?

The student failed to add couple moments that arise when forces are moved from their original points of application to the new location (correct answer)
Varignon's theorem only applies when forces are moved to their resultant's line of action, not to arbitrary points like point C
The forces must maintain their original directions when moved, but their magnitudes must be adjusted based on the new position vectors
The theorem requires that all forces be moved to the same point simultaneously, not individually as the student attempted

Explanation: When a force is moved from its original point of application to a new point, a couple moment must be added to maintain equivalence. The couple moment equals the original force times the perpendicular distance between the old and new lines of action. The student's error was neglecting these additional couple moments. Choice A correctly identifies this fundamental principle of force system equivalence. Choice B incorrectly suggests Varignon's theorem has restrictions on where forces can be moved - the theorem itself doesn't require moving forces. Choice C incorrectly suggests magnitude changes are needed. Choice D incorrectly implies the theorem requires simultaneous movement - the issue is the missing couples, not the sequence of operations.

Question 12

A force $F = 200$ N acts at point P(6, 8) m along a line making a 53° angle with the positive x-axis. Using Varignon's theorem, the force is decomposed into components $F_x = 120$ N and $F_y = 160$ N. When calculating the moment about point Q(2, 3) m, which expression correctly represents the application of Varignon's theorem?

$M_Q = 120 \times 5 + 160 \times 4 = 1240 \text{ N⋅m}$
$M_Q = 160 \times (6-2) - 120 \times (8-3) = 40 \text{ N⋅m}$ (correct answer)
$M_Q = 120 \times (8-3) - 160 \times (6-2) = -40 \text{ N⋅m}$
$M_Q = \sqrt{(120 \times 4)^2 + (160 \times 5)^2} = 894 \text{ N⋅m}$

Explanation: When you encounter moment calculations with force components, Varignon's theorem is your key tool. This theorem states that the moment of a resultant force equals the sum of moments created by its components. To find the moment about point Q(2, 3) from force components at P(6, 8), you need the perpendicular distances from Q to each component's line of action. The horizontal component

F_x = 120

N acts along a horizontal line at height

y = 8

, so its perpendicular distance from Q is

(8-3) = 5

m. The vertical component

F_y = 160

N acts along a vertical line at

x = 6

, so its perpendicular distance from Q is

(6-2) = 4

m. Using the right-hand rule for moment direction:

F_y

creates a counterclockwise (positive) moment, while

F_x

creates a clockwise (negative) moment. Therefore:

M_Q = 160 \times 4 - 120 \times 5 = 640 - 600 = 40

N⋅m. This matches option B. Option A incorrectly adds both moments as positive, ignoring the opposing directions of rotation. Option C reverses the sign convention, treating the counterclockwise moment as negative and clockwise as positive. Option D attempts to find a resultant magnitude using the Pythagorean theorem, which doesn't apply to scalar moment calculations. Remember: Varignon's theorem involves algebraic addition of moments, respecting their rotational directions. Always establish your sign convention first (typically counterclockwise = positive), then apply it consistently to each component's contribution.

Question 13

A system consists of three concurrent forces at point A: $F_1 = 80$ N at 30° above the horizontal, $F_2 = 120$ N at 150° from the positive x-axis, and $F_3 = 100$ N vertically downward. Using Varignon's theorem, if the moment of the resultant about point O (located 5 m to the left of point A) is calculated, which approach will yield an incorrect result?

Calculate the vector sum of the three forces first, then determine the moment of this resultant force about point O using the perpendicular distance method
Calculate the moment of each individual force about point O separately, then sum these moments vectorially to find the total moment
Calculate the scalar sum of the magnitudes of the moments of each individual force about point O, treating all moments as positive values (correct answer)
Resolve all forces into x and y components, calculate moments of these components about point O, then sum vectorially

Explanation: Varignon's theorem requires vector addition of moments, not scalar addition of magnitudes. The moment is a vector quantity with both magnitude and direction (sign). Choice C incorrectly treats moments as scalars and adds their magnitudes, ignoring the directional nature of moments. This violates the fundamental principle that moments can be clockwise or counterclockwise. Choices A, B, and D all properly apply Varignon's theorem through vector operations. Choice A uses the resultant force method, Choice B uses direct application of Varignon's theorem, and Choice D uses component resolution followed by vector summation.

Question 14

A force $\vec{F} = 150\hat{i} + 200\hat{j}$ N is applied at point P(4, 3) m. If this force is decomposed into components $F_x = 150$ N and $F_y = 200$ N acting separately at the same point, what is the difference between the magnitude of the moment of the original force about the origin and the sum of the magnitudes of the moments of the individual components about the origin?

Zero, because Varignon's theorem ensures the moment of the resultant equals the vector sum of component moments (correct answer)
50 N⋅m, due to cross-coupling terms between force components and position coordinates not being preserved
Zero, because both calculations yield identical scalar moment magnitudes regardless of decomposition method
100 N⋅m, because magnitude operation timing differs between resultant and component calculation methods

Explanation: Varignon's theorem states that the moment of a resultant force about any point equals the vector sum of the moments of its components about the same point. For the original force: M = r × F = (4î + 3ĵ) × (150î + 200ĵ) = 800 - 450 = 350 N⋅m (k-direction). For components: M₁ = (4î + 3ĵ) × (150î) = 450 N⋅m (k-direction), M₂ = (4î + 3ĵ) × (200ĵ) = -800 N⋅m (k-direction). Vector sum: 450 + (-800) = -350 N⋅m, which has magnitude 350 N⋅m. The difference is zero.

Question 15

A crane applies forces through three cables attached at point P. Cable tensions are $T_1 = 1000$ N, $T_2 = 800$ N, and $T_3 = 600$ N along unit vectors $\hat{u_1} = 0.6\hat{i} + 0.8\hat{j}$ , $\hat{u_2} = -0.8\hat{i} + 0.6\hat{j}$ , and $\hat{u_3} = 0\hat{i} - 1\hat{j}$ respectively. If point P is located at (3, 4) m from the origin, and Varignon's theorem is used to find the moment about the origin, which approach would yield an incorrect result due to a conceptual error?

Calculate the resultant of all three tension forces first, then find the moment of this resultant about the origin using the position vector to point P
Find the moment contribution of each tension force separately about the origin, then sum these moments vectorially
Resolve each tension force into x and y components, calculate moments of these six components about the origin, then sum algebraically
Calculate the perpendicular distance from the origin to each cable's line of action, multiply by the respective tension, then sum the resulting moment magnitudes (correct answer)

Explanation: Choice D makes the critical error of summing moment magnitudes rather than considering the vector nature of moments. Moments have direction (clockwise or counterclockwise), and simply adding magnitudes ignores the cancellation effects that occur when moments act in opposite directions. This violates the fundamental vector nature of Varignon's theorem. Choices A and B correctly apply Varignon's theorem through vector operations. Choice C correctly decomposes forces and uses algebraic summation, which properly accounts for signs (directions) of the moment contributions.

Question 16

A wrench applies a force $F = 50$ N at point P, located at position vector $\vec{r} = 0.3\hat{i} + 0.2\hat{j}$ m from bolt center O. The force makes a 45° angle with the positive x-axis. If this force is replaced by an equivalent system of a 35.36 N horizontal component and a 35.36 N vertical component at the same point, what additional constraint must be satisfied for Varignon's theorem to remain valid?

The perpendicular distances from point O to component lines of action must equal the original force distance
The vector sum of position vectors to component application points must equal the original position vector
Both force components must act at point P where the original force was applied (correct answer)
The scalar sum of component magnitudes must equal the original force magnitude for consistency

Explanation: Varignon's theorem requires that the component forces act at the same point as the original force. The theorem states that the moment of a resultant equals the vector sum of moments of components, but this is only valid when all forces act at the same point of application. Choice C correctly identifies this requirement. Choice A incorrectly focuses on perpendicular distances rather than points of application. Choice B misunderstands position vectors. Choice D incorrectly suggests scalar addition equivalence.

Question 17

In a parallel force system, three vertical forces $F_1 = 50$ N (downward), $F_2 = 80$ N (upward), and $F_3 = 30$ N (downward) act at x-coordinates 2 m, 5 m, and 8 m respectively. A student applies Varignon's theorem to find the location of the resultant by setting the moment of the resultant about the origin equal to the sum of moments of individual forces. If the student calculates $x_{resultant} = \frac{50(2) + 80(5) + 30(8)}{50 + 80 + 30} = 4.375$ m, what error has been made?

Varignon's theorem cannot be applied to parallel force systems because the forces don't have components in perpendicular directions
The calculation incorrectly treats all forces as positive in the numerator while using their net effect in the denominator (correct answer)
The moment arms should be the perpendicular distances to the lines of action, not the x-coordinates of the points of application
The forces should be decomposed into components before applying the theorem, even though they are already parallel

Explanation: When applying Varignon's theorem to find the location of a resultant force, you must maintain consistent sign conventions throughout your calculation. The theorem states that the moment of the resultant about any point equals the sum of moments of the individual forces about that same point. The correct approach requires treating downward forces as negative and upward forces as positive (or vice versa) in both the numerator AND denominator. The student's error lies in using the absolute values of all forces in the numerator while correctly accounting for directions in the denominator. The calculation should be:

x_{resultant} = \frac{(-50)(2) + (80)(5) + (-30)(8)}{-50 + 80 - 30} = \frac{-100 + 400 - 240}{0}

This reveals the calculation is undefined because the net force is zero—the forces are in equilibrium with no resultant. Looking at the wrong answers: (A) is incorrect because Varignon's theorem absolutely applies to parallel force systems and is commonly used for this purpose. (C) misses the point—in this problem, the x-coordinates are the correct moment arms since we're taking moments about the origin and forces are vertical. (D) is unnecessary since the forces are already parallel and don't need decomposition. The key insight is that (B) correctly identifies the sign convention error: mixing positive values in the numerator with signed values in the denominator violates the fundamental requirement for consistent mathematical treatment. Remember: Always maintain the same sign convention throughout your entire calculation when applying Varignon's theorem—inconsistent signs lead to meaningless results.