Statics and Dynamics Quiz: Varignons Theorem

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 and Dynamics.

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.

Question 1

1. 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. (correct answer)
2. Varignon's theorem does not apply here because the force passes through two distinct points; the theorem is only valid when a force acts at a single, clearly defined point of application, not along a line of action.
3. Varignon's theorem applies, but the moment must be computed using the midpoint of AB as the position vector; using either A or B alone would give an incorrect moment because the force is distributed along the segment AB rather than concentrated at one end.
4. Varignon's theorem applies only if $\vec{F}$ is resolved into components that are mutually perpendicular to the coordinate axes; since the direction of AB is neither horizontal nor vertical, the theorem cannot be used until a coordinate system aligned with AB is established.

Question 2

1. The student applied the wrong overall sign pattern; the correct expression negates every term: $M_O = -(80)(3) - (-60)(2) - (-50)(-1) - (40)(5) = -240 + 120 - 50 - 200 = -370\,\text{N·m}$ (clockwise).
2. The student paired each force component with the wrong coordinate, but the fix is simply to swap only the signs of the $F_x$ terms; the corrected sum is $M_O = -(80)(3) + (-60)(2) - (-50)(-1) + (40)(5) = -240 - 120 - 50 + 200 = -210\,\text{N·m}$, and the error was applying a positive sign to the clockwise-producing $F_x$ terms.
3. 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 $M_O = [(2)(-60)-(3)(80)] + [(5)(40)-(-1)(-50)] = [-120-240]+[200-50] = -360+150 = -210\,\text{N·m}$ (clockwise). (correct answer)
4. The student's expression is correct in structure but made an arithmetic error; properly evaluated, $M_O = 240 - 120 + 50 + 200 = 370\,\text{N·m}$ (counterclockwise), because all four terms are positive when CCW is the positive convention.

Question 3

1. $\bar{x} = \frac{L}{3} = 1.33\,\text{m}$; evaluating $\int_0^L x \cdot \frac{w_0}{L}x\,dx$ and dividing by R correctly yields $\frac{2L}{3}$, but since a triangle's centroid is located at one-third of the base measured from the zero-load vertex, the centroid is at $L/3$ from $x = 0$.
2. $\bar{x} = \frac{L}{2} = 2.0\,\text{m}$; Varignon's theorem applies to concurrent forces only, so for a distributed load the centroid of the load diagram must be located at the midpoint, as it would be for a uniform (rectangular) distribution.
3. $\bar{x} = \frac{L}{3} = 1.33\,\text{m}$; Varignon's theorem requires summing moments from the point of maximum load intensity at $x = L$, so the relevant integral is $\int_0^L (L - x)w(x)\,dx$, giving the distance of $\bar{x}$ measured from $x = L$; this distance from the peak is $L/3$, which is then incorrectly reported as the distance from $x = 0$.
4. $\bar{x} = \frac{2L}{3} = 2.67\,\text{m}$; 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. (correct answer)

Question 4

1. The magnitude 480 N·m is numerically correct, but the sign is wrong; the engineer's formula computes $r_y F_x - r_x F_y$, which equals the negative of the standard cross-product $r_x F_y - r_y F_x$. Because the magnitude happens to be the same, a student who only checks magnitudes would not detect the error.
2. The result is incorrect; the engineer swapped the pairing in the cross product. The correct Varignon expression is $M_B = (r_{BA,x})(F_y) - (r_{BA,y})(F_x) = (-3)(80) - (4)(60) = -240 - 240 = -480\,\text{N·m}$ (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). (correct answer)
3. The result is correct; the engineer's formula $F_x(y_A - y_B) - F_y(x_A - x_B)$ is a valid alternative form of the cross product when the moment center is not the origin, because shifting the reference point from O to B reverses the effective sign convention for the position vector components.
4. The result is incorrect because Strategy II is only valid when the moment center is the origin; for an arbitrary point B, Varignon's theorem must first be applied about O and then the moment transferred to B using a couple-moment transfer, which introduces a correction term not present in the engineer's calculation.

Question 5

1. $M_O = (120\cos30°)(-4) + (120\sin30°)(3) \approx -235.3\,\text{N·m}$; the negative sign indicates a clockwise moment about O, consistent with the geometry of the force placement relative to O. (correct answer)
2. $M_O = (120\cos30°)(4) + (120\sin30°)(3) \approx 596.2\,\text{N·m}$; the positive sign indicates a counterclockwise moment about O, but the perpendicular distances are both taken as positive, ignoring their directional contribution.
3. $M_O = (120\cos30°)(3) + (120\sin30°)(4) \approx 551.8\,\text{N·m}$; the moment is counterclockwise because the moment arms are assigned by matching each force component to its same-axis coordinate rather than the cross-product structure, which swaps the pairing.
4. $M_O = (120\cos30°)(-4) - (120\sin30°)(3) \approx -415.7\,\text{N·m}$; the negative sign on $F_y$ arises because the y-component of a force directed above the x-axis always produces a clockwise moment when the point of application is in the first quadrant.

Question 6

1. The correct value is $x_0 = 3\,\text{m}$; the student's reasoning is accidentally correct because when the position vector from $(x_0, 0)$ to P is vertical (same x-coordinate), the 45° force produces equal and opposite moment contributions from its components, which cancel exactly.
2. The correct value is $x_0 = 6\,\text{m}$; the student's error is assuming the zero-moment point coincides with P's x-coordinate. Setting $(3 - x_0)(-100) - (3)(100) + (3 - x_0)(100) \cdot 0 = 0$ is wrong; the correct formulation gives $-(3)(100) + (3 - x_0)(100) = 0$, so $x_0 = 6\,\text{m}$.
3. 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 $x = 3\,\text{m}$. (correct answer)
4. The correct value is $x_0 = 1.5\,\text{m}$; the student's error is ignoring that the resultant force at 45° has an effective moment arm equal to half the perpendicular distance from the origin to point P, and the line of action intersects the x-axis at the midpoint between O and the projection of P.

Question 7

1. $\vec{M}_Q' \neq \vec{M}_O$ in general, because Varignon's theorem requires the position vector to originate from the point of application of the force; using a different point on the line of action changes the component decomposition and therefore the resulting moment.
2. $\vec{M}_Q' = \vec{M}_O$ 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. (correct answer)
3. $\vec{M}_Q' = \vec{M}_O$ only if Q is closer to O than the original application point; if Q is farther from O, the larger position vector magnitude increases the moment, so the two moments are only equal at the unique perpendicular foot from O to the line of action.
4. $\vec{M}_Q' = \vec{M}_O$ only when $\vec{F}$ lies entirely in the xy-plane; in three-dimensional problems, the moment about O depends on which specific point on the line of action is selected, because the cross product is sensitive to the full three-dimensional position vector.

Question 8

1. Both approaches yield identical results, giving $M_O = (3)(300) - (4)(-50) = 900 + 200 = 1100\,\text{N·m}$ (counterclockwise), because the resultant $\vec{R} = 300\hat{i} - 50\hat{j}$ has its moment arms correctly assigned as $r_x$ to $R_y$ and $r_y$ to $R_x$ with the standard sign convention.
2. The two approaches yield the same result only for the x-components; the y-components must be treated separately because Varignon's theorem applies component-by-component, not to the full resultant. The correct moment is $M_O = (3)(-50) - (4)(300) = -1350\,\text{N·m}$.
3. The two approaches yield different results because the second engineer's method double-counts the contribution of forces with both x- and y-components (like $\vec{F}_3$); the correct moment uses only the resultant: $M_O = (3)(-50) - (4)(300) = -1350\,\text{N·m}$.
4. Both approaches yield identical results by Varignon's theorem: $\vec{R} = 300\hat{i} - 50\hat{j}\,\text{N}$, and $M_O = (3)(-50) - (4)(300) = -150 - 1200 = -1350\,\text{N·m}$ (clockwise), confirming that summing moments of components equals the moment of the resultant for concurrent forces. (correct answer)