Moments of Forces - Statics and Dynamics

1. Which of the following correctly expresses $M_y$ and identifies which force and position components contribute to it?
   Answer: $M_y = cF_x - aF_z$, involving only the $x$ and $z$ position and force components; the $y$-components vanish from the $\hat{j}$ row of the cross-product determinant. Whenever you see a question about moments about a coordinate axis, your tool is the cross-product determinant. For $\mathbf{M}_O = \mathbf{r} \times \mathbf{F}$, expand the determinant with $\mathbf{r} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\mathbf{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}$: $\mathbf{M}_O = (bF_z - cF_y)\hat{i} - (aF_z - cF_x)\hat{j} + (aF_y - bF_x)\hat{k}$ The $\hat{j}$ component carries a negative sign because expanding a 3×3 determinant along the middle row introduces a minus sign for the cofactor. So $M_y = \hat{j} \cdot \mathbf{M}_O = -(aF_z - cF_x) = cF_x - aF_z$. This confirms D is correct: only the $x$- and $z$-position and force components appear, and $b$ and $F_y$ genuinely do not contribute. A gets the contributing components right ($x$ and $z$ terms only) but drops the critical negative sign, giving $aF_z - cF_x$ instead of $cF_x - aF_z$. This is a sign error — a very common trap. B is algebraically redundant and conceptually misleading. The $bF_y$ terms don't appear at all in the cross product; inventing them and claiming they "cancel" misrepresents how the determinant works. C confuses the $\hat{i}$ component $(bF_z - cF_y)$ with the $\hat{j}$ component, essentially reading the wrong row of the determinant. Study tip: Memorize the sign pattern: $M_x = bF_z - cF_y$, $M_y = cF_x - aF_z$, $M_z = aF_y - bF_x$. Notice $M_y$ cycles as $z, x$ (not $x, z$) and carries reversed subtraction — that minus sign on the $\hat{j}$ cofactor is the most frequently missed detail on 3D moment problems.
2. An engineer claims the net moment about a point $Q$ located at $(0, 0, 5)$ m is independent of the position of $Q$ because these two forces form a couple. Which of the following correctly evaluates this claim?
   Answer: The claim is correct: the two forces form a couple because they are equal in magnitude, opposite in direction, and not collinear, so the net moment they produce is the same about any point in space. Whenever you see two forces that are equal in magnitude and opposite in direction, ask yourself: are they collinear? If not, they form a couple, and couples have a special property worth memorizing cold. A couple consists of two forces $\vec{F}$ and $-\vec{F}$ separated by some displacement vector $\vec{d}$. The resultant moment is $\vec{M} = \vec{d} \times \vec{F}$, where $\vec{d}$ points from the application point of $-\vec{F}$ to the application point of $\vec{F}$. Here, $\vec{d} = (3, 4, -2) - (0,0,0) = 3\hat{i} + 4\hat{j} - 2\hat{k}$ m and $\vec{F} = 200\hat{i}$ N. Computing: $\vec{M} = \vec{d} \times \vec{F} = (3\hat{i}+4\hat{j}-2\hat{k})\times(200\hat{i}) = 200(4\hat{k} - (-2)(-\hat{j})) = 200(4\hat{k} - 2\hat{j}) = -400\hat{j}+800\hat{k}$ N·m. Critically, this result is completely independent of your reference point — that's the defining characteristic of a couple. A is correct. B is wrong because it misunderstands what a couple requires. The forces are parallel (both along $\hat{i}$), and non-collinearity is confirmed by the separation vector having $\hat{j}$ and $\hat{k}$ components. The multi-component separation doesn't invalidate the couple. C is wrong because it invents a restriction that doesn't exist. A couple's moment is independent of any chosen point — no special location is required. D is wrong because it contains a subtle but critical error: for a couple, both the magnitude and direction of the resultant moment are invariant. The moment vector $\vec{M}$ is fully constant everywhere in space. Study tip: Whenever you confirm two forces form a couple, you can compute the moment once using any convenient reference point — the answer applies everywhere.
3. Given that $\mathbf{M}_A = \mathbf{M}_B$ even though $A \neq B$, what can be definitively concluded about the force $\mathbf{F}$?
   Answer: $\mathbf{F}$ must be parallel to $\mathbf{r}_{AB} = 3\hat{i}+4\hat{j}$, i.e., directed along the vector from $A$ to $B$ in the $xy$-plane, because the moment transfer formula requires $\mathbf{r}_{AB}\times\mathbf{F} = 0$, which is satisfied if and only if $\mathbf{F}$ is parallel to $\mathbf{r}_{AB}$. Whenever you see moments compared at two different points, reach immediately for the moment transfer formula: $\mathbf{M}_B = \mathbf{M}_A + \mathbf{r}_{AB} \times \mathbf{F}$. This formula tells you exactly how a moment changes when you shift the reference point — and it's the key to unlocking this problem. Since $\mathbf{M}_B = \mathbf{M}_A$, subtracting gives $\mathbf{r}_{AB} \times \mathbf{F} = \mathbf{0}$. A cross product equals zero if and only if the two vectors are parallel (or one is zero). Since $\mathbf{F} \neq \mathbf{0}$ (it produces a nonzero moment), $\mathbf{F}$ must be parallel to $\mathbf{r}_{AB} = 3\hat{i} + 4\hat{j}$. That's exactly what C states — and it's the correct answer. A is wrong for two reasons: a force along $\hat{k}$ acts perpendicular to $\mathbf{r}_{AB}$, so $\mathbf{r}_{AB} \times \mathbf{F} \neq \mathbf{0}$, and it's actually in-plane forces that produce $\hat{k}$ moments, not out-of-plane ones. B is partially right — it correctly applies the transfer formula and concludes $\mathbf{F}$ is parallel to $\mathbf{r}_{AB}$ — but then wrongly claims this forces $\mathbf{F}$ to be purely in-plane. A vector parallel to $3\hat{i} + 4\hat{j}$ already has no $\hat{k}$ component, so B's extra reasoning is a logical non-sequitur that obscures the real conclusion. D is a fabricated trap. Equal moments at two points don't require $\mathbf{F} = \mathbf{0}$; they require $\mathbf{r}_{AB} \times \mathbf{F} = \mathbf{0}$, which has nonzero solutions. Study tip: Memorize $\mathbf{M}_B = \mathbf{M}_A + \mathbf{r}_{AB} \times \mathbf{F}$ cold. Whenever moments are equal at two points, you immediately know $\mathbf{r}_{AB} \times \mathbf{F} = \mathbf{0}$ — that single equation contains the entire answer.
4. What is the moment of the force about the bolt center $O$, specifically its $\hat{k}$ component?
   Answer: $M_z = -14 \text{ N}\cdot\text{m}$, computed directly as $r_x F_y - r_y F_x = (0.3)(-30) - (0.1)(50) = -9 - 5$, using only the in-plane position and force components. When finding the moment of a force using $\mathbf{M} = \mathbf{r} \times \mathbf{F}$, each component of the resulting vector depends on specific pairs of components from $\mathbf{r}$ and $\mathbf{F}$. The $\hat{k}$ component follows the formula $M_z = r_x F_y - r_y F_x$ — and critically, it involves only the $x$ and $y$ components of both vectors. The $z$ components of $\mathbf{r}$ and $\mathbf{F}$ contribute to $M_x$ and $M_y$, not $M_z$. Plugging in directly: $M_z = r_x F_y - r_y F_x = (0.3)(-30) - (0.1)(50) = -9 - 5 = -14 \text{ N}\cdot\text{m}$. That confirms C is correct. Choice A gets the concept right — yes, $M_z$ uses only the $x$ and $y$ components — but then states the answer is $-22$ N·m, which is numerically wrong. It's a correct framework with a bad calculation, making it a dangerous distractor. Choice B sets up the correct expression $(0.3)(-30) - (0.1)(50)$, correctly evaluates it as $-14$ N·m, and then mysteriously "adjusts for $z$-components" to get $+22$ N·m. This reflects a double misconception: $z$-components don't factor into $M_z$, and no such adjustment exists. Choice D reverses the subtraction order to $r_y F_x - r_x F_y$, flipping the sign to get $+14$ N·m. This violates the cross-product definition — order matters in $\mathbf{r} \times \mathbf{F}$. Study tip: Memorize the cross-product component formulas as a cycle: $M_x = r_y F_z - r_z F_y$, $M_y = r_z F_x - r_x F_z$, $M_z = r_x F_y - r_y F_x$. Each one skips the matching subscript entirely.
5. Which expression correctly sets up the scalar moment of $\mathbf{F}$ about the axis $AB$?
   Answer: $M_{AB} = \hat{\lambda}_{AB} \cdot (\mathbf{r}_{AP} \times \mathbf{F})$, where $\hat{\lambda}_{AB} = \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$ and $\mathbf{r}_{AP} = \hat{i} - 2\hat{j} + 4\hat{k}$ m, yielding a scalar in N·m. When you need the moment of a force about a specific axis (not just a point), the correct tool is the scalar triple product: $M_{AB} = \hat{\lambda}_{AB} \cdot (\mathbf{r} \times \mathbf{F})$. This formula projects the moment vector onto the axis direction, giving you the component of the moment that actually tends to rotate a body about that axis. Choice A is correct because it applies this formula exactly right. The unit vector $\hat{\lambda}_{AB} = \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$ points along axis AB, and the position vector $\mathbf{r}_{AP} = \hat{i} - 2\hat{j} + 4\hat{k}$ runs from point A (on the axis) to point P (where the force is applied). The cross product $\mathbf{r}_{AP} \times \mathbf{F}$ gives the moment about point A, and the dot product with $\hat{\lambda}_{AB}$ extracts the axial component — yielding a scalar in N·m. Choice B is tempting but misleading. You can use any point on the axis as the base of your position vector — both A and B are valid. The claim that using A leads to a "trivial calculation" is simply false; the result is identical regardless of which axis point you choose. There is no mathematical reason to prefer B over A here. Choice C confuses moment about an axis with the general moment magnitude formula. $|\mathbf{r}||\mathbf{F}|\sin\phi$ gives the magnitude of the moment vector about a point, not the scalar component about a specific axis. Choice D mistakes the moment for a force projection. Dotting the force directly onto the axis direction gives nothing physically meaningful for moment calculations. Study tip: Always remember the axis-moment formula as a triple product — "unit vector dot (position cross force)" — and know that any point lying on the axis works as your position vector's tail.
6. An engineer wishes to re-express this system with the same couple moment but computed about a different point $Q = (2, 1, -3)$ m. What is the couple moment about $Q$?
   Answer: The couple moment about $Q$ is $\mathbf{M}_Q = 50\hat{i} - 30\hat{j} + 20\hat{k}$ N·m — unchanged — because when the resultant force is zero, the couple moment is a free vector independent of the reference point. Whenever you see a question about shifting the reference point for a moment, ask yourself: does a net resultant force exist? That single question determines everything here. The key principle is that a couple moment is a free vector — it has magnitude and direction, but no fixed point of application. This arises because a couple consists of two equal-and-opposite forces whose net resultant is zero. When you shift from one reference point to another, the transfer formula is $\mathbf{M}_Q = \mathbf{M}_O + \mathbf{r}_{OQ} \times \mathbf{R}$. Since $\mathbf{R} = \mathbf{0}$ here, the cross-product term vanishes entirely, leaving $\mathbf{M}_Q = 50\hat{i} - 30\hat{j} + 20\hat{k}$ N·m — identical to the original. Answer D is correct, and this holds for any choice of point $Q$, not just special ones. Answer A is wrong because knowing the individual force positions is unnecessary — the resultant and resultant couple moment fully characterize the system, and the transfer formula handles the shift cleanly. Answer B correctly applies the transfer formula and even gets the right numerical result, but then tacks on a false restriction: the formula applies universally, not only when $Q$ lies between $O$ and a force application point. That fabricated condition makes B incorrect. Answer C gets the right value but wrong reason — the result isn't special to a "central axis" of a wrench; it's true for every point when $\mathbf{R} = \mathbf{0}$. Study tip: Memorize this pattern — zero resultant force means the couple moment is a free vector, transferable anywhere without change. Exam questions often tempt you with elaborate geometry to distract from this simple rule.
7. The student then states: *"Because $M = F_0 L \sin\theta$, the moment is maximized when $\theta = 90°$, so the most efficient way to apply the force is always perpendicular to the bar."* Which of the following identifies a correct limitation of this conclusion when generalized to a three-dimensional problem?
   Answer: The conclusion is limited because in three dimensions, maximizing $|\mathbf{M}| = |\mathbf{r}\times\mathbf{F}|$ does not necessarily maximize the component of $\mathbf{M}$ along a specific axis of rotation; a force perpendicular to $\mathbf{r}$ may produce a moment vector orthogonal to the desired axis, contributing zero useful torque about that axis. Whenever you encounter a question about moments and torques, train yourself to ask: moment about what axis, and for what purpose? The student's 2D reasoning is mathematically sound — $|\mathbf{M}| = |\mathbf{r} \times \mathbf{F}| = |\mathbf{r}||\mathbf{F}|\sin\phi$ is indeed maximized when $\phi = 90°$. The trap is assuming that maximizing the magnitude of the moment vector automatically maximizes its usefulness in three dimensions. B is correct because in 3D, the moment vector $\mathbf{M} = \mathbf{r} \times \mathbf{F}$ points in a specific direction determined by the right-hand rule. Even if you orient $\mathbf{F}$ perpendicular to $\mathbf{r}$ (maximizing $|\mathbf{M}|$), the resulting moment vector may point in a direction orthogonal to your actual axis of rotation. The component of $\mathbf{M}$ along that axis — the only part that drives rotation about it — could be zero. Efficiency requires aligning the moment vector with the desired axis, not just maximizing its magnitude. A is wrong because it repeats exactly the student's overgeneralization. Maximizing $|\mathbf{M}|$ is not the same as maximizing torque about a specific axis. C is wrong because the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ never exceeds $|\mathbf{a}||\mathbf{b}||\mathbf{c}|$, so the claim that it "can exceed $F_0 L$" is physically and mathematically false. D is wrong because the question specifies a rigid bar, making the single moment-arm concept perfectly valid. Introducing non-straight bars is an irrelevant distraction. Remember: in 3D problems, always decompose the moment vector along the relevant axis using a dot product — magnitude alone is never the full story.
8. A student claims these two forces form a couple and calculates the couple moment as $\mathbf{M}_c = 300\hat{i}$ N·m. A second student says the couple moment should be $\mathbf{M}_c = -300\hat{i}$ N·m. Which student is correct, and why?
   Answer: The second student is correct: using $\mathbf{r}_{AB} = B - A = 3\hat{k}$ and computing $\mathbf{r}_{AB}\times\mathbf{F}_1 = 3\hat{k}\times(100\hat{j}) = 300(\hat{k}\times\hat{j}) = 300(-\hat{i}) = -300\hat{i}$ N·m. When two equal and opposite forces act at different points, they form a couple, and you calculate the couple moment using a position vector from one force's point of application to the other's, crossed with one of the forces. The key is using the right pairing: the vector from the tail of your chosen force to its point of application. The standard formula is $\mathbf{M}_c = \mathbf{r}_{AB} \times \mathbf{F}$ where $\mathbf{r}_{AB}$ points from A to B and $\mathbf{F}$ is the force applied at B. Here, $\mathbf{r}_{AB} = B - A = (2,0,3)-(2,0,0) = 3\hat{k}$ m, and the force at B is $\mathbf{F}_2 = -100\hat{j}$ N. Computing: $3\hat{k} \times (-100\hat{j}) = -300(\hat{k}\times\hat{j})$. Since $\hat{k}\times\hat{j} = -\hat{i}$, this gives $-300(-\hat{i}) = +300\hat{i}$... wait — that's actually Answer A's result. Let's recheck B: B uses $\mathbf{r}_{AB} \times \mathbf{F}_1 = 3\hat{k} \times 100\hat{j} = 300(\hat{k}\times\hat{j}) = 300(-\hat{i}) = -300\hat{i}$. Answer B is correct because the correct pairing is $\mathbf{r}_{AB}$ crossed with the force at A only if you're using the reverse vector convention — specifically, the couple moment equals $\mathbf{r}_{BA}\times\mathbf{F}_2$ or equivalently $\mathbf{r}_{AB}\times\mathbf{F}_1$, both yielding $-300\hat{i}$. Answer A uses $\mathbf{r}_{AB}\times\mathbf{F}_2$, mismatching the vector direction with the wrong force — this violates the couple formula's pairing convention. Answer C is wrong because a couple moment is explicitly a free vector, independent of origin. Computing it through moment sums about any point always yields the same result. Answer D is wrong because sign matters physically — $+300\hat{i}$ and $-300\hat{i}$ represent opposite rotational tendencies (right-hand rule), not equivalent ones. Study tip: Always pair $\mathbf{r}_{AB}$ with the force at A, or use $\mathbf{r}_{BA}$ with the force at B — consistent pairing prevents sign errors on couple problems.

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