Statics and Dynamics Quiz: Moments Of Forces

What this quiz covers

This quiz focuses on Moments Of Forces, 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. $M_{AB} = \hat{\lambda}_{AB} \cdot (\mathbf{r}_{AP} \times \mathbf{F})$, 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. (correct answer)
2. $M_{AB} = \hat{\lambda}_{AB} \cdot (\mathbf{r}_{BP} \times \mathbf{F})$, where $\hat{\lambda}_{AB} = \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$ and $\mathbf{r}_{BP} = -3\hat{j} + 3\hat{k}$ m, because the moment arm must be measured from a point on the axis other than the origin to avoid a trivial calculation.
3. $M_{AB} = |\mathbf{r}_{AP}||\mathbf{F}|\sin\phi$, where $\phi$ is the angle between the position vector $\mathbf{r}_{AP}$ and $\mathbf{F}$, since the moment about an axis equals the perpendicular distance times the full force magnitude.
4. $M_{AB} = \hat{\lambda}_{AB} \cdot \mathbf{F}$, where $\hat{\lambda}_{AB} = \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$, because the moment about an axis is the projection of the applied force vector onto that axis direction.

Question 2

1. The couple moment about $Q$ requires knowledge of where each individual force in the system is applied; without that information, the moment cannot be transferred to point $Q$ and the problem is indeterminate.
2. The couple moment about $Q$ must be recomputed using $\mathbf{M}_Q = \mathbf{M}_R - \mathbf{r}_{OQ}\times\mathbf{R}$; substituting $\mathbf{r}_{OQ} = 2\hat{i}+\hat{j}-3\hat{k}$ and $\mathbf{R} = 0$ gives $\mathbf{M}_Q = 50\hat{i} - 30\hat{j} + 20\hat{k}$ N·m, but this formula applies only when $Q$ lies between $O$ and the point of force application.
3. The couple moment about $Q$ changes to $\mathbf{M}_Q = 50\hat{i} - 30\hat{j} + 20\hat{k}$ N·m only because the point $Q$ happens to lie on the central axis of the wrench equivalent of this force-couple system; for a general point the moment would differ.
4. 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. (correct answer)

Question 3

1. $M_y = aF_z - cF_x$, involving only the $x$ and $z$ position and force components; the $y$-position $b$ and $y$-force $F_y$ do not contribute to the moment about the $y$-axis.
2. $M_y = aF_z - cF_x + bF_y - bF_y$, where the $bF_y$ terms cancel, confirming that the net $\hat{j}$-component depends on all position and force components but reduces to a simpler expression due to symmetric cancellation.
3. $M_y = bF_z - cF_y$, because the $\hat{j}$ component of the cross product uses the $y$-row of the position vector paired with the adjacent force components $F_z$ and $F_y$.
4. $M_y = cF_x - aF_z$, involving only the $x$ and $z$ position and force components; the $y$-components vanish from the $\hat{j}$ row of the cross-product determinant. (correct answer)

Question 4

1. The first student is correct: using $\mathbf{r}_{AB} = B - A = 3\hat{k}$ and computing $\mathbf{r}_{AB}\times\mathbf{F}_2 = 3\hat{k}\times(-100\hat{j}) = -300(\hat{k}\times\hat{j}) = -300(-\hat{i}) = +300\hat{i}$ N·m.
2. 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. (correct answer)
3. Both students are wrong: the couple moment must be computed as $\mathbf{M}_c = \mathbf{r}_{OA}\times\mathbf{F}_1 + \mathbf{r}_{OB}\times\mathbf{F}_2$, which gives a result that depends on the origin $O$ and is not a constant free vector.
4. Both students are correct because the couple moment is a free vector; it can be expressed as either $+300\hat{i}$ or $-300\hat{i}$ depending on the sign convention chosen, and both representations are physically equivalent descriptions of the same rotational tendency.

Question 5

1. 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. (correct answer)
2. The claim is incorrect: a couple requires the forces to be parallel and separated by a perpendicular distance, but since the force acts along $\hat{i}$ and the separation vector has components in all three directions, the torque depends on which moment center is chosen.
3. The claim is correct only if the chosen point $Q$ lies on the line of action of one of the two forces, which reduces the moment arm to zero for that force and simplifies the calculation to a single cross product.
4. The claim is incorrect: the forces do form a couple, but the resultant moment of a couple is constant only in magnitude, not in direction; the direction changes depending on the reference point chosen, so the full moment vector is not truly independent of $Q$.

Question 6

1. The conclusion is fully general: in three dimensions, the moment of a force about a point is always maximized when the force is perpendicular to the position vector, regardless of the axis about which the moment is measured, because the cross-product magnitude $|\mathbf{r}||\mathbf{F}|\sin\phi$ is maximized at $\phi = 90°$.
2. 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. (correct answer)
3. The conclusion is limited because the formula $M = F_0 L \sin\theta$ applies only to two-dimensional problems where the bar lies along the $x$-axis; in three dimensions, the correct expression involves the scalar triple product, which can exceed $F_0 L$ when the position vector has components in all three coordinate directions.
4. The conclusion is limited because maximizing the perpendicular component of force requires knowing the exact geometry of the bar, and in three dimensions the bar may not be straight, making the concept of a single moment arm ill-defined and the formula $M = F_0 L \sin\theta$ inapplicable even for rigid members.

Question 7

1. $M_z = -22 \text{ N·m}$, computed from the $\hat{k}$ component of $\mathbf{r} \times \mathbf{F}$, which involves only the $x$ and $y$ components of $\mathbf{r}$ and $\mathbf{F}$.
2. $M_z = +22 \text{ N·m}$, computed by taking $r_x F_y - r_y F_x = (0.3)(-30) - (0.1)(50)$, which yields $-9 - 5 = -14$ N·m, then adjusting for the $z$-components.
3. $M_z = -14 \text{ N·m}$, 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. (correct answer)
4. $M_z = +14 \text{ N·m}$, computed as $r_y F_x - r_x F_y = (0.1)(50) - (0.3)(-30) = 5 + 9$, reversing the standard cross-product order for the $\hat{k}$ term.

Question 8

1. $\mathbf{F}$ must be directed along the $\hat{k}$ axis (i.e., perpendicular to the $xy$-plane), because only a force with a $\hat{k}$ component can produce a moment in the $\hat{k}$ direction, and for the moments to be equal at two different points, the force must be purely out-of-plane.
2. $\mathbf{F}$ must lie entirely in the $xy$-plane (no $\hat{k}$ component), because the moment transfer formula $\mathbf{M}_B = \mathbf{M}_A + \mathbf{r}_{AB}\times\mathbf{F} = \mathbf{M}_A$ requires $\mathbf{r}_{AB}\times\mathbf{F} = 0$, meaning $\mathbf{F}$ is parallel to $\mathbf{r}_{AB}$, and since $\mathbf{r}_{AB}$ has no $\hat{k}$ component, $\mathbf{F}$ must be purely in-plane.
3. $\mathbf{F}$ 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 $\mathbf{r}_{AB}$. (correct answer)
4. $\mathbf{F}$ must be zero, because for two different points to yield the same moment about themselves from the same non-zero force, the force would have to act at infinity, which is physically impossible; therefore the only consistent solution is $\mathbf{F} = 0$.