Moment of a Couple - Statics

A torque wrench applies a couple moment of 150 N⋅m to a bolt. The wrench handle is 0.5 m long, and forces are applied perpendicular to the handle at its ends. What is the magnitude of each force in the couple?
Answer: 300 N When you encounter problems involving couple moments, remember that a couple consists of two equal and opposite forces separated by a distance, creating pure rotation without translation. A couple moment is calculated using the formula $M = F \times d$ , where M is the moment, F is the magnitude of each force, and d is the perpendicular distance between the forces. Since the forces are applied at the ends of the 0.5 m wrench handle, this distance becomes your moment arm. Given that the couple moment is 150 N⋅m and the distance is 0.5 m, you can solve for the force: $F = \frac{M}{d} = \frac{150 \text{ N⋅m}}{0.5 \text{ m}} = 300 \text{ N}$ . This confirms answer C is correct. Looking at the incorrect options: A) 75 N results from incorrectly dividing the moment by 2, perhaps confusing this with the concept that couples involve two forces. B) 150 N comes from assuming the force magnitude equals the moment magnitude, ignoring the distance factor entirely. D) 600 N suggests multiplying the moment by 4, possibly from incorrectly applying the distance twice or other algebraic errors. The key insight is that in a couple, both forces have the same magnitude but create rotation through their separation distance. Remember: for couple problems, always identify the perpendicular distance between the force lines of action, then use $M = F \times d$ directly. Don't overthink the "two forces" aspect—the formula already accounts for the couple's geometry.
A couple is formed by two forces of magnitude F acting on a rectangular plate. If the perpendicular distance between the force lines of action is d, and the plate rotates 45° about its center, what happens to the moment of the couple?
Answer: The moment remains unchanged at Fd regardless of the plate orientation The moment of a couple is invariant under rotation of the body. A couple's moment magnitude is always equal to the force magnitude times the perpendicular distance between the lines of action (M = Fd), and this value does not change when the body rotates. The couple vector rotates with the body, but its magnitude remains constant. Choice A incorrectly applies a trigonometric reduction. Choice B incorrectly suggests the moment increases. Choice D incorrectly applies a cosine factor as if dealing with moment components.
A rectangular plate measures 0.6 m by 0.8 m. A couple is formed by applying 50 N forces at the midpoints of opposite longer sides, with forces directed perpendicular to those sides. What is the moment of the couple?
Answer: 30 N⋅m The longer sides are 0.8 m, and the perpendicular distance between them (width of rectangle) is 0.6 m. Since forces are applied perpendicular to the longer sides, the moment arm equals the width: M = 50 N × 0.6 m = 30 N⋅m. Choice A uses half the width. Choice C uses the length instead of width. Choice D uses only force magnitude. Choice E multiplies force by length incorrectly.
A mechanic applies equal and opposite forces to the ends of a 0.3 m long wrench. The measured torque is 21 N⋅m. Later, the same forces are applied to a 0.5 m wrench in the same manner. Assuming the perpendicular distance scales proportionally with wrench length, what is the new couple moment?
Answer: 35 N⋅m This question tests your understanding of couple moments and how they scale with geometric changes. A couple consists of two equal and opposite parallel forces, and the couple moment equals the force magnitude times the perpendicular distance between the force lines. To solve this, you need to establish the relationship between wrench length and couple moment. From the first scenario: $M = F \times d$ , where $M = 21 \text{ N⋅m}$ and the perpendicular distance $d$ is proportional to the 0.3 m wrench length. Since the problem states that perpendicular distance scales proportionally with wrench length, when the wrench length increases from 0.3 m to 0.5 m, the distance increases by a factor of $\frac{0.5}{0.3} = \frac{5}{3}$ . With the same applied forces, the new couple moment becomes: $M_{new} = 21 \times \frac{5}{3} = 35 \text{ N⋅m}$ . Option A (12.6 N⋅m) represents the common error of dividing instead of multiplying by the scaling factor, or incorrectly calculating $21 \times \frac{3}{5}$ . Option B (21 N⋅m) suggests the misconception that couple moment is independent of geometry, ignoring the proportional scaling stated in the problem. Option D (70 N⋅m) likely results from incorrectly squaring the length ratio or misapplying the scaling factor. When working with couples in statics problems, always identify what changes and what stays constant. Here, forces remain the same while the perpendicular distance scales with length, making the couple moment directly proportional to the wrench length.
A wrench applies two equal and opposite forces of 80 N each to a bolt. The forces are applied at points 0.25 m apart along the wrench handle. If the wrench handle makes a 30° angle with the horizontal and the forces are applied vertically (one up, one down), what is the moment of the couple?
Answer: 20 N⋅m because couple moment is independent of handle orientation The moment of a couple is F × d where d is the perpendicular distance between the lines of action of the forces. Since both forces are vertical and 0.25 m apart horizontally, the perpendicular distance between their lines of action is 0.25 m, regardless of the wrench handle orientation. The couple moment = 80 × 0.25 = 20 N⋅m. Choice A incorrectly applies cos(30°). Choice B incorrectly applies sin(30°) and calculates 80 × 0.25 × 0.5 = 10 N⋅m, but shows 17.3. Choice D incorrectly inflates the value.
A wrench applies two equal and opposite forces of 80 N each to a bolt. The effective length of the wrench (perpendicular distance between force lines) is 0.25 m. If the wrench is rotated 45° about the bolt center, what happens to the couple moment?
Answer: It remains 20 N⋅m When analyzing couple moments in statics, remember that a couple is defined by two equal, opposite, and parallel forces. The key insight is that a couple moment depends only on the magnitude of the forces and the perpendicular distance between their lines of action—not the orientation of the couple in space. Let's calculate the original couple moment: $M = F \times d = 80 \text{ N} \times 0.25 \text{ m} = 20 \text{ N⋅m}$ . When you rotate the entire wrench 45° about the bolt center, you're performing a rigid body rotation. This rotation doesn't change the perpendicular distance between the force lines of action (still 0.25 m) or the force magnitudes (still 80 N each). The couple moment remains 20 N⋅m, making C correct. Choice A (14.14 N⋅m) incorrectly applies $20 \cos(45°) = 20 \times 0.707 = 14.14$ , suggesting the student mistakenly thinks rotation affects the couple moment like it would a single force component. Choice B (10 N⋅m) represents half the original moment, possibly from confusion about how rotation affects the system. Choice D (28.28 N⋅m) incorrectly applies $20/\cos(45°) = 20/0.707 = 28.28$ , showing another misunderstanding of how rotation relates to couple moments. Study tip: Couple moments are invariant under rotation—they're independent of coordinate system orientation. If you see a couple rotation problem, immediately recognize that only changes in force magnitude or perpendicular distance between force lines affect the couple moment, not spatial orientation changes.
Two parallel forces of magnitude 50 N each act in opposite directions on a rigid body. The perpendicular distance between the lines of action of these forces is 0.8 m. What is the moment of this couple?
Answer: 40 N⋅m When you encounter parallel forces acting in opposite directions on a rigid body, you're dealing with a couple – one of the fundamental loading types in statics. A couple creates pure rotation without translation, and its moment is independent of the reference point you choose. The moment of a couple is calculated using the simple formula: $M = F \times d$ , where F is the magnitude of either force and d is the perpendicular distance between the force lines of action. Here, you have forces of 50 N separated by 0.8 m, so: $M = 50 \text{ N} \times 0.8 \text{ m} = 40 \text{ N⋅m}$ Looking at the wrong answers: Choice A (25 N⋅m) might tempt you if you mistakenly used half the force magnitude, perhaps thinking you need to "average" the opposing forces. Choice C (50 N⋅m) could result from incorrectly using just the force magnitude while forgetting to multiply by the distance. Choice D (80 N⋅m) represents a common error where students add the force magnitudes (50 + 50 = 100 N) before multiplying by distance, misunderstanding that couple moment depends on one force times the separation distance, not the sum of forces. The correct answer is B (40 N⋅m). Study tip: Remember that for couples, you always use the magnitude of ONE force times the perpendicular distance between them. The fact that forces are equal and opposite is what makes it a couple – don't let that lead you to add or subtract the forces in your calculation.
A square plate has side length 0.4 m. Forces of 30 N are applied at opposite corners, directed parallel to one side of the square. What is the moment of the couple formed by these forces?
Answer: 12 N⋅m When forces are applied at opposite corners parallel to a side, the perpendicular distance between force lines equals the side length (0.4 m). The couple moment is M = 30 N × 0.4 m = 12 N⋅m. Choice A uses half the side length. Choice B uses the diagonal incorrectly (0.4√2 × 30 ÷ 2). Choice D uses the diagonal length (0.4√2 × 30). Choice E uses double the side length.
Three couples act on a rigid body simultaneously. The first couple has a moment of 15 N⋅m clockwise, the second has a moment of 25 N⋅m counterclockwise, and the third has a moment of 8 N⋅m clockwise. What is the resultant couple moment?
Answer: 2 N⋅m counterclockwise When analyzing multiple couples acting on a rigid body, you're dealing with pure moment addition. Couples are special force systems that create rotation without translation, and their moments can be directly added algebraically regardless of where they act on the body. To find the resultant couple moment, establish a sign convention first. Let's use counterclockwise as positive and clockwise as negative. Now convert each couple: - First couple: 15 N⋅m clockwise = -15 N⋅m - Second couple: 25 N⋅m counterclockwise = +25 N⋅m - Third couple: 8 N⋅m clockwise = -8 N⋅m Sum them algebraically: $M_{resultant} = -15 + 25 + (-8) = +2 \text{ N⋅m}$ Since the result is positive, the resultant is 2 N⋅m counterclockwise. Looking at the wrong answers: Choice A incorrectly assigns the clockwise direction to the 2 N⋅m result, likely from using the opposite sign convention or making an algebraic error. Choice C (17 N⋅m clockwise) appears to come from adding only the clockwise moments (15 + 8 = 23) and subtracting the counterclockwise (23 - 25 = -2, but then incorrectly stating direction). Choice D (17 N⋅m counterclockwise) might result from incorrectly adding 25 - 8 = 17 while ignoring the first couple entirely. Study tip: Always establish your sign convention clearly at the start, then stick to it throughout the calculation. Remember that couple moments add algebraically regardless of their position on the body—this is what makes couple analysis straightforward compared to general moment problems.
A couple has a moment of 45 N⋅m. If one of the forces in the couple is 150 N, and this force is replaced by a 90 N force while maintaining the same couple moment, what must be the new perpendicular distance between the forces?
Answer: 0.5 m When you encounter couple problems in statics, remember that a couple is defined by two parallel forces of equal magnitude acting in opposite directions, separated by a perpendicular distance. The key property is that the moment of a couple remains constant regardless of where you calculate it from. The moment of a couple is calculated as $M = F \times d$ , where F is the force magnitude and d is the perpendicular distance between the forces. Since the couple moment must remain constant at 45 N⋅m, you can set up an equation relating the original and new configurations. Originally: $45 = 150 \times d_1$ , so $d_1 = 0.3$ m For the new configuration with 90 N forces: $45 = 90 \times d_2$ Solving for the new distance: $d_2 = \frac{45}{90} = 0.5$ m This confirms answer C) 0.5 m is correct. Looking at the wrong answers: A) 0.2 m would give a moment of only 18 N⋅m (90 × 0.2), which is too small. B) 0.3 m represents the original distance when the force was 150 N, but this would only produce 27 N⋅m with the 90 N force. D) 0.75 m would create a moment of 67.5 N⋅m (90 × 0.75), which exceeds the required 45 N⋅m. Remember this inverse relationship: when dealing with couples, if you decrease the force magnitude, you must proportionally increase the distance to maintain the same moment. Always verify your answer by multiplying force times distance to ensure it equals the original couple moment.
An L-shaped bracket has arms of length 0.3 m and 0.4 m meeting at a right angle. Equal and opposite forces of 60 N are applied at the free ends of the arms, both forces pointing in the same direction parallel to one arm. What is the moment of the couple?
Answer: 24 N⋅m When forces at the arm ends are parallel to one arm, the perpendicular distance between force lines equals the length of the other arm. If forces are parallel to the 0.3 m arm, the perpendicular distance is 0.4 m. M = 60 N × 0.4 m = 24 N⋅m. Choice A uses the shorter arm length. Choice C uses the average arm length. Choice D uses the hypotenuse length. Choice E uses both arm lengths incorrectly.
Two parallel forces of 40 N each act in opposite directions on a beam. The forces are applied at points 1.2 m apart along the beam's length. If the forces make a 60° angle with the beam's longitudinal axis, what is the couple moment?
Answer: 41.6 N⋅m When you encounter parallel forces acting in opposite directions, you're dealing with a couple - a system that produces pure rotational effect without translation. The key insight is that couples create the same moment about any point, making the calculation straightforward. For a couple, the moment equals the force magnitude times the perpendicular distance between the force lines of action. Here, you have 40 N forces separated by 1.2 m, but they're applied at 60° to the beam's axis. The perpendicular distance between the parallel force lines is the component of the 1.2 m separation that's perpendicular to the force direction. Since the forces make 60° with the beam, the perpendicular distance is: $d_{\perp} = 1.2 \sin(60°) = 1.2 \times \frac{\sqrt{3}}{2} = 1.04 \text{ m}$ The couple moment is: $M = F \times d_{\perp} = 40 \times 1.04 = 41.6 \text{ N⋅m}$ Answer A (20.8 N⋅m) represents half the correct value - you might get this if you incorrectly used half the force magnitude. Answer B (24 N⋅m) comes from using the cosine component instead of sine: $40 \times 1.2 \times \cos(60°) = 24$ . Answer D (48 N⋅m) results from using the full 1.2 m distance without accounting for the angle: $40 \times 1.2 = 48$ . Remember: for angled couples, always use the perpendicular distance between force lines, not the distance along the beam. The sine function gives you the perpendicular component when forces are angled to the connecting line.
A couple consists of two 120 N forces separated by a distance of 0.6 m. If the angle between each force vector and the line connecting their points of application is 30°, what is the moment of the couple?
Answer: 36 N⋅m When you encounter a couple in statics, remember that it creates a pure moment that's independent of where you calculate it. A couple consists of two equal, parallel forces acting in opposite directions. For any couple, the moment is calculated as $M = F \times d_{\perp}$ , where $F$ is the magnitude of either force and $d_{\perp}$ is the perpendicular distance between the lines of action of the forces. The key insight here is finding the perpendicular distance. You're given that the forces are separated by 0.6 m, but each force makes a 30° angle with the line connecting their application points. This means the perpendicular distance between the force lines of action is $d_{\perp} = 0.6 \times \sin(30°) = 0.6 \times 0.5 = 0.3 \text{ m}$ . Therefore: $M = 120 \text{ N} \times 0.3 \text{ m} = 36 \text{ N⋅m}$ , confirming answer A. Answer B (62.35 N⋅m) likely comes from using $0.6 \times \cos(30°) = 0.519 \text{ m}$ instead of the sine—a common error when students confuse which trigonometric function gives the perpendicular component. Answer C (72 N⋅m) results from incorrectly using the full 0.6 m separation distance without accounting for the angle: $120 \times 0.6 = 72$ . Answer D (124.7 N⋅m) appears to stem from multiplying by both forces (240 N total) and using the cosine distance. Study tip: Always visualize couples by sketching the perpendicular distances between force lines of action. The moment arm is never measured along the connecting line unless the forces are perpendicular to it.
A couple moment of 36 N⋅m is required to turn a valve. If the available force that can be applied is 120 N, what is the minimum perpendicular distance required between the two forces of the couple?
Answer: 0.3 m When you encounter a couple moment problem, you're dealing with two equal and opposite forces that create pure rotation. The key relationship is that the moment of a couple equals the force magnitude times the perpendicular distance between the forces: $M = F \times d$ . Given that you need a couple moment of 36 N⋅m and can apply 120 N of force, you can solve directly for the minimum distance: $d = \frac{M}{F} = \frac{36 \text{ N⋅m}}{120 \text{ N}} = 0.3 \text{ m}$ This confirms answer B is correct. Looking at the wrong answers: A) 0.15 m would only produce half the required moment ( $120 \times 0.15 = 18$ N⋅m). This might result from incorrectly dividing the required moment by 2, perhaps confusing couple mechanics with single force moments. C) 0.6 m would create double the needed moment ( $120 \times 0.6 = 72$ N⋅m). This could come from accidentally multiplying instead of dividing, or doubling the correct answer. D) 4.32 m results from incorrectly multiplying the moment by the force ( $36 \times 120 = 4320$ , then misplacing the decimal). This shows a fundamental misunderstanding of the couple moment formula. Remember that couple problems are typically straightforward once you identify them. The moment equation $M = F \times d$ is your primary tool, and unlike single force problems, you don't need to worry about the location of the couple—it produces the same rotational effect regardless of where it's applied to the body.
A circular disk of radius 1.5 m has four forces applied tangentially at its edge. Two forces of 80 N each are applied at opposite points (180° apart), both in the clockwise direction. Two forces of 60 N each are applied at points 90° from the first pair, both in the counterclockwise direction. What is the net moment about the disk's center?
Answer: 60 N⋅m clockwise because the 80 N forces create a stronger couple When you encounter forces applied tangentially to a circular object, think about couples and moments. This question tests your understanding of how multiple couples interact to create net rotation. Each pair of opposite forces creates a couple. The 80 N forces (180° apart, both clockwise) form a couple with moment $M_1 = F \times d = 80 \text{ N} \times 3.0 \text{ m} = 240 \text{ N⋅m}$ clockwise, where the distance is the full diameter. Similarly, the 60 N forces (180° apart, both counterclockwise) create $M_2 = 60 \text{ N} \times 3.0 \text{ m} = 180 \text{ N⋅m}$ counterclockwise. The net moment is the algebraic sum: $240 \text{ N⋅m}$ clockwise minus $180 \text{ N⋅m}$ counterclockwise equals $60 \text{ N⋅m}$ clockwise. This confirms answer B is correct. Answer A incorrectly suggests the counterclockwise forces dominate, but 240 > 180, so the clockwise couple is stronger. Answer C makes an error in calculation or concept—the difference is 60, not 30, and the direction follows the stronger couple. Answer D falls into the trap of thinking opposite forces always create equilibrium, but this ignores that we have two different couples of different magnitudes working against each other. Remember: when analyzing couples, calculate each couple's moment separately using the full distance between the forces, then find the algebraic sum. The stronger couple determines both the magnitude of the net moment and its direction.
A compound couple system consists of three pairs of forces on a rectangular frame. Pair 1: 100 N forces separated by 0.5 m (clockwise). Pair 2: 80 N forces separated by 0.75 m (counterclockwise). Pair 3: 120 N forces separated by 0.4 m (clockwise). If the frame rotates 90° about its center, what is the resultant couple moment after rotation?
Answer: 38 N⋅m clockwise, same as before rotation due to couple invariance When analyzing compound couple systems, remember that couples have a unique property: their moment is independent of the reference point and remains constant under rigid body rotation. This is because a couple represents pure rotational tendency without any translational effects. To find the resultant couple moment, you simply add the individual couple moments algebraically. Let's establish a sign convention where clockwise is positive: Pair 1: $M_1 = 100 \text{ N} \times 0.5 \text{ m} = +50 \text{ N⋅m}$ (clockwise) Pair 2: $M_2 = 80 \text{ N} \times 0.75 \text{ m} = -60 \text{ N⋅m}$ (counterclockwise) Pair 3: $M_3 = 120 \text{ N} \times 0.4 \text{ m} = +48 \text{ N⋅m}$ (clockwise) Resultant: $M_{total} = 50 - 60 + 48 = 38 \text{ N⋅m}$ clockwise The 90° rotation doesn't change this result due to couple invariance. Choice A incorrectly assumes rotation changes the couple's sign or magnitude. Choice C wrongly suggests that rotation causes cancellation – couples don't cancel due to orientation changes alone. Choice D makes the false assumption that perpendicular orientation doubles the moment, which has no basis in statics theory. The correct answer is B: 38 N⋅m clockwise remains unchanged after rotation. Study tip: Remember that couples are "free vectors" – their moments remain constant regardless of rotation or translation of the rigid body. Only the individual force magnitudes and perpendicular distances matter for the calculation, not the body's orientation in space.
Two parallel forces create a couple on a beam. The first force is 120 N acting upward at x = 2 m, and the second force is 120 N acting downward at x = 5 m. A third force of 80 N acts downward at x = 3.5 m. If the couple is replaced by a single equivalent couple applied at different points, where should two 90 N forces (opposite directions) be placed to create the same moment as the original couple alone?
Answer: 4.0 m apart, positioned symmetrically about any point on the beam The original couple has moment = 120 N × (5-2) m = 360 N⋅m. To create the same moment with 90 N forces: 90 N × d = 360 N⋅m, so d = 4.0 m. Since a couple's moment is independent of its position on the body, the 90 N forces can be placed anywhere as long as they are 4.0 m apart and parallel/opposite. The 80 N force at x = 3.5 m doesn't affect the couple calculation since we're only replacing the couple portion. Choice B incorrectly requires specific positioning. Choice C incorrectly factors in the third force. Choice D unnecessarily constrains the location.
A L-shaped bracket experiences two couples simultaneously. The first couple consists of 40 N forces separated by 0.6 m, creating a clockwise moment. The second couple consists of 30 N forces separated by 0.8 m, creating a counterclockwise moment. If these couples are in equilibrium, what single equivalent couple using 60 N forces would be needed to balance an additional 36 N⋅m counterclockwise couple applied to the system?
Answer: 0.6 m separation with clockwise orientation to provide the balancing moment When analyzing coupled moments in statics, you need to first determine the net effect of existing couples, then find what's required for equilibrium with any additional applied moments. Let's work through this systematically. The first couple creates a clockwise moment of $40 \text{ N} \times 0.6 \text{ m} = 24 \text{ N⋅m}$ clockwise. The second couple creates a counterclockwise moment of $30 \text{ N} \times 0.8 \text{ m} = 24 \text{ N⋅m}$ counterclockwise. These two couples are indeed in equilibrium (24 N⋅m clockwise = 24 N⋅m counterclockwise). Now you add a 36 N⋅m counterclockwise couple to this balanced system. To restore equilibrium, you need a 36 N⋅m clockwise couple to balance it. Using 60 N forces, the required separation distance is $d = \frac{36 \text{ N⋅m}}{60 \text{ N}} = 0.6 \text{ m}$ , and it must be clockwise oriented. Looking at the wrong answers: Choice A gives the wrong orientation (counterclockwise) and wrong separation (1.2 m would create 72 N⋅m). Choice B also has the wrong orientation and would only create 24 N⋅m. Choice C has the correct orientation but wrong separation—0.8 m would create 48 N⋅m, which is too much. Choice D correctly specifies 0.6 m separation with clockwise orientation, providing exactly the 36 N⋅m clockwise moment needed. Study tip: Always calculate the moment magnitude first (Force × distance), then determine the required direction for equilibrium. Remember that couples are pure moments—their position doesn't matter, only their magnitude and rotational direction.
Three couples act simultaneously on a rigid body. Couple A produces 20 N⋅m clockwise, Couple B produces 35 N⋅m counterclockwise, and Couple C produces 12 N⋅m clockwise. If an additional couple is applied to achieve equilibrium, what must be its moment?
Answer: 3 N⋅m clockwise When analyzing couples acting on rigid bodies, remember that couples are pure moments that can be moved anywhere on the body without changing their effect. For equilibrium, the sum of all couple moments must equal zero. To solve this problem, establish a sign convention first. Let's use counterclockwise as positive and clockwise as negative. The three given couples produce: - Couple A: $-20$ N⋅m (clockwise) - Couple B: $+35$ N⋅m (counterclockwise) - Couple C: $-12$ N⋅m (clockwise) The net moment from these three couples is: $(-20) + (+35) + (-12) = +3$ N⋅m counterclockwise. For equilibrium, you need an additional couple that exactly cancels this net moment. Therefore, the equilibrium couple must be $-3$ N⋅m, or 3 N⋅m clockwise. Looking at the wrong answers: Option B gives 3 N⋅m counterclockwise, which would add to the existing counterclockwise moment rather than cancel it. Options C and D both show 32 N⋅m, which appears to come from incorrectly adding the magnitudes of all couples ( $20 + 35 + 12 = 67$ ) and then making an error, or possibly adding just two of the couples. These approaches ignore the directional nature of the moments. Study tip: Always establish a clear sign convention first, then algebraically sum all moments. The equilibrium couple must have the opposite sign of your calculated net moment to bring the total to zero.
Two 60 N forces form a couple. When measured about point P, the sum of moments of these individual forces is 18 N⋅m. What is the moment of the couple?
Answer: 18 N⋅m When you encounter problems involving couples in statics, remember that a couple has a unique property: its moment is the same about any point in the plane. This is because a couple consists of two equal, opposite, parallel forces that create pure rotation without translation. The key insight here is understanding what "the sum of moments of these individual forces about point P" means versus "the moment of the couple." When you calculate the moment of each force separately about point P and add them together, you get 18 N⋅m. However, this sum of individual moments IS the moment of the couple. The couple's moment doesn't depend on where point P is located - it would be 18 N⋅m about any point you choose. Looking at the wrong answers: Choice A (9 N⋅m) might tempt you if you incorrectly think the couple moment is half the sum of individual moments, perhaps confusing it with an average. Choice C (36 N⋅m) could result from mistakenly doubling the given value, possibly thinking you need to account for both forces somehow. Choice D suggests the problem lacks sufficient information, but this misses the fundamental principle that a couple's moment is independent of the reference point. The correct answer is B (18 N⋅m) because the moment of a couple equals the sum of moments of its constituent forces about any point. Study tip: Remember that for couples, the moment is invariant - it's the same regardless of your chosen reference point. This makes couple problems often simpler than they initially appear.

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