Statics Quiz: Units And Sign Conventions
19 questions · exam conditions
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Units And Sign ConventionsQuestion 1 of 19

A 3.0 kN·m clockwise moment and a 1500 N·m counterclockwise moment act on a body. With CCW positive, the net moment is:

-3000 N·m
-4500 N·m
-1500 N·m
+1500 N·m
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Statics Quiz

Statics Quiz: Units And Sign Conventions

Practice Units And Sign Conventions 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 Units And Sign Conventions, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

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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.

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Question 1

A 3.0 kN·m clockwise moment and a 1500 N·m counterclockwise moment act on a body. With CCW positive, the net moment is:

  1. -3000 N·m
  2. -4500 N·m
  3. -1500 N·m (correct answer)
  4. +1500 N·m
Explanation: With CCW positive, the 3.0 kN·m clockwise moment is -3000 N·m and the 1500 N·m counterclockwise moment is +1500 N·m. Adding them gives -1500 N·m. The tempting error is adding 3000 and 1500 to get 4500 N·m, but that treats the counterclockwise moment as if it had the same sense.

Question 2

With east positive, a 0.25 kN west force and a 150 N east force act on a particle. The resultant is:

  1. +100 N
  2. -250 N
  3. +150 N
  4. -100 N (correct answer)
Explanation: Convert 0.25 kN to 250 N. With east as positive, the west force is -250 N and the east force is +150 N, so the resultant is -250 + 150 = -100 N, or 100 N west. A tempting wrong answer is +100 N, which uses the correct 250 - 150 = 100 arithmetic but assigns the wrong direction; the larger westward force makes the resultant westward.

Question 3

A force F = (-0.40i + 0.30j) kN. What is its direction measured CCW from the +x axis?

  1. 143° (correct answer)
  2. 37°
  3. -37°
  4. 53°
Explanation: The x-component is negative and the y-component is positive, so the force is in the second quadrant, between 90° and 180°. The reference angle is arctan(0.30/0.40) = about 37° above the negative x-axis. Measured CCW from the +x axis, that is 180° - 37° = 143°. The tempting 37° ignores the negative x-component and treats the vector as if it were in the first quadrant.

Question 4

A pressure is 2.5 kN/m². What is this in N/mm²?

  1. 0.025 N/mm²
  2. 2500 N/mm²
  3. 0.0025 N/mm² (correct answer)
  4. 2.5 N/mm²
Explanation: Convert 2.5 kN to 2,500 N. One square metre is 1,000,000 square millimetres, so divide 2,500 N by 1,000,000 to get 0.0025 N/mm². The tempting wrong answer is 2.5 N/mm² because it treats 1 m² as 1,000 mm², but area conversion must square the linear 1,000 mm per metre.

Question 5

A reaction force is computed as -0.0045 MN with upward positive. Its actual direction and magnitude are:

  1. 4.5 kN upward
  2. 4.5 kN downward (correct answer)
  3. 0.45 kN downward
  4. 45 kN downward
Explanation: Since 1 MN = 1000 kN, -0.0045 MN = -4.5 kN. With upward positive, a negative value points downward, so the magnitude is 4.5 kN downward. Don't choose 4.5 kN upward: that drops the minus sign and reverses the direction.

Question 6

A distributed load of 2.5 kN/m acts over a 4-meter span. When this load is replaced by an equivalent point load for analysis purposes, what is the magnitude of the equivalent point load in pounds-force?

  1. 2,247 lbf (correct answer)
  2. 2,481 lbf
  3. 2,696 lbf
  4. 2,923 lbf
  5. 3,158 lbf
Explanation: When dealing with distributed loads in statics, you need to understand that any distributed load can be replaced by an equivalent point load for analysis purposes. The magnitude of this equivalent point load equals the total area under the load distribution curve, and it acts at the centroid of that area. For a uniformly distributed load, you calculate the equivalent point load by multiplying the load intensity by the length over which it acts. Here, you have 2.5 kN/m acting over 4 meters, so the equivalent point load is 2.5 kN/m×4 m=10 kN2.5 \text{ kN/m} \times 4 \text{ m} = 10 \text{ kN}. Now you must convert from kilonewtons to pounds-force. Using the conversion factor 1 kN = 224.8 lbf: 10 kN×224.8 lbf/kN=2,248 lbf10 \text{ kN} \times 224.8 \text{ lbf/kN} = 2,248 \text{ lbf}, which rounds to 2,247 lbf. This confirms answer A is correct. Answer B (2,481 lbf) likely results from using an incorrect conversion factor or calculation error. Answer C (2,696 lbf) might come from confusing kilonewtons with kilograms and using gravitational acceleration incorrectly. Answer D (2,923 lbf) represents a significant calculation error, possibly from multiplying instead of applying the proper conversion. Remember this two-step process: first find the total load by multiplying intensity by length, then convert units carefully. Always double-check your unit conversions in statics problems, as they're a common source of errors on exams.

Question 7

A force vector is expressed as F = 40î - 30ĵ + 20k̂ (in Newtons). If this force creates a moment about the origin when applied at point P(2, -1, 3) meters, what is the magnitude of the y-component of the moment vector in N·m?

  1. 40 N·m
  2. 60 N·m
  3. 80 N·m (correct answer)
  4. 100 N·m
  5. 120 N·m
Explanation: When you encounter moment problems involving force vectors and position vectors, you need to use the cross product formula: M⃗=r⃗×F⃗\vec{M} = \vec{r} \times \vec{F}, where r⃗\vec{r} is the position vector from the origin to the point of application. To find the moment vector, set up the cross product with position vector r⃗=2i^−1j^+3k^\vec{r} = 2\hat{i} - 1\hat{j} + 3\hat{k} and force vector F⃗=40i^−30j^+20k^\vec{F} = 40\hat{i} - 30\hat{j} + 20\hat{k}: Expanding the determinant:
  • My=−[2(20)−3(40)]=−[40−120]=−(−80)=80 N⋅mM_y = -[2(20) - 3(40)] = -[40 - 120] = -(-80) = 80 \text{ N⋅m}
The magnitude of the y-component is 80 N⋅m, making C correct. A) 40 N⋅m represents the x-component of the force vector, showing confusion between force components and moment components. B) 60 N⋅m could result from calculation errors in the cross product, possibly mixing up which terms to subtract or incorrect sign handling. D) 100 N⋅m might come from adding terms instead of subtracting in the cross product expansion, or from computational mistakes in the determinant evaluation. Study tip: Always remember that moments are calculated using cross products, not dot products. Practice the determinant method for 3D cross products, and pay careful attention to signs—they determine the direction of rotation. The key is methodical expansion of the determinant and careful arithmetic.

Question 8

A cantilever beam has moments applied as follows: +150 N·m (clockwise positive convention), -200 N·m, and +75 N·m. If the sign convention is changed so that counterclockwise moments are positive, what would be the algebraic sum of these same physical moments under the new convention?

  1. -25 N·m (correct answer)
  2. +25 N·m
  3. -75 N·m
  4. +75 N·m
  5. -125 N·m
Explanation: When you encounter problems involving sign conventions in statics, remember that the physical moments themselves don't change—only how we represent them mathematically changes based on our chosen convention. Let's work through this systematically. Under the original convention (clockwise positive), we have three moments: +150 N·m, -200 N·m, and +75 N·m. The algebraic sum is 150+(−200)+75=+25150 + (-200) + 75 = +25 N·m. Now, when we switch to counterclockwise positive convention, each moment's sign flips because what was previously considered positive (clockwise) is now negative, and vice versa. So our moments become: -150 N·m, +200 N·m, and -75 N·m. The new algebraic sum is (−150)+200+(−75)=−25(-150) + 200 + (-75) = -25 N·m. Looking at the choices: Answer A (-25 N·m) is correct as shown above. Answer B (+25 N·m) represents the original sum before the sign convention change—this traps students who forget that changing conventions affects the signs. Answer C (-75 N·m) might result from incorrectly applying the sign change to only some moments or making arithmetic errors. Answer D (+75 N·m) could come from various computational mistakes or misunderstanding how sign conventions work. Here's your key strategy: When sign conventions change in statics problems, the magnitude of the resultant stays the same, but the sign flips. Always calculate the sum under the original convention first, then simply change the sign of that result for the new convention.

Question 9

In a free body diagram, forces are labeled with the following sign convention: rightward forces are positive, leftward forces are negative, upward forces are positive, and downward forces are negative. If three forces act on a particle: F₁ = +150 N (horizontal), F₂ = -75 N (horizontal), and F₃ = +200 N at 45° above the positive x-axis, what is the net force in the y-direction?

  1. +141 N (correct answer)
  2. +165 N
  3. +186 N
  4. +212 N
  5. +245 N
Explanation: When analyzing forces acting on a particle, you need to break each force into its x and y components, then sum the components in each direction separately. This problem asks specifically for the net force in the y-direction, so focus on the vertical components of each force. Let's examine each force: F₁ = +150 N acts horizontally, so it has zero y-component. F₂ = -75 N also acts horizontally, contributing nothing to the y-direction. Only F₃ = +200 N at 45° above the positive x-axis has a vertical component. To find the y-component of F₃, use trigonometry: F3y=F3sin⁡(45°)=200×sin⁡(45°)=200×0.707=141.4 NF_{3y} = F_3 \sin(45°) = 200 \times \sin(45°) = 200 \times 0.707 = 141.4 \text{ N} Since F₃ points above the x-axis and upward forces are positive, this component is +141.4 N. The net force in the y-direction is therefore +141 N, which is answer choice A. The wrong answers likely come from common mistakes: B (+165 N) might result from incorrectly using cosine instead of sine, or making an error in the trigonometric calculation. C (+186 N) and D (+212 N) could come from adding horizontal force components to the vertical calculation, or from using incorrect trigonometric relationships. Remember this key strategy: always identify which forces actually contribute to the direction you're analyzing. Horizontal forces don't affect vertical equilibrium, and angled forces must be decomposed using sine for the perpendicular component and cosine for the parallel component.

Question 10

In a truss analysis, joint forces are reported as: Tension = +850 N, Compression = -650 N, and Compression = -400 N. If the analysis software uses the opposite sign convention where compression is positive and tension is negative, what would be the sum of these same physical forces as displayed by the software?

  1. -200 N
  2. +200 N (correct answer)
  3. +850 N
  4. +1,900 N
  5. -1,900 N
Explanation: When working with truss analysis, you need to understand that sign conventions for tension and compression can vary between different sources and software programs. This question tests your ability to convert between different sign conventions while keeping the physical forces unchanged. The original analysis reports three forces with the standard convention (tension positive, compression negative): +850 N (tension), -650 N (compression), and -400 N (compression). The software uses the opposite convention where compression is positive and tension is negative. To convert, you simply flip the signs of all forces: the +850 N tension becomes -850 N, the -650 N compression becomes +650 N, and the -400 N compression becomes +400 N. Now sum these converted values: −850+650+400=+200 N-850 + 650 + 400 = +200 \text{ N} Looking at the wrong answers: Choice A (-200 N) represents summing the original values without converting the sign convention first (850−650−400=−200850 - 650 - 400 = -200). Choice C (+850 N) appears to ignore the compression forces entirely and just report the original tension value. Choice D (+1,900 N) results from adding all the absolute values without considering signs at all (850+650+400=1,900850 + 650 + 400 = 1,900). The key strategy here is to remember that changing sign conventions means flipping the algebraic signs of all forces, but the physical reality remains the same. Always convert first, then perform your calculations. Watch for problems that test sign convention understanding, as they're common in structural analysis courses.

Question 11

In a coordinate system where the positive x-axis points east and positive y-axis points north, a wind force of 120 N acts from the southwest direction (225° from positive x-axis). What are the force components using the convention that eastward and northward are positive?

  1. Fx = -84.9 N, Fy = -84.9 N
  2. Fx = +84.9 N, Fy = +84.9 N (correct answer)
  3. Fx = -84.9 N, Fy = +84.9 N
  4. Fx = +84.9 N, Fy = -84.9 N
  5. Fx = +120 N, Fy = 0 N
Explanation: When analyzing force components in statics, you need to carefully distinguish between the direction a force comes from versus the direction it acts toward. This problem states the wind force "acts from the southwest," which means it originates from 225° but actually pushes in the opposite direction—toward the northeast. Since the force acts toward the northeast (opposite of 225°), it actually points at 225° - 180° = 45° from the positive x-axis. Using trigonometry to find components:
  • Fx=120cos⁡(45°)=120×0.707=+84.9 NF_x = 120 \cos(45°) = 120 \times 0.707 = +84.9 \text{ N}
  • Fy=120sin⁡(45°)=120×0.707=+84.9 NF_y = 120 \sin(45°) = 120 \times 0.707 = +84.9 \text{ N}
Both components are positive because the force pushes eastward (+x direction) and northward (+y direction). Answer A gives negative components for both directions, which would occur if you mistakenly used 225° directly without recognizing that forces act in the direction opposite to where they originate. Answer C incorrectly makes the x-component negative while keeping y positive, suggesting confusion about which quadrant the force acts in. Answer D makes the y-component negative, which would place the force in the fourth quadrant rather than the first. Study tip: Always clarify whether an angle describes where a force comes from or where it points. Wind forces, pressure forces, and similar "pushing" forces typically act in the direction opposite to their stated origin angle. Draw a quick sketch to visualize the actual force direction before calculating components.

Question 12

A torque wrench reads 45 ft·lb when tightening a bolt. If this same physical torque is measured using a meter-kilogram-force torque wrench, what would be the reading in m·kgf?

  1. 6.24 m·kgf (correct answer)
  2. 8.45 m·kgf
  3. 9.18 m·kgf
  4. 11.3 m·kgf
  5. 13.7 m·kgf
Explanation: Unit conversion problems in statics require careful attention to the relationship between different measurement systems. When you encounter torque conversions, remember that torque has units of force times distance, so you need conversion factors for both components. To convert 45 ft·lb to m·kgf, you need two key conversion factors: 1 ft = 0.3048 m and 1 lb = 0.4536 kgf. Since torque involves both length and force, you multiply by both conversion factors: 45 ft\cdotplb×0.3048mft×0.4536kgflb=6.24 m\cdotpkgf45 \text{ ft·lb} \times 0.3048 \frac{\text{m}}{\text{ft}} \times 0.4536 \frac{\text{kgf}}{\text{lb}} = 6.24 \text{ m·kgf} Answer A (6.24 m·kgf) is correct because it properly accounts for both the length and force conversions. Answer B (8.45 m·kgf) likely results from using an incorrect conversion factor, possibly confusing pounds with newtons or using an approximated conversion. Answer C (9.18 m·kgf) suggests the student may have only converted the length units (45 × 0.3048 ≈ 13.7) and made an error with the force conversion, or used completely wrong conversion factors. Answer D (11.3 m·kgf) appears to come from converting only the distance (45 ft × 0.3048 m/ft ≈ 13.7 m) while treating the pound-force incorrectly, perhaps dividing instead of multiplying by the conversion factor. Study tip: Always identify what units you're converting from and to, then systematically convert each component. Write out your conversion factors as fractions to ensure units cancel properly—this prevents the common mistake of dividing when you should multiply.

Question 13

A uniformly distributed load of 3 kips/ft acts over a 12-foot span. When converted to SI units, what is the equivalent distributed load in kN/m?

  1. 43.8 kN/m (correct answer)
  2. 48.2 kN/m
  3. 52.6 kN/m
  4. 57.1 kN/m
  5. 61.5 kN/m
Explanation: Unit conversion problems in statics require careful attention to conversion factors and dimensional analysis. When converting distributed loads from US customary units (kips/ft) to SI units (kN/m), you need to convert both the force component (kips to kN) and the length component (ft to m). To solve this, start with the given load of 3 kips/ft. First, convert kips to kilonewtons: 1 kip = 4.448 kN, so 3 kips = 3 × 4.448 = 13.344 kN. Next, convert feet to meters: 1 ft = 0.3048 m. Since you're converting a rate (force per unit length), you divide the converted force by the converted length unit: 13.344 kN0.3048 m=43.8 kN/m\frac{13.344 \text{ kN}}{0.3048 \text{ m}} = 43.8 \text{ kN/m} Answer A (43.8 kN/m) is correct using the proper conversion factors. Answer B (48.2 kN/m) likely results from using an incorrect conversion factor, possibly confusing kips with pounds or using an approximated value. Answer C (52.6 kN/m) suggests an error in the dimensional analysis, perhaps multiplying by the length conversion factor instead of dividing. Answer D (57.1 kN/m) represents a more significant calculation error, possibly combining multiple conversion mistakes. Always remember the key conversion factors for structural engineering: 1 kip = 4.448 kN and 1 ft = 0.3048 m. When converting rates or ratios, pay close attention to whether you multiply or divide by each conversion factor—force per length requires dividing the converted force by the converted length unit.

Question 14

A force system consists of F₁ = 200 N at 30° above the horizontal (positive direction) and F₂ = 150 N at 45° below the horizontal (negative direction). Using vector addition with proper sign conventions, what is the magnitude of the resultant horizontal component?

  1. 67 N
  2. 89 N
  3. 134 N
  4. 167 N
  5. 279 N (correct answer)
Explanation: When analyzing force systems in statics, you need to break each force into its horizontal and vertical components using trigonometry, then apply proper sign conventions for direction. For this problem, start with F₁ = 200 N at 30° above horizontal. Its horizontal component is F1x=200cos⁡(30°)=200×0.866=173.2 NF_{1x} = 200 \cos(30°) = 200 \times 0.866 = 173.2 \text{ N} (positive, pointing right). For F₂ = 150 N at 45° below horizontal, its horizontal component is F2x=150cos⁡(45°)=150×0.707=106.1 NF_{2x} = 150 \cos(45°) = 150 \times 0.707 = 106.1 \text{ N} (also positive, pointing right). The resultant horizontal component is: Rx=F1x+F2x=173.2+106.1=279.3 NR_x = F_{1x} + F_{2x} = 173.2 + 106.1 = 279.3 \text{ N} This value doesn't match any of the given options A through D, making the correct answer E (presumably "none of the above" or a similar option). Answer A (67 N) likely comes from incorrectly subtracting the components: 173.2 - 106.1 = 67.1 N. Answer B (89 N) might result from using sine instead of cosine for one force. Answer C (134 N) could come from taking only the larger component and making calculation errors. Answer D (167 N) is close to F₁'s horizontal component alone, suggesting someone ignored F₂ entirely. Always remember to carefully identify the direction of each force component and use the correct trigonometric function—cosine for horizontal components, sine for vertical components. Double-check your sign conventions and verify that your final answer makes physical sense.

Question 15

A couple consists of two parallel forces: +500 N at position (2, 0) m and -500 N at position (2, 3) m. If the coordinate system origin is moved to point (1, 1.5) m, what is the magnitude of the couple moment about the new origin?

  1. 0 N·m
  2. 750 N·m
  3. 1,000 N·m
  4. 1,500 N·m (correct answer)
  5. 2,000 N·m
Explanation: When you encounter a couple problem in statics, remember that a couple's moment magnitude is invariant—it remains the same regardless of where you calculate it from. This is a fundamental property that makes couples unique among force systems. A couple consists of two equal and opposite parallel forces separated by a distance. The moment of a couple equals the force magnitude times the perpendicular distance between the forces. Here, you have +500 N at (2, 0) and -500 N at (2, 3). Since both forces act along the same vertical line (x = 2), the perpendicular distance between them is simply the difference in their y-coordinates: 3 - 0 = 3 meters. The couple moment magnitude is: M=F×d=500 N×3 m=1500 N\cdotpmM = F \times d = 500 \text{ N} \times 3 \text{ m} = 1500 \text{ N·m} This value doesn't change when the origin moves to (1, 1.5) m—that's the key insight about couples. Answer A (0 N·m) incorrectly assumes the forces cancel out completely, ignoring that they create a rotational effect. Answer B (750 N·m) might result from mistakenly using half the force magnitude or half the distance. Answer C (1,000 N·m) could come from incorrectly calculating the distance as 2 m instead of 3 m, perhaps by confusing coordinate positions. Remember this golden rule for statics: couples always produce the same moment regardless of the reference point. When you see equal and opposite parallel forces, calculate the moment using force times perpendicular distance, and don't worry about coordinate system changes.

Question 16

A concentrated moment of 800 lb·in is applied to a shaft. When this moment is expressed in SI base units (N·m), and then the equivalent torque per unit length is calculated for a shaft of length 0.5 meters, what is the distributed moment in N·m/m?

  1. 181 N·m/m (correct answer)
  2. 203 N·m/m
  3. 225 N·m/m
  4. 247 N·m/m
  5. 269 N·m/m
Explanation: When you encounter problems involving unit conversions and distributed loads in statics, you need to work systematically through the conversion process before calculating the distribution. First, convert the concentrated moment from English to SI units. Using the conversion factor 1 lb·in = 0.113 N·m: 800 lb\cdotpin×0.113N\cdotpmlb\cdotpin=90.4 N\cdotpm800 \text{ lb·in} \times 0.113 \frac{\text{N·m}}{\text{lb·in}} = 90.4 \text{ N·m} Next, to find the equivalent distributed moment (torque per unit length), divide this concentrated moment by the shaft length: 90.4 N\cdotpm0.5 m=180.8 N\cdotpm/m\frac{90.4 \text{ N·m}}{0.5 \text{ m}} = 180.8 \text{ N·m/m} This rounds to 181 N·m/m, making A correct. The wrong answers represent common calculation errors. B (203 N·m/m) likely results from using an incorrect conversion factor or rounding errors in the conversion step. C (225 N·m/m) suggests someone may have used 1 lb·in = 0.141 N·m instead of the correct 0.113 N·m conversion. D (247 N·m/m) indicates a more significant error, possibly multiplying instead of dividing by the length or using a completely wrong conversion factor. Study tip: Always double-check your unit conversion factors—they're often the source of errors in statics problems. Write out the units in your calculations to ensure they cancel properly, and remember that distributed loads are always total load divided by the length over which it's distributed.

Question 17

A force of 500 N acts at coordinates (3, 4) meters relative to point O. If the perpendicular distance from O to the line of action of the force is 2.4 meters, and moments about O are calculated using M = F × d, what is the moment about point O?

  1. 1,200 N·m (correct answer)
  2. 1,500 N·m
  3. 2,000 N·m
  4. 2,500 N·m
  5. 3,000 N·m
Explanation: When you encounter moment problems in statics, remember that a moment represents the rotational effect of a force about a point. The fundamental formula is M=F×dM = F \times d, where d is the perpendicular distance from the point to the force's line of action. In this problem, you're given everything you need directly: a 500 N force and a perpendicular distance of 2.4 meters from point O to the force's line of action. The coordinates (3, 4) might seem important, but they're actually irrelevant when the perpendicular distance is already provided. Simply apply the moment formula: M=500 N×2.4 m=1,200 N\cdotpmM = 500 \text{ N} \times 2.4 \text{ m} = 1,200 \text{ N·m}, which is answer A. The wrong answers represent common calculation errors. Answer B (1,500 N·m) might result from incorrectly using 3 meters (the x-coordinate) as the moment arm. Answer C (2,000 N·m) could come from using 4 meters (the y-coordinate) as the perpendicular distance. Answer D (2,500 N·m) represents the most tempting trap - using the distance from O to the point (3,4), which is 32+42=5\sqrt{3^2 + 4^2} = 5 meters, then multiplying by 500 N. The key insight is that the perpendicular distance (moment arm) is not necessarily the same as the distance to where the force is applied. When a problem gives you the perpendicular distance explicitly, use it directly rather than trying to calculate it from coordinates. This distinction between "distance to the point of application" and "perpendicular distance to the line of action" is crucial in moment calculations.

Question 18

A moment of 250 lb·ft about point O is to be converted to SI units. Additionally, if this moment is caused by a force applied at a distance of 1.5 meters from point O, what is the magnitude of the force in Newtons?

  1. 226 N (correct answer)
  2. 248 N
  3. 272 N
  4. 294 N
  5. 315 N
Explanation: This problem tests your ability to work with moments in different unit systems and apply the fundamental moment equation. When you encounter unit conversion problems in statics, always convert systematically and then apply the relevant equations. First, convert the moment from lb·ft to N·m. Using the conversion factor: 250 lb\cdotpft×1.356N\cdotpmlb\cdotpft=339 N\cdotpm250 \text{ lb·ft} \times 1.356 \frac{\text{N·m}}{\text{lb·ft}} = 339 \text{ N·m} Now apply the moment equation: M=F×dM = F \times d, where M is the moment, F is the force, and d is the perpendicular distance. Solving for force: F=Md=339 N\cdotpm1.5 m=226 NF = \frac{M}{d} = \frac{339 \text{ N·m}}{1.5 \text{ m}} = 226 \text{ N} This confirms answer A is correct. Answer B (248 N) likely results from using an incorrect conversion factor or rounding errors during calculation. Answer C (272 N) suggests a calculation error, possibly from incorrectly converting the moment or misapplying the distance. Answer D (294 N) appears to stem from a more significant computational mistake, possibly using the wrong conversion entirely or making an algebraic error. Study tip: Always double-check your unit conversions in statics problems—use the standard conversion that 1 lb·ft = 1.356 N·m. When solving moment problems, remember that moments and forces are related through distance, so if you know two of the three quantities (M, F, d), you can always find the third using M=F×dM = F \times d.

Question 19

A rigid body equilibrium problem involves forces measured in different unit systems. The horizontal forces are 2.5 kN rightward and 400 lbf leftward. The vertical forces are 1800 N upward and 350 lbf downward. Using standard engineering sign conventions (+x right, +y up), which statement correctly describes the net force components?

  1. Net horizontal force is 721 N rightward; net vertical force is 243 N downward
  2. Net horizontal force is 721 N rightward; net vertical force is 243 N upward (correct answer)
  3. Net horizontal force is 1779 N leftward; net vertical force is 243 N upward
  4. Net horizontal force is 4279 N rightward; net vertical force is 3357 N upward
Explanation: Converting to Newtons: 400 lbf = 400 × 4.448 = 1779 N, 350 lbf = 350 × 4.448 = 1557 N. Horizontal forces: +2500 N (rightward) - 1779 N (leftward) = +721 N rightward. Vertical forces: +1800 N (upward) - 1557 N (downward) = +243 N upward. Choice A has wrong direction for vertical component. Choice C has wrong direction for horizontal component. Choice D fails to account for opposing directions and incorrectly adds all magnitudes as positive.