Statics and Dynamics Quiz: Force System Resultants
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Force System ResultantsQuestion 1 of 5

A space truss joint is subjected to four concurrent forces: F1=10i^+0j^+0k^\mathbf{F}_1 = 10\hat{i} + 0\hat{j} + 0\hat{k} kN, F2=0i^+15j^+0k^\mathbf{F}_2 = 0\hat{i} + 15\hat{j} + 0\hat{k} kN, F3=0i^+0j^−12k^\mathbf{F}_3 = 0\hat{i} + 0\hat{j} - 12\hat{k} kN, and F4=−10i^−15j^+12k^\mathbf{F}_4 = -10\hat{i} - 15\hat{j} + 12\hat{k} kN. All four forces act through the same point PP.

What is the resultant of this concurrent force system, and what is the simplest equivalent representation?

The resultant is FR=10i^+15j^−12k^\mathbf{F}_R = 10\hat{i}+15\hat{j}-12\hat{k} kN, because F4\mathbf{F}_4 acts in the opposite direction and cancels only the first three force components in magnitude.
The resultant is FR=0\mathbf{F}_R = \mathbf{0} with a nonzero couple moment at PP, because concurrent forces that cancel each other still generate rotational effects about their common point.
The resultant is FR=0\mathbf{F}_R = \mathbf{0} with no couple moment, meaning the system is in equilibrium at PP and the simplest representation is simply the null force vector.
The resultant is FR=0\mathbf{F}_R = \mathbf{0}, but it must be expressed as a pure couple because any system of four or more forces always produces a residual couple moment when reduced.
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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Force System Resultants

Practice Force System Resultants in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A space truss joint is subjected to four concurrent forces: F1=10i^+0j^+0k^\mathbf{F}_1 = 10\hat{i} + 0\hat{j} + 0\hat{k} kN, F2=0i^+15j^+0k^\mathbf{F}_2 = 0\hat{i} + 15\hat{j} + 0\hat{k} kN, F3=0i^+0j^−12k^\mathbf{F}_3 = 0\hat{i} + 0\hat{j} - 12\hat{k} kN, and F4=−10i^−15j^+12k^\mathbf{F}_4 = -10\hat{i} - 15\hat{j} + 12\hat{k} kN. All four forces act through the same point PP.

What is the resultant of this concurrent force system, and what is the simplest equivalent representation?

  1. The resultant is FR=10i^+15j^−12k^\mathbf{F}_R = 10\hat{i}+15\hat{j}-12\hat{k} kN, because F4\mathbf{F}_4 acts in the opposite direction and cancels only the first three force components in magnitude.
  2. The resultant is FR=0\mathbf{F}_R = \mathbf{0} with a nonzero couple moment at PP, because concurrent forces that cancel each other still generate rotational effects about their common point.
  3. The resultant is FR=0\mathbf{F}_R = \mathbf{0} with no couple moment, meaning the system is in equilibrium at PP and the simplest representation is simply the null force vector. (correct answer)
  4. The resultant is FR=0\mathbf{F}_R = \mathbf{0}, but it must be expressed as a pure couple because any system of four or more forces always produces a residual couple moment when reduced.
Explanation: When analyzing concurrent force systems, your first move should always be vector addition — sum all force components along each axis separately, then interpret the result physically. Adding all four forces component by component: ΣFx=10+0+0+(−10)=0 kN\Sigma F_x = 10 + 0 + 0 + (-10) = 0 \text{ kN} ΣFy=0+15+0+(−15)=0 kN\Sigma F_y = 0 + 15 + 0 + (-15) = 0 \text{ kN} ΣFz=0+0+(−12)+12=0 kN\Sigma F_z = 0 + 0 + (-12) + 12 = 0 \text{ kN} The resultant is FR=0\mathbf{F}_R = \mathbf{0}. Since all forces act through the same point PP, there are no moment arms — no force is offset from any other — so no couple moment is generated either. The system is simply in equilibrium, making C correct. A is wrong because it misreads F4\mathbf{F}_4. Notice that F4=−10i^−15j^+12k^\mathbf{F}_4 = -10\hat{i} - 15\hat{j} + 12\hat{k} exactly cancels F1+F2+F3=10i^+15j^−12k^\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 10\hat{i} + 15\hat{j} - 12\hat{k} component-for-component. There is nothing left over. B introduces a false concept. Concurrent forces — by definition — all pass through one common point, so their moment about that point is zero regardless of their magnitudes. No couple moment arises. D applies a rule that simply doesn't exist. There is no principle stating that four or more forces always produce a residual couple. A couple only appears when forces are non-concurrent and offset from each other. A useful habit: when forces are labeled as concurrent, mentally flag that rotation effects at the common point are automatically zero — your only job is the vector sum.

Question 2

A distributed line load acts along a horizontal beam of length L=6L = 6 m. The load intensity varies linearly from w=0w = 0 at x=0x = 0 to w=12w = 12 kN/m at x=6x = 6 m (a triangular distribution). The beam also carries a concentrated couple moment M0=30M_0 = 30 kN·m (counterclockwise) applied at x=2x = 2 m.

When this entire load system is reduced to a single resultant force–couple at x=0x = 0, what is the resultant force FRF_R and resultant couple moment MRM_R (taking counterclockwise as positive)?

  1. FR=36F_R = 36 kN (downward) and MR=−174M_R = -174 kN·m (clockwise), computed by placing the distributed load resultant at x=4x = 4 m from the origin and adding the concentrated couple. (correct answer)
  2. FR=36F_R = 36 kN (downward) and MR=−144M_R = -144 kN·m (clockwise), computed by placing the distributed load resultant at x=3x = 3 m (midpoint) from the origin and adding the concentrated couple.
  3. FR=36F_R = 36 kN (downward) and MR=−114M_R = -114 kN·m (clockwise), computed by placing the distributed load resultant at x=2x = 2 m from the origin and adding the concentrated couple.
  4. FR=72F_R = 72 kN (downward) and MR=−174M_R = -174 kN·m (clockwise), computed by integrating the full trapezoidal area under the load curve from x=0x = 0 to x=6x = 6 m.
Explanation: When reducing a distributed load system to a resultant force-couple at a reference point, you need two things: the magnitude of the resultant force and the moment of the entire load system about that reference point. For a triangular distributed load, the resultant force equals the area of the triangle: FR=12(12)(6)=36 kN (downward)F_R = \frac{1}{2}(12)(6) = 36 \text{ kN (downward)}. This eliminates choice D immediately — doubling to 72 kN would only be correct for a rectangular (uniform) distribution of 12 kN/m, not triangular. The critical step is locating where this resultant acts. A triangular load's resultant acts at one-third of the base from the larger end — that is, at x=23(6)=4x = \frac{2}{3}(6) = 4 m from the origin (where the load is zero). Many students mistakenly place it at the midpoint (x=3x = 3 m) or at the load's starting point. Now compute the moment about x=0x = 0: the distributed load contributes −(36)(4)=−144 kN\cdotpm-(36)(4) = -144 \text{ kN·m} (clockwise, negative), and the concentrated couple adds +30 kN\cdotpm+30 \text{ kN·m}. Total: MR=−144+30=−174 kN\cdotpmM_R = -144 + 30 = -174 \text{ kN·m}. This confirms choice A is correct. Choice B places the resultant at x=3x = 3 m (the midpoint, correct for a uniform load), giving −(36)(3)+30=−78-(36)(3) + 30 = -78 kN·m — a classic trap. Choice C incorrectly places the resultant at x=2x = 2 m, which happens to be where M0M_0 is applied but has no geometric basis for the triangular load. Study tip: Always remember the 13\frac{1}{3} rule — a triangular load's resultant sits at one-third of the length measured from the larger end, not the midpoint.

Question 3

A wrench (force–couple system) consists of a force F=6i^+8j^+0k^\mathbf{F} = 6\hat{i} + 8\hat{j} + 0\hat{k} kN and a couple moment M=24i^+32j^+60k^\mathbf{M} = 24\hat{i} + 32\hat{j} + 60\hat{k} kN·m. An engineer claims this wrench can be reduced to a single resultant force (a pure force with no couple) by moving the line of action to an appropriate point.

Is the engineer's claim correct, and what is the key reason?

  1. The claim is incorrect because the couple moment has a component parallel to the force (M⋅u^F≠0\mathbf{M} \cdot \hat{u}_F \neq 0), meaning the system is a true wrench (screw) and cannot be reduced to a single force without a remaining couple along the force's axis. (correct answer)
  2. The claim is correct because the resultant force is nonzero, and any force–couple system with a nonzero resultant force can always be shifted to eliminate the couple moment entirely by choosing the right point.
  3. The claim is incorrect because the resultant force lies entirely in the xyxy-plane, making it impossible to shift the line of action to cancel a couple moment with a k^\hat{k} component regardless of the chosen point.
  4. The claim is correct because the magnitude of the couple moment (≈72.1\approx 72.1 kN·m) exceeds the magnitude of the force (10 kN), allowing complete cancellation through a superposition of shifted force components.
Explanation: When you encounter a force–couple system and are asked whether it can reduce to a single resultant force, the first thing you should check is whether the couple moment has any component parallel to the force. This is the defining test for a true mechanical wrench (screw system). Here's why: when you shift a force's line of action to a new point, you can only cancel the component of M that is perpendicular to F. The component of M parallel to F — called the pitch component — is invariant and cannot be eliminated by any shift of the line of action. To check this, compute the unit vector along the force: u^F=6i^+8j^10=0.6i^+0.8j^\hat{u}_F = \frac{6\hat{i}+8\hat{j}}{10} = 0.6\hat{i}+0.8\hat{j}, then evaluate M⋅u^F=(24)(0.6)+(32)(0.8)+(60)(0)=14.4+25.6=40≠0\mathbf{M}\cdot\hat{u}_F = (24)(0.6)+(32)(0.8)+(60)(0) = 14.4+25.6 = 40 \neq 0. Because this dot product is nonzero, the system is a true wrench and cannot be reduced to a single force. Answer A is correct. Answer B is tempting but wrong — it confuses a necessary condition (nonzero resultant force) with a sufficient one. A nonzero force is required, but it doesn't guarantee full couple cancellation if a parallel moment component exists. Answer C incorrectly focuses on the geometry of the force's plane rather than the parallel-moment criterion; the k^\hat{k} observation is a red herring. Answer D is nonsense — magnitude comparison between M and F has no bearing on reducibility. Study tip: Always compute M⋅u^F\mathbf{M}\cdot\hat{u}_F first. If it's nonzero, the system is a true screw — a single resultant force is impossible.

Question 4

Four coplanar forces act on a body in the xyxy-plane. After computing the resultant force, an engineer finds FR=0i^+0j^\mathbf{F}_R = 0\hat{i} + 0\hat{j} N but the resultant couple moment about an arbitrary point OO is MO=50M_O = 50 N·m (counterclockwise). The engineer considers reducing this further.

Which of the following statements about further reduction of this system is correct?

  1. The system cannot be reduced further only if point OO is at the centroid of the force application points; otherwise, a different reference point could yield a zero couple moment.
  2. The system can be reduced to a single force by relocating the resultant to a new point where the couple moment vanishes, since any system with zero net force can be simplified this way.
  3. The system can be reduced to a single force, but only if the engineer first decomposes the couple into two equal-and-opposite forces and then recombines them with the zero resultant.
  4. The system cannot be reduced further; it is already in its simplest form as a pure couple, and the couple moment MO=50M_O = 50 N·m is the same about every point in the plane. (correct answer)
Explanation: When a system of forces is reduced to its simplest equivalent, the key question is: what combination of resultant force and couple moment remains? Here, the resultant force is already zero (FR=0\mathbf{F}_R = 0), leaving only a net couple moment of MO=50M_O = 50 N·m. This is the definition of a pure couple — and understanding its properties is exactly what this question tests. A pure couple has a remarkable property: its moment is independent of the reference point. Unlike the moment of a force, which changes as you shift the reference location, a couple moment remains constant regardless of where you compute it. So MO=50M_O = 50 N·m counterclockwise about point OO is equally 50 N·m counterclockwise about any point in the plane. This system is already in its simplest possible form — a pure couple — making D correct. A is wrong because it misunderstands couple moments entirely. The invariance of a couple's moment has nothing to do with centroids or reference-point selection — it's a fundamental geometric property, not something you can engineer away by choosing a clever point. B is wrong because relocating a resultant force to eliminate a couple moment only works when FR≠0\mathbf{F}_R \neq 0. When the net force is zero, there is no force to relocate, so you cannot "absorb" the couple moment into a shifted force position. C is wrong because decomposing the couple into two equal-and-opposite forces doesn't change the physics. Recombining them with a zero resultant simply reconstructs the same pure couple — no simplification is achieved. Study tip: Memorize this rule — when FR=0\mathbf{F}_R = 0 but M≠0M \neq 0, the system is a pure couple and cannot be reduced further. This is a classic exam trap.

Question 5

A machine component is subjected to the following force system in 3D: a force F=4i^+3k^\mathbf{F} = 4\hat{i} + 3\hat{k} kN at the origin, plus a couple moment MC=12i^+9k^\mathbf{M}_C = 12\hat{i} + 9\hat{k} kN·m. An engineer wishes to determine the wrench (screw) axis of this system.

What is the pitch pp of the equivalent wrench, and along what direction does the wrench axis point?

  1. p=2.4p = 2.4 m, and the wrench axis is parallel to 4i^+3k^4\hat{i}+3\hat{k}, computed by dividing only the i^\hat{i} component of MC\mathbf{M}_C by the magnitude of F\mathbf{F}: p=12/5=2.4p = 12/5 = 2.4 m.
  2. p=3.0p = 3.0 m, and the wrench axis is parallel to 4i^+3k^4\hat{i}+3\hat{k} (i.e., along u^=0.8i^+0.6k^\hat{u} = 0.8\hat{i}+0.6\hat{k}), because MC\mathbf{M}_C is entirely parallel to F\mathbf{F} and the pitch is computed as p=(MC⋅F)/∣F∣2p = (\mathbf{M}_C \cdot \mathbf{F})/|\mathbf{F}|^2. (correct answer)
  3. p=3.0p = 3.0 m, and the wrench axis is parallel to i^+k^\hat{i}+\hat{k} (equal i^\hat{i} and k^\hat{k} components), because the pitch calculation is correct but the axis direction is taken as the average of the nonzero basis directions in F\mathbf{F}.
  4. p=0p = 0 m (zero pitch), and the wrench axis is parallel to 4i^+3k^4\hat{i}+3\hat{k}, because MC\mathbf{M}_C is parallel to F\mathbf{F}, which is incorrectly interpreted as meaning there is no rotational component and the system reduces to a pure force.
Explanation: When a force system reduces to a wrench (screw), you need two things: the pitch and the axis direction. The pitch measures how much couple moment accompanies each unit of force along the axis, and the axis points along the resultant force direction. The pitch formula is p=MC⋅F∣F∣2p = \frac{\mathbf{M}_C \cdot \mathbf{F}}{|\mathbf{F}|^2}. Here, F=4i^+3k^\mathbf{F} = 4\hat{i} + 3\hat{k} gives ∣F∣2=16+9=25|\mathbf{F}|^2 = 16 + 9 = 25 kN². The dot product MC⋅F=(12)(4)+(9)(3)=48+27=75\mathbf{M}_C \cdot \mathbf{F} = (12)(4) + (9)(3) = 48 + 27 = 75 kN·m·kN. So p=75/25=3.0p = 75/25 = 3.0 m. Notice that MC=12i^+9k^=3(4i^+3k^)\mathbf{M}_C = 12\hat{i} + 9\hat{k} = 3(4\hat{i} + 3\hat{k}), meaning MC\mathbf{M}_C is exactly parallel to F\mathbf{F} — no perpendicular component exists to shift the axis. The wrench axis therefore points along u^=0.8i^+0.6k^\hat{u} = 0.8\hat{i} + 0.6\hat{k}, confirming B as correct. Choice A makes a partial calculation error: it uses only the i^\hat{i} component (12/5 = 2.4) rather than computing the full dot product, ignoring the k^\hat{k} contribution entirely. Choice C gets the pitch right but incorrectly takes the axis as i^+k^\hat{i} + \hat{k} — a "naive average" of nonzero directions that has no physical basis; the axis must align with F\mathbf{F}, not an arbitrary combination. Choice D misreads the parallel condition: MC∥F\mathbf{M}_C \parallel \mathbf{F} does not mean zero rotational component — it actually means the wrench is pure (no perpendicular moment), giving maximum pitch, not zero pitch. Your study tip: always compute pitch using the full dot product MC⋅F\mathbf{M}_C \cdot \mathbf{F}, and remember that MC∥F\mathbf{M}_C \parallel \mathbf{F} signals a pure wrench with nonzero pitch, not a degenerate case.