Force System Resultants - Statics and Dynamics

1. What is the pitch $p$ of the equivalent wrench, and along what direction does the wrench axis point?
   Answer: $p = 3.0$ m, and the wrench axis is parallel to $4\hat{i}+3\hat{k}$ (i.e., along $\hat{u} = 0.8\hat{i}+0.6\hat{k}$), because $\mathbf{M}_C$ is entirely parallel to $\mathbf{F}$ and the pitch is computed as $p = (\mathbf{M}_C \cdot \mathbf{F})/|\mathbf{F}|^2$. 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 $p = \frac{\mathbf{M}_C \cdot \mathbf{F}}{|\mathbf{F}|^2}$. Here, $\mathbf{F} = 4\hat{i} + 3\hat{k}$ gives $|\mathbf{F}|^2 = 16 + 9 = 25$ kN². The dot product $\mathbf{M}_C \cdot \mathbf{F} = (12)(4) + (9)(3) = 48 + 27 = 75$ kN·m·kN. So $p = 75/25 = 3.0$ m. Notice that $\mathbf{M}_C = 12\hat{i} + 9\hat{k} = 3(4\hat{i} + 3\hat{k})$, meaning $\mathbf{M}_C$ is exactly parallel to $\mathbf{F}$ — no perpendicular component exists to shift the axis. The wrench axis therefore points along $\hat{u} = 0.8\hat{i} + 0.6\hat{k}$, confirming B as correct. Choice A makes a partial calculation error: it uses only the $\hat{i}$ component (12/5 = 2.4) rather than computing the full dot product, ignoring the $\hat{k}$ contribution entirely. Choice C gets the pitch right but incorrectly takes the axis as $\hat{i} + \hat{k}$ — a "naive average" of nonzero directions that has no physical basis; the axis must align with $\mathbf{F}$, not an arbitrary combination. Choice D misreads the parallel condition: $\mathbf{M}_C \parallel \mathbf{F}$ 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 $\mathbf{M}_C \cdot \mathbf{F}$, and remember that $\mathbf{M}_C \parallel \mathbf{F}$ signals a pure wrench with nonzero pitch, not a degenerate case.
2. What is the resultant of this concurrent force system, and what is the simplest equivalent representation?
   Answer: The resultant is $\mathbf{F}_R = \mathbf{0}$ with no couple moment, meaning the system is in equilibrium at $P$ and the simplest representation is simply the null force vector. 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: $\Sigma F_x = 10 + 0 + 0 + (-10) = 0 \text{ kN}$ $\Sigma F_y = 0 + 15 + 0 + (-15) = 0 \text{ kN}$ $\Sigma F_z = 0 + 0 + (-12) + 12 = 0 \text{ kN}$ The resultant is $\mathbf{F}_R = \mathbf{0}$. Since all forces act through the same point $P$, 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 $\mathbf{F}_4$. Notice that $\mathbf{F}_4 = -10\hat{i} - 15\hat{j} + 12\hat{k}$ exactly cancels $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 10\hat{i} + 15\hat{j} - 12\hat{k}$ 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.
3. When this entire load system is reduced to a single resultant force–couple at $x = 0$, what is the resultant force $F_R$ and resultant couple moment $M_R$ (taking counterclockwise as positive)?
   Answer: $F_R = 36$ kN (downward) and $M_R = -174$ kN·m (clockwise), computed by placing the distributed load resultant at $x = 4$ m from the origin and adding the concentrated couple. 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: $F_R = \frac{1}{2}(12)(6) = 36 \text{ kN (downward)}$. 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 $x = \frac{2}{3}(6) = 4$ m from the origin (where the load is zero). Many students mistakenly place it at the midpoint ($x = 3$ m) or at the load's starting point. Now compute the moment about $x = 0$: the distributed load contributes $-(36)(4) = -144 \text{ kN}\cdot\text{m}$ (clockwise, negative), and the concentrated couple adds $+30 \text{ kN}\cdot\text{m}$. Total: $M_R = -144 + 30 = -174 \text{ kN}\cdot\text{m}$. This confirms choice A is correct. Choice B places the resultant at $x = 3$ m (the midpoint, correct for a uniform load), giving $-(36)(3) + 30 = -78$ kN·m — a classic trap. Choice C incorrectly places the resultant at $x = 2$ m, which happens to be where $M_0$ is applied but has no geometric basis for the triangular load. Study tip: Always remember the $\frac{1}{3}$ rule — a triangular load's resultant sits at one-third of the length measured from the larger end, not the midpoint.
4. Is the engineer's claim correct, and what is the key reason?
   Answer: The claim is incorrect because the couple moment has a component parallel to the force ($\mathbf{M} \cdot \hat{u}_F \neq 0$), 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. 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: $\hat{u}_F = \frac{6\hat{i}+8\hat{j}}{10} = 0.6\hat{i}+0.8\hat{j}$, then evaluate $\mathbf{M}\cdot\hat{u}_F = (24)(0.6)+(32)(0.8)+(60)(0) = 14.4+25.6 = 40 \neq 0$. 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 $\hat{k}$ observation is a red herring. Answer D is nonsense — magnitude comparison between M and F has no bearing on reducibility. Study tip: Always compute $\mathbf{M}\cdot\hat{u}_F$ first. If it's nonzero, the system is a true screw — a single resultant force is impossible.
5. Which of the following statements about further reduction of this system is correct?
   Answer: The system cannot be reduced further; it is already in its simplest form as a pure couple, and the couple moment $M_O = 50$ N·m is the same about every point in the plane. 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 ($\mathbf{F}_R = 0$), leaving only a net couple moment of $M_O = 50$ 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 $M_O = 50$ N·m counterclockwise about point $O$ 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 $\mathbf{F}_R \neq 0$. 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 $\mathbf{F}_R = 0$ but $M \neq 0$, the system is a pure couple and cannot be reduced further. This is a classic exam trap.

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