2D Rigid-Body Equilibrium - Statics and Dynamics

1. If the net force on the rigid body must be zero (force equilibrium satisfied), what must $\mathbf{F}_3$ be, and what moment does $\mathbf{F}_3$ alone produce about the origin?
   Answer: $\mathbf{F}_3 = (0\hat{i} + 2\hat{j})\text{ kN}$, and it produces a moment of $+2\text{ kN}\cdot\text{m}$ (counterclockwise) about the origin, computed as $r_x F_{3y} - r_y F_{3x} = 1(2) - 1(0) = +2\text{ kN}\cdot\text{m}$. When a rigid body is in force equilibrium, the vector sum of all forces must equal zero. This gives you a straightforward way to find any unknown force, and it's a separate calculation from finding the moment that force produces. Start by summing the known forces: $\mathbf{F}_1 + \mathbf{F}_2 = (3-3)\hat{i} + (-4+2)\hat{j} = (0\hat{i} - 2\hat{j})\text{ kN}$. For equilibrium, $\mathbf{F}_3$ must cancel this resultant, so $\mathbf{F}_3 = (0\hat{i} + 2\hat{j})\text{ kN}$. Once you have $\mathbf{F}_3$, its moment about the origin is computed using the 2D cross product formula: $M = r_x F_y - r_y F_x$, where $(r_x, r_y) = (1, 1)$ is the point of application. Plugging in: $M = (1)(2) - (1)(0) = +2\text{ kN}\cdot\text{m}$, which is counterclockwise. Choice B captures both results correctly. Choice A finds the right $\mathbf{F}_3$ but claims the moment is zero, inventing a false "self-canceling" rule. No such rule exists — a vertical force applied off the origin always produces a moment unless its line of action passes directly through the reference point. Choice C gets the force backwards ($-2\hat{j}$ instead of $+2\hat{j}$), which would actually double the imbalance rather than restore equilibrium, and its moment calculation follows from that initial error. Choice D finds the correct force and even computes $+2\text{ kN}\cdot\text{m}$ correctly, but then contradicts itself by reassigning the sign to satisfy a separate moment-equilibrium condition — a trap that confuses *finding $\mathbf{F}_3$'s individual moment* with checking overall moment balance. Key tip: Always treat force equilibrium and moment calculations as two independent steps. Finding $\mathbf{F}_3$ uses $\sum \mathbf{F} = 0$; computing its moment is a separate cross product. Never adjust a calculated moment to force a global equilibrium condition — that's a different equation entirely.
2. What is the horizontal component of the pin reaction at A?
   Answer: $A_x = 80\text{ N}$ directed to the left, because the roller at C provides no horizontal resistance, so applying $\sum F_x = 0$ to the whole frame requires the pin at A to supply all $80\text{ N}$ of horizontal resistance. When analyzing rigid frames, your first instinct should be to apply global equilibrium to the entire structure treated as a single free body. Identify every external reaction — only then does the internal force distribution matter. Here, the frame's external supports are the pin at A (which can provide both horizontal and vertical reactions) and the vertical roller at C (which provides only a vertical reaction — no horizontal component by definition). The single external horizontal force is $H = 80\text{ N}$ applied to the right. Applying global horizontal equilibrium: $\sum F_x = 0: \quad A_x + H = 0 \implies A_x = -80\text{ N}$ The negative sign means $A_x$ acts to the left. Since the roller at C cannot resist any horizontal load, the pin at A must single-handedly balance the entire $80\text{ N}$. That confirms C is correct. Choice A reaches the right magnitude but the wrong direction — claiming A "pushes in the same direction as H" violates equilibrium entirely; the reactions must oppose applied loads, not reinforce them. Choice B applies a false symmetry argument. Symmetry in load position might affect vertical reactions, but horizontal equilibrium isn't about sharing — it's about which supports are even capable of providing horizontal resistance. The roller at C cannot, so A carries all of it. Choice D misunderstands structural continuity. A rigid frame transmits forces in all directions through its joints; an internal connection at B passes both horizontal and vertical components to the column, so horizontal equilibrium of the column absolutely requires a horizontal reaction at A. Study tip: Always catalog support types first — pins resist forces in any direction, rollers resist in only one. Then apply $\sum F_x = 0$ globally before analyzing individual members.
3. For $w_0 = 6\text{ kN/m}$, what are the vertical reactions at A and B, and what is the correct reasoning for their values?
   Answer: $R_A = R_B = 12\text{ kN}$, because the symmetric tent distribution has its resultant at the midpoint of the beam, and by symmetry both reactions must be equal and each carries half the total load of $24\text{ kN}$. Whenever you see a distributed load problem on a simply supported beam, your first instinct should be to find the resultant force and its location, then apply equilibrium. For a symmetric load distribution, symmetry alone can do most of the work. Here, the tent-shaped load is perfectly symmetric about the midpoint at $x = 4\text{ m}$. The resultant of the entire load equals the total area under the distribution: two triangles, each with base $4\text{ m}$ and height $w_0 = 6\text{ kN/m}$, giving $P_{total} = 2 \times \frac{1}{2}(4)(6) = 24\text{ kN}$. Because the load is symmetric, the resultant acts exactly at $x = 4\text{ m}$ — the midpoint of the 8 m beam. Summing moments about either support confirms $R_A = R_B = 12\text{ kN}$, which is answer A, and the reasoning is sound: symmetry places the resultant at mid-span, so each support carries exactly half. Answer B is a classic trap — it correctly locates the centroid of the left triangle alone at $x = \frac{8}{3}\text{ m}$, but ignores the right triangle entirely. You must account for the combined resultant of the full load, not just one piece of it. Answer C arrives at the right numbers but for a completely wrong reason. The roller's inability to resist horizontal forces has no bearing on how vertical reactions split; that split depends solely on moment equilibrium. Answer D introduces a fictional concept — support rigidity does not appear in static equilibrium. Statics assumes rigid supports; relative stiffness only matters in statically indeterminate (structural analysis) problems. Study tip: Always combine all load components into one resultant before taking moments — splitting a symmetric load and analyzing only half is a common and costly mistake.
4. Which of the following statements correctly characterizes the reactions that link AB and link CD can each exert on the plate, and why?
   Answer: Each link is a two-force member and can only exert a force along its own axis; link AB exerts a force along its $45°$ direction and link CD exerts a purely horizontal force. Together with the applied $F$, these three concurrent forces must satisfy $\sum F_x = 0$ and $\sum F_y = 0$ for equilibrium of the plate. Whenever you see links (or bars) that are pin-connected at both ends with no load applied between the pins, recognize them immediately as two-force members. A two-force member can only exert a force directed along the line connecting its two pins — no moment, no transverse force, just a single axial force (tension or compression). That principle is the key to this problem. Link AB is pinned at A and B with nothing in between, so it can only push or pull along its 45° axis. Link CD is pinned at C and D and oriented horizontally, so it can only exert a purely horizontal force. Both forces pass through the plate at known directions, giving you two unknowns (the magnitude of each link force). With the applied $F = 100\text{ N}$ at 30° below horizontal, you have three concurrent forces, and equilibrium requires $\sum F_x = 0$ and $\sum F_y = 0$ — exactly two equations for two unknowns. The system is statically determinate. D is correct. Choice A is wrong because being pin-connected to a rigid plate does not allow a link to develop a transverse reaction — that would require the link itself to resist bending, which a two-force member (pin-pin, no intermediate load) cannot do. Choice B contains a true statement about rigid-body equilibrium having a moment equation, but it incorrectly implies this creates an over-constraint problem. When forces are concurrent, $\sum M = 0$ about the concurrent point is automatically satisfied and adds no new information. Choice C confuses pin-connected links with fixed-support members. Pins carry no moment, so stiffness is irrelevant — the reactions are statically determinate, not indeterminate. Study tip: Always check first whether a member is a two-force member (pin-pin, no intermediate loads). If it is, the reaction direction is known immediately, which dramatically simplifies your free-body diagram.
5. A student claims that the equilibrium equations alone (without compatibility) yield a unique solution for all three wire tensions. Which statement best identifies the error in this claim and the correct approach?
   Answer: The claim is incorrect: with three unknowns ($T_1, T_2, T_3$) and only two independent equilibrium equations ($\sum F_y = 0$ and $\sum M = 0$), the system is indeterminate, and a compatibility condition based on the bar's deformation geometry must supplement equilibrium to solve for all three tensions uniquely. Whenever you encounter a problem with multiple unknown support reactions, your first instinct should be to count unknowns versus independent equilibrium equations — this tells you immediately whether the system is statically determinate or indeterminate. Here, you have three unknown wire tensions: $T_1$, $T_2$, and $T_3$. For a rigid bar in a plane with only vertical forces, you get exactly two independent equilibrium equations: $\sum F_y = 0$ and $\sum M = 0$. Two equations, three unknowns — the system is statically indeterminate to the first degree. To solve it, you must introduce a compatibility condition: since the bar is rigid, its deformed shape is a straight line (it can only translate and rotate). This means the elongations of the three wires are geometrically constrained, giving you a third equation that, combined with each wire's force-deformation relationship ($\delta = TL/AE$), closes the system. Answer A correctly captures this entire reasoning chain. Answer B is wrong because horizontal translation is irrelevant — all forces and wires are vertical, so $\sum F_x = 0$ is trivially satisfied and introduces no new unknowns. The system is indeterminate to the first degree, not the second. Answer C is a tempting trap: you might assume the geometry forces wire 3 slack, but that conclusion itself requires a compatibility analysis to verify. You cannot assume slackness from load position alone without checking deformations. Answer D confuses flexibility of members with the need for plastic analysis. Wires being tension-only members doesn't invalidate equilibrium equations — plastic (limit) analysis applies to ductile structures at collapse, not to elastic indeterminate systems. Study tip: Always count unknowns minus equilibrium equations first. If the difference is n, you need exactly n compatibility equations — one for each degree of indeterminacy.

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