Concurrent Force Systems - Statics and Dynamics

1. Which of the following correctly gives the tension in Cable A ($T_A$) and Cable B ($T_B$)?
   Answer: $T_A = \frac{W \cos50°}{\sin70°}$ and $T_B = \frac{W \cos20°}{\sin70°}$, where $W = 1962 \text{ N}$, because applying Lami's theorem to the three concurrent forces at the junction yields these expressions directly. When a point is held in equilibrium by three concurrent forces — two cable tensions and a weight — your first instinct should be to write equilibrium equations by resolving forces horizontally and vertically. At the junction, Cable A pulls up-left at 20° and Cable B pulls up-right at 50°, while weight $W = 200 \times 9.81 = 1962 \text{ N}$ pulls straight down. The two equilibrium equations are: Horizontal: $T_A \cos20° = T_B \cos50°$ Vertical: $T_A \sin20° + T_B \sin50° = W$ Substituting the horizontal equation into the vertical one and applying the sine addition identity $\sin(20°+50°) = \sin70°$, you arrive at: $T_A = \frac{W\cos50°}{\sin70°}, \quad T_B = \frac{W\cos20°}{\sin70°}$ This confirms A is correct — and it's also exactly what Lami's theorem produces when you apply it to the three concurrent forces, so both methods agree. B is wrong because it swaps cosines for sines. The horizontal balance equation uses cosine projections, not sine. Plugging in sine values satisfies neither equilibrium condition. C is wrong on two counts: the tensions are not equal (the angles differ), and the expression $\sin20° + \sin50°$ doesn't arise from the correct simultaneous equations — it ignores the horizontal constraint entirely. D is wrong because it treats each cable as if it independently supports the full weight projected onto its own direction, which ignores that both cables act simultaneously and must be solved as a coupled system. Study tip: Always write both equilibrium equations (horizontal and vertical) before solving. A common trap is assuming you can find each tension independently — you almost always need simultaneous equations when multiple cables support a single load.
2. If the system is in equilibrium, what is the ratio $P : R$?
   Answer: $P : R = 1 : 1$, because for any symmetric three-force concurrent system with one force at 120° and another at 240°, all three magnitudes must be equal to satisfy both horizontal and vertical equilibrium simultaneously. When three concurrent coplanar forces are in equilibrium, their vector sum must equal zero — meaning both the x-components and y-components must independently sum to zero. That's your entry point here. Notice the geometry: $P$ acts at $0°$, $Q = 600\text{ N}$ at $120°$, and $R$ at $240°$. These three angles are evenly spaced by exactly $120°$ around the origin. This symmetry is the key insight. Writing the equilibrium equations: $\sum F_x = P + Q\cos120° + R\cos240° = P - \frac{Q}{2} - \frac{R}{2} = 0$ $\sum F_y = Q\sin120° + R\sin240° = \frac{\sqrt{3}}{2}Q - \frac{\sqrt{3}}{2}R = 0$ The y-equation immediately gives $Q = R$, so $R = 600\text{ N}$. Substituting into the x-equation: $P = \frac{Q}{2} + \frac{R}{2} = \frac{600}{2} + \frac{600}{2} = 600\text{ N}$. Therefore $P = Q = R$, confirming $P : R = 1 : 1$, answer D. Choice A incorrectly claims $P : R = 1 : 2$ based on vague geometric reasoning without actually solving the equilibrium equations. Choice B introduces a $\sqrt{3}$ ratio that would arise if the angles were asymmetric — it confuses sine and cosine projections. Choice C reverses the correct ratio and mischaracterizes how Q interacts with P and R in the x-direction. Study tip: Whenever three concurrent forces are separated by equal $120°$ intervals, they must all be equal in magnitude — this is a classic result worth memorizing. Always check for angular symmetry before grinding through algebra; it can save you significant time on exam day.
3. After finding $\mathbf{F}_4$ for equilibrium, a fifth force $\mathbf{F}_5 = (1\hat{i} - 3\hat{j}) \text{ kN}$ is added to the system. What single additional force $\mathbf{F}_6$ must be applied at P to restore equilibrium, and what is its magnitude?
   Answer: $\mathbf{F}_6 = (-1\hat{i} + 3\hat{j}) \text{ kN}$ with magnitude $\sqrt{10} \approx 3.16 \text{ kN}$, because restoring equilibrium after adding $\mathbf{F}_5$ requires only canceling $\mathbf{F}_5$ itself, since the original four forces were already balanced. Whenever you see a concurrent force equilibrium problem, anchor yourself to one core principle: a system in equilibrium has a zero resultant. Any force added to a balanced system disturbs equilibrium by exactly its own vector value — and restoring equilibrium requires canceling precisely that disturbance. Here's the key insight: once $\mathbf{F}_4$ is chosen so that $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + \mathbf{F}_4 = \mathbf{0}$, those four forces are balanced. Adding $\mathbf{F}_5 = (1\hat{i} - 3\hat{j})$ kN creates a net resultant of exactly $\mathbf{F}_5$ at point P. To restore equilibrium, you need one force that cancels this resultant — the equilibrant of $\mathbf{F}_5$, which is simply its negative: $\mathbf{F}_6 = (-1\hat{i} + 3\hat{j})$ kN. Its magnitude is $\sqrt{(-1)^2 + (3)^2} = \sqrt{10} \approx 3.16$ kN. This confirms A is correct. B is wrong because it reverses only the x-component while keeping the y-component positive — $(1\hat{i} + 3\hat{j})$ does not cancel $\mathbf{F}_5$; it would actually increase the y-imbalance. C describes a force in the same direction as $\mathbf{F}_5$, which would amplify the imbalance rather than cancel it. The concept of a "couple neutralizing" a single concurrent force is also physically incorrect here. D introduces unnecessary complexity by claiming $\mathbf{F}_4$ must be "recomputed." Since the original four forces remain unchanged and were already balanced, $\mathbf{F}_4$ needs no revision. Study tip: When equilibrium is broken by adding one force, the restoring force is always the negative of that added force — no need to re-examine the previously balanced forces.
4. Setting up the two equilibrium equations for joint A, which system of equations is correct?
   Answer: $500\cos35° + F_{AB}\cos70° + F_{AC}\cos160° + N\cos105° = 0$ and $500\sin35° + F_{AB}\sin70° + F_{AC}\sin160° + N\sin105° = 0$, treating all forces with their actual direction angles from the positive x-axis so that all sign information is carried by the trigonometric functions. When analyzing equilibrium at a pin joint, your most reliable strategy is to express every force using its actual direction angle measured from the positive x-axis, then let the trigonometric functions carry all sign information automatically. This avoids the common mistake of manually flipping signs mid-equation. For joint A, you have four forces with direction angles: $P$ at $35°$, $F_{AB}$ at $70°$, $F_{AC}$ at $160°$ (the actual direction the force acts on the pin, already accounting for compression), and $N$ at $105°$. Applying $\sum F_x = 0$ and $\sum F_y = 0$ consistently gives the equations in B — every term uses $\cos\theta$ or $\sin\theta$ with the true angle, so $\cos160°$ and $\sin160°$ are both negative and positive respectively, naturally encoding the correct sign. This is the correct approach. A is wrong because it uses the member's geometric angle ($-20°$) rather than the actual force direction on the pin ($160°$). The compression was already handled when you identified the force direction as $160°$; using $-20°$ and subtracting double-counts the sign reversal. C is wrong for a subtler reason: it correctly uses $160°$ for the x-component but then incorrectly negates only the y-component of $F_{AC}$. There is no physical justification for treating x and y components differently — compression affects the entire force vector, not just one component. D is wrong because it ignores the surface inclination. A roller reaction is perpendicular to its contact surface, not necessarily to the horizontal. A surface inclined at $15°$ from horizontal produces a normal at $90° + 15° = 105°$, not $90°$. Study tip: Always convert every force to its actual pin-force direction angle first, then write both equilibrium equations mechanically using $\cos\theta$ and $\sin\theta$. Never manually insert extra minus signs — let the angles do the work.
5. A student claims: *'If $\alpha$ is doubled while $W$ and $\beta$ remain constant, then $T_1$ (tension in Rope 1) will also approximately double for small angles.'* Which response most precisely evaluates this claim?
   Answer: The claim is incorrect, because even for small $\alpha$, the equilibrium equations couple $T_1$ and $T_2$ through both $\alpha$ and $\beta$; doubling $\alpha$ changes both $T_1$ and $T_2$, and the net effect on $T_1$ alone is not a simple doubling due to the coupled system of equations. When a system of forces is in equilibrium, every unknown tension is determined simultaneously by the full set of equilibrium equations — not by any single equation alone. That interdependence is the heart of this question. For the ring, horizontal and vertical equilibrium give: $T_1 \sin\alpha = T_2 \sin\beta \quad \text{and} \quad T_1 \cos\alpha + T_2 \cos\beta = W$ To find $T_1$, you must solve this coupled system. Doing so yields: $T_1 = \frac{W \sin\beta}{\sin(\alpha + \beta)}$ Now apply the small-angle approximation ($\sin\theta \approx \theta$, $\cos\theta \approx 1$): $T_1 \approx \frac{W\beta}{\alpha + \beta}$ Doubling $\alpha$ gives $T_1 \approx \frac{W\beta}{2\alpha + \beta}$, which is smaller than the original — not double. The result depends on both $\alpha$ and $\beta$ together, so C is correct: the coupling through $\beta$ prevents any simple proportionality. Answer A is tempting because the small-angle linearization step is valid, but it stops too early. Linearizing doesn't make $T_1$ proportional to $\alpha$ alone — the denominator $(\alpha + \beta)$ still contains $\alpha$, so the relationship is not linear in $\alpha$. Answer B incorrectly claims the relationship is "always nonlinear regardless of angle magnitude," which contradicts the well-established small-angle approximation. Answer D correctly notes $\cos\alpha \approx 1$ but then wrongly concludes $T_1 \approx W - T_2$, ignoring that $T_2$ itself changes when $\alpha$ changes — the same coupling error, in reverse. Study tip: Whenever you have a coupled system of equations, always solve for the target variable completely before drawing conclusions about proportionality. Partial substitution is the most common trap in equilibrium problems.

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