Checking Solutions - Statics and Dynamics

1. A student applies the work-energy theorem to a system and derives the final speed of a block after it slides a distance $d$ down a rough incline of angle $\theta$, starting from rest, as: $v = \sqrt{2gd(\sin\theta + \mu_k\cos\theta)}$ The student performs a unit check and confirms the expression has units of m/s. Which of the following best identifies whether this answer should be accepted or rejected, and why?
   Answer: The answer should be rejected. The friction force opposes motion, so kinetic friction removes energy from the system; the work done by friction is negative. The correct expression is $v = \sqrt{2gd(\sin\theta - \mu_k\cos\theta)}$. The plus sign inside the radical causes the formula to predict a speed greater than the frictionless case, which violates physical reasonableness — friction cannot accelerate the block. Whenever you apply the work-energy theorem on an incline, your first instinct after getting an answer should be a physical reasonableness check, not just a unit check. Ask yourself: does this result make sense compared to a simpler, known case? Here, the frictionless result is $v = \sqrt{2gd\sin\theta}$. Friction opposes motion — it removes mechanical energy from the system, so the block must arrive at the bottom slower than the frictionless case. The work done by kinetic friction is negative: $W_f = -\mu_k mg\cos\theta \cdot d$. Applying the work-energy theorem correctly gives $\frac{1}{2}mv^2 = mgd\sin\theta - \mu_k mgd\cos\theta$, yielding $v = \sqrt{2gd(\sin\theta - \mu_k\cos\theta)}$. The student's plus sign instead produces a speed larger than the frictionless result — friction is effectively accelerating the block, which is physically impossible. C correctly identifies both the sign error and the physical violation, making it the right choice. A is wrong because it mischaracterizes the problem. The formula is incorrect at all angles, not just where $\mu_k\cos\theta > \sin\theta$; the flaw is the wrong sign, period. Pointing to a subset of angles doesn't identify the root error. B is wrong because it treats a passing unit check as sufficient validation. Units confirm dimensional consistency, but they cannot detect sign errors — a formula and its negative can have identical units. D is wrong because physical laws don't become "irrelevant" at steep angles. A sign error is a sign error regardless of which term dominates numerically. Strategy tip: Always benchmark your answer against a simpler limiting case. If removing friction makes the result slower than your derived formula, your signs are wrong — no unit check will catch that.
2. Using limiting cases and physical reasoning to check the reported minimum speed, which of the following conclusions is correct?
   Answer: The reported minimum speed $v_{\min} = 0$ is incorrect. At $t^* = v_0\sin\theta/g$ (the apex), the vertical velocity component is zero but the horizontal component $v_0\cos\theta$ remains nonzero (for $\theta \neq 90^\circ$), so $v_{\min} = v_0\cos\theta > 0$. The student's error is using $t^* = v_0/g$ instead of $t^* = v_0\sin\theta/g$, and conflating zero vertical velocity with zero total speed. Whenever you see a question about projectile motion extrema, train yourself to analyze each velocity component separately before drawing conclusions about total speed. The speed can only equal zero if every component simultaneously equals zero — a much stricter condition than just one component vanishing. At the apex of projectile motion, the vertical component $v_y = v_0\sin\theta - gt$ equals zero, giving the correct apex time $t^* = \frac{v_0\sin\theta}{g}$. However, the horizontal component $v_x = v_0\cos\theta$ is constant throughout the flight (no air resistance means no horizontal acceleration). So the minimum speed is $v_{\min} = \sqrt{(v_0\cos\theta)^2 + 0^2} = v_0\cos\theta$, which is only zero when $\theta = 90^\circ$. This confirms C is correct: the student made two linked errors — using $t^* = v_0/g$ (missing the $\sin\theta$ factor) and then incorrectly concluding that zero vertical velocity implies zero total speed. A is wrong because the condition for zero total speed has nothing to do with whether $\theta \geq 45^\circ$. The horizontal component persists at any angle except $90^\circ$, so $v_{\min} = 0$ fails for all non-vertical launches. B is wrong on two counts: $\sin\theta$ does not cancel when differentiating $v(t)$ — it remains in the apex time formula — and $t^* = v_0/g$ is only correct when $\theta = 90^\circ$. D is wrong because $gt$ carries units of m/s (acceleration × time = velocity), so $(gt)^2$ has units of $\text{m}^2/\text{s}^2$, perfectly consistent with the other squared-velocity terms. The formula is dimensionally sound. Study tip: Always check limiting cases — set $\theta = 90^\circ$ and $\theta = 0^\circ$ to instantly expose errors in projectile motion formulas.
3. A more experienced engineer reviews the solution and argues that the student's three checks are necessary but not sufficient, and identifies a fourth check that could reveal an error the student missed. Which of the following is the most rigorous additional check available without re-solving the entire problem?
   Answer: Apply the method of sections by cutting through member AB and two other members, then write equilibrium equations for the resulting free body. Agreement with 85 kN in tension corroborates the result using an independent equation set; a disagreement reveals a computational error that the three original checks could not detect. Whenever you encounter a question about verifying structural analysis results, ask yourself: does this check use a truly independent method, or does it just repackage the same assumptions? The method of joints produces equilibrium equations at each joint individually. Its weakness is that a computational error at one joint can propagate invisibly through the solution — and the three checks the student performed (unit consistency, sign convention, dimensional geometry) all operate within that same framework. None of them catch arithmetic mistakes or incorrect force directions in the equilibrium equations themselves. The method of sections, choice B, is the correct answer because it constructs an entirely separate free-body diagram by cutting through the truss — including member AB — and writes independent equilibrium equations for that cut section. If this independent calculation also yields $F_{AB} = 85\text{ kN}$ in tension, you have genuine corroboration. If it disagrees, you've detected an error that no internal consistency check could have found. This is what makes it rigorous: it's a mathematically independent verification. Choice A is a material strength check, not a mechanics verification. Whether 85 kN exceeds a cross-section's capacity tells you nothing about whether the equilibrium analysis was performed correctly — it's a design consideration, not a structural analysis check. Choice C is based on a false premise. Member forces in a truss do not sum to zero globally; that statement confuses internal forces with external reactions. This "check" would never reveal a real error. Choice D misapplies Maxwell's reciprocal theorem, which relates displacements under reciprocal loads, not directly member forces under a specific loading case. It cannot straightforwardly verify a single member force value. Study tip: On statics questions asking for "independent verification," always favor methods that use a completely different free-body diagram and equation set — that's the gold standard for catching computational errors.
4. A student solves for the magnitude of the resultant of two concurrent forces $F_1 = 50\text{ N}$ and $F_2 = 120\text{ N}$ acting at an angle $\alpha$ to each other, obtaining $R = 140\text{ N}$. As a check, the student notes that when $\alpha = 0^\circ$ the formula gives $R = 170\text{ N}$, and when $\alpha = 180^\circ$ it gives $R = 70\text{ N}$. Which statement about the original answer is most justified by these limiting-case checks?
   Answer: The answer is plausible: $R = 140\text{ N}$ lies strictly between the minimum of 70 N (forces opposing) and maximum of 170 N (forces aligned), which is a necessary condition for any valid resultant at an intermediate angle. When checking a physics answer, it helps to distinguish between necessary conditions (things that must be true) and sufficient conditions (things that guarantee correctness). The parallelogram (or cosine) law tells you the resultant of two concurrent forces must lie between $|F_2 - F_1|$ and $F_1 + F_2$, corresponding to the extremes of $\alpha = 180°$ and $\alpha = 0°$ respectively. Here, those bounds are $|120 - 50| = 70\text{ N}$ and $120 + 50 = 170\text{ N}$. Since $R = 140\text{ N}$ falls strictly inside this range, the answer passes a meaningful sanity check — it is physically plausible. That's exactly what A captures, making it the most justified statement. B is the critical trap here. Passing both limiting-case checks tells you the formula structure behaves correctly at the boundaries, but it says nothing about whether you substituted $\alpha$ correctly or avoided an arithmetic error in the middle. Boundary checks are necessary, not sufficient — do not confuse the two. C contains a real physics insight ($R_{90°} = \sqrt{50^2 + 120^2} = 130\text{ N}$), but its conclusion is wrong. If $140\text{ N} > 130\text{ N}$, that simply means $\alpha < 90°$, which is perfectly valid. Nothing in the problem forbids that angle, so there is no reason to "reject" the answer. D misapplies the triangle inequality. The correct bounds are $|F_2 - F_1| \leq R \leq F_1 + F_2$, which gives $70\text{ N}$ to $170\text{ N}$ — not $50\text{ N}$ to $120\text{ N}$. The 50 N and 120 N are the individual force magnitudes, not the resultant's bounds. Study tip: On statics problems, always run the $\alpha = 0°$ and $\alpha = 180°$ checks first — but remember they bound your answer, not prove it. A result inside the bounds is plausible; only a complete derivation confirms it.
5. Which of the following represents the best solution-checking practice for evaluating this reported result?
   Answer: The dimensional check is valid and confirms the formula structure, but the reported negative value $a_n = -8.5\text{ m/s}^2$ is physically unreasonable. Because $a_n = v^2/\rho$ is a magnitude — with both $v^2$ and $\rho$ strictly positive — a negative result indicates a sign or direction-convention error in the computation that the unit check cannot detect. Whenever you encounter a problem involving curvilinear motion, you need to distinguish between formula validation (what a units check confirms) and physical reasonableness (what a units check cannot confirm). These are two separate layers of verification. The formula $a_n = v^2/\rho$ involves only squared speed and radius of curvature. Since $v^2 \geq 0$ always and $\rho > 0$ by definition (it's a geometric length), the ratio is strictly non-negative. Reporting $a_n = -8.5 \text{ m/s}^2$ as a magnitude violates this constraint — the negative sign signals a computation error, likely a direction-convention mistake or an algebraic sign error, that slipped through undetected. Answer A correctly identifies both points: the units check is legitimate and confirms formula structure, but it operates independently of sign validity and cannot catch this class of error. Answer B introduces a plausible-sounding but subtly wrong idea. In a signed coordinate framework, vector components can be negative, but $a_n = v^2/\rho$ specifically defines the magnitude of the normal acceleration — it is not a signed component expression. Mixing these concepts is a common trap. Answer C overstates the power of dimensional analysis. Units confirm that your formula is dimensionally consistent; they say nothing about whether numerical values, signs, or physical constraints are satisfied. A units check passing is necessary but never sufficient. Answer D invents a constraint that does not exist. There is no geometric or physical rule bounding $|a_n|$ relative to $|a_t|$; these components are independent. Study tip: Always layer your checks — units first, then sign/direction conventions, then order-of-magnitude reasonableness. A clean units check is just the starting point, not the finish line.
6. A student solves a statics problem involving a simply supported beam and reports the reaction at the left support as $R_A = 450\text{ N}$ and at the right support as $R_B = 200\text{ N}$ for a single downward point load of $P = 600\text{ N}$ applied somewhere along the beam. The student checks moment equilibrium about one point and confirms it is satisfied. Which statement best identifies the most important additional check the student should perform and its implication?
   Answer: The student should verify force equilibrium in the vertical direction: $R_A + R_B = P$ requires $450 + 200 = 650\text{ N} \neq 600\text{ N}$. This check reveals that the answers are wrong, since confirming moment equilibrium about a single point does not guarantee force equilibrium is also satisfied — both independent equations must hold simultaneously. Whenever you solve a statics problem with a simply supported beam, remember that equilibrium requires two independent conditions to be satisfied simultaneously: the sum of all forces equals zero and the sum of all moments equals zero. Checking only one condition never fully validates your solution. Here, the student reports $R_A = 450\text{ N}$ and $R_B = 200\text{ N}$ for a single downward load $P = 600\text{ N}$. The critical vertical force check is straightforward: $\sum F_y = 0$ requires $R_A + R_B = P$, meaning $450 + 200 = 650\text{ N}$, but the applied load is only $600\text{ N}$. Since $650 \neq 600$, the solution is wrong, and no amount of moment-checking around a single point can rescue it. This is why D is correct — it identifies the violated equation and explains precisely why the error went undetected. Choice A is tempting but flawed: confirming the load's geometric position is a reasonableness check, not a mathematical verification. A solution can seem geometrically plausible while still violating equilibrium. Choice B contains a subtle misconception: for a simply supported beam with a single point load, force equilibrium and the two moment equations are not all independent — in fact, checking moments about both supports plus force equilibrium gives three equations for two unknowns, where any two imply the third. Force equilibrium is never truly redundant; it's the simplest and most direct check. Choice C is incomplete — positive values confirm assumed directions but say nothing about whether the magnitudes are actually correct. Your strategy: always run both checks — $\sum F = 0$ and $\sum M = 0$ — before accepting any reaction solution. The arithmetic check $R_A + R_B = P$ takes five seconds and catches errors that moment equations alone will miss.
7. The student performs a unit check and confirms the expression has units of rad/s. Which of the following most completely and correctly evaluates whether the formula should be accepted?
   Answer: The formula should be accepted with caution. The unit check passes, and the limiting cases $\theta \to 0$ and $d \to 0$ both correctly give $\omega \to 0$. These checks are consistent with the formula, but they cannot validate the coefficient $4/3$ — an incorrect coefficient such as $1$ or $2$ would pass the same limiting-case tests. Independent verification via energy methods confirms the coefficient is correct, but that requires re-deriving the result. When evaluating a physics formula, you should think in layers: dimensional analysis first, then limiting cases, then coefficient verification. Passing one layer doesn't mean passing all of them — and understanding what each layer can and cannot tell you is exactly what this question tests. The formula $\omega = \sqrt{\frac{4gd\sin\theta}{3R^2}}$ can be verified by energy conservation. Setting potential energy loss equal to total kinetic energy (translational plus rotational, using $I = \frac{1}{2}mR^2$ for a disk and the no-slip condition $v_c = \omega R$) yields exactly this expression with coefficient $4/3$. So the formula is correct. Choice B is right because it honestly acknowledges what the checks do and don't prove: the unit check and limiting cases ($\theta \to 0$, $d \to 0$ both giving $\omega \to 0$) are consistent with the formula but cannot distinguish the correct coefficient from an incorrect one like $1$ or $2$. Full confidence requires re-deriving the result independently. Choice A is wrong because it invents a false rule. The coefficient $4/3$ appears in the denominator under the radical when you work through the energy algebra correctly — there is no requirement that such coefficients be less than 1. Choice C is wrong because it misinterprets what a formula's domain of validity means. The formula models rolling without slipping; applying it at $\theta = 90°$ gives a mathematically finite answer, but the physical model simply breaks down there. That's a modeling limitation, not a derivation error — the formula isn't required to self-destruct at its boundary. Choice D introduces a false dimensional conflict. The variable $d$ is an independent displacement along the incline, not a function of $R$, so there is no conflation of variables. Strategy tip: On problems asking you to evaluate a formula, always ask: "What can this test actually detect, and what can it miss?" Limiting cases and unit checks are necessary but not sufficient — they can never validate a specific numerical coefficient.

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