Statics and Dynamics Quiz: Checking Solutions

What this quiz covers

This quiz focuses on Checking Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The formula should be rejected. Although the unit check passes, the coefficient $4/3$ violates the physical constraint that all energy-method coefficients for rolling bodies must be rational numbers strictly less than 1, since the rolling kinetic energy is always a fraction of the total kinetic energy. This indicates an error in the moment-of-inertia calculation for the disk.
2. 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. (correct answer)
3. The formula should be rejected because the limiting case $\theta \to 90^\circ$ yields $\omega = \sqrt{4gd/(3R^2)}$, but rolling without slipping on a vertical surface is physically impossible since there is no normal force to generate the required friction. A valid formula must give an undefined or zero result at this limit, so the formula's finite prediction reveals a fundamental modeling error.
4. The formula should be rejected on dimensional grounds. Applying the no-slip constraint $v_c = \omega R$ requires that $\omega$ scale as $R^{-1}$, but inspection of the formula shows it scales as $R^{-1}$ only if $d$ is independent of $R$. Since the problem does not specify this independence, the formula conflates geometric and kinematic variables and cannot be accepted without additional clarification.

Question 2

1. Verify that $F_{AB} = 85\text{ kN}$ does not exceed the allowable tensile capacity of the member's cross-section. A force exceeding the material strength would be physically unreasonable for a stable structure, indicating that an error was made somewhere in the equilibrium analysis.
2. 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. (correct answer)
3. Confirm that the sum of all reported member forces in the truss equals zero, since global force balance requires tensile and compressive member forces to cancel when summed across all members, providing a system-level consistency check that is independent of the method of joints.
4. Apply Maxwell's reciprocal theorem, which relates member forces under reciprocal unit loads, to verify that the force in AB due to the applied loading equals the force predicted by reciprocal loading at joint B — providing an independent analytical cross-check of the reported 85 kN value.

Question 3

1. The answer should be rejected, but only for angles where $\mu_k\cos\theta > \sin\theta$, because in that case the expression under the radical becomes larger than $2gd\sin\theta$, yielding a speed greater than the frictionless result — a physically impossible outcome that proves the sign is wrong for those specific angles.
2. The answer should be accepted. The unit check confirms dimensional consistency, and the presence of both $\sin\theta$ and $\cos\theta$ terms is structurally correct since both the gravitational component and the normal force (which drives friction) depend on the angle; the sign of the friction term cannot be determined without knowing the direction convention.
3. 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. (correct answer)
4. The answer should be accepted for $\theta > 45^\circ$ because at steep angles $\sin\theta > \cos\theta$, making the gravitational term dominant and rendering the sign of the friction term physically irrelevant to the reasonableness check.

Question 4

1. 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. (correct answer)
2. The answer must be correct: passing both the $\alpha = 0^\circ$ and $\alpha = 180^\circ$ limiting cases is sufficient to confirm that the formula and the specific numerical result are free of errors.
3. The answer is suspicious: for $\alpha = 90^\circ$ the resultant should be $\sqrt{50^2 + 120^2} = 130\text{ N}$, and since 140 N > 130 N, the forces must be acting at an angle less than 90°. Because the problem does not specify $\alpha$, the result 140 N is outside the acceptable range and should be rejected.
4. The answer is suspicious: the triangle inequality requires the resultant to satisfy $|F_2 - F_1| \leq R \leq F_1 + F_2$, and since 140 N lies outside the range 50 N to 120 N, the result violates this inequality and is physically impossible.

Question 5

1. The reported minimum speed $v_{\min} = 0$ is incorrect only for $\theta < 45^\circ$; for $\theta \geq 45^\circ$ the horizontal and vertical components are equal at the apex, allowing the total speed to reach zero, confirming the student's result for steeper launch angles.
2. The reported minimum speed is correct for the special case $\theta = 90^\circ$ (vertical launch), and the formula $t^* = v_0/g$ is valid for all launch angles because $\sin\theta$ cancels out when differentiating $v(t)$ to find the minimum.
3. 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. (correct answer)
4. The formula for $v(t)$ fails dimensional analysis because the argument of the square root mixes squared velocity terms with a squared acceleration-time product $g^2t^2$, making the expression dimensionally inconsistent and invalidating any conclusions drawn from it.

Question 6

1. 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. (correct answer)
2. The negative value is acceptable and physically meaningful: while the scalar magnitude $v^2/\rho$ is positive by definition, in a signed coordinate system the normal acceleration component can be expressed as negative when the center of curvature lies in the negative normal direction, so the sign simply reflects the chosen axis orientation.
3. The dimensional check is the most critical verification step; since it confirms units of m/s², both the formula and the numerical result $-8.5\text{ m/s}^2$ are validated. For intermediate-level curvilinear-motion problems, a successful unit check is generally sufficient to conclude the computation is correct.
4. The result should be further checked by confirming that $|a_n|$ does not exceed the tangential acceleration $a_t$ in magnitude, because the normal component is geometrically bounded by the tangential component for any physically realizable curved path at non-constant speed.

Question 7

1. The student should verify that the ratio $R_A/R_B = 450/200 = 2.25$ is physically reasonable by confirming that the load $P$ is located closer to support $A$ than to support $B$, since a larger reaction at $A$ requires the load to be nearer to $A$; if this geometric condition holds, the solution is confirmed.
2. The student should verify moment equilibrium about the other support point to obtain a second independent equation, which when combined with the first moment equation uniquely confirms both reactions; checking force equilibrium is redundant because it is algebraically dependent on the two moment equations.
3. The student should check that both reactions are positive, confirming the assumed directions are correct; since $R_A = 450\text{ N} > 0$ and $R_B = 200\text{ N} > 0$, both directions are confirmed and the solution is physically consistent with a simply supported beam under a downward load.
4. 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. (correct answer)