Question 1

Statics – Resolution of Forces and Resultant In engineering mechanics, the resolved component of the resultant of two forces (inclined at an angle θ to each other) taken along a specified direction is equal to which of the following?

Accepted Answer

Introduction / Context: This question tests the fundamental resolution principle in statics. When combining multiple forces, any single component of the resultant along a chosen axis equals the algebraic sum of the corresponding components of the individual forces along that same axis. This principle underpins free-body diagram analysis, equilibrium checks, and vector addition in mechanics. Given Data / Assumptions: Two forces act at an angle θ to one another. A specific reference direction (axis) is chosen. Standard right-hand sign convention for components. Concept / Approach: The vector resultant R of forces F1 and F2 is obtained by vector addition. Components are linear operations on vectors. Therefore, the component of R along any direction is simply the component-wise sum: R_dir = F1_dir + F2_dir. This holds regardless of the angle θ between forces or their magnitudes, provided signs (algebraic sense) are respected. Step-by-Step Solution: Choose a direction (say x-axis) and resolve each force: F1x = F1 * cos α1, F2x = F2 * cos α2.Compute resultant component: Rx = F1x + F2x (algebraic sum).Repeat for any direction; linearity of components guarantees additivity. Verification / Alternative check: Construct a vector diagram or use Cartesian components: R = (F1x + F2x) i + (F1y + F2y) j. Projection of R on any axis equals the sum of projections of F1 and F2 on that axis. Why Other Options Are Wrong: (b) Ignores direction and signs; not generally valid. (c) Applies only to a special projection relationship and even then not universally. (d) and (e) combine magnitudes with trigonometric factors incorrectly; components must be taken force-by-force before summing. Common Pitfalls: Mixing scalar magnitudes with vector components; forgetting to use algebraic signs for directions; using a single common angle θ when individual force angles to the reference differ. Final Answer: Algebraic sum of the resolved components of the two forces along that direction

Question 2

Suspended Cable Under Self-Weight – Identify the Correct Curve Shape A perfectly flexible, inextensible cable supporting only its own weight (uniformly distributed along its length) hangs freely between two supports. The resulting shape of the cable…

Accepted Answer

Introduction / Context: Recognizing the natural curve of a cable under self-weight is essential in bridge engineering (suspension systems), power-line design, and structural form-finding. The equilibrium of a flexible line under a continuous gravity load leads to a well-known transcendental curve called the catenary. Given Data / Assumptions: Cable is perfectly flexible and inextensible. Only load is its own weight (uniform per unit length measured along the cable). No uniformly distributed load per horizontal span is assumed here. Concept / Approach: For a cable whose load is uniformly distributed along its length, the differential equilibrium leads to the catenary equation: y = a cosh(x/a) − a, where a = H/w, H is horizontal tension and w is weight per unit length. If the load is instead uniform per horizontal projection (as for a level deck with hangers), the cable takes a parabolic shape. Distinguishing these loading modes is the key concept. Step-by-Step Solution: Model the cable as a funicular for the given loading (self-weight along length).Set up equilibrium of a differential element → tension components vary with slope.Solve the governing equation to obtain the catenary form y = a cosh(x/a) − a. Verification / Alternative check: Compare with the parabola case (uniform load per horizontal span). Real cables supporting their own weight only match the catenary closely; when they carry a uniform deck load, the profile becomes nearly parabolic over typical spans. Why Other Options Are Wrong: (a) Parabola corresponds to uniform load per horizontal projection, not self-weight along length. (b) Circle and (d) ellipse are not solutions of the governing equilibrium. (e) A straight line is impossible in equilibrium with gravity; 'small sag' parabolic approximations are only approximations, not the exact curve for self-weight. Common Pitfalls: Confusing uniform deck load with self-weight distribution; assuming parabola for all suspended curves. Final Answer: Catenary

Question 3

Simple Harmonic Motion – Identifying the Quantity Measured in Cycles Per Second For a body executing simple harmonic motion (SHM), the number of cycles completed per second is called:

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Introduction / Context: Terminology in vibrations is precise and must not be confused. In simple harmonic motion, the key measurable quantities are amplitude, period, and frequency. Frequency dictates how many complete cycles occur per unit time and is fundamental in structural dynamics, seismology, and machine design. Given Data / Assumptions: Sinusoidal or harmonic motion about an equilibrium position. Constant amplitude and frequency for ideal SHM. No damping or external forcing considered. Concept / Approach: Frequency f is defined as cycles per second (hertz). The period T is the time for one complete cycle, with f = 1 / T. Amplitude A is the maximum displacement from equilibrium, independent of f in ideal linear SHM. Beats arise from superposition of close frequencies, not from a single harmonic oscillator. Step-by-Step Solution: Recognize that the asked quantity is 'cycles per second' → by definition, frequency.Recall relation f = 1 / T and units: hertz (s^-1).Exclude amplitude and beat, which are distinct concepts. Verification / Alternative check: In a mass–spring system with natural circular frequency ω_n, frequency f = ω_n / (2π), and period T = 2π / ω_n. The interpretation remains consistent for pendulums and other linear SHM systems for small oscillations. Why Other Options Are Wrong: (a) 'Oscillation' refers to the motion itself, not a rate. (b) Amplitude is a displacement magnitude. (c) Period is time per cycle, not cycles per second. (d) Beats occur from superposition of two different frequencies. Common Pitfalls: Interchanging period and frequency; using 'oscillation' as a unit-bearing quantity. Final Answer: Frequency

Question 4

Projectile Motion – Finding the Angle of Projection from Range–Height Ratio A projectile is launched with speed u at an angle θ to the horizontal on level ground. If its horizontal range R is 2.5 times its greatest height H, determine the angle of p…

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Introduction / Context: Projectile motion relationships connect range, maximum height, and launch angle. Recognizing and using standard formulas allows quick back-calculation of the launch angle when given a ratio such as R/H, a common exam style in engineering mechanics and physics. Given Data / Assumptions: Horizontal range: R = (u^2 / g) * sin 2θ. Greatest height: H = (u^2 * sin^2 θ) / (2 g). Given R = 2.5 * H. Neglect air resistance; level launch and landing elevations. Concept / Approach: Form the ratio R/H to eliminate u and g, leaving a relation purely in θ. Solve for θ using trigonometric identities, specifically sin 2θ = 2 sin θ cos θ, which simplifies the ratio to a function of cot θ. Step-by-Step Solution: R/H = [(u^2/g) * sin 2θ] / [ (u^2 sin^2 θ)/(2g) ]= 2 * sin 2θ / sin^2 θ = 2 * (2 sin θ cos θ) / sin^2 θ = 4 * cot θGiven R/H = 2.5 ⇒ 2.5 = 4 * cot θ ⇒ cot θ = 0.625θ = arctan(1 / 0.625) = arctan(1.6) ≈ 58° Verification / Alternative check: Compute with a calculator: arctan(1.6) ≈ 58.0°. A quick sanity check: if θ = 60°, cot θ ≈ 0.577, giving R/H ≈ 2.308; slightly low, confirming 58° is more accurate for 2.5. Why Other Options Are Wrong: 57° and 59° are near but yield R/H values of about 2.56 and 2.44 respectively. 60° gives 2.308. 62° deviates further. Common Pitfalls: Using H = u^2 sin θ / (2g) by mistake (missing the square); misapplying sin 2θ identity; forgetting that R/H must be dimensionless and u cancels. Final Answer: 58°

Question 5

Suspension Bridges – Eliminating Bending at the Pier Base In a suspension bridge, how is bending at the base of a pier minimized or avoided by the detailing of the main suspension and anchor cables?

Accepted Answer

Introduction / Context: Pier detailing in suspension bridges is critical to transfer cable forces efficiently. If cable forces introduce large moments, the pier base experiences undesirable bending. Properly designed saddles and bearings help align the resultant through the pier's centroidal core, reducing bending demands. Given Data / Assumptions: Main suspension cable carries vertical loads to the piers and anchors. Anchor cables balance horizontal components. Roller-mounted saddles allow relative movement and load alignment. Concept / Approach: By seating both suspension and anchor cables over a saddle on rollers, the cable can adjust position and transmit primarily compressive resultant through the pier. The rollers allow small longitudinal movements due to temperature and live-load effects while aligning the cable's line of action with the pier centerline, minimizing bending moments at the base. Step-by-Step Solution: Consider force components from the cable at the pier top.Use a roller saddle to permit horizontal translation and rotation of the cable seat.Ensure the resultant passes within the kern of the pier section to avoid tension and minimize bending. Verification / Alternative check: Bridge practice shows roller or rocker saddles are standard at pylons to reduce fixed-end moments; finite element models confirm markedly lower base moments compared to rigid cable fixity. Why Other Options Are Wrong: (a) and (b) impose rigidity, inducing bending from temperature/load shifts. (d) contradicts efficient design philosophy. (e) is structurally infeasible for a suspension system. Common Pitfalls: Confusing anchor saddles at abutments with pylon saddles; assuming fixed connections are beneficial for stability. Final Answer: Attach suspension and anchor cables to a saddle mounted on rollers at the pier top so that the line of action passes through the pier and horizontal movement is accommodated

Question 6

Low-Earth Orbit Perturbations up to ~720 km Altitude – Dominant Effects Which of the following physical effects measurably perturb a satellite's orbit up to an altitude of about 720 km above Earth's surface?

Accepted Answer

Introduction / Context: Low-Earth Orbits (LEO) are subject to several non-Keplerian perturbations. Mission design and orbit maintenance require understanding of which effects dominate at various altitudes. Around a few hundred kilometers up to roughly 700–800 km, multiple perturbations act concurrently with different time scales and magnitudes. Given Data / Assumptions: Altitude regime: up to approximately 720 km. Typical LEO satellites with modest area-to-mass ratios. Keplerian two-body baseline with perturbations superposed. Concept / Approach: Earth’s gravity field (J2 and higher): Causes node regression and apsidal rotation; effects are significant throughout LEO. Third-body forces (Sun/Moon): Though weaker than J2 and drag, they produce secular and periodic changes in orbital elements, especially inclination and right ascension of ascending node. Aerodynamic drag: Residual atmosphere below ~800–1000 km induces semi-major axis decay and reduces perigee height; magnitude depends on solar activity and ballistic coefficient. Step-by-Step Solution: Identify perturbations relevant to LEO: J2+, third bodies, atmospheric drag.Assess altitude dependence: drag prominent below ~700–800 km; J2 significant across LEO; third-body always present but smaller.Select the comprehensive choice covering all real effects. Verification / Alternative check: Standard orbit determination models (e.g., SGP4/SDP4 families) include J2 and drag for LEO; high-fidelity propagators include Sun/Moon gravitation even at LEO altitudes. Why Other Options Are Wrong: (a), (b), (c) alone each omit other non-negligible contributors in this altitude band. (e) contradicts observed decay and nodal drift behavior of LEO satellites. Common Pitfalls: Assuming drag is negligible above 400–500 km regardless of solar cycle; ignoring J2-driven precessions that are pivotal for Sun-synchronous orbit design. Final Answer: All of the above

Question 7

Elevator Dynamics – Apparent Weight on a Scale During Descending Motion A man stands on a spring scale in a lift. At rest, the scale reads 60 kg. When the lift begins to descend with constant acceleration equal to g/5, what reading will the scale in…

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Introduction / Context: Apparent weight measured by a scale is the normal reaction between the person and the scale, not the true weight. In accelerating frames like elevators, Newton’s second law modifies the normal reaction depending on the direction and magnitude of the elevator’s acceleration. Given Data / Assumptions: Scale reading at rest = 60 kg → mass m = 60 kg (since scale is calibrated in kg as m = W/g). Lift accelerates downward with a = g/5. Downward is taken positive for acceleration; neglect cable mass and friction. Concept / Approach: For a person of mass m in a descending elevator, the normal reaction N equals m(g − a). A scale calibrated in kilograms effectively displays m_eff = N/g. Therefore, the indicated mass reduces by the factor (1 − a/g) for downward acceleration. Step-by-Step Solution: m = 60 kg; a = g/5.N = m * (g − a) = 60 * (g − g/5) = 60 * (4g/5) = 48 g.Scale reading (in kg) = N / g = 48. Verification / Alternative check: Limiting cases: if a = 0 → reading 60 kg (matches rest). If a → g (free fall) → reading 0 kg, consistent with weightlessness. Our a = g/5 gives an intermediate reduction to 48 kg. Why Other Options Are Wrong: (b) ignores acceleration. (c) would require upward acceleration. (d) and (e) correspond to different accelerations (g/10 and g/6 respectively), not the given value. Common Pitfalls: Confusing mass with weight units; forgetting that the scale reads the normal reaction; sign errors when the elevator moves downward. Final Answer: 48 kg

Question 8

Lami’s theorem in coplanar concurrent force systems: Which precise statement correctly expresses Lami’s theorem for three coplanar forces acting at a point in equilibrium?

Accepted Answer

Introduction / Context: Lami’s theorem is a core tool in engineering mechanics for solving problems with three coplanar, concurrent forces in equilibrium. It provides a direct proportionality between each force and the sine of the angle between the other two forces, avoiding component resolution when only three forces act at a point. Given Data / Assumptions: Three forces are coplanar and concurrent at a point. The system is in static equilibrium (resultant equals zero). Angles are the included angles between the directions of the other two forces. Concept / Approach: Lami’s theorem states: F1 / sin(α) = F2 / sin(β) = F3 / sin(γ), where α is the angle between F2 and F3, β between F3 and F1, and γ between F1 and F2. This relation arises from the equilibrium polygon (triangle of forces) and the sine rule applied to that triangle. Step-by-Step Solution: Recognize three-force equilibrium at a point → triangle of forces is valid.Apply the sine rule to the force triangle → each force ∝ sine of opposite included angle.Conclude the correct statement is the proportionality to the sine of the angle between the other two forces. Verification / Alternative check: Resolving each force along two perpendicular axes and imposing ΣFx = 0 and ΣFy = 0 leads to the same sine relationships after elimination, confirming Lami’s theorem. Why Other Options Are Wrong: (a) is too broad; equilibrium is not automatic. (b) uses a triangle condition but is not the precise theorem. (d) claims inverse proportionality, which is incorrect. (e) is not applicable. Common Pitfalls: Forgetting the forces must be exactly three, coplanar, and concurrent; misidentifying the included angles; mixing angle definitions. Final Answer: Three coplanar forces in equilibrium are each proportional to the sine of the angle between the other two

Question 9

Simple pendulum — identify the statement that is NOT applicable: For small oscillations of a simple pendulum, which one of the following statements is NOT valid?

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Introduction / Context: The simple pendulum is a classic model used to illustrate simple harmonic motion for small angles. Its period depends primarily on pendulum length and gravitational acceleration. This question asks you to detect a false statement among commonly quoted properties for small oscillations. Given Data / Assumptions: Small angular displacement approximation (sin θ ≈ θ in radians). Simple pendulum: point mass, massless string, no air resistance, frictionless pivot. Standard time period formula applies. Concept / Approach: For a simple pendulum under the small-angle approximation, the time period is T = 2π √(l / g). From this, T ∝ √l and T ∝ 1/√g, and T does not depend on mass. It is also commonly stated that T is independent of amplitude for small angles. Any claim that contradicts these proportionalities is not applicable. Step-by-Step Solution: Write the governing relation: T = 2π √(l / g).Infer proportionalities: T ∝ √l and T ∝ 1/√g.Note independence from mass and (approximately) from amplitude for small angles.Identify the incorrect statement: “T ∝ l” is false. Verification / Alternative check: Dimensional analysis: [T] must scale as √(length/acceleration). A linear dependence on length violates dimensional consistency unless g changes accordingly. Why Other Options Are Wrong: (a) and (e) are standard properties under small-angle assumptions. (b) and (c) directly match the formula. Only (d) contradicts the square-root dependence. Common Pitfalls: Applying the conclusions to large amplitudes where T increases slightly with amplitude; ignoring air drag or mass distribution (compound pendulum effects). Final Answer: The time period is proportional to l (directly proportional to length)

Question 10

Equilibrium of two forces at a point: If two forces act at a point and the system is in equilibrium, what must be true about their magnitudes and directions?

Accepted Answer

Introduction / Context: Equilibrium under two forces is a foundational result in statics. When only two forces act on a rigid body and it remains in equilibrium, those forces must be strictly collinear, equal in magnitude, and opposite in direction so that both resultant force and resultant moment are zero. Given Data / Assumptions: Rigid body assumption. Only two forces act on the body. No couple acting separately. Concept / Approach: For ΣF = 0 in vector form with only two forces F1 and F2, we must have F1 + F2 = 0 → F2 = −F1, which implies equal magnitudes and opposite directions along the same line. Also, ΣM about any point is then zero because their lines of action are coincident (no net moment arm). Step-by-Step Solution: Assume only F1 and F2 act; impose ΣF = 0 → F2 = −F1.Conclude magnitudes are equal, directions opposite, and line of action the same.Check ΣM = 0: collinearity ensures no net couple is produced. Verification / Alternative check: Non-collinear equal and opposite forces produce a couple, violating equilibrium; perpendicular forces cannot cancel each other fully. Why Other Options Are Wrong: (a) same sense cannot balance; (c) perpendicular forces cannot yield zero resultant; (d) incorrect; (e) requires an additional force to balance the components. Common Pitfalls: Confusing “same line” with “parallel but not collinear”; forgetting that equal and opposite but non-collinear forces create a pure moment. Final Answer: Equal in magnitude and along the same line in opposite sense

Question 11

Static friction — identify the statement that is NOT a law: Which of the following is NOT a valid law of limiting static friction between two dry solid surfaces?

Accepted Answer

Introduction / Context: For dry, rigid bodies at impending motion, classical Coulomb friction provides empirical “laws” for limiting static friction. These guide quick calculations in statics and dynamics. This question asks you to identify a statement that does not belong to these laws. Given Data / Assumptions: Dry, clean, nominally rigid surfaces in contact. Impending motion (limiting static friction), not kinetic friction. No significant adhesion, temperature, or surface contamination effects. Concept / Approach: Classical laws: (1) The frictional force opposes impending motion; (2) Limiting friction F_lim is proportional to the normal reaction N, with coefficient μ_s roughly constant for given surfaces; (3) F_lim is largely independent of apparent area of contact for clean, rigid bodies; (4) At the instant of impending motion, speed is zero and speed effects are irrelevant (unlike kinetic friction at higher speed). Hence any statement claiming dependence on apparent area is not a law in this idealization. Step-by-Step Solution: Recall F_lim = μ_s * N.Note that μ_s depends on surface pair (roughness/material), not on apparent area.Therefore the claim that limiting friction depends on apparent area is NOT a law. Verification / Alternative check: Experiments with varying block footprint but same normal load show nearly unchanged limiting friction in the ideal regime, corroborating area-independence. Why Other Options Are Wrong: (a), (c), and (d) match Coulomb laws; (e) is consistent at the threshold since sliding speed is zero. Common Pitfalls: Confusing real microscopic contact area (which may affect μ_s) with apparent geometric area; mixing static and kinetic friction behaviors. Final Answer: Limiting friction depends on the apparent area of contact

Question 12

Seconds pendulum — beats (oscillations) per second: A seconds pendulum has a time period of 2 s. How many oscillations (beats) per second does it execute?

Accepted Answer

Introduction / Context: A seconds pendulum is defined so that one complete oscillation takes 2 seconds. Converting period to frequency (oscillations per second) is a straightforward application of simple harmonic motion relations. Given Data / Assumptions: Time period T = 2 s. Small-angle approximation so the motion is nearly simple harmonic. One “beat” here means one full oscillation (to and fro). Concept / Approach: The relationship between period and frequency is f = 1 / T. Therefore, for T = 2 s, the frequency is 0.5 Hz, which is 0.5 oscillations (beats) per second. Step-by-Step Solution: Write f = 1 / T.Substitute T = 2 s → f = 1 / 2 = 0.5 s^-1.Interpret s^-1 as “per second” → 0.5 beat per second. Verification / Alternative check: In 10 seconds, number of oscillations = time / period = 10 / 2 = 5. That is consistent with 0.5 per second. Why Other Options Are Wrong: (b)–(e) correspond to shorter periods not matching a seconds pendulum. Common Pitfalls: Counting half-oscillations as beats; mixing up time period and frequency units. Final Answer: 0.5 beat per second

Question 13

Elastic collision with a wall — time ratio and coefficient of restitution: A smooth marble ball rolls on a horizontal floor toward a vertical wall. The time taken to return to the launch point is twice the time taken to reach the wall. What is the c…

Accepted Answer

Introduction / Context: Coefficient of restitution e relates the speeds of separation and approach in a one-dimensional impact. For a ball striking a fixed wall normally, e equals the ratio of rebound speed to approach speed. Timing information can be used to infer e when distances are equal and rolling resistance is negligible. Given Data / Assumptions: Straight-line normal impact on a fixed wall. Horizontal floor, negligible rolling resistance and air drag. Outbound and inbound path lengths are equal. Return time is twice the outbound time. Concept / Approach: Let distance to the wall be D, approach speed be u, and rebound speed be v. Then time to reach the wall t1 = D / u; time to return t2 = D / v. Given t2 = 2 t1, we obtain v = u / 2. For a ball–wall impact, e = rebound speed / approach speed = v / u. Step-by-Step Solution: t1 = D / u, t2 = D / v.Given t2 = 2 t1 → D / v = 2 D / u → v = u / 2.Compute e = v / u = (u / 2) / u = 0.5. Verification / Alternative check: If the ball took, say, 1 s to reach the wall, it would take 2 s to return, totaling 3 s, consistent with speeds u and u/2 on equal paths. Why Other Options Are Wrong: (a) and (c) correspond to v = 0.25 u or 0.75 u, contradicting the time ratio. (d) e = 1 implies perfectly elastic impact with equal times out and back. (e) would mean the ball sticks (no rebound), which is inconsistent. Common Pitfalls: Using distance ratio incorrectly; forgetting that for a fixed wall, e is speed ratio, not including sign; assuming oblique impact when the motion is normal. Final Answer: 0.50

Question 14

Apparent weight in an accelerating lift (elevator): A man’s apparent weight is less than his real weight when the lift is moving downward under which condition?

Accepted Answer

Introduction / Context: Apparent weight is the normal reaction N exerted by the lift floor on the person. In a non-inertial frame undergoing vertical acceleration, N differs from the true weight W = m * g, and this is often tested in elevator physics questions. Given Data / Assumptions: Mass m of the person; gravitational acceleration g. Lift accelerates vertically with magnitude a. Positive upward direction convention. Concept / Approach: Equation of motion: ΣF = m * a. For downward acceleration (a negative with upward positive), we have N − m * g = m * a → N = m * (g + a). If the acceleration is downward, a = −|a|, hence N = m * (g − |a|) < m * g. Thus apparent weight decreases. For upward acceleration, N = m * (g + |a|) > m * g. At uniform speed (a = 0), N = m * g (no change). Step-by-Step Solution: Write N − m * g = m * a.For downward acceleration: a = −|a| → N = m * (g − |a|).Therefore N < m * g → apparent weight less than real weight. Verification / Alternative check: Free-body diagram of the person confirms direction of forces and the resulting normal reaction magnitude under different acceleration signs. Why Other Options Are Wrong: (a) Uniform speed implies a = 0, so no change. (c) Momentum alone does not determine N; acceleration does. (d) and (e) correspond to upward acceleration, which increases apparent weight. Common Pitfalls: Confusing velocity with acceleration; taking downward retardation as “less weight” when it actually means upward acceleration and greater apparent weight. Final Answer: With downward acceleration

Question 15

Two-body system with friction — block on table and hanging mass: A 6.6 kg block A rests on a horizontal platform and is connected over a smooth pulley to a hanging 3.2 kg mass. The coefficient of friction between block A and the platform is 1/3. Wha…

Accepted Answer

Introduction / Context: This is a standard two-mass system: a block on a horizontal surface connected to a hanging mass over a frictionless pulley. The acceleration depends on the net driving force (weight of the hanging mass minus frictional resistance on the block) divided by the total mass of the system. Given Data / Assumptions: Mass on table m1 = 6.6 kg; hanging mass m2 = 3.2 kg. Coefficient of friction μ = 1/3 between m1 and table. Pulley and string are light and frictionless; motion is impending in the direction pulled by m2. Take g = 9.8 m/s^2. Concept / Approach: Friction force opposing motion on the table is F_f = μ * N = μ * m1 * g. The driving force is m2 * g − F_f. The acceleration of the connected system is a = (driving force) / (m1 + m2), provided the hanging weight exceeds friction so that motion occurs. Step-by-Step Solution: Compute friction: F_f = (1/3) * 6.6 * 9.8 = 2.2 * 9.8 N.Compute hanging weight: W2 = 3.2 * 9.8 N.Net driving force: W2 − F_f = (3.2 − 2.2) * 9.8 = 1.0 * 9.8 N.Total mass: m_total = 6.6 + 3.2 = 9.8 kg.Acceleration: a = (1.0 * 9.8) / 9.8 = 1.00 m/s^2. Verification / Alternative check: Check motion condition: m2 * g > μ * m1 * g → 3.2 > 2.2 (true). Therefore system accelerates with the computed value. Why Other Options Are Wrong: (a) and (b) underestimate net driving force; (d) and (e) overestimate by ignoring friction magnitude or miscalculating total mass. Common Pitfalls: Using mass difference directly without subtracting friction; forgetting that friction depends on normal reaction m1 * g; rounding g too crudely. Final Answer: 1.00 m/s^2

Question 16

Resultant of concurrent forces on a particle (engineering mechanics) A number of forces act simultaneously on a single particle of a rigid body. According to the principle of composition of concurrent forces, state whether they can be replaced by an…

Accepted Answer

Introduction / Context: When several forces act simultaneously on a single particle (i.e., their lines of action intersect at one point), their combined effect can be represented by a single resultant. This core idea underpins free-body diagrams, equilibrium checks, and vector reduction in engineering mechanics. Given Data / Assumptions: All forces are concurrent at a single point (particle idealization). Rigid-body effects such as moments from offset forces are not present because lines of action meet at the particle. Standard vector algebra applies to force addition. Concept / Approach: The principle of composition states that two or more concurrent forces can be replaced by their vector sum. The result is a single force (resultant) whose magnitude and direction are obtained by vector addition, and whose line of action passes through the same point as the original forces. Step-by-Step Solution: Represent each force Fi by components along chosen axes.Compute Rx = ΣFi,x and Ry = ΣFi,y.Magnitude R = √(Rx^2 + Ry^2) and direction θ = atan2(Ry, Rx).Place the resultant as a single force R through the common point (the particle). Verification / Alternative check: Graphical polygon-of-forces or parallelogram construction gives the same resultant as analytical components, confirming equivalence. Why Other Options Are Wrong: (b) contradicts vector composition; (c) centre of gravity is irrelevant for a particle problem; (d) a pure couple arises only from equal and opposite non-concurrent forces. Common Pitfalls: Confusing particle mechanics (concurrent forces) with rigid-body mechanics where non-concurrency introduces resultant force plus a couple. Final Answer: May be replaced by a single force acting at the particle (resultant exists).

Question 17

Dynamic equilibrium form of Newton’s second law Let P be the net applied force on a mass m producing acceleration f. The relation P − m f = 0 is called what in mechanics?

Accepted Answer

Introduction / Context: Newton’s second law, P = m f, is often rewritten for analysis as P − m f = 0 by transferring inertia to the left-hand side. This rearrangement is widely known as the equation of dynamic equilibrium or D’Alembert’s principle, enabling dynamic problems to be treated as if in static equilibrium with an added inertial force. Given Data / Assumptions: Net applied force P acts on a body of mass m. Resulting acceleration is f (vector). Classical (nonrelativistic) mechanics. Concept / Approach: D’Alembert’s principle introduces an inertial force F_in = − m f, and imposes equilibrium: ΣF + F_in = 0. This converts a dynamics problem to a statics form, simplifying free-body analysis and allowing the use of equilibrium methods (ΣF = 0, ΣM = 0) with the inertial force included. Step-by-Step Solution: Start with Newton: P = m f.Rearrange: P − m f = 0.Interpret m f as an inertia term; −m f is the inertial force added to external forces to achieve equilibrium. Verification / Alternative check: Applying the method to a translating block or rotating link yields the same accelerations as direct second-law application, confirming equivalence. Why Other Options Are Wrong: (a) names the original Newtonian form, not the rearranged equilibrium form; (c) applies only when accelerations are zero; (d) and (e) do not describe this specific mechanical identity. Common Pitfalls: Forgetting the inertial force is fictitious and used for analysis convenience; mixing signs when moving m f across the equality. Final Answer: Equation of dynamic equilibrium (D’Alembert’s form)

Question 18

Conditions of equilibrium for a rigid body in a plane State the necessary conditions that must be satisfied simultaneously for a coplanar force system to be in static equilibrium.

Accepted Answer

Introduction / Context: To ensure a rigid body in a plane does not translate or rotate, three independent equilibrium equations are required. These are the foundation of structural analysis, machine design, and statics problem solving. Given Data / Assumptions: Coplanar forces acting on a rigid body. Classical statics (no acceleration). Any reference point may be chosen for taking moments. Concept / Approach: Equilibrium demands zero net translation and zero net rotation. Translational equilibrium in 2D requires ΣH = 0 and ΣV = 0. Rotational equilibrium requires ΣM = 0 about any point (if it is zero about one point and ΣH = ΣV = 0, it is zero about all points). Step-by-Step Solution: Impose ΣFx = 0 → no acceleration along x.Impose ΣFy = 0 → no acceleration along y.Impose ΣM_O = 0 about an arbitrary point O → no angular acceleration. Verification / Alternative check: Choosing different moment centers yields consistent results when ΣH = ΣV = 0; otherwise, contradictions appear, revealing imbalance. Why Other Options Are Wrong: (a), (b), or (c) alone do not prevent the other type of motion; (e) ignores translational equilibrium. Common Pitfalls: Forgetting to include couple moments; assuming ΣM = 0 about one special point is sufficient without verifying forces sum to zero. Final Answer: All three: ΣH = 0, ΣV = 0, and ΣM = 0.

Question 19

Principle of transmissibility of a force (rigid-body mechanics) According to the principle of transmissibility, the mechanical effect of a force on a rigid body remains what when the point of application is shifted along its line of action?

Accepted Answer

Introduction / Context: The principle of transmissibility enables engineers to move a force along its own line of action without changing the external effect on a rigid body. This simplifies free-body diagrams and statics reductions, provided the body can be treated as perfectly rigid. Given Data / Assumptions: Rigid-body assumption (no deformation-induced stress redistribution considered). Same line of action maintained during shifting. Moments from couples are treated separately. Concept / Approach: Forces that share the same line of action are statically equivalent in their external effect on a rigid body. Moving the point of application along the line of action does not change the resultant force or net moment about any external point; internal stress distributions can differ, but statics problems focus on external effects. Step-by-Step Solution: Identify the force vector F and its line of action.Shift F along the same line so that position vector r changes only along the line.Moment M_O = r × F remains unchanged for any external point O when the shift is collinear with F. Verification / Alternative check: Compare net force and net moment before and after shifting: both are identical; hence, external equilibrium and reactions are unchanged. Why Other Options Are Wrong: (a) and (d) incorrectly tie effectiveness to C.G.; (b) contradicts the principle; (e) is inapplicable. Common Pitfalls: Applying transmissibility to deformable bodies where internal effects matter; confusing shifting along the line of action with moving off the line, which introduces a couple. Final Answer: The same at every point on its line of action (unchanged effect)

Question 20

Railway/Highway superelevation formulation If G is the track gauge (distance between rail centrelines), v the vehicle speed, g gravitational acceleration, and r the curve radius, the equilibrium superelevation e (height by which outer rail is raised…

Accepted Answer

Introduction / Context: Superelevation (cant) is provided on curved tracks and highways to balance the lateral acceleration and reduce reliance on side friction. The design relation links the required rail height difference to speed, radius, and gauge/width. Given Data / Assumptions: Track gauge (or effective wheelbase width) G. Vehicle travels at speed v on a circular curve of radius r. Gravity g; steady-state condition (no transition dynamics). Equilibrium condition neglecting side friction (pure balancing of centripetal acceleration). Concept / Approach: At equilibrium, the resultant of weight and centrifugal inertia is perpendicular to the track plane. With track banked at angle θ, tan θ equals the required centripetal acceleration divided by g, i.e., tan θ = v^2/(g r). The rail height difference e relates to θ and gauge G by tan θ ≈ e/G for small angles, yielding the design formula. Step-by-Step Solution: Centripetal ratio: v^2/(g r).For small θ, tan θ ≈ θ ≈ e/G.Equate: e/G = v^2/(g r).Solve for e: e = (G * v^2) / (g * r). Verification / Alternative check: Dimensional check: G [L], v^2/(g r) dimensionless → e has length [L], consistent. For larger r or smaller v, e reduces as expected. Why Other Options Are Wrong: (b) omits G; (c)–(e) invert variables incorrectly and are dimensionally inconsistent for length. Common Pitfalls: Confusing gauge G with lane width; using degrees instead of radians for tan θ small-angle approximation. Final Answer: e = (G * v^2) / (g * r)

Applied Mechanics Questions