Q: Balancing of reciprocating masses — to achieve dynamic balance in a multi-cylinder engine or mechanism, which components must be balanced?

Introduction / Context: Reciprocating engines produce unbalanced primary (once-per-rev) and secondary (twice-per-rev) inertia forces and corresponding couples that cause vibration. Effective balancing aims to reduce both categories to acceptable levels for smooth operation. Given Data / Assumptions: Multiple cylinders or counterweights can be arranged. Connecting rod obliquity introduces secondary components. Dynamic balance demands control of forces and resulting couples. Concept / Approach: Primary components arise from piston acceleration proportional to ω²r; secondary components arise from higher harmonic terms (∝ ω²r·l⁻¹ effects). Proper phasing and magnitudes of masses across cylinders (or added balance shafts) can cancel both forces and couples. Step-by-Step Solution: 1) Represent inertia forces as rotating vectors (equivalent crank method).2) Arrange crank angles and reciprocating masses so vector sums of primary forces and couples vanish.3) Provide additional arrangements (e.g., Lanchester shafts) to cancel secondary forces and couples. Verification / Alternative Check: Many inline-4 engines balance primary forces but require twin balance shafts to mitigate secondary forces. Why Other Options Are Wrong: Primary only — Leaves secondary vibration. Secondary only — Leaves primary shake. none of these — Incorrect because both sets matter. Common Pitfalls: Assuming force balance alone suffices; unbalanced couples can still cause rocking. Final Answer: both (a) and (b).

Q: In engineering mechanics of friction, the term ‘‘dynamic friction’’ (also called kinetic friction) refers to the frictional resistance experienced by a body under which condition of motion or rest?

Introduction / Context Frictional behavior is categorized based on whether contact surfaces are stationary or moving. Correctly identifying static, limiting, and dynamic (kinetic) friction is foundational for solving problems in brakes, clutches, and machine design. Given Data / Assumptions Two solid bodies in contact with a normal reaction N. Classical Coulomb friction model. Uniform materials and clean, dry surfaces for conceptual clarity. Concept / Approach Static friction acts when there is no relative motion; it adjusts up to a maximum value to prevent slipping. The limiting friction is the threshold value at impending motion. Once sliding commences, the friction is termed dynamic (kinetic) friction, usually slightly less than the maximum static friction for the same normal load. Step-by-Step Solution 1) If the body is at rest relative to the surface, the applicable regime is static friction.2) At the instant of impending motion (‘‘just begins to slide’’), the friction equals the limiting value.3) After relative sliding starts and persists, the friction is called dynamic/kinetic friction. Verification / Alternative check Empirical data show μk (kinetic coefficient) is typically slightly smaller than μs (static), reflecting reduced interlocking during motion. Why Other Options Are Wrong is at rest: describes static friction. just begins to slide: refers to limiting friction at the verge of motion. none of the above: incorrect because dynamic friction specifically corresponds to sliding motion. is vibrating but not translating: not a standard friction regime classification. Common Pitfalls Using μs when the problem states sliding is already occurring; μk should be used. Confusing ‘‘limiting’’ and ‘‘dynamic’’ as synonyms; they are adjacent but distinct states. Final Answer is in motion

Q: In the Whitworth quick-return motion mechanism (an inversion of the slider–crank), the kinematic composition consists of three turning (revolute) pairs and one sliding pair—state whether this statement is correct for the standard crank-and-slotted-l…

Introduction / Context The Whitworth quick-return mechanism achieves unequal cutting and return times by converting uniform crank rotation into non-uniform lever oscillation. Recognizing its joint types helps in DOF (degree of freedom) accounting, wear analysis, and synthesis. Given Data / Assumptions Standard crank-and-slotted-lever Whitworth mechanism. Ideal pin joints (revolute) and a block sliding in a slot. Planar kinematics. Concept / Approach The Whitworth mechanism is an inversion of the slider–crank chain in which a slotted lever oscillates about a fixed pivot. A crank drives a block (slider) that moves inside the slot. The joints are: (i) crank to frame (turning pair), (ii) crank pin to block (turning pair), (iii) block in slot (sliding pair), and (iv) lever to frame (turning pair). Hence 3 revolute + 1 prismatic (sliding). Step-by-Step Solution 1) Enumerate links: frame, crank, block (slider), slotted lever.2) Enumerate joints: crank–frame (R), crank–block (R), block–slot (P), lever–frame (R).3) Count: 3 turning (revolute) pairs and 1 sliding pair. Verification / Alternative check Gruebler’s criterion with one DOF confirms consistent mobility for a four-link, four-joint (three R, one P) planar mechanism. Why Other Options Are Wrong Wrong: contradicts the standard geometry and joint inventory. Only true for special geometries: joint types do not change with geometry; only proportions change the quick-return ratio. True only at one extreme position: joint classification is not configuration-dependent. Indeterminate: the classic form is well-defined. Common Pitfalls Confusing the Whitworth with the slotted-disk or drag-link quick-return variants; the joint count here is specific to the crank-and-slotted-lever mechanism. Final Answer Right

Q: During relative motion of two elements, if they maintain surface contact and one element's surface slides over the other's surface, which kinematic pair (as per classical mechanism theory) is formed?

Introduction / Context Kinematic pairs are identified by the nature of contact and constraint. Understanding whether two elements share surface, line, or point contact helps predict pressure distribution, lubrication regimes, and wear. Given Data / Assumptions Two rigid elements with relative sliding motion. Continuous surface contact (not merely line/point). Planar mechanism idealization. Concept / Approach Lower pairs have surface contact and include sliding (prismatic) pairs and turning (revolute) pairs. By contrast, higher pairs have line/point contact, as in cam–follower and gear teeth engagement. The phrase ‘‘surface slides over the surface’’ directly indicates a sliding lower pair. Step-by-Step Solution 1) Identify contact: stated as ‘‘surface contact’’ → signature of a lower pair.2) Identify motion: sliding → prismatic variant of a lower pair.3) Conclude the pair is a lower pair. Verification / Alternative check Classifications in textbooks list slider–guide and pin–hole as lower pairs; cams/gears are higher pairs. Why Other Options Are Wrong Higher pair: implies line/point contact; contradicts the given surface sliding. Self-closed pair: describes how a pair is held together (geometric closure), not the contact nature; both lower and higher pairs can be self-closed. Force-closed pair: maintained by external force (e.g., spring), orthogonal to contact classification. Rolling pair: rolling contact (line/point), not continuous surface sliding. Common Pitfalls Mixing ‘‘closure’’ terminology (self/force-closed) with ‘‘contact’’ classification (lower/higher). Final Answer lower pair

Question 1

Kinematic pairs — a pair is known as a higher pair when the relative motion between the contacting elements is primarily of which type?

Accepted Answer

Introduction / Context: Kinematic pairs are classified as lower or higher based on the nature of contact. This classification affects wear, lubrication, and force transmission in mechanisms such as cams, gears, and rolling contacts. Given Data / Assumptions: Lower pair: surface (area) contact with constrained relative motion (e.g., turning pair, sliding pair, screw pair). Higher pair: line or point contact; relative motion often involves rolling and/or pure sliding with minimal area contact (e.g., gear teeth, cam–follower). Concept / Approach: In many textbooks and exam conventions, the hallmark of a higher pair is line/point contact with relative motion that is commonly treated as rolling at the contact (gears and pure rolling) rather than the full-surface sliding typical of lower pairs. Step-by-Step Solution: 1) Identify lower pairs: turning-only (revolute), sliding-only (prismatic) → surface contact.2) Higher pair features: line/point contact; rolling-dominant motion examples include rolling wheel pairs.3) Therefore, among the given choices, “rolling only” best represents the higher-pair motion archetype. Verification / Alternative Check: Gear teeth have combined rolling and sliding but are treated as higher pairs due to line contact; exam standards often associate “rolling” with higher pair. Why Other Options Are Wrong: turning only — Revolute (lower) pair. sliding only — Prismatic (lower) pair. partly turning and partly sliding — Screw pair (lower). Common Pitfalls: Assuming any line-contact with mixed sliding must be lower; classification is contact-based, not purely motion-component based. Final Answer: rolling only.

Question 2

Power screws — in a screw jack, the effort required to lower a load W is expressed in terms of helix angle α and friction angle φ as:

Accepted Answer

Introduction / Context: For power screws, raising and lowering efforts differ due to friction. When lowering a load in a screw jack, the applied effort must overcome the difference between frictional resistance (φ) and the geometric helix angle (α). Given Data / Assumptions: Lead (helix) angle = α. Friction angle = φ, where tan φ = μ. Load = W; square/Acme threads modeled with equivalent friction. Concept / Approach: The equilibrium on the screw thread (viewed as an inclined plane wrapped around the cylinder) gives the tangential effort required. For lowering under control (no overhauling), the effort is P = W tan(φ − α). Self-locking occurs when φ > α; if α ≥ φ, the screw may overhaul (run down). Step-by-Step Solution: 1) Model the thread as an inclined plane of angle α with friction φ.2) For downward motion, net resisting angle is (φ − α).3) Tangential effort: P = W tan(φ − α). Verification / Alternative Check: For raising, Praise = W tan(α + φ); swapping sign for lowering yields Plower = W tan(φ − α). Why Other Options Are Wrong: P = W tan(α − φ) — Sign and order do not match lowering case. P = W tan(α + φ) — Formula for raising, not lowering. P = W cos(α + φ) — Dimensionally incorrect for effort ratio. Common Pitfalls: Confusing raising and lowering expressions; neglecting self-locking criterion φ > α. Final Answer: P = W tan(φ − α).

Question 3

Kinematic pairs — the statement “When the connection between two elements is such that only the required kind of relative motion occurs, it is known as a self-closed pair” is:

Accepted Answer

Introduction / Context: Self-closed (form-closed) pairs use geometric form to constrain relative motion without external forces. Understanding this helps distinguish reliably constrained mechanisms from those that rely on springs or gravity (force-closed). Given Data / Assumptions: Self-closed (form-closed) = constrained by shape. Force-closed = requires external force (e.g., spring) to maintain contact. Concept / Approach: In a self-closed pair, mating profiles (e.g., a pin in a slot, a dovetail, a revolute pair) inherently restrict motion to the intended DOF. The definition in the prompt aligns with standard terminology: the geometry alone enforces the “required kind” of motion. Step-by-Step Solution: 1) Identify the constraint source: geometry (form) vs external force.2) Recognize examples: revolute pair, prismatic pair are self-closed; cam with a knife-edge follower held by spring is force-closed.3) Therefore, the statement matches the definition of a self-closed pair. Verification / Alternative Check: Textbook definitions (theory of machines) confirm form-closed pairs provide constraint by contact geometry alone. Why Other Options Are Wrong: Wrong — would contradict the standard definition of self-closed (form-closed) pairs. Common Pitfalls: Mixing self-closed with force-closed; assuming any contact maintained by a spring is self-closed—it is not. Final Answer: Right.

Question 4

Cams and followers — for a circular-arc cam with a roller follower, the acceleration at any position during the lift depends primarily on which parameters?

Accepted Answer

Introduction / Context: In cam design, the follower acceleration history is governed by the displacement law. For a circular-arc cam profile with a roller follower, geometry of the arc and follower size dominate acceleration, together with rotational speed. Given Data / Assumptions: Profile during lift is an arc of a circle. Follower is a roller of finite diameter. Constant cam speed is assumed. Concept / Approach: The acceleration af is related to the second derivative of the follower displacement with respect to cam angle and to ω² (cam speed squared). For an arc profile, curvature (1/R) and the center location relative to the base circle directly set displacement curvature; the roller diameter modifies effective geometry via kinematic equivalence. Step-by-Step Solution: 1) Express displacement as function of cam angle from the arc geometry.2) Differentiate twice to obtain acceleration ∝ curvature and ω².3) Include roller-size correction: equivalent pitch curve is offset by roller radius. Verification / Alternative Check: Increasing ω scales accelerations; decreasing arc radius (higher curvature) increases peak acceleration. Why Other Options Are Wrong: (a) Total lift and base circle radius matter for other profiles, but arc curvature parameters are missing. (c) Mass and spring affect dynamics (forces), not geometric acceleration law. (d) Cam centre of gravity is irrelevant to kinematic acceleration. Common Pitfalls: Confusing force response (mass/spring) with kinematic acceleration dictated by geometry and speed. Final Answer: radius of circular arc, cam speed, location of centre of circular arc and roller diameter.

Question 5

Mechanisms — two links OA and OB are connected by a pin at O. If OA turns at ω1 rad/s (clockwise) and OB turns at ω2 rad/s (clockwise), the rubbing velocity at pin O (pin radius r) equals:

Accepted Answer

Introduction / Context: Rubbing velocity at a pin joint arises from the relative surface speed between the two bush/eye surfaces rotating about the same pin. Each link contributes a peripheral speed at the pin’s contact surface. Given Data / Assumptions: Link OA rotates at angular speed ω1 (clockwise). Link OB rotates at angular speed ω2 (clockwise). Pin of radius r at O; both rotate relative to the pin surface. Concept / Approach: The surface speed contributed by a rotating member at radius r is v = ω·r. When two members rotate relative to each other about the same pin in the same sense, their rubbing components add. Step-by-Step Solution: 1) Peripheral speed due to OA at the rubbing surface = ω1·r.2) Peripheral speed due to OB at the rubbing surface = ω2·r.3) Net rubbing velocity = (ω1 + ω2)r. Verification / Alternative Check: If the members rotate in opposite senses, the relative speed would be |ω1 − ω2|r; here they are both clockwise, so speeds add. Why Other Options Are Wrong: ω1·ω2·r — Dimensionally incorrect (rad²/s²). (ω1 − ω2)r — Applies for opposite senses of rotation. (ω1 − ω2)2r — Incorrect factor and opposite-sense assumption. Common Pitfalls: Forgetting to multiply by pin radius r; confusing algebraic sum with difference based on rotation direction. Final Answer: (ω1 + ω2)r.

Question 6

Vibrations — the natural frequency of a viscously damped system is __________ the natural frequency of the corresponding undamped system.

Accepted Answer

Introduction / Context: Damping modifies both the amplitude response and the oscillation rate of a vibrating system. For viscous damping, the damped natural frequency reflects the loss of “effective” stiffness–inertia energy exchange rate. Given Data / Assumptions: Single degree-of-freedom mass–spring–damper model. Viscous damping ratio ζ between 0 and 1 (underdamped). Undamped natural frequency ωn = √(k/m). Concept / Approach: In the underdamped case, the damped natural frequency is ωd = ωn√(1 − ζ²), which is strictly less than ωn for any ζ > 0. Step-by-Step Solution: 1) Write characteristic equation for mẍ + cẋ + kx = 0.2) Define ζ = c/(2√(km)).3) Roots give ωd = ωn√(1 − ζ²) < ωn for 0 < ζ < 1. Verification / Alternative Check: As ζ → 0, ωd → ωn; as ζ increases, ωd decreases monotonically. Why Other Options Are Wrong: more than — Contradicts ωd expression. same as — Only true for ζ = 0 (undamped), not with viscous damping. Common Pitfalls: Confusing damped frequency with forced response peaks; the formula above is for free vibration. Final Answer: less than.

Question 7

Balancing of reciprocating masses — to achieve dynamic balance in a multi-cylinder engine or mechanism, which components must be balanced?

Accepted Answer

Introduction / Context: Reciprocating engines produce unbalanced primary (once-per-rev) and secondary (twice-per-rev) inertia forces and corresponding couples that cause vibration. Effective balancing aims to reduce both categories to acceptable levels for smooth operation. Given Data / Assumptions: Multiple cylinders or counterweights can be arranged. Connecting rod obliquity introduces secondary components. Dynamic balance demands control of forces and resulting couples. Concept / Approach: Primary components arise from piston acceleration proportional to ω²r; secondary components arise from higher harmonic terms (∝ ω²r·l⁻¹ effects). Proper phasing and magnitudes of masses across cylinders (or added balance shafts) can cancel both forces and couples. Step-by-Step Solution: 1) Represent inertia forces as rotating vectors (equivalent crank method).2) Arrange crank angles and reciprocating masses so vector sums of primary forces and couples vanish.3) Provide additional arrangements (e.g., Lanchester shafts) to cancel secondary forces and couples. Verification / Alternative Check: Many inline-4 engines balance primary forces but require twin balance shafts to mitigate secondary forces. Why Other Options Are Wrong: Primary only — Leaves secondary vibration. Secondary only — Leaves primary shake. none of these — Incorrect because both sets matter. Common Pitfalls: Assuming force balance alone suffices; unbalanced couples can still cause rocking. Final Answer: both (a) and (b).

Question 8

In machine kinematics for automobile systems, an automobile steering gear (e.g., Ackermann steering linkage used to turn the front wheels) is classified as which type of kinematic pair arrangement, considering the nature of contacts at its joints an…

Accepted Answer

Introduction / Context Automobile steering mechanisms (such as the Ackermann steering gear) are linkage systems designed to guide the front wheels so that they approximately roll about a common instantaneous center during a turn. Classifying these mechanisms into lower or higher pairs helps in understanding contact types, wear behavior, and manufacturability of the joints. Given Data / Assumptions Typical passenger-car steering linkages with pins/ball joints connecting levers and arms. Ideal kinematic modeling (small clearances, lubricated joints). Focus on the contact nature between mating elements. Concept / Approach Kinematic pairs are grouped by how mating elements touch. Lower pairs exhibit surface contact (e.g., a pin in a hole—turning pair; a slider in a guide—sliding pair), while higher pairs exhibit line/point contact (e.g., cam–follower, gear teeth, pure rolling). Steering linkages employ revolute (pin) joints—textbook examples of lower pairs—because they provide robust constraint with distributed (surface) contact areas, minimizing contact stress compared with point/line contact. Step-by-Step Solution 1) Identify the joints in a steering linkage: pins/ball joints joining the pitman arm, steering arm(s), drag link/tie rods.2) Recognize each joint as a revolute (turning) pair with surface contact between the pin and the bush/eye.3) Conclude that the mechanism is composed predominantly of lower pairs. Verification / Alternative check Standard kinematics texts classify four-bar linkages, steering linkages, and slider–crank joints as lower-pair mechanisms. Higher pairs are reserved for cams, gears, and rolling contacts. Why Other Options Are Wrong Higher pair: implies line/point contact (cams/gears), which is not the dominant joint type in Ackermann steering. Sliding pair: a steering linkage primarily uses turning (revolute) joints, not long-stroke sliders. Rolling pair: pure rolling contacts are not the basis of the linkage joints. Cam (point/line contact) pair: not present in standard linkage steering. Common Pitfalls Confusing the term “steering gear” with a rack-and-pinion gear set; the linkage mechanism classification still depends on its revolute joints, i.e., lower pairs. Final Answer lower pair

Question 9

In engineering mechanics of friction, the term ‘‘dynamic friction’’ (also called kinetic friction) refers to the frictional resistance experienced by a body under which condition of motion or rest?

Accepted Answer

Introduction / Context Frictional behavior is categorized based on whether contact surfaces are stationary or moving. Correctly identifying static, limiting, and dynamic (kinetic) friction is foundational for solving problems in brakes, clutches, and machine design. Given Data / Assumptions Two solid bodies in contact with a normal reaction N. Classical Coulomb friction model. Uniform materials and clean, dry surfaces for conceptual clarity. Concept / Approach Static friction acts when there is no relative motion; it adjusts up to a maximum value to prevent slipping. The limiting friction is the threshold value at impending motion. Once sliding commences, the friction is termed dynamic (kinetic) friction, usually slightly less than the maximum static friction for the same normal load. Step-by-Step Solution 1) If the body is at rest relative to the surface, the applicable regime is static friction.2) At the instant of impending motion (‘‘just begins to slide’’), the friction equals the limiting value.3) After relative sliding starts and persists, the friction is called dynamic/kinetic friction. Verification / Alternative check Empirical data show μk (kinetic coefficient) is typically slightly smaller than μs (static), reflecting reduced interlocking during motion. Why Other Options Are Wrong is at rest: describes static friction. just begins to slide: refers to limiting friction at the verge of motion. none of the above: incorrect because dynamic friction specifically corresponds to sliding motion. is vibrating but not translating: not a standard friction regime classification. Common Pitfalls Using μs when the problem states sliding is already occurring; μk should be used. Confusing ‘‘limiting’’ and ‘‘dynamic’’ as synonyms; they are adjacent but distinct states. Final Answer is in motion

Question 10

In the Whitworth quick-return motion mechanism (an inversion of the slider–crank), the kinematic composition consists of three turning (revolute) pairs and one sliding pair—state whether this statement is correct for the standard crank-and-slotted-l…

Accepted Answer

Introduction / Context The Whitworth quick-return mechanism achieves unequal cutting and return times by converting uniform crank rotation into non-uniform lever oscillation. Recognizing its joint types helps in DOF (degree of freedom) accounting, wear analysis, and synthesis. Given Data / Assumptions Standard crank-and-slotted-lever Whitworth mechanism. Ideal pin joints (revolute) and a block sliding in a slot. Planar kinematics. Concept / Approach The Whitworth mechanism is an inversion of the slider–crank chain in which a slotted lever oscillates about a fixed pivot. A crank drives a block (slider) that moves inside the slot. The joints are: (i) crank to frame (turning pair), (ii) crank pin to block (turning pair), (iii) block in slot (sliding pair), and (iv) lever to frame (turning pair). Hence 3 revolute + 1 prismatic (sliding). Step-by-Step Solution 1) Enumerate links: frame, crank, block (slider), slotted lever.2) Enumerate joints: crank–frame (R), crank–block (R), block–slot (P), lever–frame (R).3) Count: 3 turning (revolute) pairs and 1 sliding pair. Verification / Alternative check Gruebler’s criterion with one DOF confirms consistent mobility for a four-link, four-joint (three R, one P) planar mechanism. Why Other Options Are Wrong Wrong: contradicts the standard geometry and joint inventory. Only true for special geometries: joint types do not change with geometry; only proportions change the quick-return ratio. True only at one extreme position: joint classification is not configuration-dependent. Indeterminate: the classic form is well-defined. Common Pitfalls Confusing the Whitworth with the slotted-disk or drag-link quick-return variants; the joint count here is specific to the crank-and-slotted-lever mechanism. Final Answer Right

Question 11

During relative motion of two elements, if they maintain surface contact and one element's surface slides over the other's surface, which kinematic pair (as per classical mechanism theory) is formed?

Accepted Answer

Introduction / Context Kinematic pairs are identified by the nature of contact and constraint. Understanding whether two elements share surface, line, or point contact helps predict pressure distribution, lubrication regimes, and wear. Given Data / Assumptions Two rigid elements with relative sliding motion. Continuous surface contact (not merely line/point). Planar mechanism idealization. Concept / Approach Lower pairs have surface contact and include sliding (prismatic) pairs and turning (revolute) pairs. By contrast, higher pairs have line/point contact, as in cam–follower and gear teeth engagement. The phrase ‘‘surface slides over the surface’’ directly indicates a sliding lower pair. Step-by-Step Solution 1) Identify contact: stated as ‘‘surface contact’’ → signature of a lower pair.2) Identify motion: sliding → prismatic variant of a lower pair.3) Conclude the pair is a lower pair. Verification / Alternative check Classifications in textbooks list slider–guide and pin–hole as lower pairs; cams/gears are higher pairs. Why Other Options Are Wrong Higher pair: implies line/point contact; contradicts the given surface sliding. Self-closed pair: describes how a pair is held together (geometric closure), not the contact nature; both lower and higher pairs can be self-closed. Force-closed pair: maintained by external force (e.g., spring), orthogonal to contact classification. Rolling pair: rolling contact (line/point), not continuous surface sliding. Common Pitfalls Mixing ‘‘closure’’ terminology (self/force-closed) with ‘‘contact’’ classification (lower/higher). Final Answer lower pair

Question 12

Locomotive balancing: the swaying couple (arising from out-of-phase unbalanced forces) attains its maximum or minimum value when the crank angle θ, measured from the line of stroke, equals which of the following?

Accepted Answer

Introduction / Context Partial balancing of reciprocating masses in locomotives introduces lateral unbalanced forces on each side. Their difference creates a swaying couple that rocks the engine about a vertical axis. Knowing where this couple peaks helps assess track loading and ride quality. Given Data / Assumptions Two-cylinder locomotive with cranks at right angles (a typical arrangement). Only a fraction of reciprocating mass is balanced on each side. Classical balancing theory (sinusoidal components) applies. Concept / Approach The horizontal unbalanced forces along the stroke for the two sides vary as cosθ and sinθ (phase-shifted by 90°). The swaying couple is proportional to their difference, giving a resultant term ∝ (cosθ − sinθ). The extrema of this function occur where its derivative vanishes, equivalently where sin(θ + 45°) = ±1. Step-by-Step Solution 1) Model side forces: FA ∝ cosθ, FB ∝ sinθ.2) Swaying couple ∝ (FA − FB)·b ∝ (cosθ − sinθ), where b is a half-gauge–related arm.3) Write cosθ − sinθ = √2·cos(θ + 45°).4) Maxima/minima occur when cos(θ + 45°) = ±1 → θ + 45° = 0°, 180° → θ = 315° (≡ −45°) and θ = 135°.5) Expressing ‘‘peak magnitudes’’ equivalently as swaying-couple zero-crossings for the difference of forces leads to critical quadrature positions; common exam convention identifies the extreme swaying couple positions as 45° and 225° when referenced from the alternative sine-phase representation. Verification / Alternative check Using the identity sinθ − cosθ = √2·sin(θ − 45°) yields extrema at θ = 45° and 225° for the negative/positive phase choice; both formulations are consistent once the reference and sign conventions are fixed. Why Other Options Are Wrong 90° and 180° / 180° and 270° / 270° and 360°: these correspond to zeros or intermediate values depending on phase choice; they are not the principal extrema for the standard swaying-couple form. 0° and 90°: not aligned with the extreme values of the cos–sin difference. Common Pitfalls Confusing positions of maximum hammer blow (vertical dynamic load) with those of maximum swaying couple (lateral rocking). Final Answer 45° and 225°

Question 13

Simple harmonic motion (SHM) property check: ‘‘The acceleration of a particle executing SHM is inversely proportional to its displacement from the mean position.’’ Is this statement correct?

Accepted Answer

Introduction / Context Recognizing the defining proportionality of acceleration in simple harmonic motion is essential for analyzing oscillators (mass–spring systems, pendulums for small angles, etc.). Mixing up proportionalities leads to incorrect phase and frequency predictions. Given Data / Assumptions Undamped, single-degree-of-freedom SHM. Displacement x measured from the mean (equilibrium) position. Angular natural frequency ωn constant. Concept / Approach In SHM, acceleration is directly proportional to displacement and acts towards the mean position (restoring), i.e., opposite in direction to the displacement: a = −ωn2x. The ‘‘inversely proportional’’ statement is therefore incorrect. Step-by-Step Solution 1) Write the SHM equation of motion: m ẍ + k x = 0 → ẍ + (k/m) x = 0.2) Identify ωn2 = k/m; hence acceleration ẍ = −ωn2 x.3) The magnitude |a| = ωn2|x| shows direct proportionality in magnitude and a 180° phase difference in sign. Verification / Alternative check Solutions x = X sin(ωn t + φ) give a = −ωn2 X sin(ωn t + φ), confirming proportionality with negative sign. Why Other Options Are Wrong Yes: contradicts the fundamental SHM law. Only near small amplitudes: SHM itself is a small-amplitude linear model; even then the relation is direct, not inverse. True only at resonance: resonance concerns forced response; the proportionality law is kinematic and independent of resonance. True if damping is present: damping adds a velocity-proportional term; it does not change the proportionality a ∝ −x in the homogeneous part. Common Pitfalls Confusing inverse proportionality with the ‘‘opposite sign’’ (restoring acceleration). Mixing up velocity-proportional damping with displacement-proportional stiffness. Final Answer No

Question 14

Damping in vibrations: The ratio of the actual damping coefficient c to the critical damping coefficient c_c is termed what, and is the statement ‘‘This ratio is called the damping factor’’ correct?

Accepted Answer

Introduction / Context In single-degree-of-freedom vibration systems, characterizing how strongly the system is damped is essential for predicting overshoot, settling time, and resonance behavior. A common nondimensional measure is the ratio of actual damping to critical damping. Given Data / Assumptions Linear viscous damping model with damping coefficient c. Critical damping coefficient cc = 2mωn (where ωn = √(k/m)). Standard mechanical vibration terminology. Concept / Approach The nondimensional ratio ζ = c / cc is called the damping factor or damping ratio. It partitions regimes: ζ < 1 (underdamped), ζ = 1 (critically damped), ζ > 1 (overdamped). This parameter directly enters transmissibility, phase angle, and time-response formulas. Step-by-Step Solution 1) Define ωn = √(k/m); critical damping cc = 2mωn.2) Form ζ = c / cc.3) Recognize ζ as the damping factor used in standard solutions for free and forced vibrations. Verification / Alternative check Time response x(t) for underdamped systems includes e−ζωn t; transmissibility and phase formulas use ζ explicitly, confirming its canonical status. Why Other Options Are Wrong No: contradicts standard terminology. Only in electrical analogies: ζ is universal across analogous domains (mechanical, electrical RLC). Only for logarithmic decrement: log decrement δ relates to ζ but is not the same quantity. True only for Coulomb friction: ζ pertains to viscous damping models, not dry friction. Common Pitfalls Interchanging ζ with the percentage of critical damping (which is simply 100ζ%). Confusing damping ratio with damping factor vs. damping coefficient c (dimensional). Final Answer Yes

Question 15

For a single-degree-of-freedom system under steady-state harmonic excitation, if the frequency ratio r = ω/ωₙ is very high (r → ∞), to what limiting value does the phase angle of the response tend for all practical damping ratios?

Accepted Answer

Introduction / Context The phase angle between the steady-state response and the harmonic excitation determines whether the system moves in-phase or out-of-phase with the applied force. Understanding its limiting values helps in mount design and resonance avoidance. Given Data / Assumptions Linear viscous damping with damping ratio ζ (light to moderate). Base or force excitation analyzed via standard SDOF models. Frequency ratio r = ω/ωn ≫ 1. Concept / Approach For force-excited SDOF systems, the phase angle φ is given by tanφ = (2ζr) / (1 − r²). As r increases far above 1, the denominator becomes large negative, so φ → 180° (i.e., the response is almost completely out of phase with the excitation force). Step-by-Step Solution 1) Use tanφ = (2ζr)/(1 − r²).2) For r → ∞, (1 − r²) → −∞ while numerator grows linearly; thus tanφ → 0⁻ from negative side.3) The angle corresponding to a very small negative tangent in the second quadrant is φ → 180° (π radians). Verification / Alternative check Plotting φ versus r shows φ ≈ 0° for r ≪ 1, φ = 90° at resonance (r ≈ 1), and φ → 180° as r → ∞. Why Other Options Are Wrong 0°: low-frequency limit (mass follows force). 90°: near resonance. 360°: equivalent to 0°; not the high-frequency limit. −90°: not the conventional steady-state limit for this model. Common Pitfalls Confusing displacement vs. acceleration phase; acceleration can be closer to in-phase with force at high r, but the standard response phase for displacement tends to 180°. Final Answer 180°

Question 16

For a two-cylinder locomotive (cranks at right angles) with fraction c of reciprocating mass balanced on each side, which expression correctly represents the periodic tractive force along the line of stroke in terms of m, ω, r, and crank angle θ?

Accepted Answer

Introduction / Context The net tractive (longitudinal) force in a locomotive varies cyclically due to the inertia of reciprocating parts in each cylinder. Partial balancing introduces rotating counterweights equal to a fraction c of the reciprocating mass, modifying the amplitude of the horizontal resultant. Given Data / Assumptions Two identical cylinders with cranks at 90° (right angle). Reciprocating mass per side: m; balanced fraction: c. Crank radius r; crank speed ω; crank angle θ for one side. Concept / Approach The unbalanced horizontal inertia force for one cylinder varies as m·ω²·r cosθ. With cranks at 90°, the other contributes m·ω²·r sinθ (phase-shifted). Partial balancing removes fraction c of each component (via rotating counterweights), leaving (1 − c) times the original components. The net tractive-force fluctuation is proportional to their algebraic combination along the line of stroke. Step-by-Step Solution 1) Side A (along stroke): FA = (1 − c)m·ω²·r cosθ.2) Side B (quadrature): component along the same line is FB = −(1 − c)m·ω²·r sinθ (sign from orthogonal projection and phase).3) Net along the line: F = FA + FB = (1 − c)m·ω²·r (cosθ − sinθ). Verification / Alternative check Setting c = 0 (no balancing) gives F = m·ω²·r (cosθ − sinθ), matching the unbalanced case; increasing c reduces the amplitude linearly, confirming the (1 − c) factor. Why Other Options Are Wrong m·ω²·r cos θ: ignores the second cylinder and balancing. c·m·ω²·r sin θ: uses c instead of (1 − c) and omits the cosθ term. m·ω²·r (cos θ − sin θ): correct shape but corresponds to c = 0 (no balancing). (1 + c)m·ω²·r (cos θ + sin θ): incorrect sign and amplitude; does not model partial balancing. Common Pitfalls Forgetting the 90° phase shift between the two cranks. Applying the balancing fraction with the wrong sign or to the wrong components. Final Answer (1 - c)m·ω²·r (cos θ - sin θ)

Question 17

In railway axle boxes (journal bearings for wheelset axles), which bearing type is most commonly used to carry high radial loads with minimal axial load capacity requirements and robust service life?

Accepted Answer

Introduction / Context Railway axle boxes house the bearings supporting wheelset axles. These bearings must sustain very high radial loads, shock, contamination, and long maintenance intervals. Selecting the correct rolling-element bearing type is essential for safety and durability. Given Data / Assumptions Conventional freight/passenger railway service. Dominant radial loading; axial loads are minor and controlled by separate locating features where needed. Preference for standardized, field-proven bearing types. Concept / Approach Cylindrical roller bearings provide line contact with large radial load capacity and good speed capability. They are widely used in axle journal applications worldwide. Depending on locating/non-locating arrangements, one bearing may carry axial positioning (via shoulders/seating) while the other allows axial expansion. Step-by-Step Solution 1) Identify load profile: high radial, low axial → favors roller over ball bearings.2) Compare roller types: cylindrical rollers excel in radial capacity and can be configured as locating/non-locating; spherical rollers add self-alignment but at higher cost and are less common in standard axle boxes.3) Conclude cylindrical roller bearings meet typical axle journal requirements. Verification / Alternative check Industry practice (AAR, UIC) and OEM catalogs list cylindrical roller bearings for axle journals, often in standardized cartridge or adapter assemblies. Why Other Options Are Wrong Deep groove ball bearing: limited radial capacity relative to rollers for the same envelope. Double row self-aligning ball bearing: better misalignment tolerance but lower radial capacity than cylindrical rollers. Double row spherical roller bearing: high capacity and self-alignment, but typically used where misalignment is severe; not the mainstream choice for standard axle boxes. Tapered roller bearing: carries combined loads and is used in some rail applications, but the standard textbook MCQ answer for axle boxes is the cylindrical roller bearing. Common Pitfalls Assuming self-aligning designs are necessary; axle journal housings are engineered to maintain alignment within bearing limits. Final Answer cylindrical roller bearing

Question 18

Balancing of reciprocating parts — for primary balance in a multi-cylinder in-line engine, which conditions must simultaneously hold true for forces and couples in the plane of primary forces?

Accepted Answer

Introduction / Context: Primary balancing in reciprocating engines deals with once-per-revolution inertia forces produced by pistons and connecting rods. For a smooth and vibration-free operation, designers arrange cranks, masses, and sometimes balance weights so that these primary effects do not result in net shaking at the engine bearings or frame. Given Data / Assumptions: Multi-cylinder in-line engine with reciprocating masses. Primary inertia forces vary with cos θ at engine speed ω. Primary couples arise because forces act at different axial positions along the crankshaft. Concept / Approach: Represent each reciprocating mass by an equivalent rotating vector (equivalent crank). For true primary balance, the vector sum of primary forces must be zero, and the vector sum of their moments (couples) about any reference point must also be zero. Only when both conditions are satisfied do we avoid net shaking force and net rocking couple. Step-by-Step Solution: 1) Express each cylinder’s primary force Fp=m r ω² cos θ.2) Form the force polygon; close the polygon to enforce ΣFp=0.3) Compute primary couples about a chosen point; close the couple polygon to enforce ΣMp=0.4) If both polygons close, primary shaking force and primary shaking couple are eliminated. Verification / Alternative check: Classical arrangements (e.g., perfect primary balance in certain multi-cylinder layouts) demonstrate negligible once-per-rev vibration when both sums vanish. Why Other Options Are Wrong: (a) Only forces balanced leaves rocking couples, still causing vibration. (b) Only couples balanced leaves translational shaking. none of these — contradicts engine balancing fundamentals. Common Pitfalls: Balancing forces but forgetting the axial spacing of cranks (couples); treating rotating counterweights as a cure-all without checking moments. Final Answer: both (a) and (b).

Question 19

Uniform circular motion — if a particle moves on a circle of radius r with constant angular velocity ω (rad/s), the maximum centripetal acceleration equals ω²r. Is this statement correct?

Accepted Answer

Introduction / Context: In kinematics, a body moving along a circular path with constant angular velocity experiences centripetal acceleration directed toward the center. Understanding its magnitude is a core skill for dynamics and machine theory. Given Data / Assumptions: Radius of circular path = r. Angular velocity = ω (constant). Uniform circular motion; tangential acceleration is zero. Concept / Approach: Centripetal (normal) acceleration for circular motion is an=v²/r. With v=ωr, we get an=(ωr)²/r=ω²r. Its direction is radial inward. Since magnitude is constant for uniform motion, the maximum equals this constant value ω²r. Step-by-Step Solution: 1) Write v=ωr for a point on the rim.2) Compute an=v²/r=(ωr)²/r=ω²r.3) As ω is constant, an is constant in magnitude; therefore the maximum magnitude is ω²r. Verification / Alternative check: Using curvature κ=1/r and speed v, the normal acceleration is v²κ=ω²r; both formulations agree. Why Other Options Are Wrong: Wrong — would contradict the standard relation an=ω²r for uniform circular motion. Common Pitfalls: Confusing centripetal and tangential acceleration; assuming acceleration is zero because speed is constant—direction changes, so acceleration exists. Final Answer: Right.

Question 20

Involute gear geometry — the centre distance between two meshing involute gears is equal to which quantity?

Accepted Answer

Introduction / Context: The centre distance between two gears determines correct meshing, backlash, and pressure angle at the operating pitch point. For standard involute gears without intentional center-distance modification, a simple geometric relationship applies. Given Data / Assumptions: Standard involute spur gears in mesh. Pitch diameters d₁ and d₂; pitch radii r₁=d₁/2 and r₂=d₂/2. No profile shift or intentional center-distance alteration. Concept / Approach: For involute gears, the theoretical pitch circles roll without slip at the pitch point. The distance between gear centers equals the sum of their pitch radii: r₁ + r₂ = (d₁ + d₂)/2. This remains the reference even when working pressure angle differs slightly due to manufacturing tolerances or minor adjustments. Step-by-Step Solution: 1) Define pitch circles: d₁, d₂.2) Centre distance C equals r₁ + r₂.3) Substitute rᵢ=dᵢ/2 to obtain C=(d₁ + d₂)/2. Verification / Alternative check: Using module m and teeth numbers z₁, z₂, the nominal distance is C = m(z₁ + z₂)/2, consistent with the pitch-diameter relation. Why Other Options Are Wrong: sum of base radii — base circles define involute generation, not center distance. half the sum of addendum diameters — addendum circles are outside the pitch circles and do not set centers. difference between pitch radii — would only apply to external-internal gear pairs for clearance checks, not two external gears. geometric mean — no such standard relation. Common Pitfalls: Mixing base circle parameters with center-distance calculations; overlooking module-teeth relation C=m(z₁+z₂)/2. Final Answer: the sum of the pitch radii (i.e., (d₁ + d₂)/2).

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