Question 1

Mass moment of inertia — thin rod comparison  How does the mass moment of inertia of a thin uniform rod about one end compare with that about its midpoint?

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Introduction / Context:Designers frequently switch reference axes using parallel-axis relations. For a slender rod, knowing standard mass moments of inertia allows quick dynamic and strength calculations for beams, pendulums, and linkages.Given Data / Assumptions:Uniform thin rod of length L and mass m.Axis 1: through the centroid (midpoint) and perpendicular to the rod.Axis 2: through one end and perpendicular to the rod.Concept / Approach:Standard results: I_centroid = (1/12) * m * L^2. Using the parallel-axis theorem to shift to the end: I_end = I_centroid + m * (L/2)^2 = (1/12) m L^2 + (1/4) m L^2 = (1/3) m L^2. The ratio I_end / I_centroid = (1/3) / (1/12) = 4.Step-by-Step Solution:Write I_centroid = (1/12) m L^2.Apply parallel-axis: I_end = (1/12) m L^2 + m * (L/2)^2.Compute: I_end = (1/12 + 1/4) m L^2 = (1/3) m L^2.Form ratio: I_end / I_centroid = (1/3) / (1/12) = 4 → “four times”.Verification / Alternative check:Dimensional consistency: both are in kg·m^2 and differ only by the dimensionless factor 4, confirming correctness.Why Other Options Are Wrong:Same/twice/thrice/half: Do not match the computed factor from standard formulas and the parallel-axis theorem.Common Pitfalls:Forgetting to square the shift distance in the parallel-axis term m * d^2 or using area moments instead of mass moments.Final Answer:Four times

Question 2

Vibrations — definition of frequency  Does “frequency of vibrations” mean the number of cycles per second (measured in hertz, Hz)?

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Introduction / Context:In vibration analysis, frequency quantifies how often a periodic motion repeats. Whether analyzing machine imbalance, structural resonance, or signal processing, the definition is consistent across domains.Given Data / Assumptions:Periodic motion with a well-defined cycle.SI units are used: 1 hertz (Hz) = 1 cycle per second.Amplitude or damping does not redefine frequency.Concept / Approach:Frequency f is the reciprocal of the period T: f = 1 / T. It counts complete cycles per unit time and is independent of amplitude. In angular terms, ω = 2π f (rad/s). This definition applies to mechanical vibrations, acoustics, and electrical waveforms alike.Step-by-Step Solution:Identify the repeating pattern (one full cycle).Measure the time for one cycle → T seconds.Compute frequency: f = 1 / T cycles per second = Hz.Verification / Alternative check:For a rotating shaft at 1200 rpm: f = 1200 / 60 = 20 Hz. The same calculation is used for sound waves and electrical signals, confirming universality of the definition.Why Other Options Are Wrong:No / only for sound/electrical: Restricting to specific domains is incorrect; the definition is universal.Depends on amplitude: Frequency is independent of amplitude; changing amplitude does not change cycles per second.Common Pitfalls:Confusing angular frequency ω (rad/s) with f (Hz). They are related by ω = 2π f but are not the same unit.Final Answer:Yes

Question 3

Pendulum clock taken underground  If a pendulum clock is taken 1 km below the Earth's surface (into a mine), it will _____ time compared with at the surface.

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Introduction / Context:Pendulum clocks use gravitational acceleration g to set their period. Moving the clock to locations where g changes will affect its timekeeping. Underground, g is slightly smaller than at the surface, and this influences the oscillation period.Given Data / Assumptions:Pendulum length L is constant.Small-angle approximation holds: simple pendulum model.At a depth of 1 km, effective g is reduced compared with surface g.Concept / Approach:The period of a simple pendulum is T = 2π * sqrt(L / g_eff). When g_eff decreases, the period increases (oscillations take longer). A clock that counts oscillations will then register fewer cycles per true second, meaning it loses time relative to a standard.Step-by-Step Solution:At the surface: T_surface = 2π * sqrt(L / g_surface).In the mine: T_mine = 2π * sqrt(L / g_mine) with g_mine Therefore T_mine > T_surface (slower oscillation rate).Thus the clock falls behind the true time: it loses time.Verification / Alternative check:Numerically, even a small percentage decrease in g produces a proportional increase in period because T ∝ g^(−1/2). Precision clocks correct for local gravity; pendulum regulators are adjusted when moved geographically or in elevation.Why Other Options Are Wrong:Gain: Would require larger g (e.g., at the poles vs equator, not underground).Same rate: Contradicts T dependence on g.Stop / alternate faster-slower: No physical basis under steady conditions.Common Pitfalls:Thinking gravity is stronger underground. Inside a uniform Earth model, gravitational pull decreases with depth due to the shell theorem.Final Answer:lose

Question 4

Area moments — square about its base  State whether the following is true or false: “The second moment of area (area moment of inertia) of a square of side a about its base is a^4 / 3.”

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Introduction / Context:Area moments of inertia are essential in bending stress and deflection calculations. Recognizing standard formulas helps engineers set up beam problems quickly and accurately.Given Data / Assumptions:Plane area: square cross-section of side a.Axis: the base line of the square (edge) lying in the plane of the area.Use standard area-moment formulas.Concept / Approach:For a rectangle of breadth b and height h, the area moment about its base is I_base = b * h^3 / 3. A square is a special rectangle with b = a and h = a, so I_base = a * a^3 / 3 = a^4 / 3.Step-by-Step Solution:Start with rectangle formula: I_base = b * h^3 / 3.Substitute b = a and h = a (square).Compute: I_base = a * a^3 / 3 = a^4 / 3.Therefore, the statement is true.Verification / Alternative check:Using centroidal value I_centroid = b * h^3 / 12 = a^4 / 12 and the parallel-axis theorem to the base at distance a/2: I_base = I_centroid + A * (a/2)^2 = a^4/12 + a^2 * a^2 / 4 = a^4/12 + a^4/4 = a^4/3.Why Other Options Are Wrong:False / mass moment only / centroidal axis: Confuse area vs mass moments or axes; the given base-axis formula is correct.Common Pitfalls:Mixing up centroidal vs base axes and area moments vs mass moments. Always specify the reference axis.Final Answer:True

Question 5

Elevator dynamics — comparing cable tensions  A lift (mass m) moves with the same magnitude of acceleration a in two cases: upward and downward. If the cable tension while moving downward is half of the tension while moving upward, what is the accel…

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Introduction / Context:Elevator problems illustrate Newton’s second law with attention to sign conventions. Cable tension depends on both gravity and acceleration direction, and comparing tensions in two scenarios allows solving for the unknown acceleration.Given Data / Assumptions:Lift mass = m (constant).Upward motion with upward acceleration magnitude a → T_up = m(g + a).Downward motion with downward acceleration magnitude a → T_down = m(g − a).Given: T_down = (1/2) * T_up.Concept / Approach:Apply Newton’s second law along the vertical with the chosen positive direction upward. Write expressions for tension in both cases and use the given ratio to solve for a in terms of g.Step-by-Step Solution:T_up = m(g + a).T_down = m(g − a).Given: m(g − a) = (1/2) * m(g + a).Cancel m and solve: g − a = 0.5g + 0.5a → 0.5g = 1.5a → a = (1/3) g.Verification / Alternative check:Numerical check with g = 9.9 m/s^2 (approx.): a = 3.3 m/s^2. Then T_up = m(13.2), T_down = m(6.6), indeed T_down = 0.5 T_up.Why Other Options Are Wrong:g/2, g/4, 3g/4: Do not satisfy the given tension ratio when substituted.None of these: Incorrect because a consistent value exists: g/3.Common Pitfalls:Using the same sign for acceleration in both cases or forgetting that T_down decreases when accelerating downward.Final Answer:g/3

Question 6

Statics — coplanar concurrent forces  Which description correctly defines coplanar concurrent forces?

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Introduction / Context:Classifying force systems is essential for selecting the right equilibrium equations. “Coplanar concurrent” is a common case in 2D statics where ΣF_x, ΣF_y are sufficient and moments can be taken about the point of concurrency.Given Data / Assumptions:All forces considered act in a single geometric plane.All lines of action intersect at a single point (concurrent).Rigid-body assumptions for statics.Concept / Approach:“Coplanar” means contained in one plane. “Concurrent” means all lines of action pass through one common point. In such systems, the resultant is obtained by vector addition, and equilibrium conditions reduce to ΣF = 0 at that point.Step-by-Step Solution:Check planarity: all forces lie in the same plane → coplanar.Check intersection: all meet at one point → concurrent.Therefore, the correct definition is forces that meet at one point and are coplanar.Verification / Alternative check:Examples include cables meeting at a ring joint in a 2D truss node; solving requires balancing only two scalar components.Why Other Options Are Wrong:Not coplanar / not concurrent: These describe spatial systems or non-concurrent cases requiring moment equations about arbitrary points.Parallel forces in different planes: Neither concurrent nor coplanar.Common Pitfalls:Confusing concurrent with parallel (parallel lines do not meet), or mixing spatial (3D) with planar cases.Final Answer:Forces that meet at one point and their lines of action also lie on the same plane

Question 7

Impact mechanics — kinetic energy loss  In a direct (head-on) impact between two bodies, the loss of kinetic energy _____ on the value of the coefficient of restitution (e).

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Introduction / Context:The coefficient of restitution e measures how “bouncy” a collision is. It directly influences post-impact relative speed and therefore the change in kinetic energy. Understanding this dependence is crucial in collision analysis in mechanics and vehicle safety.Given Data / Assumptions:Straight-line, head-on impact.Coefficient of restitution e in the range 0 ≤ e ≤ 1.No external impulses during the short impact interval.Concept / Approach:For two masses m1 and m2 with initial relative speed u_rel and final relative speed v_rel, restitution states: v_rel = e * u_rel (opposite direction). Total kinetic energy change depends on both masses and e. Perfectly elastic (e = 1) collisions conserve kinetic energy; perfectly inelastic (e = 0) collisions maximize kinetic energy loss for the given masses and initial speeds.Step-by-Step Solution:Use momentum conservation for the system to get final velocities in terms of e.Compute initial and final kinetic energies: K_i and K_f.Energy loss ΔK = K_i − K_f is a function of e (larger loss for smaller e).Verification / Alternative check:Special cases: e = 1 → ΔK = 0 (no loss). e = 0 → bodies stick together (maximum loss). Intermediate values yield intermediate losses, confirming dependence on e.Why Other Options Are Wrong:Does not depend / always zero: Contradicted by the limiting cases above.Independent of masses or relative speed: ΔK also depends on masses and initial relative speed; e is one of several parameters.Common Pitfalls:Assuming momentum conservation implies energy conservation. Momentum is always conserved for isolated impacts; kinetic energy is not unless e = 1.Final Answer:depends

Question 8

Fundamentals of units (SI system) What is the SI unit of power used in physics and engineering measurements?

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Introduction / Context:Power quantifies the rate at which work is done or energy is converted per unit time. This question checks your recall of the correct SI (International System of Units) unit for power used in science and engineering.Given Data / Assumptions:No numerical data are needed.Use standard SI base and derived units.Work is measured in joules; time is measured in seconds.Concept / Approach:By definition, power P = work / time. In SI, work (or energy) is the joule (J) and time is the second (s). Therefore, the SI derived unit of power is joule per second, which has the special name watt (W).Step-by-Step Solution:Start from definition: P = W / t.Replace symbols with SI units: [P] = J / s.Recognize the named derived unit: 1 W = 1 J/s.Therefore, the correct SI unit for power is the watt.Verification / Alternative check:Electrical power also uses watts via P = V * I, where V (volt) * I (ampere) yields joule per second, confirming consistency across mechanical and electrical domains.Why Other Options Are Wrong:horsepower: a non-SI unit (1 hp ≈ 746 W).joule: unit of energy, not power.kg·m: incomplete and not a power unit (and lacks per-second term).Common Pitfalls:Confusing energy (joules) with power (watts). Power always involves “per unit time”.Final Answer:watt

Question 9

Centre of gravity of a right-angled triangle A statement claims: “The centre of gravity (centroid) of a right-angled triangle lies at its geometrical centre.” Determine whether this statement is correct or incorrect.

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Introduction / Context:In statics and engineering mechanics, the centroid (centre of gravity in a uniform field) of standard plane figures is a frequently used result. Right-angled triangles are common in structural and area-moment problems.Given Data / Assumptions:Uniform thickness and density so CG coincides with geometric centroid.Right triangle with legs of lengths b and h meeting at the right angle.Concept / Approach:The centroid of a triangle lies at the intersection of its medians. For a right-angled triangle, measured from the right-angled vertex along the legs, the centroid coordinates are x̄ = b/3 and ȳ = h/3. This is not the “geometrical centre” in the sense of the midpoint of a bounding rectangle or the triangle’s circumcenter.Step-by-Step Solution:Locate medians from each vertex to the midpoint of the opposite side.Their intersection point (centroid) divides each median in a 2:1 ratio from the vertex.For a right triangle with legs b and h: centroid is at (b/3, h/3) from the right-angled vertex.This is not the “centre” of the figure in a symmetric sense (triangles are not centrally symmetric), so the statement is incorrect.Verification / Alternative check:Using area-integration: x̄ = (1/A)∫x dA over the triangular region yields x̄ = b/3 and similarly ȳ = h/3, confirming the standard centroid location.Why Other Options Are Wrong:“Correct” would imply a symmetric centre; a right triangle lacks central symmetry.Common Pitfalls:Confusing centroid with circumcenter or incenter; these are distinct triangle centres and lie at different locations.Final Answer:Incorrect

Question 10

Effect of changing gravitational acceleration on weight If the local gravitational acceleration doubles (g becomes 2g), how does the weight of a given body change? Assume mass remains constant.

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Introduction / Context:Weight is the gravitational force acting on a body. This question tests the relationship between weight, mass, and gravitational acceleration, a foundational concept in mechanics.Given Data / Assumptions:Mass m of the body is constant.Original gravitational acceleration = g; new acceleration = 2g.Weight W is defined as m * g.Concept / Approach:Weight depends linearly on gravitational acceleration. Doubling g doubles the weight for the same mass.Step-by-Step Solution:Original weight: W_old = m * g.New weight: W_new = m * (2g) = 2 * (m * g).Hence W_new = 2 * W_old.Verification / Alternative check:Units remain force (newtons): N = kg * m/s^2. Doubling the acceleration term doubles the numerical value of the force for fixed mass.Why Other Options Are Wrong:Half or same: would require g to be halved or unchanged.Four times: would require g to become 4g, not 2g.Common Pitfalls:Mixing up mass (invariant in this scenario) with weight (force that changes with g). Mass is measured in kilograms; weight is measured in newtons.Final Answer:twice its original weight

Question 11

Collinear but opposite forces (P > Q) Two forces P and Q act along the same straight line but in opposite directions, with P greater than Q. What is the magnitude of the resultant force?

Accepted Answer

Introduction / Context:Resultant of collinear forces is found by algebraic addition with appropriate signs. Here, the forces are opposite in sense along the same line.Given Data / Assumptions:P and Q act along one line.P > Q and directions are opposite.We seek only the magnitude of the resultant.Concept / Approach:Assign a positive direction; take the algebraic sum. If P is chosen positive and Q acts in the negative sense, the resultant R = P − Q. The direction will be that of the larger force, i.e., along P.Step-by-Step Solution:Choose +x along P. Then P = +P, Q = −Q.Algebraic sum: R = P + (−Q) = P − Q.Since P > Q, R > 0, so resultant acts along P.Verification / Alternative check:Vector diagram: two arrows in opposite directions; the smaller cancels part of the larger, leaving difference in the larger’s sense.Why Other Options Are Wrong:P + Q: would apply if both forces acted in the same direction.P/Q or Q/P: dimensionally incorrect for a force resultant.Common Pitfalls:Confusing magnitude with signed result; here only the magnitude is requested, which is the absolute difference.Final Answer:P - Q

Question 12

Projectile motion on a horizontal plane – time of flight For a projectile launched with initial speed u at an angle θ above the horizontal (neglect air resistance), the total time of flight t back to the launch level is:

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Introduction / Context:The time of flight of a projectile returning to its original elevation depends on the vertical component of its initial velocity and gravity. This is a staple result in kinematics.Given Data / Assumptions:Initial speed u, launch angle θ above horizontal.Constant gravitational acceleration g downward.Projectile lands at the same vertical level as launch point.Concept / Approach:Analyze vertical motion. The vertical component of the initial velocity is u sin θ. The projectile rises until vertical velocity becomes zero, then descends symmetrically, making the total time twice the time to reach the peak.Step-by-Step Solution:Vertical velocity at any time: v_y = u sin θ − g t.At the apex: v_y = 0 ⇒ t_up = (u sin θ) / g.Time of flight (symmetric): t = 2 * t_up = (2 u sin θ) / g.Verification / Alternative check:Use displacement equation y = u sin θ * t − (1/2) g t^2 and set y = 0 (besides t = 0). Solving yields t = (2 u sin θ)/g, confirming the result.Why Other Options Are Wrong:(u sin θ)/g: only the ascent time.(u^2 sin 2θ)/g: related to horizontal range, not time.(2 u cos θ)/g: uses the horizontal component incorrectly.Common Pitfalls:Forgetting the factor of 2 or using sin 2θ (a range formula) in the time expression.Final Answer:t = (2 u sin θ) / g

Question 13

Definition of a couple in planar mechanics Two equal and opposite parallel forces whose lines of action are different form a couple. Assess the statement.

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Introduction / Context:Couples are essential in statics because they create pure rotation without net translation. Recognizing when forces constitute a couple is a core skill.Given Data / Assumptions:Two forces F and −F, parallel, separated by a perpendicular distance d.Lines of action do not coincide (nonzero arm).Concept / Approach:A couple is a free vector characterized by its moment M = F * d (about any point). The net force is zero (equal and opposite), but the net moment is nonzero because of the arm between the forces.Step-by-Step Solution:Compute resultant force: F + (−F) = 0.Compute moment about an arbitrary point O: M_O = F * d (direction given by right-hand rule).Since M_O ≠ 0 for d ≠ 0, the system produces rotation only ⇒ a couple.Verification / Alternative check:Shift the reference point; a couple’s moment is invariant with respect to the choice of origin (free vector property), confirming the identification.Why Other Options Are Wrong:“Incorrect” would imply either unequal magnitudes or coincident lines of action; the statement explicitly rules those out.Common Pitfalls:Confusing a couple with a single force pair where lines of action coincide (which would cancel with no moment).Final Answer:Correct

Question 14

Components of acceleration in circular motion When a particle moves along a circular path, does its acceleration have both normal (centripetal) and tangential components?

Accepted Answer

Introduction / Context:Nonlinear motion is analyzed by decomposing acceleration into components aligned with the path (tangential) and toward the center of curvature (normal or centripetal). Understanding this decomposition is vital in dynamics.Given Data / Assumptions:Particle follows a circular trajectory of radius r.Speed may be constant or varying with time.Concept / Approach:Acceleration a is resolved as a = a_t + a_n, where a_t = dv/dt (along the direction of motion) and a_n = v^2 / r (directed toward the center). For circular motion, a_n exists whenever v ≠ 0. If speed varies, a_t is nonzero; if speed is constant, a_t = 0 but a_n remains.Step-by-Step Solution:Normal component: a_n = v^2 / r.Tangential component: a_t = dv/dt.Therefore, both components may be present in general; at least the normal component exists for any nonzero speed.Verification / Alternative check:Uniform circular motion: dv/dt = 0 ⇒ a_t = 0; still a_n = v^2 / r. Nonuniform: both a_t and a_n nonzero, confirming the general statement.Why Other Options Are Wrong:“False” would deny the normal component which is fundamental to circular motion for v ≠ 0.Common Pitfalls:Equating acceleration solely with change in speed; change in direction also constitutes acceleration via the normal component.Final Answer:True

Question 15

Moment of inertia of a solid sphere about a tangential axis Find the area moment of inertia (second moment of mass) of a solid sphere of mass m and radius r about an axis tangent to its surface.

Accepted Answer

Introduction / Context:Moments of inertia for standard solids are key in rotational dynamics. The tangent-axis value can be obtained using the parallel-axis theorem from the centroidal value.Given Data / Assumptions:Solid (uniform) sphere of mass m and radius r.Known centroidal MOI of a solid sphere: I_center = 2 m r^2 / 5 about any diameter.Axis sought is tangent to the sphere (parallel to a diameter and shifted by r).Concept / Approach:Apply the parallel-axis theorem: I_tangent = I_center + m d^2, where d is the perpendicular distance between axes. Here, d = r.Step-by-Step Solution:Start with I_center = 2 m r^2 / 5.Distance from center to tangent axis: d = r.Use parallel-axis theorem: I_tangent = I_center + m r^2.Compute: I_tangent = (2/5) m r^2 + 1 * m r^2 = (2/5 + 5/5) m r^2 = 7 m r^2 / 5.Verification / Alternative check:Dimensional check: units are kg·m^2. Value exceeds the centroidal MOI, as expected when shifting the axis away from the center.Why Other Options Are Wrong:2 m r^2 / 5: centroidal value only, not tangent.2 m r^2 / 3 and 7 m r^2 / 3: incorrect coefficients; no standard derivation yields these.Common Pitfalls:Forgetting to add m r^2 when moving to a tangent axis; mixing up solid versus hollow sphere formulas.Final Answer:7 m r^2 / 5

Question 16

Systems of pulleys – third system (textbook machine theory) For the third system of pulleys with n pulleys, what is the velocity ratio (VR) expressed in terms of n?

Accepted Answer

Introduction / Context:Velocity ratio (VR) is a key performance parameter of simple machines. For classical systems of pulleys (first, second, and third systems), VR depends on how the ropes are reeved and how many pulleys support the moving block.Given Data / Assumptions:Third system of pulleys as defined in standard machine theory texts.Number of pulleys = n.Ideal kinematics without slip; friction ignored for VR.Concept / Approach:In the third system, the free (moving) block is supported by 2n rope segments effectively sharing the load (for the canonical arrangement), which makes the distance moved by the effort twice the distance moved by the load for each supporting segment count n, producing VR = 2n.Step-by-Step Solution:Count effective supporting rope segments under the moving block: 2n.By kinematics, VR = (distance moved by effort) / (distance moved by load).For 2n supporting segments, VR = 2n.Verification / Alternative check:Compare with other systems: first system VR = n; second system VR = 2^n. The third system’s linear dependence VR = 2n contrasts with the exponential dependence of the second system.Why Other Options Are Wrong:n: corresponds to the first system.n^2: not a standard VR in these classic arrangements.2n − 1: sometimes arises in variant counting errors; not the canonical result.Common Pitfalls:Miscounting supporting rope segments or confusing the three classical systems’ formulas.Final Answer:2n

Question 17

Systems of pulleys – first system (textbook machine theory) For the first system of pulleys containing n pulleys, what is the velocity ratio (VR)?

Accepted Answer

Introduction / Context:The velocity ratio (VR) of a machine indicates the ideal mechanical advantage in distance terms. In the first system of pulleys, several pulleys are combined so that the number of supporting segments equals the number of pulleys in the moving block.Given Data / Assumptions:First system of pulleys as defined in standard machine theory.Number of pulleys = n.Ideal ropes (no slip, negligible stretch) and friction ignored for VR.Concept / Approach:For the first system, the moving block is supported by n rope segments. The effort end must move n units of length for the load to rise 1 unit, so VR equals the number of supporting segments, i.e., VR = n.Step-by-Step Solution:Count supporting rope segments under the moving block: n.By definition, VR = distance moved by effort / distance moved by load.Therefore, VR = n.Verification / Alternative check:Check limiting cases: n = 1 gives VR = 1 (a single fixed pulley changes direction only). For n = 2, VR = 2, and so on—matching textbook examples.Why Other Options Are Wrong:n^2: not applicable in first system; confusion with other compound mechanisms.2n or 2n − 1: correspond to other arrangements or counting mistakes.Common Pitfalls:Confusing the first system with the second (VR = 2^n) or third (VR = 2n) systems; always identify the reeving pattern first.Final Answer:n

Question 18

Triangle of forces for a rigid body — equivalent couple: Three non-concurrent forces act on a rigid body and are represented (in magnitude, direction, and line of action) by the three sides of a triangle taken in order. These forces are not in simpl…

Accepted Answer

Introduction / Context:This question tests the triangle (polygon) law of forces for a rigid body when lines of action are non-concurrent. When three forces form a closed triangle taken in order, their vector sum is zero; however, if the forces do not meet at a single point, the system is not a null-resultant but a pure couple. The magnitude of this couple connects directly to the geometry (area) of the force triangle.Given Data / Assumptions:Three coplanar forces with specified magnitudes, directions, and lines of action.The forces correspond to the three sides of a triangle taken sequentially (head-to-tail), so the polygon closes.Forces do not share a common point of concurrency on the rigid body.Concept / Approach:For a particle, a closed force polygon implies equilibrium (net force = 0). For a rigid body, additionally all moments must cancel. If lines of action are not concurrent, the translational resultant may be zero while a non-zero free couple remains. Classical statics shows that the moment of this couple equals twice the area of the triangle of forces (with an appropriate force–length scale).Step-by-Step Solution:Construct the triangle of forces by placing the three vectors head-to-tail in the given order; the triangle closes, indicating ΣF = 0.Because lines of action are distinct (non-concurrent), translate forces to a common reference and evaluate net moment; a residual couple remains.By geometric/statics proof, the moment of this couple M equals 2 * Δ, where Δ is the signed area of the triangle of forces in the chosen scale.Therefore, the equivalent system is a pure couple of moment equal to twice the triangle’s area.Verification / Alternative check:Take any two sides of the triangle as equal-and-opposite forces separated by a perpendicular distance. Their moment equals force * arm. A symmetric construction around the third side recovers M = 2Δ consistently, independent of the chosen pair, confirming uniqueness of the couple.Why Other Options Are Wrong:area of the triangle / half the area: Underestimates the standard result; derivation yields a factor of 2.none of these: Incorrect, because a well-known closed-form exists.Common Pitfalls:Confusing particle equilibrium (ΣF = 0) with rigid-body equilibrium (ΣF = 0 and ΣM = 0).Assuming concurrency just because the force polygon closes. Concurrency concerns lines of action on the body, not the drawn polygon.Final Answer:twice the area of the triangle

Question 19

Reversible vs. non-reversible machines — definition check: A machine which is capable of doing work in the reversed direction after the effort is removed is called a non-reversible machine. Is this statement correct?

Accepted Answer

Introduction / Context:In machine theory, it is important to distinguish between reversible and non-reversible (self-locking) machines. This affects safety, back-driving behavior, and efficiency calculations for hoists, jacks, and geared mechanisms.Given Data / Assumptions:The statement claims that a machine which can reverse and do work after the effort is removed is “non-reversible”.Standard definitions of reversible and non-reversible machines apply.Concept / Approach:A reversible machine is one that, if the load is allowed to act (with effort removed), will back-drive the mechanism and return energy to the source. A non-reversible (self-locking) machine will not move when the effort is removed; friction or geometry prevents back-driving. Therefore, the statement in the stem contradicts the standard definition.Step-by-Step Solution:Definition: Reversible machine ⇒ capable of back-driving under load, delivering work in reverse.Definition: Non-reversible machine ⇒ cannot be back-driven (self-locking) when effort is removed.The statement describes back-driving ability, hence it matches “reversible,” not “non-reversible.”Therefore, the statement is incorrect.Verification / Alternative check:Examples: A low-friction pulley system can back-drive when the load pulls downward (reversible). A screw jack with large friction angle is self-locking and will not descend without applied effort (non-reversible). The behavior clearly differentiates the two cases.Why Other Options Are Wrong:Yes: Accepts a definition that is opposite to standard terminology.Common Pitfalls:Confusing “reversible” with “reversible thermodynamics.” Here it is purely mechanical back-drive capability.Assuming any high-efficiency machine is non-reversible; in fact, higher efficiency often makes back-driving easier.Final Answer:No

Question 20

Polygon law (vector method) for resultant forces — terminology check: The vector (graphical) method of finding the resultant of several coplanar forces is also known as the polygon law of forces. Is this statement correct?

Accepted Answer

Introduction / Context:Engineers frequently use graphical methods—like the triangle or polygon law—to combine multiple forces. Correct terminology matters in communicating design steps and justifications in reports and exams.Given Data / Assumptions:Multiple coplanar forces acting on a body.Resultant obtained by head-to-tail vector addition.Concept / Approach:The polygon law states: if several forces acting on a particle are represented in magnitude and direction by the sides of a polygon taken in order, then the resultant is represented by the closing side of the polygon taken in the opposite order. This is exactly what the vector (graphical) method constructs.Step-by-Step Solution:Draw the first force vector to scale.From its head, draw the second force vector; continue sequentially for all forces.The resultant vector is obtained by joining the tail of the first vector to the head of the last—this is the polygon law.Hence, calling the vector method the “polygon law of forces” is correct and standard.Verification / Alternative check:For two forces, the polygon becomes a parallelogram; the parallelogram law is the special case of the polygon law for n = 2. Consistency across n confirms the terminology.Why Other Options Are Wrong:Incorrect: Would contradict long-established statics nomenclature used in textbooks and codes.Common Pitfalls:Mixing up the direction of the closing side; it must be taken in the opposite order to represent the resultant.Not drawing to scale, which invalidates numerical interpretation.Final Answer:Correct

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