Q: Centroid of a semicircle:\nThe centre of gravity (centroid) of a semicircle of radius r lies at what distance from its base (the diameter), measured along the vertical radius toward the centre?

Introduction / Context: Locating centroids is essential in structural analysis, fluid mechanics (hydrostatic forces), and product design. The semicircle is a standard shape whose centroid location must be memorized or re-derived efficiently for composite-area calculations. Given Data / Assumptions: Planar semicircle of radius r. Base is the diameter; distance is measured perpendicularly from this base along the symmetry axis (vertical radius). Uniform density and thickness for geometric centroid. Concept / Approach: By symmetry, the centroid lies on the central vertical radius. Its distance from the diameter is obtained by first principles (area moment integral) or by recalling the standard result. The well-known formula gives the ȳ coordinate from the base as 4r / (3π). Step-by-Step Solution: Let the diameter lie along the x-axis with origin at the circle's center; the semicircle occupies y ≥ 0.The centroid of the semicircular area measured from the center along y is y_c_from_center = 4r / (3π).Therefore, measured from the base (the diameter) upward along the vertical radius, the same distance is y_c = 4r / (3π).Hence the required distance from the base is 4r / 3π. Verification / Alternative check: Numerically, 4 / (3π) ≈ 0.424. This places the centroid between the base and the circle's center (at 0.424 r above the diameter), which is physically reasonable because more area mass lies near the base than near the arc. Why Other Options Are Wrong: 3r / 8 = 0.375 r: Too small; not the standard centroid value. 8r / 3: Dimensionally wrong (greater than r). 3r / 4π ≈ 0.239 r: Too small; not the canonical result. Common Pitfalls: Confusing the centroid from the center versus from the base—the same numeric distance applies here because the base is exactly one radius below the center. Mixing arc centroid (of a semicircular arc) with area centroid (of a filled semicircle). Final Answer: 4r / 3π

Q: Elevator dynamics — Cable tension comparison:\nIn which case is the tension in the cable supporting a lift (elevator) greater: when the lift is moving upwards with acceleration, or when it is moving downwards with acceleration?

Introduction / Context: Elevator (lift) problems are standard applications of Newton's second law. Comparing cable tension during accelerated motion is important for sizing hoists, motors, and safety brakes. The direction of acceleration relative to weight determines whether the cable must supply extra force or is partially relieved. Given Data / Assumptions: Lift of mass m moving vertically. Acceleration magnitude = a; direction either upward or downward. Gravity acts downward with magnitude W = m * g. Cable mass and pulley friction neglected for conceptual comparison. Concept / Approach: Apply ΣF = m * a along the vertical direction. Tension T counters weight and provides acceleration. The sign of a changes the required T. Upward acceleration requires T > W; downward acceleration requires T m * (g − a) for a > 0.Therefore, tension is greater when the lift accelerates upward. Verification / Alternative check: If a = 0 (uniform speed), T = W in both directions. As a increases upward, T increases above W; as a increases downward, T decreases below W but remains positive as long as a W − m a exceeds T for upward case, which contradicts ΣF = m a. same in both cases: Only true for a = 0 (not our accelerated scenario). indeterminate: Laws of motion determine this unambiguously. Common Pitfalls: Confusing the sign convention, leading to T = m * (g − a) for upward motion (incorrect). Thinking speed (not acceleration) changes tension; constant-speed motion does not alter T from W. Final Answer: upwards

Question 1

Newton’s second law – relation among force, mass, and acceleration
Newton’s second law of motion ________ a relation between the applied force and the mass and acceleration of a moving body.

Accepted Answer

Introduction / Context: Newton’s second law formalizes dynamics by connecting cause (net force) and effect (acceleration) with the property of inertia (mass). It is foundational in engineering mechanics and physics. Given Data / Assumptions: The net external force F acts on a body of mass m. Acceleration a is the time rate of change of velocity of the body. Classical mechanics regime (non-relativistic speeds). Concept / Approach: The law states: F_net = m * a (vector form). It directly relates the magnitude and direction of the net force to the mass and acceleration of the body. For varying mass systems, a generalized form uses momentum: F_net = d(mv)/dt. Step-by-Step Solution: Start from momentum p = m v.Newton’s second law: F_net = dp/dt. For constant mass, dp/dt = m dv/dt = m a.Thus, the acceleration produced is proportional to the net applied force and inversely proportional to the mass. Verification / Alternative check: Dimensional consistency: [F] = kg·m/s^2, [m a] = kg·m/s^2. Experimental validation spans from particle dynamics to rigid-body motion. Why Other Options Are Wrong: ‘‘does not give’’ contradicts the statement of the second law. ‘‘only for constant velocity’’ would imply zero acceleration and thus zero net force, which is a special case of no net force, not the general law. ‘‘only for zero mass’’ is nonsensical; the law is meaningful for finite mass. Common Pitfalls: Confusing net force with a single applied force while ignoring opposing forces, or forgetting vector directions in multi-force systems. Final Answer: gives

Question 2

Area moments – triangular section about its base
The second moment of area (area moment of inertia) of a triangular section with base b and height h about its base is:

Accepted Answer

Introduction / Context: Area moments of inertia are central to bending stress and deflection analyses. For common shapes, standard formulas are used to speed design and checks. Given Data / Assumptions: Plane area is a triangle of base b and height h. Axis of interest is along the base line of the triangle (in-plane axis through the base). Small-deflection, linear elasticity context (typical beam theory usage). Concept / Approach: The area moment of inertia about the base for a triangle is derived by integrating y^2 dA from the base toward the apex, using a linear variation of width with height. Step-by-Step Solution: Let the triangle have base along y = 0 and apex at y = h.Width at a distance y from base varies linearly: w(y) = (b/h) (h − y).Differential area: dA = w(y) dy.Area moment about base: I_base = ∫ y^2 dA = ∫_0^h y^2 w(y) dy = ∫_0^h y^2 (b/h)(h − y) dy.Compute: (b/h) ∫_0^h (h y^2 − y^3) dy = (b/h) [ h * (h^3 / 3) − (h^4 / 4) ] = (b/h) (h^4/3 − h^4/4) = (b/h) (h^4/12) = b h^3 / 12. Verification / Alternative check: Parallel-axis transformation from centroidal axis (I_cg = b h^3 / 36) to base: I_base = I_cg + A d^2 with d = h/3 and A = b h / 2. This also yields b h^3 / 12. Why Other Options Are Wrong: b h^3 / 4 and b h^3 / 8 are too large; they correspond to different shapes/axes. b h^3 / 36 is the centroidal moment for a triangle about a base-parallel centroidal axis, not about the base itself. Common Pitfalls: Mixing centroidal and base axes or forgetting linear width variation. Final Answer: b h^3 / 12

Question 3

Internal stress without external motion – effect of applied equilibrating forces
If a given force (or system of forces) acting on a body ______ the position of the body, yet keeps it in overall equilibrium, its effect is to produce internal stress …

Accepted Answer

Introduction / Context: Stress can exist in a body even when there is no external motion. Self-equilibrating or balanced external loads may leave the body at rest while inducing internal forces to maintain equilibrium. Given Data / Assumptions: External forces act but net force and net moment are zero (static equilibrium). No change in the macroscopic position or orientation of the body. Continuum mechanics assumptions apply (internal force transmission through sections). Concept / Approach: Equilibrium of the whole body does not imply zero internal forces. Internal stress arises to balance the applied loads over internal cross-sections so that every part is in equilibrium. Step-by-Step Solution: Consider a prismatic bar in axial tension with two equal and opposite end forces. The bar remains at rest (no change in position) but carries internal axial stress σ = P / A.Similarly, a beam under equal and opposite couples remains in equilibrium but develops internal bending stresses and shear.Hence, when external loads do not change the body’s overall position yet are present, they produce internal stress to satisfy local equilibrium. Verification / Alternative check: Cut-free-body approach: section the body and draw free-body diagrams of each part; internal forces and moments appear to balance external loads, proving the presence of stress fields. Why Other Options Are Wrong: ‘‘change’’ implies motion; the statement emphasizes stress without overall motion. Options about rotation or acceleration contradict static equilibrium condition. Common Pitfalls: Assuming ‘‘no motion’’ means ‘‘no stress’’—internal forces can be substantial while the body remains at rest. Final Answer: does not change

Question 4

Coefficient of restitution – expression for collinear impact (same direction)
Two bodies move in the same straight line with initial velocities u1 and u2 (u1 > u2) before impact, and velocities v1 and v2 after impact, along the same direction. The …

Accepted Answer

Introduction / Context: The coefficient of restitution (e) quantifies how ‘‘bouncy’’ a collision is along the line of impact. It is defined as the ratio of the relative speed of separation to the relative speed of approach. Given Data / Assumptions: One-dimensional (collinear) impact. Before impact: velocities u1 and u2 in the same direction, with u1 > u2 so that collision occurs. After impact: velocities v1 and v2 in the same line. Standard sign convention: speeds taken along the same positive direction. Concept / Approach: By definition, e = (relative velocity of separation) / (relative velocity of approach), taken along the line of impact with consistent direction signs. Step-by-Step Solution: Relative approach speed = u1 − u2 (since u1 > u2 and both along same direction).Relative separation speed after collision = v2 − v1 (if both continue along the same positive direction).Therefore, e = (v2 − v1) / (u1 − u2).For perfectly elastic impact, e = 1; for perfectly inelastic (sticking), e = 0. Verification / Alternative check: Check dimensions: the ratio is dimensionless as required. Test the limits: if the bodies stick (v1 = v2), numerator is zero → e = 0; if they exchange speeds in an elastic case, numerator equals denominator → e = 1. Why Other Options Are Wrong: Signs reversed in (a) and (c) produce negative or inverted ratios. (d) uses sums of velocities, not relative speeds, violating the definition. ‘‘None’’ is incorrect since (b) is valid. Common Pitfalls: Mixing direction signs or using speeds instead of signed velocities; always apply the definition consistently along the line of impact. Final Answer: e = (v2 − v1) / (u1 − u2)

Question 5

Engineering Mechanics — Concurrent forces:
Which of the following best describes concurrent forces in statics, i.e., forces acting on a body whose lines of action share a common geometric condition?

Accepted Answer

Introduction / Context: In engineering mechanics and statics, force systems are classified by the geometry of their lines of action. Understanding what makes forces concurrent is fundamental for applying equilibrium conditions and simplifying complex systems into resultants acting through specific points. Given Data / Assumptions: We are dealing with ideal, rigid-body statics. “Concurrent” refers to the geometry of lines of action only, not magnitudes. Options contrast collinear, coplanar, and point-intersecting arrangements. Concept / Approach: Forces are concurrent when their lines of action intersect at a single common point, irrespective of the angle between them or whether the forces are coplanar or spatial. This permits replacing the system by a single resultant through that intersection when appropriate. Step-by-Step Solution: Identify the definition: “Concurrent” means meeting at a point.Collinear forces lie on the same straight line (a special case of concurrency but overly restrictive).Coplanar forces lie in the same plane; they may or may not meet at a single point.Therefore, the precise descriptor is “meet at one point.” Verification / Alternative check: A common test is to draw multiple non-parallel lines. If you can pass them through a single point, the system is concurrent; if they merely share a plane, they are coplanar; if they share a line, they are collinear (which is a subset, not the general definition). Why Other Options Are Wrong: lie on the same line: Too restrictive; that is collinearity, not general concurrency. meet on the same plane: Coplanarity does not guarantee intersection at one point. none of these: Incorrect because “meet at one point” is correct. Common Pitfalls: Confusing coplanar with concurrent. Assuming parallel forces can be concurrent; strictly, parallel distinct lines never meet, so they are non-concurrent unless collinear. Final Answer: meet at one point

Question 6

Machines — Velocity ratio of a Weston differential pulley block:
If D and d are the diameters of the larger and the smaller rigidly-connected pulleys, respectively, what is the velocity ratio (ideal conditions, no slip, no friction)?

Accepted Answer

Introduction / Context: The Weston differential pulley (chain block) is a classic lifting machine. Its kinematics yield a characteristic velocity ratio (VR) that relates the distance moved by the effort chain to the distance moved by the load. Knowing VR is essential for estimating mechanical advantage and sizing equipment under ideal and real conditions. Given Data / Assumptions: Larger pulley diameter = D; smaller pulley diameter = d. Pulleys are rigidly connected to rotate together (common angular speed). Idealized behavior: no slip, no friction, inextensible chain. Velocity ratio VR = distance moved by effort / distance moved by load. Concept / Approach: Because the two pulleys rotate together, the chain segments over the larger and smaller pitch circles move with different linear speeds. The net motion of the load depends on the differential take-up between these two circumferences. Under ideal conditions, the VR for a Weston differential pulley is expressed using the diameters (or radii) as a function of their difference. Step-by-Step Solution: Let circumferential speed on larger pulley = v_L; on smaller pulley = v_S.For the same angular speed ω: v_L = ω * (D/2) and v_S = ω * (d/2).Differential take-up per unit time = v_L − v_S = ω * (D − d)/2.The load is supported by two segments, so load speed = (v_L − v_S)/2 = ω * (D − d)/4.Effort chain speed corresponds to motion around the larger pulley (two strands engaged), effectively ω * D/2 on the hauling side; combining the standard derivation gives VR = (effort speed) / (load speed) = (ω * D/2) / (ω * (D − d)/4) = 2D / (D − d). Verification / Alternative check: Dimensionally, as D approaches d, (D − d) becomes small and VR grows very large, which matches the intuition that closer diameters give higher theoretical advantage but very slow load motion. Why Other Options Are Wrong: D / (D − d): Misses the factor of 2 from two supporting strands. (D + d) / (D − d): Incorrect combination; does not follow from differential take-up. D / d: Belongs to simple belt speed ratios, not the Weston differential block VR. 2d / (D − d): Uses the smaller diameter in the numerator; not supported by derivation. Common Pitfalls: Forgetting the two-strand support halves the load speed. Confusing torque/MA expressions with kinematic VR. Final Answer: 2D / (D − d)

Question 7

Kinematics — Identify vector quantities:
Which of the following physical quantities are vectors in elementary mechanics?

Accepted Answer

Introduction / Context: Physics distinguishes scalar and vector quantities. Scalars have magnitude only, whereas vectors have both magnitude and direction and follow vector addition laws (e.g., triangle or parallelogram rule). Correctly classifying displacement, velocity, and acceleration is crucial for solving motion problems and applying kinematic equations consistently. Given Data / Assumptions: Motion is along straight or curved paths in Newtonian mechanics. Terms refer to standard definitions: linear displacement, linear velocity, linear acceleration. We assume three-dimensional vectors unless a 1D restriction is explicitly stated. Concept / Approach: Ask whether the quantity requires direction for full specification and obeys vector algebra. Displacement is the vector difference between final and initial position; velocity is the time rate of change of displacement; acceleration is the time rate of change of velocity. Each inherently includes direction. Step-by-Step Solution: Displacement: position_final − position_initial, direction matters → vector.Velocity: v = ds/dt, derived from displacement → vector.Acceleration: a = dv/dt, derived from velocity → vector.Therefore, all three listed quantities are vectors. Verification / Alternative check: If you reverse the direction of motion, displacement, velocity, and acceleration change sign (or direction), which is a hallmark of vector behavior. Their magnitudes alone would be insufficient to describe motion fully. Why Other Options Are Wrong: Linear displacement: True but incomplete to select alone. Linear velocity: True but incomplete to select alone. Linear acceleration: True but incomplete to select alone. Common Pitfalls: Confusing displacement (vector) with distance (scalar). Assuming “linear” implies scalar; “linear” here refers to translational motion, not the absence of direction. Final Answer: all of these

Question 8

Simple harmonic motion (SHM) — Maximum acceleration:
A particle performs SHM of amplitude r and angular frequency ω. What is the maximum magnitude of its acceleration?

Accepted Answer

Introduction / Context: In simple harmonic motion, acceleration is proportional to displacement and directed toward the equilibrium (restoring) position. Being able to relate maximum acceleration to amplitude and angular frequency is essential for vibration analysis and design (e.g., springs, pendulums, machinery). Given Data / Assumptions: SHM with displacement x(t) = r * sin(ωt + φ) or r * cos(ωt + φ). Amplitude = r (maximum displacement from equilibrium). Angular frequency = ω (radians per second). Concept / Approach: For SHM, the defining relation is a = −ω^2 * x. The maximum magnitude of a occurs when |x| is maximum, i.e., at the extremes where |x| = r. Therefore, |a|max depends on ω^2 times r. Step-by-Step Solution: Start from acceleration–displacement relation: a = −ω^2 * x.Maximum |x| = r at the ends of motion.Thus maximum |a| = ω^2 * r.Therefore the correct expression is ω^2r. Verification / Alternative check: Differentiation method: x = r cos(ωt) ⇒ v = dx/dt = −rω sin(ωt); a = dv/dt = −rω^2 cos(ωt) = −ω^2 x. Maximum |a| occurs at cos(ωt) = ±1 → |a| = rω^2. Why Other Options Are Wrong: ω: Has units of 1/s, not acceleration. ωr: Has units of speed, not acceleration. ω / r: Units inconsistent; not from SHM relations. Common Pitfalls: Confusing maximum speed (ωr) with maximum acceleration (ω^2r). Dropping the square on ω in a = −ω^2 x. Final Answer: ω^2r

Question 9

Machines — Ideal machine relationship:
In an ideal machine (no friction, no losses), the mechanical advantage (MA) is __________ the velocity ratio (VR).

Accepted Answer

Introduction / Context: Mechanical advantage (MA) and velocity ratio (VR) are cornerstone measures in machine theory. For ideal machines, energy is conserved perfectly, forging a simple relationship between input motion and output force that informs performance estimation and design trade-offs. Given Data / Assumptions: Ideal machine: no friction, perfect rigidity, no elastic losses. MA = Load / Effort. VR = Distance moved by effort / Distance moved by load. Efficiency η = MA / VR. Concept / Approach: Under ideal conditions, η = 1. Therefore MA = VR. Real machines have η < 1 due to friction and other losses, hence MA < VR in practice. The question explicitly specifies an ideal machine, so equality applies. Step-by-Step Solution: Define efficiency: η = (Work output) / (Work input).For machines, η = MA / VR.Ideal machine ⇒ η = 1.Thus MA = VR. Verification / Alternative check: Consider a simple lever or pulley block with ideal bearings and inextensible ropes. The kinematic ratio (VR) exactly predicts the achievable load amplification (MA), confirming equality when losses are absent. Why Other Options Are Wrong: less than: True for real machines, not ideal. greater than: Violates energy conservation. undefined relative to: Relationship is well-defined via efficiency. Common Pitfalls: Using real-world intuition (friction) for an idealized theoretical question. Mixing up definitions of MA and VR. Final Answer: equal to

Question 10

Centroid of a semicircle:
The centre of gravity (centroid) of a semicircle of radius r lies at what distance from its base (the diameter), measured along the vertical radius toward the centre?

Accepted Answer

Introduction / Context: Locating centroids is essential in structural analysis, fluid mechanics (hydrostatic forces), and product design. The semicircle is a standard shape whose centroid location must be memorized or re-derived efficiently for composite-area calculations. Given Data / Assumptions: Planar semicircle of radius r. Base is the diameter; distance is measured perpendicularly from this base along the symmetry axis (vertical radius). Uniform density and thickness for geometric centroid. Concept / Approach: By symmetry, the centroid lies on the central vertical radius. Its distance from the diameter is obtained by first principles (area moment integral) or by recalling the standard result. The well-known formula gives the ȳ coordinate from the base as 4r / (3π). Step-by-Step Solution: Let the diameter lie along the x-axis with origin at the circle's center; the semicircle occupies y ≥ 0.The centroid of the semicircular area measured from the center along y is y_c_from_center = 4r / (3π).Therefore, measured from the base (the diameter) upward along the vertical radius, the same distance is y_c = 4r / (3π).Hence the required distance from the base is 4r / 3π. Verification / Alternative check: Numerically, 4 / (3π) ≈ 0.424. This places the centroid between the base and the circle's center (at 0.424 r above the diameter), which is physically reasonable because more area mass lies near the base than near the arc. Why Other Options Are Wrong: 3r / 8 = 0.375 r: Too small; not the standard centroid value. 8r / 3: Dimensionally wrong (greater than r). 3r / 4π ≈ 0.239 r: Too small; not the canonical result. Common Pitfalls: Confusing the centroid from the center versus from the base—the same numeric distance applies here because the base is exactly one radius below the center. Mixing arc centroid (of a semicircular arc) with area centroid (of a filled semicircle). Final Answer: 4r / 3π

Question 11

Elevator dynamics — Cable tension comparison:
In which case is the tension in the cable supporting a lift (elevator) greater: when the lift is moving upwards with acceleration, or when it is moving downwards with acceleration?

Accepted Answer

Introduction / Context: Elevator (lift) problems are standard applications of Newton's second law. Comparing cable tension during accelerated motion is important for sizing hoists, motors, and safety brakes. The direction of acceleration relative to weight determines whether the cable must supply extra force or is partially relieved. Given Data / Assumptions: Lift of mass m moving vertically. Acceleration magnitude = a; direction either upward or downward. Gravity acts downward with magnitude W = m * g. Cable mass and pulley friction neglected for conceptual comparison. Concept / Approach: Apply ΣF = m * a along the vertical direction. Tension T counters weight and provides acceleration. The sign of a changes the required T. Upward acceleration requires T > W; downward acceleration requires T < W (assuming nonzero a). Step-by-Step Solution: Case 1 (accelerating upward): Take upward positive. ΣF = T − W = m * a ⇒ T = W + m * a = m * (g + a).Case 2 (accelerating downward): Take upward positive. ΣF = T − W = −m * a ⇒ T = W − m * a = m * (g − a).Compare: m * (g + a) > m * (g − a) for a > 0.Therefore, tension is greater when the lift accelerates upward. Verification / Alternative check: If a = 0 (uniform speed), T = W in both directions. As a increases upward, T increases above W; as a increases downward, T decreases below W but remains positive as long as a < g, which is typical in normal operation. Why Other Options Are Wrong: downwards: Would imply T > W − m a exceeds T for upward case, which contradicts ΣF = m a. same in both cases: Only true for a = 0 (not our accelerated scenario). indeterminate: Laws of motion determine this unambiguously. Common Pitfalls: Confusing the sign convention, leading to T = m * (g − a) for upward motion (incorrect). Thinking speed (not acceleration) changes tension; constant-speed motion does not alter T from W. Final Answer: upwards

Question 12

Engineering Mechanics – Area Moment of Inertia
What is the second moment of area (area moment of inertia) of a square plate of side a about one of its diagonals passing through the centroid?

Accepted Answer

Introduction / Context: This question tests knowledge of second moment of area (also called area moment of inertia) for standard plane figures used in strength of materials and structural analysis. For a square plate, common reference axes are the centroidal x–y axes (parallel to the sides) and the centroidal axes along the diagonals. Because a square has fourfold symmetry, several centroidal axes become principal axes with equal moments. Given Data / Assumptions: Plane figure: square of side a. Axis: along a diagonal, through the centroid (in-plane area moment). Homogeneous thin plate; standard engineering definitions apply. Concept / Approach: For a square of side a, the centroidal area moments about axes parallel to the sides are Ix = Iy = a^4 / 12. For a square, the principal centroidal axes also include the two diagonals; by symmetry, the in-plane moment about any centroidal principal axis equals a^4 / 12 as well. Another quick route is to use Mohr’s circle for area moments with Ix = Iy and product of inertia Ixy = 0, which implies every centroidal axis is principal with the same value. Step-by-Step Solution: Known: Ix = Iy = a^4 / 12 for a square. For a square, diagonal axes are principal centroidal axes because of symmetry. Therefore, Id (about any centroidal diagonal) = a^4 / 12. Verification / Alternative check: Using Mohr’s circle: center = (Ix + Iy)/2 = a^4/12, radius = sqrt(((Ix − Iy)/2)^2 + Ixy^2) = 0, so all centroidal orientations have the same value = a^4/12. Why Other Options Are Wrong: a^2/8, a^3/12: wrong dimensions; area moments scale with length^4. a^4/16 or a^4/24: do not match the principal value for a square's centroidal axes. Common Pitfalls: Confusing polar moment J = Ix + Iy = a^4/6 with an in-plane moment about a single axis. Assuming diagonal gives a different value; for a square it does not. Final Answer: a^4 / 12

Question 13

Statics – Law of Moments under Equilibrium
According to the law of moments, if several coplanar forces act on a particle and the system is in equilibrium, which statement is correct?

Accepted Answer

Introduction / Context: In engineering statics, the law of moments (Varignon’s theorem) and equilibrium conditions are fundamental. For a system of coplanar forces in equilibrium, both net force and net moment about any arbitrary point in the plane must vanish. Given Data / Assumptions: Forces are coplanar and act on a particle/system. The system is in static equilibrium. Concept / Approach: Equilibrium of coplanar forces requires: ΣFx = 0, ΣFy = 0, and ΣM(any point) = 0. Varignon’s theorem also states that the moment of a resultant equals the sum of moments of the component forces about the same point. If the system is in equilibrium, the resultant is zero; hence the algebraic sum of moments must be zero about any point. Step-by-Step Solution: Equilibrium implies resultant R = 0. By Varignon: M_R about any point O = ΣM_O(forces). Since R = 0 ⇒ M_R = 0 ⇒ ΣM_O(forces) = 0. Verification / Alternative check: Choosing different moment centers does not change the conclusion; if equilibrium holds, the algebraic sum of moments about any point is zero. Why Other Options Are Wrong: 'Their algebraic sum is zero': True for forces (ΣF = 0) but the law of moments specifically concerns moments, not merely the force sum. Equal distances: has no general relevance to equilibrium. 'Sum of moments equals moment of resultant': The theorem is true, but in equilibrium the resultant is zero; the specific equilibrium statement is that the sum of moments is zero. Moment of inertia zero: unrelated to static equilibrium of forces. Common Pitfalls: Confusing the necessary condition ΣF = 0 with the moment condition ΣM = 0; both are required. Final Answer: The algebraic sum of their moments about any point in the plane is zero

Question 14

Dynamics – Apparent Weight in an Accelerating Lift
A lift (elevator) moves downward with acceleration 9.8 m/s^2. The normal reaction/pressure exerted by a man on the lift floor becomes zero. Is this statement true?

Accepted Answer

Introduction / Context: Apparent weight in elevators is a standard dynamics concept. The normal reaction (N) from the floor is perceived as 'weight' on a scale. When the elevator accelerates, N differs from the true weight mg. Given Data / Assumptions: Lift acceleration a = 9.8 m/s^2 downward (approximately g). Man of mass m standing on the floor; vertical motion only. Neglect air resistance; consider ideal contact. Concept / Approach: For downward acceleration a, the equation of motion is: N + (−mg) = m(−a). Rearranged, N = m(g − a). Apparent weight equals N. If a = g, then N = 0, meaning weightlessness (free fall). Step-by-Step Solution: Write vertical force balance taking upward positive. Sum of forces: N − mg = −m a. Thus N = m(g − a). Given a = g ⇒ N = m(g − g) = 0. Verification / Alternative check: This is the classic 'free-fall' condition: all bodies inside accelerate equally, so the contact force vanishes. Why Other Options Are Wrong: 'False': Incorrect because the derived formula gives N = 0 at a = g. 'Only if the man jumps' or 'mass is zero': Unnecessary; the result follows from dynamics. 'Air resistance' is not part of the contact-force calculation here. Common Pitfalls: Using N = mg ± ma without sign care; always derive N = m(g − a) for downward acceleration. Final Answer: True

Question 15

Machine Theory – Maximum Efficiency of a Screw Jack
For a square-thread screw jack with friction angle φ and lead angle λ, what is the maximum theoretical efficiency and the condition for it?

Accepted Answer

Introduction / Context: A screw jack's efficiency depends on its lead angle λ and the friction angle φ. Designers often seek the condition that maximizes efficiency, subject to self-locking and strength constraints. Given Data / Assumptions: Square-thread screw; standard friction model with friction angle φ. Lead angle λ is the helix angle measured on the mean radius. General efficiency formula: η = tan λ / tan(λ + φ). Concept / Approach: To find the maximum, differentiate η with respect to λ and set the derivative to zero. Using a trigonometric identity via the derivative of ln η, the optimality condition emerges as 2λ + φ = 90°. Substituting this into the efficiency formula yields a compact expression in terms of φ alone. Step-by-Step Solution: Start with η = tan λ / tan(λ + φ). Set d(ln η)/dλ = 0 ⇒ sec^2λ/tanλ − sec^2(λ+φ)/tan(λ+φ) = 0. This simplifies to sin(2λ) = sin(2λ + 2φ) ⇒ 2λ + φ = 90° (non-trivial solution). At 2λ + φ = 90°, tan(λ + φ) = tan(90° − λ) = cot λ, so η_max = tan^2 λ. Using λ = 45° − φ/2, tan λ = tan(45° − φ/2) ⇒ η_max simplifies to (1 − sin φ)/(1 + sin φ). Verification / Alternative check: Deriving η_max directly from tangent half-angle identities also leads to η_max = (1 − sin φ)/(1 + sin φ). Why Other Options Are Wrong: (b) Inverts the correct ratio. (c) Gives the general η expression, not the maximized value or correct condition. (d) Uses an incorrect condition and value. (e) Ignores dependence on λ and gives an incorrect constant value. Common Pitfalls: Confusing lead angle λ with friction angle φ. Assuming maximum efficiency occurs at maximum λ without proof. Final Answer: η_max = (1 − sin φ) / (1 + sin φ), attained when 2λ + φ = 90°

Question 16

Resultant of Two Forces – Magnitude Formula
Two forces P and Q act at an angle θ to each other. What is the magnitude of their resultant R?

Accepted Answer

Introduction / Context: The parallelogram (or triangle) law of forces gives the resultant magnitude and direction when two vectors act at an angle θ. This is foundational in statics and dynamics. Given Data / Assumptions: Two vectors (forces) of magnitudes P and Q. Included angle between them is θ. Concept / Approach: Resolve vectors or use the cosine rule on the triangle formed by P, Q, and the resultant R. The cosine rule for sides a, b with included angle C gives c^2 = a^2 + b^2 − 2ab cos C; adapting signs based on geometry yields the standard resultant formula. Step-by-Step Solution: Construct the parallelogram of forces with sides P and Q. Diagonal from the common point represents R. By geometry: R^2 = P^2 + Q^2 + 2 P Q cos θ (when θ is the included angle between P and Q tail-to-tail). Thus R = sqrt(P^2 + Q^2 + 2 P Q cos θ). Verification / Alternative check: Component method: Rx = P + Q cos θ, Ry = Q sin θ ⇒ R = sqrt((P + Q cos θ)^2 + (Q sin θ)^2) which simplifies to the same expression. Why Other Options Are Wrong: Linear sum P + Q + θ has wrong dimensions. The minus-sign cosine form corresponds to the side opposite the included angle in a different arrangement (effectively P and −Q). Other algebraic forms do not follow from vector addition. Common Pitfalls: Confusing included angle and supplementary angle; always base on the geometry used. Final Answer: R = sqrt(P^2 + Q^2 + 2 P Q cos θ)

Question 17

Newtonian Mechanics – Identify the Statement
The total linear momentum of a system of masses in any direction remains constant unless acted upon by an external force in that direction. This statement is known as:

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Introduction / Context: Conservation laws form the backbone of mechanics. Momentum conservation is especially important for collisions, rockets, and systems with internal interactions but negligible external influence. Given Data / Assumptions: Closed system or net external force along the direction considered is zero. Classical (Newtonian) mechanics regime. Concept / Approach: Newton's second law in vector form is ΣF = d( p )/dt. If ΣF = 0 in a given direction, then dp/dt = 0, implying momentum p is constant in that direction. This is the formal basis of the principle of conservation of momentum. Step-by-Step Solution: Start: ΣF_dir = d( p_dir )/dt. If ΣF_dir = 0 ⇒ d( p_dir )/dt = 0. Therefore p_dir = constant over time. This is termed the conservation of momentum in that direction. Verification / Alternative check: Applies to multi-body systems with internal forces obeying action–reaction; internal forces cancel in total momentum change, leaving only external forces to alter system momentum. Why Other Options Are Wrong: First law: inertia principle (no change in state without net force), not a conservation statement for a system’s momentum. Second law: defines relation between force and rate of change of momentum. Energy conservation: different physical quantity. D’Alembert’s principle: transforms dynamics into statics with inertial forces, not a conservation law. Common Pitfalls: Confusing energy and momentum conservation; they are distinct and context-dependent. Final Answer: Principle of conservation of momentum

Question 18

Units & Definitions – Definition of the Newton (N)
A newton is defined as the magnitude of force which, acting on a mass of 1 kg, produces an acceleration of 1 m/s^2 in the direction of the force. Is this definition correct?

Accepted Answer

Introduction / Context: SI base and derived units are core to engineering problem solving. The newton (N) is the SI unit of force, defined via Newton’s second law. Given Data / Assumptions: SI units throughout: mass in kilogram, length in meter, time in second. Linear acceleration scenario. Concept / Approach: Newton’s second law states F = m * a. Setting m = 1 kg and a = 1 m/s^2 gives F = 1 N by definition. Step-by-Step Solution: Use F = m a. Let m = 1 kg, a = 1 m/s^2. Then F = 1 kg * 1 m/s^2 = 1 N. Verification / Alternative check: Dimensional analysis: [N] = kg·m/s^2. Why Other Options Are Wrong: 'No': contradicts the SI definition. 'Only in cgs': cgs uses dyne, not newton. Rotational motion: torque unit is N·m, not N; definition remains linear and exact. Common Pitfalls: Mixing SI and cgs units; always keep consistency. Final Answer: Yes

Question 19

Lifting Machines – Reversibility and Efficiency
A machine lifts a load of 1000 N by an effort of 80 N. If the velocity ratio (VR) is 20, classify the machine.

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Introduction / Context: In lifting machines, reversibility indicates whether the machine will move backward under the load when effort is removed. Efficiency helps determine this behavior. Given Data / Assumptions: Load W = 1000 N. Effort P = 80 N. Velocity ratio VR = 20. Mechanical advantage MA = W / P. Concept / Approach: Efficiency η = MA / VR. A practical rule: if η > 50%, the machine is reversible; if η ≤ 50%, it is non-reversible (self-locking). Ideal machines have η = 100%. Step-by-Step Solution: Compute MA = W / P = 1000 / 80 = 12.5. Compute η = MA / VR = 12.5 / 20 = 0.625 = 62.5%. Since η > 50%, the machine is reversible. Verification / Alternative check: If the efficiency is this high, frictional losses are not sufficient to prevent back-driving under load when effort is removed. Why Other Options Are Wrong: Non-reversible: incorrect because η > 50%. Ideal: would require η = 100% (not the case). 'Self-acting only on lowering' and 'Indeterminate': not supported by the data. Common Pitfalls: Confusing mechanical advantage with velocity ratio; remember η = MA / VR. Final Answer: Reversible (efficiency > 50%)

Question 20

Force Systems – Definition Check
Non-coplanar concurrent forces are best described as forces that:

Accepted Answer

Introduction / Context: Classifying force systems (concurrent, parallel, coplanar, non-coplanar) is fundamental in engineering mechanics for selecting the correct equilibrium equations and solution strategies. Given Data / Assumptions: We are distinguishing between concurrency (meeting at a point) and coplanarity (lying in a single plane). Concept / Approach: Concurrent forces: all lines of action intersect at a single point. Coplanar forces: all lines of action lie within the same geometric plane. Non-coplanar concurrent forces: concurrent but spanning three-dimensional space (not confined to one plane). Step-by-Step Solution: Identify concurrency: forces meet at one point. Check coplanarity: if lines do not lie on one plane, the system is non-coplanar. Thus: 'Meet at one point, but not in the same plane' is the correct description. Verification / Alternative check: Examples include three or more forces acting along directions in 3D that all intersect at a common joint, such as cable forces meeting at a node in space trusses. Why Other Options Are Wrong: Non-concurrent and non-coplanar: describes a general spatial system, not concurrent. Concurrent and coplanar: that is coplanar concurrent, not non-coplanar. Non-concurrent but coplanar: describes a different class. Always parallel: unrelated to concurrency or coplanarity. Common Pitfalls: Mixing up 'concurrent' with 'coplanar'; they are independent properties. Final Answer: Meet at one point, but their lines of action do not lie in the same plane

Engineering Mechanics Questions