Question 1

Centroid of a shaded parabolic area (figure not provided): Identify the centroid location of the shaded region under a parabola from the y-axis as indicated in the missing figure.

Accepted Answer

Data insufficient to determine from the given text

Question 2

Principle of virtual work: When a body in equilibrium is subjected to an infinitesimally small imagined displacement, the corresponding imagined work is called what?

Accepted Answer

Introduction / Context: The principle of virtual work is a powerful energy method in statics. It asserts that for a system in equilibrium, the total virtual work done by external forces during any kinematically admissible virtual displacement is zero. This approach often simplifies calculations of reactions and internal forces when direct force balance is cumbersome. Given Data / Assumptions: Body/system is in static equilibrium. A virtual displacement is infinitesimally small and consistent with constraints. No change in actual loads; only an imagined compatibility-preserving motion is considered. Concept / Approach: Virtual work is not physical work performed; instead, it is a mathematical device. If δr denotes a virtual displacement, the virtual work δW = Σ(F · δr). For equilibrium, ΣδW = 0 for all permissible δr. This allows solving for unknown reactions without explicitly writing all component equilibrium equations. Step-by-Step Solution: Define a small, compatible virtual displacement consistent with supports and constraints.Compute the virtual work of each external force using F · δr (or M · δθ for moments).Set the algebraic sum of virtual works equal to zero and solve for the unknown(s). Verification / Alternative check: Results obtained via virtual work can be checked against conventional ΣF = 0 and ΣM = 0 equations in simple cases, ensuring consistency. Why Other Options Are Wrong: “Imaginary work” is a colloquial phrase but not a defined mechanics term. “Negative work” refers to work opposite to displacement direction, not to the method. “Complementary work” is used in elastic energy methods, not this principle. Common Pitfalls: Using virtual displacements that violate constraints (e.g., fixed supports allowed to move). Confusing virtual work (mathematical) with actual energy transfer. Final Answer: virtual work

Question 3

Time-dependent kinematics: A particle starts from rest and moves on a straight line with x(t) = t^3 - 3 t^2 + 5. What is the ratio of accelerations a(5 s) : a(3 s)?

Accepted Answer

Introduction / Context: This exercise reinforces the relationship between position, velocity, and acceleration through differentiation. With position given as a time polynomial, differentiating once gives velocity, and differentiating twice gives acceleration. Ratios at specific times test accurate substitution and algebra. Given Data / Assumptions: Position: x(t) = t^3 - 3 t^2 + 5. We need accelerations at t = 5 s and t = 3 s. One-dimensional motion; units are consistent. Concept / Approach: Velocity v(t) = dx/dt; acceleration a(t) = dv/dt = d^2x/dt^2. Use power-rule differentiation term-by-term. Step-by-Step Solution: Compute velocity: v(t) = d/dt(t^3 - 3 t^2 + 5) = 3 t^2 - 6 t.Compute acceleration: a(t) = d/dt(3 t^2 - 6 t) = 6 t - 6.Evaluate: a(5) = 65 - 6 = 24; a(3) = 63 - 6 = 12.Form ratio: a(5) : a(3) = 24 : 12 = 2 : 1 → 2. Verification / Alternative check: Second derivative check: differentiating x twice should remove the constant term and reduce polynomial degree properly; the result 6 t - 6 satisfies this. Why Other Options Are Wrong: 3, 4, 5 do not match the evaluated ratio. 1 would imply equal accelerations at 5 s and 3 s, which is false for a(t) linear in t. Common Pitfalls: Stopping after the first derivative (giving velocity instead of acceleration). Arithmetic mistakes when substituting t values. Final Answer: 2

Question 4

Effect of changing gravitational acceleration: If the local gravitational acceleration at a place is doubled, what happens to the weight of a body at that location (mass constant)?

Accepted Answer

Introduction / Context: Weight is the gravitational force acting on a body. While mass is an intrinsic property and does not change with location, weight depends on the local value of gravitational acceleration g. This question probes the direct proportionality between weight and g. Given Data / Assumptions: Mass m of the body is constant. Local gravitational acceleration g is doubled. No buoyancy or other forces considered. Concept / Approach: By definition, weight W = m * g. If g becomes 2g, the new weight W′ = m * (2g) = 2 * (m * g) = 2W. Hence, weight doubles when g doubles. Step-by-Step Solution: Write the relation: W = m * g.Replace g with 2g: W′ = m * 2g.Conclude W′ = 2W → doubled. Verification / Alternative check: Dimensional consistency: weight has units of newtons (N), proportional to g; doubling g must double W if m is unchanged. Why Other Options Are Wrong: Half or zero contradicts direct proportionality. “Not affected” confuses weight with mass. “None of these” is incorrect since a clear outcome exists. Common Pitfalls: Mixing up mass and weight; only weight depends on g. Final Answer: be doubled

Question 5

Kinematics of plane motion: The locus of the instantaneous centre of rotation (IC) of a moving rigid body in plane motion is called what?

Accepted Answer

Introduction / Context: In planar kinematics, every instant of a general plane motion can be seen as a pure rotation about an instantaneous centre (IC). Tracking the path of the IC is useful in mechanism analysis and gear tooth geometry. Given Data / Assumptions: Rigid body undergoing plane motion relative to a fixed frame. The instantaneous centre exists at each instant (excluding singular configurations). Concept / Approach: For pure rolling without slip of one curve over another, the locus of the point of contact (which acts as the instantaneous centre) on a base circle forms an involute. This connection underlies the common use of involute tooth profiles in gears: the line of action corresponds to the base circle’s involute. More generally, the locus of ICs relative to the fixed frame is called the fixed centrode; when the body itself is considered, the locus is the moving centrode. In many textbook contexts, the IC traced by rolling of a curve on another is described by an involute of the base curve. Step-by-Step Solution: Recognize that a plane motion can be generated by pure rolling of centrodes.For a circle base, the IC path associated with unwinding (rolling without slip) is an involute.Select the curve classically linked to instantaneous centres in rolling—an involute. Verification / Alternative check: Gear fundamentals: the common tangent at the point of contact is related to the involute geometry of the base circle, reaffirming the role of the involute as an IC locus in rolling generation. Why Other Options Are Wrong: Straight line and spiral are special cases not generally representing IC loci for rolling generation. Centroid is a mass/area property, not a curve of motion. Cycloid describes rolling of a circle along a line, not the generic IC locus in these classical contexts. Common Pitfalls: Confusing centroid (a point property) with centrode/involute (path/curve concepts). Final Answer: involute

Question 6

Simple machine performance: A 500 kgf load is lifted through 13 cm by an effort of 25 kgf moving 650 cm. What is the efficiency of the lifting machine (in %)?

Accepted Answer

Introduction / Context: Efficiency compares useful output to input in any simple machine. For lifting devices, output work equals load times load displacement, whereas input work equals effort times effort displacement. Expressing all quantities in consistent force units (e.g., kgf) and distance units (e.g., cm) keeps the ratio dimensionless and directly convertible to percent. Given Data / Assumptions: Load = 500 kgf. Load is lifted by 13 cm. Effort = 25 kgf moves through 650 cm. Neglect elastic energy storage and other losses beyond what efficiency captures. Concept / Approach: Efficiency η = Output work / Input work. For vertical lifting at constant speed, Output = Load * Load distance; Input = Effort * Effort distance. No unit conversion is needed if the same force and length units are used in numerator and denominator. Step-by-Step Solution: Compute output: W_out = 500 * 13 = 6500 (kgf·cm).Compute input: W_in = 25 * 650 = 16250 (kgf·cm).Calculate efficiency: η = 6500 / 16250 = 0.40.Express as percentage: η% = 0.40 * 100 = 40%. Verification / Alternative check: Mechanical advantage MA = Load / Effort = 500 / 25 = 20. Velocity ratio VR = Effort distance / Load distance = 650 / 13 = 50. Efficiency = MA / VR = 20 / 50 = 0.40 → 40%, confirming the result. Why Other Options Are Wrong: 30%, 50%, 55%, 60% do not match the computed ratio of works. Common Pitfalls: Mixing units (e.g., metres with centimetres) between numerator and denominator. Using load distance and effort distance in the wrong places. Final Answer: 40%

Question 7

Two-force resultants at 120°: Two forces act at 120°. The larger force is 50 kgf. If the resultant is perpendicular to the smaller force, determine the magnitude of the smaller force.

Accepted Answer

Introduction / Context: Vector addition of two forces can be simplified using components. Here, the resultant's orthogonality to one of the forces (the smaller) provides a clean dot-product condition that determines the unknown magnitude without trigonometric law-of-cosines algebra. Given Data / Assumptions: Forces F_small = P (unknown) and F_large = 50 kgf. Angle between them = 120°. Resultant R is perpendicular to the smaller force. Planar (2D) force system. Concept / Approach: Let the smaller force lie along the +x axis: S = (P, 0). The 50 kgf force at 120° has components L = 50 (cos 120°, sin 120°) = (−25, 25√3). Then R = S + L = (P − 25, 25√3). The perpendicularity condition is S · R = 0. Step-by-Step Solution: Write vectors: S = (P, 0); L = (−25, 25√3).Compute resultant: R = (P − 25, 25√3).Orthogonality: S · R = P(P − 25) + 0 * 25√3 = 0.Solve: P(P − 25) = 0 → nonzero solution P = 25 kgf. Verification / Alternative check: Law of cosines approach: impose R ⟂ S ⇒ R^2 = L^2 + S^2 (since R is perpendicular to S, triangle is right-angled at S). With L = 50 and S = 25, R^2 = 50^2 + 25^2 = 3125, consistent with the component result. Why Other Options Are Wrong: 20, 30, 35, 40 kgf do not satisfy the perpendicularity (dot-product) condition for a 120° spacing with a 50 kgf companion force. Common Pitfalls: Using 60° instead of 120° for components, which flips the sign of the x-component. Forgetting that cos 120° = −1/2 and sin 120° = √3/2. Final Answer: 25 kg

Question 8

Static indeterminacy (incomplete figure) — the vertical reaction at support A of the shown structure is to be determined from the given diagram (insufficient in text).

Accepted Answer

2 t

Question 9

Parallelogram Law of Forces — when two forces acting simultaneously on a particle are represented (in magnitude and direction) by two adjacent sides of a parallelogram, their resultant is represented by:

Accepted Answer

Introduction / Context: The Parallelogram Law of Forces is a foundational result in engineering mechanics for adding two concurrent forces. It allows engineers to replace two forces by a single equivalent resultant that acts through the same point with a direction and magnitude obtained geometrically. This law underpins graphical statics, vector addition, and equilibrium checks for pin-jointed structures and particles. Given Data / Assumptions: Two forces act simultaneously on a single particle (concurrent at one point). Each force is represented by a directed segment equal in magnitude and direction to the real force. The two segments are drawn as adjacent sides of a parallelogram. Concept / Approach: Construct a parallelogram with the two force vectors as adjacent sides. The diagonal drawn from the common tail (the point of concurrency) gives the vector sum of the two forces. This diagonal's length represents the magnitude of the resultant, and its orientation gives the resultant's direction. Analytically, this matches vector addition F_R = F_1 + F_2, including the cosine rule for magnitude and the law of sines for direction if needed. Step-by-Step Solution: At the point of application, place the tails of both vectors together as adjacent sides.Complete the parallelogram using lines parallel to each vector.Draw the diagonal from the common tail through the parallelogram to the opposite corner.Interpret this diagonal as the resultant in both magnitude and direction. Verification / Alternative check (if short method exists): Resolve the two forces into perpendicular components and add components algebraically; the combined vector equals the geometrical diagonal. Computed magnitude via the cosine rule agrees with the diagonal length. Why Other Options Are Wrong: Longer/shorter side are not generally aligned with the resultant. The non-intersecting diagonal does not start from the concurrency point, so it cannot represent the resultant of the two applied forces. The average of diagonals lacks physical meaning for vector sum. Common Pitfalls (misconceptions, mistakes): Drawing the diagonal from the wrong corner; forgetting that the law applies only to concurrent forces; mixing up magnitude scaling when using non-uniform drawing scales. Final Answer: the diagonal of the parallelogram which passes through the point of intersection of the forces

Question 10

Units of force in engineering systems — identify the correct interpretation of engineer’s force units among the choices provided.

Accepted Answer

Introduction / Context: Engineering problems use several unit systems. In mechanics, the SI unit of force is the newton (an absolute unit based on fundamental dimensions), while the CGS unit is the dyne (also absolute). Many curricula and exam questions group these as “engineer’s units” to emphasize familiarity with both, even though gravitational units (kgf, lbf) are also used in practice. Given Data / Assumptions: Newton (N) is the SI absolute unit (1 N = 1 kgm/s^2). Dyne is the CGS absolute unit (1 dyne = 1 gcm/s^2). The options list absolute units familiar to engineers. Concept / Approach: Since both the newton and the dyne are absolute force units that engineers must convert between (1 N = 10^5 dynes), and each individual statement naming these as absolute units is correct, the most inclusive choice “All the above” is appropriate for this set-based question format. Step-by-Step Solution: Recognize newton and dyne as absolute force units in SI and CGS.Note that the question asks which statements are correct among the provided options.Since (a), (b), and (c) each state true facts, select the aggregate option.Acknowledge that kgf/lbf are gravitational units but are not in the listed correct subset. Verification / Alternative check (if short method exists): Dimensional analysis: both units reduce to mass*length/time^2; conversion 1 N = 10^3 g * 10^2 cm/s^2 = 10^5 dynes supports consistency. Why Other Options Are Wrong: Choosing only one unit omits the other recognized absolute unit. “Kilogram-force only” is a gravitational unit and not an absolute unit; it is outside the scope of the listed correct statements. Common Pitfalls (misconceptions, mistakes): Confusing absolute units with gravitational units; forgetting the CGS–SI conversion factor of 10^5 between dyne and newton. Final Answer: All the above.

Question 11

Satellite launch geometry — to directly attain a geostationary (synchronous) equatorial orbit, the preferred launch site latitude is:

Accepted Answer

Introduction / Context: Geostationary (synchronous) orbits are circular, equatorial, and prograde with zero inclination relative to Earth’s equator. Achieving such an orbit most efficiently requires launching into an equatorial plane to avoid costly plane-change maneuvers at high velocity. Given Data / Assumptions: Target orbit is geostationary: inclination i = 0°, period = sidereal day. Plane-change maneuvers in orbit are expensive in delta-v, especially at GEO speeds (~3.07 km/s). Earth rotation provides an eastward velocity boost greatest at the equator. Concept / Approach: Launching from the equator aligns the initial orbital plane with the Earth’s equator, eliminating inclination. Any non-equatorial site injects into an inclined orbit requiring a plane change later, increasing fuel requirements. Additionally, equatorial sites maximize the rotational speed assist, improving payload to orbit. Step-by-Step Solution: Define target: equatorial, zero-inclination GEO.Select launch azimuth due east at equator to match target plane.Avoid inclination that would otherwise require large plane-change burns.Therefore, choose an equatorial site for direct GEO insertion (e.g., Kourou, Sea Launch on equator). Verification / Alternative check (if short method exists): Compute plane-change delta-v: Δv_pc ≈ 2vsin(i/2); at GEO speeds even a few degrees imply substantial propellant. i = 0° makes Δv_pc = 0. Why Other Options Are Wrong: Latitudes 30°, 45°, 60°, or poles inject into inclined planes that need correction. Polar launches are most inefficient for GEO plane matching. Common Pitfalls (misconceptions, mistakes): Assuming any latitude is equivalent because plane changes can be done “later”; underestimating the large Δv penalty of inclination changes at high orbital velocities. Final Answer: on the equator

Question 12

Elevator dynamics — how does the cable tension compare with the weight when a lift (elevator) is moving?

Accepted Answer

Introduction / Context: The apparent tension in an elevator cable depends on the elevator’s acceleration. Understanding this relation helps in sizing hoists and in interpreting load-sensor readings and safety-brake triggers. Given Data / Assumptions: Elevator car mass m, weight W = mg. Upward acceleration a > 0 when speeding up upward or slowing down downward; downward acceleration a > 0 when speeding up downward or slowing down upward. Neglect cable mass and friction for the basic relation. Concept / Approach: Apply Newton’s second law along the vertical. For upward acceleration: T − W = ma ⇒ T = W + ma > W. For downward acceleration: W − T = ma ⇒ T = W − m*a < W. Thus, tension increases above weight when accelerating upward and decreases below weight when accelerating downward. Step-by-Step Solution: Identify acceleration direction from motion state (speeding up/down).Write force balance with sign convention upward positive.Solve for T to compare with W for each case.Conclude: tension is less when moving downwards (accelerating downward), and more when moving upwards (accelerating upward). Verification / Alternative check (if short method exists): Elevator scale thought experiment: a person’s apparent weight is lower when the car accelerates downward and higher when it accelerates upward, mirroring cable tension behavior. Why Other Options Are Wrong: Options claiming constant tension ignore acceleration. Saying “more when moving downwards” reverses the sign. “Less when moving upwards” contradicts Newton’s law for upward acceleration. Common Pitfalls (misconceptions, mistakes): Confusing direction of motion with direction of acceleration (e.g., moving up while slowing down implies downward acceleration and reduced tension). Final Answer: it is less when the lift is moving downwards

Question 13

Friction on an incline — the angle an inclined plane makes with the horizontal when a body placed on it is just about to move down the plane is called the:

Accepted Answer

Introduction / Context: At impending motion on a rough inclined plane, the balance between gravitational component down the slope and limiting friction defines a characteristic angle. This angle is central to heap stability, conveying, and slope design problems in mechanical and geotechnical engineering. Given Data / Assumptions: Static friction model with coefficient μ between block and plane. Impending motion downward (limiting equilibrium). Plane angle θ with the horizontal. Concept / Approach: At the verge of sliding down, the component of weight along the plane balances the limiting friction: mgsin θ = μmgcos θ ⇒ tan θ = μ. The angle θ_r at which this occurs is the angle of repose. The angle of friction φ is defined by tan φ = μ; therefore θ_r = φ at impending motion. Step-by-Step Solution: Resolve weight into components along and normal to the plane.Set along-plane equilibrium at limiting friction: mgsin θ = μmgcos θ.Obtain tan θ = μ and identify θ as the angle of repose.Recognize equality with the angle of friction under limiting conditions. Verification / Alternative check (if short method exists): Plot friction cone or use friction circle; impending motion occurs when resultant reaction makes an angle φ with the normal such that tan φ = μ, again implying θ_r = φ. Why Other Options Are Wrong: Angle of inclination is a generic geometric term. Angle of friction is a material property equal to the angle of repose only at impending motion, but the named condition asked is specifically the angle of repose. Common Pitfalls (misconceptions, mistakes): Confusing steady sliding (kinetic friction) with impending motion (static friction); mixing degrees and radians when computing θ from μ. Final Answer: angle of repose

Question 14

Law of Polygon of Forces — choose the correct statement that defines the polygon method for the resultant and equilibrium of coplanar concurrent forces.

Accepted Answer

Introduction / Context: The polygon law generalizes the parallelogram law to more than two concurrent forces. It provides a graphical method to determine the single resultant of several forces acting at a point, and to check equilibrium when the force polygon closes. Given Data / Assumptions: Forces are coplanar and concurrent. A consistent scale is used to represent magnitudes. Vectors are drawn head-to-tail in sequence. Concept / Approach: Arrange force vectors one after another, preserving magnitude and direction. The straight line joining the start of the first vector to the end of the last vector (the “closing side”) represents the single resultant. If the polygon closes (end meets start), the resultant is zero and the system is in equilibrium. Step-by-Step Solution: Select a force and draw it to scale.From its head, draw the next force to scale in its given direction.Continue for all forces, forming a polygon.Draw the closing side from the tail of the first to the head of the last; this is the resultant (zero if the polygon closes naturally). Verification / Alternative check (if short method exists): Component-wise vector addition of all forces yields the same resultant as the closing side in the graphical construction. Why Other Options Are Wrong: (a) describes the special equilibrium case only, not the general resultant construction. (c) contains distorted wording; while “opposite order” appears in some texts, option (b) states the standard result clearly. (d) again addresses only the equilibrium outcome without the general construction. Common Pitfalls (misconceptions, mistakes): Using inconsistent scales; not preserving directions; confusing the meaning of a closed polygon (equilibrium) versus an open polygon (nonzero resultant). Final Answer: If forces acting on a point can be represented in magnitude and direction by the sides of a polygon taken in order, then the resultant is represented in magnitude and direction by the closing side of the polygon.

Question 15

Kinematics of a triangular three-link system — three rigid rods hinged together to form a triangle undergo general motion (rotation + translation). How many instantaneous centres of rotation exist for the triangle?

Accepted Answer

Introduction / Context: Instantaneous centres (ICs) are points about which a link has a pure rotation at a given instant. For mechanisms, the number and location of ICs aid in velocity analysis without calculus. The Kennedy–Aronhold theorem governs the count and collinearity of ICs among multiple links in plane motion. Given Data / Assumptions: The system consists of three rigid links forming a triangle with pin joints. Plane motion is considered (combination of translation and rotation). No slipping at pins; ideal hinges. Concept / Approach: For a mechanism with n links in plane motion, the number of instantaneous centres is N = n*(n − 1)/2. With three links, N = 32/2 = 3. Kennedy’s theorem further states that the three ICs for any three bodies in plane motion are collinear, aiding construction and checks in velocity diagrams. Step-by-Step Solution: Identify n = 3 links: two moving links plus the ground (if treated as a link) — still total three for a closed triangular set.Compute N = n(n − 1)/2 = 3.Enumerate the ICs: one for each pair of links.Use collinearity (Kennedy) when constructing the ICs geometrically. Verification / Alternative check (if short method exists): Alternative velocity analysis using relative velocities will reveal three pairwise relative rotations consistent with three ICs. Why Other Options Are Wrong: Values 1, 2 undercount; 4, 5 overcount for n = 3 in planar motion. Common Pitfalls (misconceptions, mistakes): Forgetting to include the frame as a link in open-chain analyses; assuming one IC suffices for composite rigid-body motion without considering relative motions between links. Final Answer: 3

Question 16

Head-on impact (incomplete data) — two balls have masses in ratio 2:1 and equal-and-opposite momenta before collision (speeds in ratio 1:2 opposite). With coefficient of restitution e (not specified), the velocities of separation are stated to be:

Accepted Answer

original velocity in the opposite direction

Question 17

Ring on a two-segment string — a 90 cm string is tied between points A and B at the same level, 60 cm apart. A 120 g ring slides on the string. A horizontal force P is applied so that the ring is vertically below B in equilibrium. Find P (in gram-fo…

Accepted Answer

Introduction / Context: This statics problem couples geometry with equilibrium of a particle acted on by string tensions, weight, and an external horizontal force. Such configurations model pulley/string support systems and help build skill in resolving concurrent forces using geometry-derived direction cosines. Given Data / Assumptions: String length L = 90 cm connects points A and B at the same level, AB = 60 cm. A ring of weight W = 120 g-force slides frictionlessly on the string. A horizontal force P acts on the ring so that its final position is vertically below B. String is light and inextensible; tensions are equal in both segments (T). Concept / Approach: Let the ring be vertically below B at distance y. One segment (B–ring) is vertical; the other (A–ring) is slanted. Use the length constraint to find y and the slanted segment length. Then write horizontal and vertical equilibrium of the ring with two equal tensions T, the horizontal force P, and weight W. Step-by-Step Solution: Place A at (0, 0) m, B at (0.60, 0) m, ring at (0.60, −y) m.String length: y + sqrt(0.60^2 + y^2) = 0.90 m ⇒ y = 0.25 m (25 cm).Length of slanted segment A–ring = sqrt(0.36 + 0.25^2) = 0.65 m (65 cm).Direction cosines for A–ring: horizontal component ratio = 0.60/0.65 = 12/13; vertical = 0.25/0.65 = 5/13.Let common tension be T. Vertical equilibrium: T (up from B–ring) + T*(5/13) (up from A–ring) = W ⇒ (18/13) T = 120 ⇒ T = 120 * 13 / 18 = 86.667 g.Horizontal equilibrium: P (to the right) balances leftward component of A–ring tension ⇒ P = T*(12/13) = 86.667 * 12/13 = 80 g. Verification / Alternative check (if short method exists): Use geometry first to get y = L − AB′ where AB′ is the projected length; the 5–12–13 direction ratios confirm exact arithmetic, giving P = 80 g without rounding errors. Why Other Options Are Wrong: 40 g and 60 g underestimate the necessary balance of the horizontal component. 100 g overestimates; 120 g equals weight, not the required horizontal balance. Common Pitfalls (misconceptions, mistakes): Assuming unequal tensions in a single frictionless string; forgetting that the B–ring segment is vertical and contributes no horizontal component; mixing centimetres and metres inconsistently. Final Answer: 80 g

Question 18

Simple harmonic motion (SHM) application: For a body executing SHM with amplitude 150 mm and frequency 2 vibrations/s, what is the maximum (peak) speed attained during the motion?

Accepted Answer

Introduction / Context: This problem checks recall and correct use of simple harmonic motion (SHM) kinematics. In SHM, the displacement, velocity, and acceleration are sinusoidal functions, and the peak (maximum) speed depends on the angular frequency and the amplitude. Engineers use this relation for designing vibration isolators, cams, and reciprocating systems. Given Data / Assumptions: Amplitude A = 150 mm = 0.150 m. Frequency f = 2 vibrations/s. Angular frequency ω = 2 * π * f. Ideal SHM with no damping. Concept / Approach: The SHM velocity varies as v(t) = ω * A * cos(ωt + φ). The maximum magnitude occurs when |cos(…)| = 1, hence v_max = ω * A. Convert frequency to angular frequency and multiply by amplitude. Step-by-Step Solution: Compute ω: ω = 2 * π * f = 2 * π * 2 = 4π ≈ 12.566 rad/s.Apply v_max formula: v_max = ω * A.Substitute values: v_max = 12.566 * 0.150 = 1.8849… m/s.Round sensibly: v_max ≈ 1.885 m/s. Verification / Alternative check: Using energy: at mid-position, all energy is kinetic: (1/2) m v_max^2 = (1/2) k A^2 with k = m * ω^2, giving v_max = ω * A again. This independently confirms the result. Why Other Options Are Wrong: 0.18845 m/s and 0.01885 m/s are off by factors of 10 and 100 due to unit/frequency mistakes. 18.85 m/s and 188.5 m/s are 10× and 100× too large, likely from using A in mm without converting to metres or squaring π incorrectly. Common Pitfalls: Forgetting to convert amplitude from mm to m. Using frequency f instead of angular frequency ω in v_max = ω * A. Final Answer: 1.885 m/s

Question 19

Statically determinate beam (figure not provided): The beam is supported by a hinge at A and a roller at B. Determine the reaction at support A (units: t).

Accepted Answer

3.5 t

Question 20

Centre of gravity of a triangle in engineering mechanics: The centroid is located at the intersection of which set of internal lines?

Accepted Answer

Introduction / Context: The centroid (centre of gravity for a uniform lamina) of a triangle is a foundational result in statics and structural analysis. It is used to compute moments, locate resultant forces, and determine support reactions for triangular load distributions. Given Data / Assumptions: Plane triangular lamina of uniform density. Standard Euclidean geometry applies. No skewed mass distribution or thickness variation. Concept / Approach: A triangle has several notable concurrent line sets: medians, angle bisectors, perpendicular bisectors, and altitudes. The point where medians meet is called the centroid (also denoted G). Each median connects a vertex to the midpoint of the opposite side. The centroid divides each median in a 2:1 ratio measured from the vertex. Step-by-Step Solution: Identify the three medians by locating midpoints of each side and joining them to the opposite vertices.Recognize their concurrency at a single point G due to triangle geometry theorems.Recall property: centroid divides medians in the ratio 2:1 from vertex to base side. Verification / Alternative check: Coordinate geometry check: For vertices (x1,y1), (x2,y2), (x3,y3), the centroid is at ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3), confirming the intersection of medians interpretation. Why Other Options Are Wrong: Perpendicular bisectors meet at the circumcenter, not the centroid. Angle bisectors meet at the incenter (centre of inscribed circle). Altitudes meet at the orthocenter. “None of these” is incorrect because medians are the right answer. Common Pitfalls: Confusing different triangle centers: centroid, incenter, circumcenter, and orthocenter. Final Answer: medians of the triangle

Applied Mechanics Questions