Question 1

Acceleration of a body sliding down a smooth incline A block slides down a frictionless plane inclined at angle θ to the horizontal. What is its acceleration component along the plane?

Accepted Answer

Introduction / Context:Resolving gravitational acceleration into components along and perpendicular to an inclined plane is a fundamental step in many statics and dynamics problems. With no friction, only the component of weight along the plane accelerates the block.Given Data / Assumptions:Inclination angle θ to the horizontal.Frictionless (smooth) plane.Gravity g acts vertically downward.Concept / Approach:Resolve weight W = m g into components: perpendicular to plane (m g cos θ) and along plane (m g sin θ). With no friction, the equation of motion along the plane is m a = m g sin θ, leading directly to a = g sin θ.Step-by-Step Solution:Choose axes: x along plane downward; y normal to plane.Along y: N − m g cos θ = 0 ⇒ N = m g cos θ.Along x: m a = m g sin θ ⇒ a = g sin θ.Verification / Alternative check:Limiting cases: θ = 0° ⇒ a = 0; θ = 90° ⇒ a = g. Both agree with intuition.Why Other Options Are Wrong:g is the full vertical acceleration, not along the plane; g cos θ is the normal component; g tan θ is not a component of gravitational acceleration; zero would require θ = 0° or friction balancing, which is not the case here.Common Pitfalls:Using g cos θ for the along-plane component; forgetting to define axes parallel and perpendicular to the plane.Final Answer:g sin θ

Question 2

Practical (SI) unit of work in mechanics Which is the standard practical unit of work used in engineering mechanics?

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Introduction / Context:Units must be clearly distinguished between force, work/energy, and power. In the International System of Units (SI), work and energy share the same unit, used universally in engineering calculations and specifications.Given Data / Assumptions:Work W = Force * Displacement along the line of action.Use SI base and derived units.Concept / Approach:The SI unit of work is the Joule (J), defined as 1 Newton of force acting over 1 meter of displacement: 1 J = 1 N·m. The Newton (N) is the unit of force, the Watt (W) is power (J/s), and the Dyne and Erg are CGS units.Step-by-Step Solution:Work definition: W = F * s (when F and s are colinear).Units: [W] = N·m = (kg·m/s^2)·m = kg·m^2/s^2.Name of unit: Joule (J).Verification / Alternative check:Dimensional analysis confirms [work] equals [energy], reinforcing that Joule is also the unit of energy.Why Other Options Are Wrong:Erg and Dyne belong to the CGS system (1 erg = 1 dyne·cm); Newton is force; Watt is power, not work.Common Pitfalls:Confusing Newton-meter as torque only; in SI, N·m is the dimension for both torque and work, but work is a scalar energy transfer, while torque is a vector moment.Final Answer:Joule

Question 3

Centroid of a standard T-section (10 × 15 × 5 cm) A T-section has flange 10 cm wide and 5 cm thick, with a web (stem) 5 cm thick and 10 cm deep beneath the flange (overall depth 15 cm). Measured from the bottom of the section, locate the centroid (d…

Accepted Answer

Introduction / Context:Locating the centroid (center of gravity for uniform density) of built-up sections such as T-sections is a staple of structural analysis. It involves dividing the shape into simple rectangles, using the first moment of area about a reference line, and then finding the combined centroidal location.Given Data / Assumptions:Flange: width b_f = 10 cm, thickness t_f = 5 cm.Web (stem): thickness b_w = 5 cm, depth h_w = 10 cm below flange.Bottom reference at the base of the stem; overall depth = 15 cm.Uniform density and thickness; ignore fillets.Concept / Approach:Use composite areas. The centroid measured from the chosen reference is ȳ = Σ(A_i y_i) / ΣA_i, where A_i are component areas and y_i are their centroidal distances from the reference. Take two rectangles: web and flange.Step-by-Step Solution:Area of web A_w = b_w * h_w = 5 * 10 = 50 cm^2.Centroid of web from bottom y_w = h_w/2 = 5 cm.Area of flange A_f = b_f * t_f = 10 * 5 = 50 cm^2.Centroid of flange from bottom y_f = h_w + t_f/2 = 10 + 2.5 = 12.5 cm.Total area ΣA = 50 + 50 = 100 cm^2.First moment Σ(A y) = 505 + 5012.5 = 250 + 625 = 875 cm^3.ȳ = Σ(A y) / ΣA = 875 / 100 = 8.75 cm.Verification / Alternative check:Symmetry about the vertical axis through the stem confirms x̄ is centered; the ȳ value lies between 5 and 12.5 cm, which is reasonable for equal flange and web areas.Why Other Options Are Wrong:5.00 cm and 7.50 cm are too low; 7.85 cm is also low; “None” is incorrect because the calculation yields 8.75 cm exactly for the stated geometry.Common Pitfalls:Misinterpreting the given 10 × 15 × 5 cm as a solid rectangle; forgetting to reference centroid distances consistently from the same datum; arithmetic slips in Σ(A y).Final Answer:8.75 cm

Question 4

Simple harmonic motion (SHM): A particle executes simple harmonic motion. At which position in its path does it attain the maximum speed, and why?

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Introduction / Context:In simple harmonic motion (SHM), a particle oscillates about a mean (equilibrium) position under a restoring force proportional to displacement. Understanding where speed is maximum is fundamental to vibration theory and helps connect energy concepts with kinematics in oscillatory systems.Given Data / Assumptions:Displacement x varies sinusoidally about the mean position.Amplitude A is the maximum displacement on either side.Angular frequency is ω (constant).No damping or external forcing; pure SHM.Concept / Approach:For SHM, velocity v is given by v = ω * sqrt(A^2 - x^2). Alternatively, v = dx/dt when x = A sin(ωt + φ). Maximum speed occurs when the square root term is largest, i.e., when x = 0. Energy view: at the mean position, potential energy (stored in the spring or restoring element) is minimum and kinetic energy is maximum, so speed peaks there.Step-by-Step Solution:Position dependence: v(x) = ω * sqrt(A^2 - x^2).Maximize v ⇒ maximize (A^2 - x^2) ⇒ x = 0.At x = 0 (mean position), v_max = ωA.Verification / Alternative check:At extremes x = ±A, v = 0, matching the observation that the particle turns around at the endpoints. Between extreme and mean, v increases; it is largest exactly at the centre.Why Other Options Are Wrong:Extreme point: speed is zero there (turning points).Half amplitude: speed is non-zero but not maximum.“None of these” contradicts the SHM velocity relation.Common Pitfalls:Confusing maximum displacement with maximum speed; in SHM they do not coincide. Also, mixing angular frequency ω with linear frequency f without conversion.Final Answer:through the mean position

Question 5

Area moment of inertia of a square lamina: For a square of side b, what is the second moment of area (area moment of inertia) about an axis through its centroid and parallel to one side?

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Introduction / Context:The area moment of inertia (second moment of area) quantifies a cross-section’s resistance to bending about a specified axis. For standard shapes like a square, knowing the centroidal moments is essential in beam design and structural analysis.Given Data / Assumptions:Plane area is a square lamina of side b.Axis passes through the centroid (geometric centre).Axis is parallel to one side of the square (i.e., centroidal x- or y-axis).We are dealing with area (not mass) moment of inertia.Concept / Approach:Standard formula for a rectangle of breadth b and depth h about its centroidal axis parallel to breadth is I = bh^3/12. For a square b = h, so I = bb^3/12 = b^4/12. This holds for either principal centroidal axis due to symmetry (I_x = I_y).Step-by-Step Solution:Start with rectangle formula: I_c = bh^3/12.Set h = b for a square: I_c = bb^3/12.Therefore, I_c = b^4/12 about a centroidal axis parallel to a side.Verification / Alternative check:Polar moment at centroid J_o = I_x + I_y = b^4/12 + b^4/12 = b^4/6, consistent with symmetry and standard tables.Why Other Options Are Wrong:b^3/4: wrong dimension (length^3 instead of length^4).b^4/3 or b^4/8: too large; would overestimate stiffness.b^4/36: centroidal about a diagonal is not b^4/36; this value does not correspond to a common centroidal axis for a square about a side-parallel axis.Common Pitfalls:Confusing area moment (length^4) with mass moment of inertia; using parallel-axis formula unnecessarily when the axis is through the centroid.Final Answer:b^4 / 12

Question 6

Centre of gravity (definition and meaning): For a homogeneous rigid body, the centre of gravity is the point at which the entire ____ is considered to act.

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Introduction / Context:The centre of gravity (CG) is a foundational concept in statics and dynamics. It allows engineers to replace the distributed gravitational forces on a body by a single resultant acting at a specific point for equilibrium and motion analysis.Given Data / Assumptions:Rigid body with uniform gravitational field.Homogeneous body means constant density, but the definition does not require homogeneity.We are interested in the definition—where the resultant weight acts.Concept / Approach:The CG is the point at which the resultant of gravitational forces (body weight) is taken to act. For uniform gravity fields, the centre of gravity coincides with the centre of mass. This point simplifies calculations of balance, support reactions, and rotational dynamics under gravity.Step-by-Step Solution:Replace the distributed weight density ρg over the volume with a single force W = mg.Locate the point G where moments of the distributed weight about any axis equal the moment of W about that axis.By definition, G is the centre of gravity and W is considered to act there.Verification / Alternative check:For symmetric uniform bodies, G lies at the geometric centre (e.g., the centroid for laminae). Lever-balance experiments also identify the point where net moment of weight is zero in all orientations.Why Other Options Are Wrong:Volume/area concentrated (a or b): useful for centroid definitions of volume or area, but CG specifically refers to weight (force) concentration.“All the above” mixes geometric centroids with gravitational resultant, which is imprecise.Common Pitfalls:Confusing centre of mass, centroid, and centre of gravity—these coincide only under uniform gravity and homogeneity.Final Answer:weight of the body

Question 7

Composite centroid along an axis: From a solid right circular cylinder (height 8 cm, radius 4 cm) a right circular cone of the same height and base is removed (scooped) from the top. Find the height of the centre of gravity of the remaining solid me…

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Introduction / Context:Finding the centre of gravity (CG) of a composite body is a standard statics task. When a cone is removed from a cylinder (same base and height), symmetry keeps the CG on the common axis; its height is determined by the principle of moments of volumes (or weights) about the base.Given Data / Assumptions:Cylinder: radius r = 4 cm, height h = 8 cm; CG at h/2 = 4 cm from base.Removed cone: same base and height; CG at h/4 = 2 cm from base (measured along the axis from the base).Uniform density; use volumes as weights.Concept / Approach:Use the composite-body formula for centroid: ȳ = (Σ V_i * y_i) / (Σ V_i), with the removed part contributing negative volume. Volumes: V_cyl = π r^2 h, V_cone = (1/3) π r^2 h.Step-by-Step Solution:Compute volumes: V_cyl = π4^28 = 128π, V_cone = (1/3)128π = (128/3)π.Locate centroids: y_cyl = 4 cm, y_cone = 2 cm.Composite centroid: ȳ = (V_cyly_cyl − V_coney_cone) / (V_cyl − V_cone).Substitute: ȳ = (128π4 − (128/3)π*2) / (128π − (128/3)π).Simplify numerator = 512π − (256/3)π = (1280/3)π; denominator = (256/3)π.Therefore ȳ = (1280/3)/(256/3) = 5.0 cm.Verification / Alternative check:Dimensional and symmetry checks: result lies between 4 cm (solid cylinder) and 6 cm (if material concentrated toward the top), consistent with removing more mass near the top than the bottom.Why Other Options Are Wrong:4.5, 5.25, 5.5 cm: Do not satisfy the exact composite centroid calculation.Common Pitfalls:Using the wrong cone centroid location (it is at h/4 from the base, not from the apex) and forgetting to subtract the removed volume’s moment.Final Answer:5.0 cm

Question 8

Velocity as a physical quantity: How should the velocity of a moving body be classified in mechanics?

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Introduction / Context:Velocity is central to kinematics and dynamics. Distinguishing it from speed and understanding its directional nature are essential for correctly applying Newton’s laws and for resolving motions into components.Given Data / Assumptions:Velocity combines magnitude and direction.Speed is the magnitude of velocity.Motion may be uniform or variable; classification should be general.Concept / Approach:A physical quantity that requires both magnitude and direction for complete specification is a vector. Velocity fits this definition: it specifies how fast and in what direction position is changing with time (v = dr/dt). Scalars like speed retain only magnitude, losing directional information.Step-by-Step Solution:Definition: velocity v is the time derivative of position vector r.Therefore v has the same directional character as r; it is a vector.Speed = |v| is scalar; it cannot specify direction.Verification / Alternative check:Vector addition rules apply to velocity (e.g., relative velocity problems). Component-wise, v = v_x i + v_y j (+ v_z k), confirming vector nature.Why Other Options Are Wrong:Scalar quantity: describes speed, not velocity.Constant quantity: velocity may change in time; “constant” is not an inherent classification.“None of these” is incorrect because velocity is definitively a vector.Common Pitfalls:Interchanging “speed” and “velocity”; overlooking directional changes in circular motion where speed can be constant but velocity is not.Final Answer:a vector quantity

Question 9

Impacts of smooth spheres: identify the always-correct statement Which one of the following statements is always true for collisions of smooth spheres (head-on or oblique) under classical impact theory?

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Introduction / Context:Classical impact theory for smooth spheres separates velocities into components along and perpendicular to the line of centres. The coefficient of restitution e acts only along the line of impact; perpendicular components remain unchanged. Identifying universally valid statements requires careful use of these rules.Given Data / Assumptions:Spheres are smooth (no tangential impulse).Masses can be equal or unequal; impacts can be direct (collinear) or oblique.Coefficient of restitution 0 ≤ e ≤ 1 acts along the line of centres.Concept / Approach:For a head-on elastic impact (e = 1) of equal masses, momentum conservation and restitution give a neat result: the two bodies exchange velocities along the line of centres. For other statements, conditions or numerical factors matter, making them not always true.Step-by-Step Solution:Equal masses, e = 1, direct hit: u₁ → v₂ and u₂ → v₁ (interchange), hence option (a) is always true.For option (b): when one equal sphere is initially at rest, the fraction of kinetic energy lost is (1 − e^2)/2, not (1 − e^2). Hence (b) is false in general.Option (c) is true for the struck sphere’s component along the line of centres, but with oblique impacts its overall motion includes unchanged perpendicular components; the statement as written can be misread as exclusive line-of-centres motion, so it is not universally accurate.Option (d) is only true for special geometries; not guaranteed for all right-angle approaches.Verification / Alternative check:Resolve velocities into normal and tangential components; apply restitution only to the normal component and confirm the interchange in the special case of equal masses and e = 1.Why Other Options Are Wrong:See analysis above; each relies on extra conditions or has incorrect numeric factors.Common Pitfalls:Applying restitution to tangential components; assuming energy loss fraction without considering mass equality and initial states.Final Answer:If two equal and perfectly elastic smooth spheres impinge directly, they interchange their velocities.

Question 10

Units for moment (moment of force) in the MKS gravitational system: What is the conventional unit for a moment in the MKS gravitational system?

Accepted Answer

Introduction / Context:The unit of a moment (moment of force) depends on the force unit used. In the SI coherent system, moment is expressed in newton–metre (N·m). In the older MKS gravitational system commonly used in many basic mechanics texts, force was expressed in kilogram-force (kgf), leading to the moment unit kgf·m, often abbreviated as kgm.Given Data / Assumptions:Gravitational MKS uses kilogram-force (kgf) as the force unit.Moment = Force * Distance (perpendicular).Distance measured in metres.Concept / Approach:Dimensionally, a moment is the product of a force and a length. Thus, in gravitational MKS: moment unit = kgf * m (written as kgm in many problems). In SI: 1 kgf ≈ 9.80665 N, so kgf·m can be converted to N·m when needed.Step-by-Step Solution:Moment M = F * d.Using gravitational MKS: F in kgf and d in m ⇒ M unit = kgf·m (abbrev. kgm).Therefore the correct choice is kgm.Verification / Alternative check:In SI, replace kgf by newton: 1 kgf·m ≈ 9.80665 N·m. The structure “force × length” remains the same across systems.Why Other Options Are Wrong:kg/m^2, kg/s^2, kg/s: not “force × length”.N·m^2: contains an extra metre; the standard SI unit is N·m.Common Pitfalls:Confusing mass with force; forgetting the “perpendicular distance” notion does not change the unit—still force times length.Final Answer:kgm

Question 11

Velocity ratio (VR) of an ideal inclined plane: For lifting a load along a smooth plane inclined at angle θ to the horizontal, what is the velocity ratio?

Accepted Answer

Introduction / Context:Velocity ratio (VR) is a measure of kinematic advantage in a machine: VR = distance moved by effort / distance moved by load. For an ideal (frictionless) inclined plane, VR links the plane geometry to mechanical advantage and effort calculations.Given Data / Assumptions:Smooth (frictionless) plane inclined at angle θ.Load moves a vertical height h while effort moves along the plane length l.Effort is applied parallel to the plane.Concept / Approach:For an inclined plane, l and h are related by sin θ = h/l. Therefore VR = distance moved by effort / distance moved by load = l / h = 1 / sin θ = cosec θ. This result is purely geometric and independent of load magnitude (for an ideal plane).Step-by-Step Solution:Geometry: sin θ = opposite/hypotenuse = h/l.Hence l/h = 1/sin θ.Therefore VR = 1/sin θ = cosec θ.Verification / Alternative check:Mechanical advantage MA (ideal) equals VR. For θ = 30°, VR = 2, matching the familiar result that a 30° plane halves the required ideal effort (ignoring friction).Why Other Options Are Wrong:sin θ, cos θ, tan θ, sec θ: do not match l/h for the plane’s rise and run.Common Pitfalls:Using cos θ instead of sin θ (confusing run and rise), or mixing VR with mechanical advantage in the presence of friction (MA Final Answer:cosec θ

Question 12

Period of simple harmonic motion (relationship with angular frequency): How does the time period of an SHM vary with its angular velocity ω?

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Introduction / Context:The time period T of a simple harmonic oscillator specifies the time for one complete oscillation. Connecting T with the angular frequency ω is central to vibration analysis, pendulum approximations, and spring–mass systems.Given Data / Assumptions:Pure SHM with angular frequency ω.Linear restoring law; no damping or forcing.Standard kinematic relations of SHM hold.Concept / Approach:By definition, angular frequency ω relates to the time period by ω = 2π / T, or equivalently T = 2π / ω. Thus, T varies inversely with ω: doubling ω halves T, etc.Step-by-Step Solution:Start from x = A sin(ωt + φ); one full cycle when argument increases by 2π.Hence ωT = 2π ⇒ T = 2π/ω.Therefore T ∝ 1/ω (inverse proportionality).Verification / Alternative check:For a spring–mass system, ω = sqrt(k/m) ⇒ T = 2π sqrt(m/k), again showing inverse relation to ω.Why Other Options Are Wrong:Direct or square relations contradict T = 2π/ω.Inverse square would imply T ∝ 1/ω^2, which is incorrect.Common Pitfalls:Confusing angular frequency ω (rad/s) with cyclic frequency f (Hz); note that T = 1/f and ω = 2πf.Final Answer:inversely proportional to its angular velocity

Question 13

Balancing two like parallel forces by adjusting one magnitude: For two parallel forces of 20 kg and 15 kg separated by a fixed distance, by how much should the 20 kg force be reduced so that the distance of the new resultant from the 20 kg force equ…

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Introduction / Context:Locating the line of action of a resultant of two like parallel forces is routine in statics. For forces F₁ and F₂ separated by distance L, the distance of the resultant from the F₁ line is x = F₂L/(F₁ + F₂). This problem asks for a change in F₁ such that a symmetry of distances before and after the change is achieved.Given Data / Assumptions:Initial forces: F₁ = 20 kg, F₂ = 15 kg (like, parallel).Separation = L (constant but unknown; it will cancel).After adjustment: F₁′ Condition: distance from the 20 kg line after change equals the original distance from the 15 kg line.Concept / Approach:Use the centroid-like formula for the resultant position for two forces. Set the “new-from-20 kg” distance equal to the “old-from-15 kg” distance and solve for F₁′. Notice that L cancels, so the numerical answer is independent of spacing.Step-by-Step Solution:Original distance from 15 kg line: x_old_from15 = F₁L/(F₁ + F₂) = 20L/35 = (4/7)L.New distance from 20 kg line: x_new_from20 = F₂*L/(F₁′ + F₂) = 15L/(F₁′ + 15).Set equality: 15L/(F₁′ + 15) = (4/7)L ⇒ 15/(F₁′ + 15) = 4/7.Solve: 105 = 4F₁′ + 60 ⇒ 4F₁′ = 45 ⇒ F₁′ = 11.25 kg.Reduction = 20 − 11.25 = 8.75 kg.Verification / Alternative check:With F₁′ = 11.25 kg, new resultant location from 20 kg line is 15L/26.25 = (4/7)L, exactly the previous distance from the 15 kg line.Why Other Options Are Wrong:5.5, 6.25, 10.5 kg: do not satisfy the balance equation; they yield incorrect positions for the resultant.Common Pitfalls:Reversing the “from which force” reference or using the wrong formula; forgetting that L cancels out and trying to assign a value to L.Final Answer:8.75 kg

Question 14

Highway vehicle dynamics on a horizontal curve  A vehicle of weight w (kgf equivalent) negotiates a horizontal circular curve of radius r. The centre of gravity is at height h above road level, and the distance between the wheel centres (track width…

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Introduction / Context:In highway and vehicle dynamics, overturning on a level horizontal curve occurs when the lateral inertial effect shifts the resultant of weight and centrifugal force to pass through the outer wheel contact. This question asks for the limiting speed that prevents overturning, given geometry and center-of-gravity height.Given Data / Assumptions:Vehicle weight = w (use m * g; it cancels in the ratio).Radius of curve = r.Height of centre of gravity above road = h.Track width = 2a (distance between wheel contact lines).Level road (no banking), no side slip; quasi-static analysis.Concept / Approach:At speed v, the lateral inertial force at the centre of gravity is m * v^2 / r, acting horizontally. Overturning about the outer wheels occurs when the overturning moment due to lateral force equals the restoring moment due to gravity. Use moment equilibrium about the outer wheel contact line to find the limiting velocity.Step-by-Step Solution:Overturning moment about outer wheel: M_over = (m * v^2 / r) * h.Restoring moment about outer wheel: M_resist = m * g * a.At limit of overturning: (m * v^2 / r) * h = m * g * a.Cancel m and solve: v^2 = (g * r * a) / h → v_max = √(g r a / h).Verification / Alternative check:Dimensional check: g [m/s^2] * r [m] * a [m] / h [m] gives m^2/s^2 → square root is m/s (correct). Increasing h lowers v_max; increasing track half-width a or radius r raises v_max, which matches practical intuition.Why Other Options Are Wrong:√(g r h / a): Inverts a and h; predicts higher v for higher CG—physically wrong.√(g a / (r h)) and √(g r / (a h)): Have incorrect dependence on r and a.g r a / h: Misses square root; wrong units.Common Pitfalls:Confusing overturning with skidding (the latter depends on friction). Also mixing up track half-width a with full track 2a. Always take moment about outer wheel contact for overturning on level curves.Final Answer:v_max = √(g r a / h)

Question 15

Equilibrium of a point under concurrent forces  A particle is subjected to several forces acting through a point. Which condition ensures equilibrium of the point in a plane?

Accepted Answer

Introduction / Context:For concurrent forces in a plane, equilibrium requires the vector sum of all forces to be zero. In practical computation we ensure this by resolving the forces along two perpendicular axes and setting both component sums to zero.Given Data / Assumptions:Forces are concurrent at a point (no couple-only system).Planar problem (two dimensions).Rigid-body effects of moments are irrelevant for concurrency at the same point.Concept / Approach:Vector equilibrium of concurrent forces demands ΣF = 0. In components this is equivalent to ΣF_x = 0 and ΣF_y = 0 for any orthogonal axes. These two scalar equations completely specify equilibrium in the plane for concurrent systems.Step-by-Step Solution:Choose axes (x, y) at right angles.Resolve each force F_i into components (F_ix, F_iy).Impose ΣF_x = 0 and ΣF_y = 0.Since all lines of action pass through the same point, moment equilibrium is automatically satisfied for any reference point.Verification / Alternative check:Graphical polygon of forces: if the head-to-tail polygon closes, the component sums are zero; the two methods are equivalent.Why Other Options Are Wrong:“Algebraic sum of forces is zero” is ambiguous (forces are vectors, not scalars).“Resolved parts are equal” does not enforce zero; equilibrium needs zero resultant, not equality.“Only moments zero” is insufficient; forces could still have a net vector.Common Pitfalls:Mixing vector and scalar language; forgetting that any two perpendicular directions suffice, not specifically horizontal/vertical.Final Answer:The sums of the resolved components in any two mutually perpendicular directions are both zero

Question 16

Centre of gravity of common plane laminas  For which of the following planar shapes is the centre of gravity not at its simple geometric centre (i.e., not at a point of central symmetry)?

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Introduction / Context:The centre of gravity (centroid) of symmetric shapes lies at their geometric centre. Shapes lacking central symmetry do not have a geometric centre coincident with the centroid. This concept is widely used in calculating bending stresses and deflections.Given Data / Assumptions:Uniform thickness, homogeneous lamina.Plane shapes considered: circle, square, rectangle, equilateral triangle, right-angled triangle.Concept / Approach:Shapes with full central symmetry (circle, square, rectangle) have the centroid at the intersection of their symmetry axes (geometric centre). An equilateral triangle has threefold symmetry; its centroid is at the intersection of medians, also the geometric centre of that symmetric triangle. A right-angled triangle lacks central symmetry; its centroid lies at the intersection of medians at distances b/3 and h/3 from the right-angled vertex, not at any “centre” of symmetry.Step-by-Step Solution:Circle/Rectangle/Square: multiple axes of symmetry intersect at geometric centre → centroid there.Equilateral triangle: medians, angle bisectors, and perpendicular bisectors concur—this point is the geometric centre.Right-angled triangle: only line symmetries (if isosceles right triangle) may exist; general right triangle lacks central symmetry → centroid not at a simple geometric centre.Verification / Alternative check:Compute centroid coordinates for a right-angled triangle with legs along axes: (x̄, ȳ) = (b/3, h/3) from the right-angle vertex. This is not the geometric centre of the bounding rectangle or of the triangle.Why Other Options Are Wrong:Circle, square, rectangle: clear central symmetry → centroid at geometric centre.Equilateral triangle: high symmetry implies centroid is the intuitive centre point.Common Pitfalls:Confusing “any triangle” with “equilateral triangle.” Only the equilateral has a central point that could be called a geometric centre coincident with the centroid.Final Answer:Right-angled triangle

Question 17

Area moment (polar) of a circular lamina  Find the second moment of area of a thin circular lamina of diameter d about an axis perpendicular to its plane and passing through its centre.

Accepted Answer

Introduction / Context:The polar second moment of area of a circle about its centre is used in torsion (for circular shafts) and in combined bending-torsion problems. For a thin lamina, it is purely a geometric property of the area.Given Data / Assumptions:Circular lamina, diameter d (radius r = d/2).Axis: through the centre, perpendicular to the plane (polar axis).Thin, homogeneous area (area property only).Concept / Approach:For a circle of radius r, the polar second moment J about the centre is J = I_x + I_y, where I_x = I_y = (π r^4) / 4. Hence J = (π r^4) / 2. Substitute r = d/2 to express in diameter form.Step-by-Step Solution:Given r = d/2 → r^4 = d^4 / 16.J = (π r^4) / 2 = (π / 2) * (d^4 / 16) = (π d^4) / 32.Verification / Alternative check:Rectangular components: I_x = I_y = (π r^4) / 4. Adding gives (π r^4)/2, consistent with the polar formula.Why Other Options Are Wrong:(π d^4)/64: This equals I_x or I_y expressed in d, not the polar J.(π d^4)/16, (π d^4)/8, (π d^4)/4: Overestimate J; incorrect factors.Common Pitfalls:Confusing polar J with planar I_x or I_y; forgetting that J = I_x + I_y for axes through the same point.Final Answer:J = (π d^4) / 32

Question 18

Direction of the resultant of two forces  Two coplanar forces P and Q act at an angle θ to each other. If the resultant makes an angle α with the direction of P, what is tan α in terms of P, Q, and θ?

Accepted Answer

Introduction / Context:When combining two forces by the parallelogram rule, the direction of the resultant relative to one of the forces is obtained by resolving components along and perpendicular to that force. This is a standard vector-resolution result.Given Data / Assumptions:Two forces: P (reference direction) and Q at angle θ to P.Resultant makes angle α with P.Coplanar, concurrent forces.Concept / Approach:Resolve Q into components parallel and perpendicular to P. The resultant R has components R_parallel = P + Q cos θ and R_perp = Q sin θ. Therefore tan α is the ratio of the perpendicular to parallel components.Step-by-Step Solution:Along P: component = P + Q cos θ.Perpendicular to P: component = Q sin θ.Hence tan α = (Q sin θ) / (P + Q cos θ).Verification / Alternative check:Construct parallelogram and measure α; the component formula matches the geometry. Dimensionless ratio is consistent.Why Other Options Are Wrong:Swaps P and Q or swaps sine and cosine incorrectly.Reciprocal form gives cot α, not tan α.Some include wrong signs or denominators.Common Pitfalls:Using θ from the wrong side; forgetting which component is parallel/perpendicular to P.Final Answer:tan α = (Q sin θ) / (P + Q cos θ)

Question 19

Horizontal pull in a parabolic cable (light suspension bridge)  A light suspension bridge has span l and central dip y. Each cable carries a total uniformly distributed load w (total over the span on that cable). What is the horizontal pull H at eac…

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Introduction / Context:The horizontal component of cable tension in a parabolic cable under uniform load controls anchor design and sag selection. The classic relationship links span, sag (dip), and load.Given Data / Assumptions:Each cable carries a total load w over the span.Parabolic cable approximation under uniform horizontal projection of load.Span = l, central dip (sag) = y.Statics of a flexible, inextensible cable.Concept / Approach:For a parabolic cable with uniform load per unit horizontal length w_h, the horizontal pull is H = w_h * l^2 / (8 y). If the problem states w as the total load on a cable over the span, then w_h = w / l, giving H = (w / l) * l^2 / (8 y) = (w l) / (8 y).Step-by-Step Solution:Start from standard relation: H = w_h l^2 / (8 y).Given w is total → w_h = w / l.Substitute: H = (w / l) * l^2 / (8 y) = (w l) / (8 y).Verification / Alternative check:Units: w is force, l/y is dimensionless, so H has units of force—consistent. If w were given per unit length instead, H would be w l^2 / (8 y), which corresponds to a different interpretation.Why Other Options Are Wrong:(w l^2)/(8 y): Treats w as load per unit length, not as total load.(w y)/(8 l) and (8 w y)/l^2: Wrong dependence on y (predicts larger H for larger sag, opposite of reality).w l: Ignores sag entirely.Common Pitfalls:Confusing w as total versus per-unit-length; remember to convert to w_h when applying the parabolic formula.Final Answer:H = (w l) / (8 y)

Question 20

Second moment of area of a triangular section  For a triangle of base b and height h, find the second moment of area about the centroidal axis parallel to the base.

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Introduction / Context:The centroidal second moment of area of a triangle about an axis parallel to its base is a standard result used in beam bending calculations and deflection analysis.Given Data / Assumptions:Triangle of base b and height h.Axis: through centroid, parallel to base.Homogeneous thin area (geometric property).Concept / Approach:Starting from the base axis, the area moment about the base is I_base = b h^3 / 12. The centroidal axis is located at a distance h/3 from the base. Use the parallel-axis theorem to move to the centroidal axis.Step-by-Step Solution:I_base = b h^3 / 12 (about base).Distance from base to centroidal axis = h/3.Apply parallel-axis theorem: I_base = I_centroid + A * (h/3)^2, where A = (1/2) b h.Solve for I_centroid: I_centroid = I_base − A * (h^2 / 9) = (b h^3 / 12) − (1/2 b h) * (h^2 / 9) = (b h^3 / 12) − (b h^3 / 18) = b h^3 (1/12 − 1/18) = b h^3 / 36.Verification / Alternative check:Dimension check yields length^4. The value is smaller than I about the base, as expected.Why Other Options Are Wrong:b h^3/12: About the base, not centroid.b h^3/24, /48, /6: Incorrect constants from misapplication of the parallel-axis theorem.Common Pitfalls:Using base-axis value directly for centroid; forgetting that A = (1/2) b h for a triangle.Final Answer:I = b h^3 / 36

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