Question 1

Area moments – rectangle about centroidal axis versus base For a rectangle (width b, height h), what is the ratio of its second moment of area about the centroidal horizontal axis to that about its base?

Accepted Answer

Introduction / Context: Second moment of area (area moment of inertia) determines bending stiffness and deflection of beams. For rectangles, knowing values about different axes is crucial for quick checks and design sizing in structural engineering. Given Data / Assumptions: Rectangle width b and height h. Centroidal horizontal axis (through the centroid parallel to base). Base axis coincident with the rectangle's base edge. Concept / Approach: Standard results: I_centroid = (b * h^3) / 12 about the centroidal horizontal axis. Using the parallel-axis theorem to the base gives I_base = I_centroid + A * d^2, where A = b * h and d = h/2. Thus I_base = (b * h^3)/12 + (b * h) * (h/2)^2 = (b * h^3)/3. The requested ratio is (1/12) / (1/3) = 1/4. Step-by-Step Solution: I_centroid = (b * h^3) / 12.I_base = (b * h^3) / 3.Ratio = I_centroid / I_base = [(b * h^3)/12] / [(b * h^3)/3] = 1/4. Verification / Alternative check: Dimensional consistency: both moments scale with b * h^3, so the ratio is dimensionless and independent of size. Why Other Options Are Wrong: (b), (c), (d), and (e) do not match the known values derived from standard formulas and the parallel-axis theorem. Common Pitfalls: Mixing up centroidal and base axes; forgetting the parallel-axis term A * d^2; confusing I_x and I_y orientations. Final Answer: 1/4

Question 2

Free fall from height h – find the impact speed at ground Neglecting air resistance, a body is dropped from rest from a height h above level ground. What is the correct expression for its speed v just before impact?

Accepted Answer

Introduction / Context: Impact speed from free fall is a fundamental kinematics result used across engineering—estimating drop tests, safety barriers, and potential energy conversion. It follows from constant-acceleration motion under gravity without air drag. Given Data / Assumptions: Initial velocity u = 0 from height h. Constant acceleration a = g downward. No air resistance; motion is vertical. Concept / Approach: Use the kinematic energy–speed relation or the third equation of motion. Either approach removes time and relates velocity directly to displacement under constant acceleration. Step-by-Step Solution: From v^2 = u^2 + 2 a s with u = 0 and s = h: v^2 = 2 g h.Therefore v = sqrt(2 g h). Take the positive root for downward speed magnitude.Energy method cross-check: m g h converts to kinetic energy (1/2) m v^2 → v = sqrt(2 g h). Verification / Alternative check: Dimensions: g has L/T^2 and h has L; 2 g h has L^2/T^2; its square root has L/T, the correct unit of speed. Why Other Options Are Wrong: (b) and (d) are missing the square root and have wrong units. (c) inverts the relationship. (e) has incorrect dependence on h and g. Common Pitfalls: Forgetting the square root; using time-dependent formulas unnecessarily; sign mistakes when taking displacement as negative. Final Answer: v = sqrt(2 g h)

Question 3

Wheel kinematics – find angular velocity from rim speed A wheel has diameter 10 m. If a point on its rim moves with linear speed 50 m/s, what is the angular velocity ω of the wheel?

Accepted Answer

Introduction / Context: Relating linear speed at a rim to angular speed is a routine conversion in machinery, conveyors, turbines, and vehicle dynamics. The relation v = r * ω directly connects geometry to rotation rate. Given Data / Assumptions: Diameter D = 10 m → radius r = 5 m. Linear speed at the rim v = 50 m/s. Pure rolling rotation without slip at the rim point relative to the wheel. Concept / Approach: Use v = r * ω, where v is the tangential speed and r the radius to the point of interest. Solve for ω = v / r. Step-by-Step Solution: Compute radius: r = D / 2 = 10 / 2 = 5 m.Compute ω: ω = v / r = 50 / 5 = 10 rad/s. Verification / Alternative check: Units: v/r = (m/s) / m = 1/s = rad/s, consistent for angular speed. Double the radius at same v would halve ω, matching intuition. Why Other Options Are Wrong: 20, 15, and 5 rad/s correspond to incorrect r or arithmetic; 25 rad/s would require v = 125 m/s at r = 5 m. Common Pitfalls: Using diameter instead of radius in v = r * ω; mixing rpm with rad/s without converting. Final Answer: 10 rad/s

Question 4

Centripetal acceleration – choose the correct expression in terms of v, ω, and r For motion along a circular path of radius r with linear speed v and angular speed ω, which expression correctly gives the centripetal (radial) acceleration a_c?

Accepted Answer

Introduction / Context: Centripetal acceleration quantifies the inward acceleration necessary to maintain circular motion. It is pivotal for analyzing curves in highways, rotor dynamics, and cable tensions in rotating machinery. Given Data / Assumptions: Uniform circular motion of radius r. Linear speed v and angular speed ω are related by v = r * ω. Acceleration is directed toward the center (radially inward). Concept / Approach: Standard relations are a_c = v^2 / r and, equivalently using v = r * ω, a_c = r * ω^2. Both forms are correct; selecting one consistent expression answers the question. Dimensional analysis provides a quick validation. Step-by-Step Solution: Start from v = r * ω.Use a_c = v^2 / r → a_c = (r^2 * ω^2) / r = r * ω^2 (equivalent form).Therefore, a correct expression is a_c = v^2 / r. Verification / Alternative check: Units: v^2 / r has (m^2/s^2) / m = m/s^2. Likewise, r * ω^2 → m * (1/s^2) = m/s^2. Why Other Options Are Wrong: (b) r * ω has units of velocity, not acceleration. (c) and (d) have incorrect functional forms and units. (e) v * ω^2 gives m/s^3 units, not acceleration. Common Pitfalls: Dropping the square on v or ω; confusing tangential acceleration (if speed changes) with centripetal acceleration (directional change only). Final Answer: a_c = v^2 / r

Question 5

Resultant of two perpendicular forces in engineering mechanics: If two forces of 3 kg and 4 kg act at right angles to each other, what is the magnitude of their single resultant force (use the parallelogram/Pythagorean relation and keep the given “k…

Accepted Answer

Introduction / Context: When two forces act at right angles (perpendicular) to each other, their resultant is obtained by vector addition using the Pythagorean relation. This basic statics concept appears throughout engineering mechanics, machine design, and structural analysis. Given Data / Assumptions: Magnitude of first force = 3 kg (treated as kilogram-force as stated). Magnitude of second force = 4 kg. The angle between the forces = 90 degrees. Forces are coplanar and concurrent at a point. Concept / Approach: For two perpendicular vectors of magnitudes P and Q, the resultant R satisfies R^2 = P^2 + Q^2. This is the parallelogram (or triangle) law of forces specialized to a right angle, reducing to the Pythagorean theorem. Step-by-Step Solution: Let P = 3 and Q = 4.Compute R^2 = P^2 + Q^2 = 3^2 + 4^2 = 9 + 16 = 25.Hence R = √25 = 5. Verification / Alternative check: The 3–4–5 right triangle is a classic integer triple; it frequently appears in mechanics problems. Using vector components in orthogonal axes yields the same result. Why Other Options Are Wrong: 7 kg (a) incorrectly sums magnitudes without direction; 1 kg (b) and 1/7 kg (d) are arbitrary; “None of these” (e) is unnecessary because 5 kg exactly matches the computed resultant. Common Pitfalls: Adding magnitudes directly without recognizing direction; forgetting that the resultant of orthogonal vectors uses the square root of the sum of squares. Final Answer: 5 kg

Question 6

Friction terminology in engineering mechanics: The maximum frictional force that acts just when a body is about to start sliding over another surface is referred to as what?

Accepted Answer

Introduction / Context: Frictional behavior transitions from static to kinetic as a body begins to move. The peak value attained by static friction immediately before motion begins is crucial for design of brakes, clutches, and anchorage systems. Given Data / Assumptions: Dry contact between two solid surfaces. No lubrication or adhesion effects beyond the classical model. Body is on the verge of impending motion (just about to slide). Concept / Approach: Static friction adjusts up to a maximum called the limiting friction, F_lim = μ_s * N, where μ_s is the coefficient of static friction and N is the normal reaction. Once motion starts, kinetic (sliding) friction, typically smaller, acts. Rolling friction is a different phenomenon arising from deformation and contact hysteresis of rolling bodies. Step-by-Step Solution: Identify the regime: “just begins to slide” → at the threshold of motion.At this point, static friction has reached its maximum value.This maximum value is by definition the limiting friction. Verification / Alternative check: Experimental graphs of friction force vs. applied tangential load show a rising line up to a peak (limiting value), then a drop to a lower constant value for kinetic friction. Why Other Options Are Wrong: Sliding friction (a) acts after motion starts; rolling friction (b) pertains to rolling bodies; “None of these” (d) and “Static drag” (e) do not match standard terminology. Common Pitfalls: Confusing the peak static value with kinetic friction; assuming friction remains constant across regimes. Final Answer: Limiting friction

Question 7

Instantaneous centre (IC) of rotation — construction rule: For a moving rigid member with known directions of motion at its two ends, the instantaneous centre lies at the intersection of lines drawn at each end that are inclined at which angle to th…

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Introduction / Context: The instantaneous centre (IC) of rotation is a kinematics concept used to analyze planar motion of rigid bodies. Locating the IC allows velocities to be obtained quickly without full vector calculus, which is valuable in mechanisms and machine theory. Given Data / Assumptions: Planar motion of a rigid member. Directions of velocity (tangents to paths) at two distinct points on the member are known. No constraints beyond planar rigidity. Concept / Approach: At any instant, the velocity at a point is perpendicular to the radius from the IC to that point. Hence, drawing through each known point a line perpendicular (normal) to the velocity direction gives two lines that must pass through the IC. Their intersection is the instantaneous centre. Step-by-Step Solution: At each end, draw a line at 90° to the known velocity direction.Extend these two normals until they intersect.The intersection point is the IC for that instant. Verification / Alternative check: With the IC known, compute velocity of any point as V = ω * r, where r is the distance from the IC. Cross-check with the original known velocities to confirm consistency. Why Other Options Are Wrong: 30°, 45°, and 60° (a–c) do not satisfy the orthogonality condition between velocity and radius to the IC. 0° (e) would place the line along the velocity direction, which cannot pass through the IC. Common Pitfalls: Drawing lines along the velocity direction instead of the normal; mixing up normal and tangent; using approximate sketches that miss the true intersection. Final Answer: 90°

Question 8

Vehicle on a level circular track — simultaneous slip and overturn: Let μ be the coefficient of friction. A wheeled vehicle on a level circular track will slip and overturn at the same limiting condition if the ratio of wheel base distance (track wi…

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Introduction / Context: While cornering on a level circular track, a vehicle may either skid (slip) due to inadequate lateral friction or overturn due to the resultant line of action shifting beyond the outer wheel. The limiting design often checks both criteria simultaneously to determine safe speed and proportions. Given Data / Assumptions: Level circular track, no superelevation. Coefficient of friction μ. Track width = distance between wheel contact lines = 2b (so half-track b). Height of centroid (centre of gravity) above ground = h. Concept / Approach: Skid (slip) limit occurs when lateral acceleration a_y satisfies a_y / g = μ. Overturn limit occurs when the overturning moment due to lateral inertia equals restoring moment: (m * a_y * h) = (m * g * b) → a_y / g = b / h. Simultaneous slip and overturn means these two nondimensional ratios are equal. Step-by-Step Solution: Slip limit: a_y / g = μ.Overturn limit: a_y / g = b / h.Equate: μ = b / h → (2b) / h = 2μ.Hence, wheel distance / height of centroid = 2μ. Verification / Alternative check: Express in terms of v^2 / (g r). At both limits v^2 / (g r) equals μ and equals b / h; equality confirms the ratio above. Why Other Options Are Wrong: μ (a) omits the factor 2 from wheel distance = 2b; 3μ (c), μ/2 (d), and 1/μ (e) have no basis in the moment balance with the standard track-width definition. Common Pitfalls: Using half-track instead of full wheel distance in the requested ratio; confusing skid and overturn criteria; neglecting that the question asks for the ratio using the full wheel distance. Final Answer: 2μ

Question 9

Projectile from top and bottom of a tower striking the same target simultaneously: Two shots are fired at the same time from the top and the bottom of a vertical tower toward a ground target 1000 m away (horizontally). The elevation angles are 30° f…

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Introduction / Context: This kinematics problem links horizontal range relations with vertical motion for two projectiles launched simultaneously from different elevations. It tests comfort with time-of-flight equality and trigonometric components of motion under uniform gravity. Given Data / Assumptions: Horizontal distance to target = 1000 m. Top shot elevation = 30° above horizontal; bottom shot elevation = 45°. Both hit the same ground target at the same time. Neglect air resistance; take g ≈ 9.8 m/s^2. Concept / Approach: Let common flight time be t. For the bottom (ground) shot at 45°, use vertical motion to get t, then use it in the top-shot vertical displacement to find tower height H. Horizontal components provide speeds in terms of t but cancel neatly. Step-by-Step Solution: Bottom shot: horizontal → u_b cos45 * t = 1000.Bottom shot: vertical → 0 = u_b sin45 * t − (1/2) g t^2 → u_b sin45 * t = (1/2) g t^2.Divide the two: (u_b sin45 * t) / (u_b cos45 * t) = [(1/2) g t^2] / 1000 → tan45 = (1/2) g t^2 / 1000.Since tan45 = 1 → t^2 = 2000 / g → t ≈ 14.28 s.Top shot: horizontal → u_t cos30 * t = 1000 → u_t = 1000 / (t cos30).Top shot: vertical ground drop H → −H = u_t sin30 * t − (1/2) g t^2.Compute u_t sin30 * t = (1000 / (t cos30)) * (1/2) * t = 500 / cos30 ≈ 577.35.Compute (1/2) g t^2 ≈ 0.5 * 9.8 * 2000/9.8 ≈ 1000 → −H = 577.35 − 1000 → H ≈ 422.65 m ≈ 425 m. Verification / Alternative check: Reversing the calculation with H = 425 m yields consistent times for both shots to reach the target. Why Other Options Are Wrong: 350 m, 375 m, and 400 m underestimate the required drop from the top shot; 450 m slightly overestimates compared to the computed 422–423 m. Common Pitfalls: Using equal speeds for both shots (not required); forgetting that equal arrival time fixes t via the 45° ground shot; sign mistakes in vertical displacement from the top. Final Answer: 425 m

Question 10

Time of flight of a projectile (standard flat ground): If a particle is projected with initial speed u at an elevation angle α above the horizontal on level ground, what is the total time of flight until it returns to the same level (neglect air res…

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Introduction / Context: The time of flight is a fundamental result of projectile motion under uniform gravity on level ground. It depends on the initial vertical component of velocity and gravity but not on the horizontal component directly. Given Data / Assumptions: Initial speed = u; projection angle = α (above horizontal). Landing level equals launch level (flat ground). Acceleration due to gravity = g downward; air resistance neglected. Concept / Approach: Vertical motion controls flight duration. With initial vertical component u_y = u sin α, the time to rise to the top is u_y / g, and the time to descend back to the original level is the same, giving the well-known factor of 2. Step-by-Step Solution: Vertical equation: y(t) = u sin α * t − (1/2) g t^2.Set y(T) = 0 for the nontrivial root T > 0.Solve: 0 = u sin α * T − (1/2) g T^2 → T = (2 u sin α) / g. Verification / Alternative check: Symmetry of ascent and descent under constant g gives T = 2 * (u sin α / g) immediately. Why Other Options Are Wrong: (a) is only the ascent time; (c) and (d) incorrectly use the horizontal component; (e) is a range expression, not a time expression. Common Pitfalls: Confusing range and time formulas; forgetting the factor of 2; using degrees/radians inconsistently when computing sines numerically. Final Answer: (2 u sin α) / g

Question 11

Maximum height of a projectile (given u = 4 m/s): A particle is projected from a horizontal plane with initial speed 4 m/s at an angle α above the horizontal. What is the maximum height reached, expressed in terms of α and g?

Accepted Answer

Introduction / Context: Maximum height depends on the initial vertical component of velocity and gravity. With a fixed launch speed u, the height varies with sin^2 α because vertical kinetic energy converts to gravitational potential energy at the apex where vertical speed becomes zero. Given Data / Assumptions: Initial speed u = 4 m/s. Projection angle = α (above horizontal). Uniform gravity g; no air resistance. Concept / Approach: Standard formula for maximum height on level ground: H = (u^2 sin^2 α) / (2 g). Substituting u = 4 gives a simplified expression. Step-by-Step Solution: Write H = (u^2 sin^2 α) / (2 g).Compute u^2 = 4^2 = 16.Hence H = 16 sin^2 α / (2 g) = (8 / g) sin^2 α. Verification / Alternative check: Energy method: (1/2) m (u sin α)^2 converts to m g H → H = u^2 sin^2 α / (2 g), confirming the kinematic derivation. Why Other Options Are Wrong: (b) involves sin 2α (related to range, not height); (c) misses the factor 2 in the denominator after substitution; (d)–(e) use cos^2 α which pertains to horizontal component, not maximum height. Common Pitfalls: Confusing maximum height with range expressions; forgetting to square the sine of the launch angle; arithmetic slips when substituting u = 4 m/s. Final Answer: (8 / g) sin^2 α

Question 12

Units — force in the C.G.S. system: Identify the correct name of the force unit in the centimetre–gram–second (C.G.S.) system of units.

Accepted Answer

Introduction / Context: Unit systems matter when converting measurements and interpreting formulas. The C.G.S. system uses centimetre, gram, and second as base units, leading to a distinct force unit compared with the S.I. system (M.K.S.). Given Data / Assumptions: C.G.S. base units: cm, g, s. Force defined via Newton’s second law F = m * a. Compare with S.I. base units: m, kg, s. Concept / Approach: In C.G.S., 1 dyne is the force that produces an acceleration of 1 cm/s^2 in a mass of 1 g. In S.I., the corresponding unit is the newton (1 N = 1 kg·m/s^2). The symbol “kg” is mass, not a force unit, and “Pascal” is pressure, not force. Step-by-Step Solution: Apply F = m * a with m in grams and a in cm/s^2.Unit obtained is g·cm/s^2 defined as 1 dyne.Therefore, the C.G.S. force unit is the dyne. Verification / Alternative check: Conversion: 1 N = 10^5 dynes (since 1 kg = 10^3 g and 1 m = 10^2 cm). Why Other Options Are Wrong: Newton (b) is the S.I. unit; kg (c) is not a force unit; “All of the above” (d) cannot be true; Pascal (e) is a pressure unit equal to N/m^2. Common Pitfalls: Confusing mass with force; mixing S.I. and C.G.S. units; treating pascal as a force instead of pressure. Final Answer: Dyne

Question 13

Resultant needed to neutralize two equal perpendicular forces In engineering mechanics, two concurrent forces each of magnitude T act at right angles to one another. If a single third force is applied along the internal bisector of the right angle, …

Accepted Answer

Introduction / Context: This problem checks the composition of two perpendicular concurrent forces and the idea of replacing them by a single equal-and-opposite balancing force applied along the angle bisector. It is a standard vector addition question in statics/dynamics. Given Data / Assumptions: Two forces of equal magnitude T act at right angles and are concurrent. The balancing (neutralizing) force acts along the internal bisector of the angle formed by the two forces, opposite to their resultant. Planar vector addition applies; the body is treated as a particle (no couples). Concept / Approach: The vector resultant R of two perpendicular forces F1 and F2 of equal magnitude T is obtained from the Pythagorean relation. The required single balancing force must be equal in magnitude to R and opposite in direction, acting along the same line (the bisector for equal perpendicular forces). Step-by-Step Solution: Let F1 = T î and F2 = T ĵ.Resultant R = F1 + F2 = T î + T ĵ.Magnitude |R| = √(T^2 + T^2) = T√2.Direction of R is along the angle bisector of the two forces; therefore, the neutralizing single force must be equal to T√2 and opposite to R. Verification / Alternative check: A head-to-tail triangle of forces with equal perpendicular sides length T forms an isosceles right triangle, whose hypotenuse is T√2, confirming the result geometrically. Why Other Options Are Wrong: 2T and 3T overestimate the required balancing magnitude; T/2 underestimates it; 'None of these' is incorrect because a precise value exists. Common Pitfalls: Forgetting the right-angle relation and adding magnitudes arithmetically (T + T) instead of using the Pythagorean theorem; not aligning the balancing force along the bisector opposite the resultant. Final Answer: T√2

Question 14

Minimum effort to just move a block up a rough inclined plane (force applied parallel to plane) A block of weight W rests on a plane inclined at angle θ to the horizontal. A force P is applied parallel to the plane and just sufficient to move the bl…

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Introduction / Context: For impending upward motion on a rough incline, friction acts down the plane and adds to the component of weight along the plane. This classic statics question tests correct identification of directions and use of friction-limited equilibrium at the verge of motion. Given Data / Assumptions: Weight W on a plane with inclination θ. Coefficient of friction μ, or friction angle φ with μ = tanφ. Force P applied parallel to the plane to just start upward motion (impending slip up the plane). Rigid block, Coulomb friction model. Concept / Approach: Resolve forces along and normal to the plane. At impending upward motion, friction direction is down the plane with magnitude F = μN. Use equilibrium of forces along the plane to find the minimum P that balances the downhill components (weight component plus friction). Step-by-Step Solution: Normal reaction N = W cosθ.Friction F = μ N = μ W cosθ (down the plane).Component of weight along plane = W sinθ (down the plane).For impending upward motion: P − (W sinθ + μ W cosθ) = 0.Therefore, P_min = W(sinθ + μ cosθ). Equivalently, using μ = tanφ, P = W sin(θ + φ)/cosφ. Verification / Alternative check: Check limiting cases: μ = 0 ⇒ P = W sinθ (smooth plane). For steep θ approaching 90°, expression approaches W, consistent with vertical lift. Why Other Options Are Wrong: (b) uses the wrong sign for friction; (c) and (d) mix sine/cosine placements; (e) does not match the correct parallel-to-plane loading form for impending upward motion. Common Pitfalls: Reversing friction direction; using N = W instead of N = W cosθ; forgetting friction acts at its limiting value only at impending motion. Final Answer: P = W(sinθ + μ cosθ)

Question 15

Velocity ratio of a simple screw jack (one full turn) A simple screw jack has screw pitch p = 9 mm. The operating handle length is L = 45 cm. For one complete turn of the handle, what is the velocity ratio (distance moved by effort / distance moved …

Accepted Answer

Introduction / Context: The velocity ratio (VR) of a lifting machine relates the motion of the effort point to the motion of the load. For a screw jack, the effort moves around a circle while the load rises by one pitch per revolution. This problem reinforces the geometric meaning of VR for screw mechanisms. Given Data / Assumptions: Pitch p = 9 mm = 0.9 cm per turn. Handle length L = 45 cm; effort travels a circumference per turn. One full rotation of the handle. Concept / Approach: VR = distance moved by effort in one turn / distance moved by load in one turn. Effort path is circumference 2πL; load rise is one pitch p. Use consistent units (cm). Step-by-Step Solution: Effort distance = 2 * π * L = 2 * π * 45 cm = 90π cm.Load distance = pitch p = 0.9 cm.VR = (90π) / 0.9 = 100π ≈ 314. Verification / Alternative check: Dimensional reasonableness: a fine pitch and long handle imply large mechanical advantage and thus large VR, consistent with ≈314. Why Other Options Are Wrong: 1.5, 5, and 25 are far too small; 100 omits the π factor. Common Pitfalls: Forgetting to convert mm to cm; using handle diameter instead of radius; using π incorrectly when computing circumference. Final Answer: 314

Question 16

Definition of the SI unit of force What is the name of the SI unit of force defined as the force which produces an acceleration of 1 m/s^2 when applied to a mass of 1 kg?

Accepted Answer

Introduction / Context: Units in mechanics must be clearly distinguished between force, work/energy, and pressure. The International System of Units (SI) defines the unit of force via Newton’s second law. Given Data / Assumptions: Mass m = 1 kg. Acceleration a = 1 m/s^2. Use SI base units and derived units. Concept / Approach: From Newton’s second law, F = m * a. The derived SI unit of force is the Newton (N), where 1 N = 1 kg * 1 m/s^2. Other named units belong to different systems (CGS) or different physical quantities (energy or pressure). Step-by-Step Solution: Start with F = m * a.For m = 1 kg and a = 1 m/s^2, F = 1 kg * 1 m/s^2 = 1 N. Verification / Alternative check: Dimensional analysis: [Force] = M L T^-2; substituting kg, m, s reproduces the Newton unit dimensions. Why Other Options Are Wrong: Dyne and erg are CGS units (1 dyne = 1 g·cm/s^2; 1 erg = 1 dyne·cm). Joule is SI unit of work/energy, not force. Pascal is SI unit of pressure (N/m^2). Common Pitfalls: Confusing energy (Joule) with force; mixing SI with CGS unit names. Final Answer: Newton

Question 17

Intrinsic equation of a catenary (relation between arc length and slope angle) For a catenary defined by parameter c, which of the following is the intrinsic (s–ψ) relation linking arc length S to the angle ψ that the tangent makes with the horizont…

Accepted Answer

Introduction / Context: A uniform flexible cable hanging under its own weight forms a catenary. Different equivalent forms exist: Cartesian (x–y), parametric, and intrinsic (relating arc length to slope angle). Recognizing these is important in cable and suspension-bridge analysis. Given Data / Assumptions: Standard catenary with parameter c (ratio T0/w where T0 is horizontal tension and w is weight per unit length). ψ is the angle between the tangent and the horizontal. Use known identities for catenary geometry and statics. Concept / Approach: In intrinsic form, the relation between arc length S measured from the lowest point and the slope angle ψ satisfies S = c tan ψ. Cartesian form through the lowest point is y = c cosh(x/c). Other expressions listed mix forms or variables incorrectly. Step-by-Step Solution: For a catenary, tan ψ = sinh(x/c) and y = c cosh(x/c).Arc length from lowest point obeys S = c sinh(x/c) = c tan ψ.Therefore, the intrinsic relation is S = c tan ψ. Verification / Alternative check: Differential geometry of a catenary gives dS = √(1 + (dy/dx)^2) dx; using dy/dx = sinh(x/c), integrate to get S = c sinh(x/c); combine with tan ψ = sinh(x/c) to obtain S = c tan ψ. Why Other Options Are Wrong: (b) is the correct Cartesian equation, not intrinsic; (c) and (d) assign ψ incorrectly as the argument to hyperbolic functions of y; (e) is another form for S but in terms of x, not ψ. Common Pitfalls: Confusing ψ with x/c; mixing up sinh and cosh roles; forgetting that intrinsic form eliminates x and y in favor of S and ψ. Final Answer: S = c tan ψ

Question 18

Projectile condition where range equals free-fall distance to reach the same launch speed For a projectile launched with initial speed u, let h be the vertical distance a particle must freely fall to acquire speed u (so h = u^2/(2g)). For what launc…

Accepted Answer

Introduction / Context: This problem links kinematics of free fall to the range formula of projectile motion. It asks for the projection angle that makes the range equal to the vertical distance required to reach the same speed u by gravity alone. Given Data / Assumptions: Initial speed u. Free-fall distance to reach speed u: h = u^2/(2g). Projectile launched and landing at the same elevation; no air resistance; constant g. Concept / Approach: On level ground, the horizontal range is R = (u^2/g) * sin(2θ). Set R = h and solve for θ. The equation reduces to sin(2θ) = 1/2, which has principal acute solution 2θ = 30°, giving θ = 15°. Step-by-Step Solution: Free-fall distance: h = u^2/(2g).Range: R = (u^2/g) * sin(2θ).Set R = h ⇒ (u^2/g) * sin(2θ) = u^2/(2g).Cancel u^2/g on both sides ⇒ sin(2θ) = 1/2.Therefore, 2θ = 30° (acute solution) ⇒ θ = 15°. Verification / Alternative check: Numerical check: choose u = 10 m/s; h = 10^2/(2*9.8) ≈ 5.10 m. For θ = 15°, R = (100/9.8)sin 30° ≈ 10.200.5 ≈ 5.10 m, confirming equality. Why Other Options Are Wrong: 30°, 45°, 60°, 75° correspond to sin(2θ) values of √3/2, 1, √3/2, and about 0.5 at different doubles; only θ = 15° makes sin(2θ) = 1/2 with 0° < θ < 45°. Common Pitfalls: Misreading the condition as maximum range; forgetting that the range contains sin(2θ), not sin θ; using degrees vs radians inconsistently. Final Answer: 15°

Question 19

Mechanical advantage of a lifting machine from load–effort data A 500 kg load is lifted through 13 cm by an effort of 25 kg that moves through 650 cm. Assuming ideal measurement in kgf–cm, what is the mechanical advantage (MA) of the machine?

Accepted Answer

Introduction / Context: Mechanical advantage (MA) quantifies how much a machine multiplies force. In simple lifting problems stated in kilogram-force (kgf) units, MA is the ratio of load to effort and is independent of distances. The work distances are used for velocity ratio and efficiency calculations. Given Data / Assumptions: Load = 500 kgf (using kg as kgf as customary in elementary machine problems). Effort = 25 kgf. Load rise = 13 cm; Effort movement = 650 cm (for context only). No slippage assumed in reading MA; efficiency not requested. Concept / Approach: Mechanical advantage MA = Load / Effort. The given distances are not needed to compute MA directly; they would relate to the velocity ratio (VR = effort distance / load distance) and efficiency (η = MA/VR), but here only MA is sought. Step-by-Step Solution: MA = Load / Effort = 500 / 25.Compute: 500 / 25 = 20. Verification / Alternative check: VR = 650 / 13 = 50. If efficiency = MA/VR = 20/50 = 0.4 (40%), the numbers are self-consistent for a real machine (optional check). Why Other Options Are Wrong: 15, 18, 26, and 25 do not match Load/Effort; 20 is the only correct ratio. Common Pitfalls: Mistaking VR for MA; mixing kg (mass) and kgf (force); ignoring that MA is a pure ratio. Final Answer: 20

Question 20

Defining characteristic of simple harmonic motion (SHM) In simple harmonic motion, the instantaneous acceleration of the particle is directly proportional to which quantity?

Accepted Answer

Introduction / Context: Simple harmonic motion (SHM) underlies oscillations of springs, pendulums (small angles), and many vibration problems. Its defining feature links acceleration directly to displacement and fixes the motion’s sinusoidal nature. Given Data / Assumptions: One-dimensional motion about a stable equilibrium (mean position). Linear restoring mechanism (e.g., Hooke’s law spring). No damping or external forcing. Concept / Approach: For SHM, acceleration a is proportional to displacement x and is directed towards the equilibrium: a = −ω^2 x, where ω is the angular frequency. The negative sign indicates the restoring (opposite) direction, causing oscillatory motion. Step-by-Step Solution: Postulate linear restoring force: F = −k x.Use Newton’s law: m a = −k x ⇒ a = −(k/m) x.Identify ω^2 = k/m ⇒ a = −ω^2 x (proportional to displacement, towards mean). Verification / Alternative check: Solution x(t) = A sin(ω t + φ) differentiates to a(t) = −ω^2 x(t), confirming proportionality and phase opposition. Why Other Options Are Wrong: (a) 'Rate of change of velocity' is acceleration itself, not a proportionality statement; (c) would imply linear drag-like behavior; (d) lacks magnitude information; (e) contradicts the SHM definition. Common Pitfalls: Omitting the direction (restoring sense); confusing SHM with uniform circular motion’s projection without noting the acceleration law. Final Answer: Displacement from the mean position (directed towards it)

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