Question 1

Elevator dynamics and apparent weight A 49 kg person stands on a spring scale in an elevator. Starting from rest, during the first 5 seconds the scale reading is 69 kg. Assuming constant acceleration, determine the elevator’s velocity at t = 5 s.

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Introduction / Context: Apparent weight on a scale in an accelerating elevator reflects the normal reaction. From the reading we can back-calculate acceleration and then velocity. This is a standard inverse dynamics and kinematics application. Given Data / Assumptions: Actual mass m = 49 kg. Scale shows apparent mass 69 kg during acceleration. Elevator starts from rest; duration t = 5 s; take g ≈ 9.8 m/s^2. Assume constant upward acceleration. Concept / Approach: Apparent weight N equals m(g + a) when accelerating upward. Scales calibrated in kilograms display N/g. Set N/g = 69 kg and solve for a. Then use v = u + a t with u = 0 to get velocity at 5 s. Step-by-Step Solution: N/g = 69 ⇒ N = 69 g.But N = m(g + a) ⇒ 69 g = 49 (g + a).Divide by g: 69 = 49 (1 + a/g) ⇒ a/g = (69 − 49)/49 = 20/49.Compute a ≈ (20/49) * 9.8 ≈ 4.0 m/s^2.Velocity after 5 s: v = a t = 4.0 * 5 = 20 m/s (upward). Verification / Alternative check: Using g = 9.81 yields a ≈ 4.00 m/s^2; velocity remains 20.0 m/s to three significant digits. Why Other Options Are Wrong: 10 and 15 m/s underestimate, 25 and 30 m/s overestimate given the computed a = 4 m/s^2 for 5 s. Common Pitfalls: Confusing mass display (kg) with force; forgetting to divide by g; assuming downwards acceleration when the reading is higher than true weight. Final Answer: 20 m/s

Question 2

Equal horizontal ranges with different projection angles Two projectiles have projection angles 64° and 45°. If the first has initial speed 10 m/s, what initial speed must the second have to obtain the same horizontal range (neglect air resistance)?

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Introduction / Context: The range on level ground depends on u^2 sin 2θ / g. For equal ranges, speeds and double-angles must satisfy a proportionality. This problem checks manipulation of the range formula and trigonometric identities. Given Data / Assumptions: u1 = 10 m/s at θ1 = 64°. u2 = ? at θ2 = 45°. Same launch and landing elevations, no air resistance, constant g. Concept / Approach: Level-ground range: R = u^2 sin 2θ / g. For R1 = R2, set u1^2 sin 2θ1 = u2^2 sin 2θ2 and solve for u2. Use sin 90° = 1 and the identity sin(180° − x) = sin x. Step-by-Step Solution: u1^2 sin 2θ1 = u2^2 sin 2θ2.θ1 = 64° ⇒ 2θ1 = 128°; sin 128° = sin 52° ≈ 0.7880.θ2 = 45° ⇒ 2θ2 = 90°; sin 90° = 1.u2 = u1 * √(sin 128° / 1) = 10 * √0.7880 ≈ 10 * 0.8889 = 8.889 m/s.Rounded to one decimal → u2 ≈ 8.9 m/s. Verification / Alternative check: Back-substitute: R ∝ u^2 sin 2θ; with u2 = 8.9 m/s and sin 90° = 1, the ranges match within rounding. Why Other Options Are Wrong: 7.8 and 6.5 m/s are too small; 9.8 and 10.0 m/s are too large for the given sin 128° factor. Common Pitfalls: Using θ instead of 2θ in the sine term; rounding too early leading to larger error. Final Answer: 8.9 m/s

Question 3

Projectile motion on level ground: time of flight For a projectile launched with initial speed u at an angle α above the horizontal (same launch and landing elevation), the time of flight T equals:

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Introduction / Context: The time of flight of a projectile on level terrain depends only on the vertical component of the initial velocity. This is a fundamental kinematics result used to compute range and maximum height. Given Data / Assumptions: Initial speed u; projection angle α. Uniform gravitational acceleration g; no air drag. Landing height equals launch height. Concept / Approach: Vertical motion governs the ascent and descent: time to rise to peak equals time to descend back to the same elevation. Use v_y(t) = u sin α − g t and y(t) = 0 condition at landing. Step-by-Step Solution: Vertical displacement at landing: 0 = u sin α * T − (1/2) g T^2.Factor T: T (u sin α − (1/2) g T) = 0 → nontrivial root T = 2u sin α / g.Thus T depends linearly on u sin α. Verification / Alternative check: At α = 90°, T = 2u/g (pure vertical throw), matching the standard vertical result. Why Other Options Are Wrong: (b) mixes range identity; (c) and (d) use cos instead of sin; (e) is dimensionally incorrect (u^2/g is time only when accompanied by angle functions differently). Common Pitfalls: Applying the formula when landing elevation differs from launch; forgetting to double the ascent time. Final Answer: T = (2u sin α) / g

Question 4

Centres of gravity (C.G.) of common plane figures Identify the incorrect statement about the location of the centre of gravity (centroid) of the listed shapes.

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Introduction / Context: Remembering centroid locations of standard shapes is crucial for area moments, composite sections, and structural design. One of the statements below is intentionally wrong to test precise recall. Given Data / Assumptions: Plane laminae of uniform thickness and density. Standard geometric shapes: circle, triangle, rectangle, semicircle, ellipse. Concept / Approach: For symmetric figures, the centroid lies at the intersection of symmetry axes. For semicircular area, the centroid lies along the axis of symmetry at a known distance from the circle’s centre, but it is not r/2. Step-by-Step Solution: (a) Circle: centroid at centre → correct.(b) Triangle: centroid at medians’ intersection → correct.(c) Rectangle: centroid at diagonal intersection → correct.(d) Semicircle: the correct distance from the circle centre to the area centroid is 4r/(3π) ≈ 0.424 r, not r/2 = 0.5 r → statement is incorrect.(e) Ellipse: centroid at intersection of major and minor axes → correct. Verification / Alternative check: Using tabulated centroid formulae, the semicircle’s area centroid from the base is 4r/(3π) relative to the circle centre along the symmetry line, confirming (d) is wrong. Why Other Options Are Wrong: Here only (d) is wrong; the others match standard results. Common Pitfalls: Confusing the centroid of a semicircular arc (2r/π) with that of a semicircular area (4r/(3π)); mixing distances from base versus from centre. Final Answer: The C.G. of a semicircle is at a distance of r/2 from the centre.

Question 5

Second moment of area of a hollow circular section Find the moment of inertia about the centroidal axis of a hollow circular section with external diameter 8 cm and internal diameter 6 cm (use standard engineering approximation).

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Introduction / Context: The second moment of area (area moment of inertia) about the centroidal axis is fundamental for bending stress and deflection calculations. Circular tubes are common in shafts, columns, and frames, so knowing the formula is essential. Given Data / Assumptions: External diameter D = 8 cm, internal diameter d = 6 cm. Centroidal (through the centre) bending axis. Use the standard thin/solid ring formula without LaTeX. Concept / Approach: The centroidal second moment for a circular annulus is I = (π/64) * (D^4 − d^4). Plug in diameters in consistent units (cm) to get cm^4. Step-by-Step Solution: Compute D^4 = 8^4 = 4096.Compute d^4 = 6^4 = 1296.Difference = 4096 − 1296 = 2800.I = (π/64) * 2800 = 43.75 * π cm^4 ≈ 137.44 cm^4.Rounded to the nearest tenth → 137.5 cm^4. Verification / Alternative check: Compare with solid circle of D = 8 cm: I_solid = (π/64)4096 ≈ 201.1 cm^4; removing inner core of d = 6 cm removes (π/64)1296 ≈ 63.7 cm^4; difference ≈ 137.4 cm^4, confirming. Why Other Options Are Wrong: Other values correspond to incorrect arithmetic or missing π factor; 437.5 cm^4 is roughly 43.7510, not 43.75π. Common Pitfalls: Using radii instead of diameters in the D^4 form; mixing units leading to m^4 instead of cm^4. Final Answer: 137.5 cm^4

Question 6

Velocity ratio (VR) of a lifting machine: A load of 500 kg is lifted through 13 cm while the effort of 25 kg moves through 650 cm. Based on the definition VR = distance moved by effort / distance moved by load, determine the velocity ratio of the ma…

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Introduction / Context: Velocity ratio (VR) is a fundamental performance parameter of simple machines such as winches, pulleys, and screw jacks. It relates how far the effort moves compared to how far the load is lifted, independent of losses. Understanding VR helps estimate mechanical advantage and efficiency for practical lifting systems. Given Data / Assumptions: Load movement (rise) = 13 cm. Effort movement (pull) = 650 cm. VR is defined using distances only, not forces. No need to convert cm to m since it is a ratio of like units. Concept / Approach: For any ideal machine, VR = distance moved by effort / distance moved by load. This geometric ratio reflects the kinematics of the mechanism (e.g., number of rope segments supporting the load, screw lead vs. circumference, etc.). Step-by-Step Solution: VR = effort distance / load distanceVR = 650 / 13VR = 50 Verification / Alternative check: Dividing both distances by 13 confirms that the effort travels 50 times farther than the load, consistent with a high-advantage mechanism (at the cost of larger effort travel). Why Other Options Are Wrong: 55, 60, 65, 70: these arise only if one misreads distances or mixes units. The exact integer division 650/13 equals 50. Common Pitfalls: Confusing VR with mechanical advantage (MA). VR uses distances; MA uses forces (load/effort). Efficiency is MA/VR. Final Answer: 50

Question 7

Synchronous (geostationary) satellite speed check: A satellite in a synchronous (geostationary) orbit moves at an altitude of about 36,000 km above Earth’s surface. What is its approximate orbital speed in km/h?

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Introduction / Context: A geostationary satellite completes one revolution in 24 hours and appears fixed over the equator. Estimating its orbital speed reinforces the link between orbital radius, period, and tangential speed for circular motion in orbital mechanics. Given Data / Assumptions: Altitude ≈ 36,000 km above Earth. Orbital period ≈ 24 hours. Earth radius ≈ 6,371 km (used for context; standard speed memory aid: ~3.07 km/s). Concept / Approach: For circular orbits, v = 2 * π * r / T. For GEO, v is widely known to be roughly 3.07 km/s. Converting to km/h multiplies by 3600 s/h. Step-by-Step Solution: Recall v ≈ 3.07 km/s for GEO.Convert to km/h: v ≈ 3.07 * 3600 ≈ 11,052 km/h.Closest tabulated value ≈ 11,000 km per hour. Verification / Alternative check: Using r ≈ 42,164 km from Earth’s center (≈ 6,371 + 35,786) and T = 24 h yields a compatible speed near 11,000 km/h. Why Other Options Are Wrong: 7,000 to 10,000 km/h: underestimate GEO speed. Common Pitfalls: Confusing altitude (above surface) with orbital radius (from Earth’s center) or mixing km/s and km/h. Final Answer: 11,000 km per hour

Question 8

Coefficient of restitution for perfectly elastic impact: For perfectly elastic bodies undergoing a direct impact, what is the value of the coefficient of restitution (dimensionless)?

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Introduction / Context: The coefficient of restitution e quantifies the ratio of relative speeds after and before impact along the line of impact. It is key to collision problems in mechanics, from bouncing balls to vehicle crash idealizations. Given Data / Assumptions: Ideal, perfectly elastic collision. Line of impact is well defined; no energy loss. Concept / Approach: By definition, for perfectly elastic impact there is no loss of kinetic energy associated with the normal component of relative velocity. Therefore, the separation speed equals the approach speed along the line of impact, giving e = 1. Step-by-Step Solution: e = (relative speed of separation) / (relative speed of approach)Perfectly elastic → numerator = denominator → e = 1.0 Verification / Alternative check: In energy terms, ideal elasticity conserves kinetic energy (along the normal), consistent with e = 1. Real materials have 0 < e < 1. Why Other Options Are Wrong: 0 or 0.5: imply partially or fully inelastic collisions. “Between 0 and 1”: applies to real, non-ideal materials, not perfectly elastic ones. > 1: physically unrealistic for passive impacts (would add energy). Common Pitfalls: Confusing tangential effects or rotation with the normal restitution definition. Final Answer: 1.0

Question 9

Sag (central dip) of a lightly dipped trolley wire between poles: A trolley wire weighs 1 kg per metre (uniform) and spans 20 m between poles. If the horizontal tension is 1000 kgf, estimate the central dip (assume small sag, parabolic approximation…

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Introduction / Context: Light conductors and trolley wires often have small sag. For small sags, the catenary may be approximated by a parabola. The central dip depends on span, weight per unit length, and the horizontal component of tension. Given Data / Assumptions: Uniform weight per horizontal length w = 1 kgf/m. Span L = 20 m (level supports). Horizontal tension H = 1000 kgf (constant). Small sag → parabolic formula valid. Concept / Approach: For a parabolic cable with small sag, central dip y at midspan is given by y = w * L^2 / (8 * H). Units are consistent when w and H are in kgf and L in metres; result y is in metres. Step-by-Step Solution: Compute L^2 = 20^2 = 400 m^2.Compute numerator: w * L^2 = 1 * 400 = 400.Compute denominator: 8 * H = 8 * 1000 = 8000.y = 400 / 8000 = 0.05 m = 5 cm. Verification / Alternative check: If the tension doubled, sag halves, consistent with inverse dependence on H. If span doubled, sag increases fourfold, consistent with L^2 dependence. Why Other Options Are Wrong: 2–4 cm: underestimate; they would require a higher tension or shorter span. 8 cm: overestimates sag for the given high tension. Common Pitfalls: Using total tension instead of its horizontal component; mixing N and kgf inconsistently; forgetting the factor of 8. Final Answer: 5 cm

Question 10

Planar rigid bar kinematics (instantaneous angular motion): A 5 m rod moves in a vertical plane. When the rod is inclined at 60° to the horizontal, its lower end moves horizontally at 3 m/s, while the upper end moves purely vertically. What is the s…

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Introduction / Context: This is a classic rigid-body planar kinematics problem. A straight bar undergoing pure rotation at an instant has point velocities related by the cross product of angular velocity with position vectors. When one end has a known velocity and the other end is constrained along a perpendicular direction, we can solve for the instantaneous angular speed and the unknown end speed. Given Data / Assumptions: Rod length L = 5 m; rod makes 60° with the horizontal. Lower end A velocity: v_A = 3 m/s horizontally. Upper end B moves vertically (its instantaneous horizontal component is zero). Planar motion; rigid bar constraint; no deformation. Concept / Approach: Use velocity relation v_B = v_A + ω × r_BA. With r_BA = (L cos60°, L sin60°) = (2.5, 4.330) m from A to B, and ω about the out-of-plane axis, we write component equations and solve for ω and vertical speed at B. Step-by-Step Solution: Let v_A = (3, 0) m/s.For 2D rotation, ω × r = (−ω y, ω x).v_Bx = 3 − ω * 4.330 = 0 ⇒ ω ≈ 3 / 4.330 ≈ 0.693 rad/s.v_By = ω * 2.5 ≈ 0.693 * 2.5 ≈ 1.73 m/s. Verification / Alternative check: An equivalent constraint approach demands equal components of the two-end velocities along the rod (no change in length), which also yields v_B ≈ 1.73 m/s. The computed value rounds to the nearest listed option. Why Other Options Are Wrong: 0.5, 1.0 m/s: too small; would imply unrealistically low ω for the given 3 m/s at the lower end. 2.5, 3.0 m/s: too large; would require a higher ω inconsistent with B’s zero x-velocity constraint. Common Pitfalls: Forgetting that v_Bx must be zero; or using L instead of its components (x, y) with the angle; or assuming pure translation. Final Answer: 1.5 m/s

Question 11

Comparing momentum change on impact: rubber ball vs. lead ball A rubber ball rebounds from a wall, but a lead ball of the same mass and speed strikes the same wall and then falls down without rebounding. Which statement is correct regarding the chan…

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Introduction / Context: Momentum change in collisions depends on whether the object rebounds or stops. The rebound reverses the normal component of velocity, increasing the total change in momentum compared to a case where the body merely comes to rest against the surface. Given Data / Assumptions: Equal masses m and equal incident speeds for both balls. Both strike the same rigid wall head-on (consider the normal component). Rubber ball rebounds (e > 0), lead ball effectively does not rebound (e ≈ 0). Concept / Approach: Change in momentum Δp along the line of impact equals m * (v_after − v_before). If the body stops, v_after ≈ 0 → Δp ≈ −m * v_before (magnitude m * v). If it rebounds with nearly equal speed in the opposite direction, v_after ≈ −v_before → Δp ≈ −2 m * v_before (magnitude 2 m * v), which is larger. Step-by-Step Solution: Lead ball: v_after ≈ 0 → |Δp_lead| = m * v.Rubber ball: v_after ≈ −k v, with 0 < k ≤ 1 → |Δp_rubber| = m * v * (1 + k) ≥ m * v.For nearly elastic rebound (k close to 1), |Δp_rubber| ≈ 2 m * v > |Δp_lead|. Verification / Alternative check: Impulse J equals change in momentum. A larger rebound speed implies a larger impulse from the wall, consistent with higher Δp for the rubber ball. Why Other Options Are Wrong: Equal changes: false unless both either stop or rebound identically. “Momentum of rubber ball is less than lead ball” before impact: masses and speeds are equal, so initial momenta are equal. “Lead ball change greater”: contradicts the stop vs. rebound analysis. Common Pitfalls: Comparing momenta instead of changes in momentum; ignoring direction when evaluating Δp vectors. Final Answer: Change in momentum suffered by the lead ball is less than that of the rubber ball

Question 12

Coefficient of restitution from bounce heights: A ball is dropped from 2.25 m onto a smooth floor and rises to 1.00 m after the bounce. Assuming vertical impact and negligible air resistance, compute the coefficient of restitution between the ball a…

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Introduction / Context: The coefficient of restitution e relates rebound speed to impact speed along the line of impact. For vertical drops, energy relations yield a simple formula in terms of drop and rebound heights, which is widely used in sports engineering and material testing. Given Data / Assumptions: Drop height h1 = 2.25 m. Rebound height h2 = 1.00 m. Neglect air resistance; vertical motion only. Concept / Approach: For vertical impacts without air drag, e = (rebound speed) / (impact speed). Using v = √(2 g h), we get e = √(h2 / h1). Step-by-Step Solution: Compute ratio h2/h1 = 1.00 / 2.25 ≈ 0.4444.Take square root: e = √0.4444 ≈ 0.6667.Rounded to two decimal places: e ≈ 0.67. Verification / Alternative check: If h2 = h1 (perfectly elastic), e = 1. If h2 = 0 (perfectly plastic), e = 0. Our result lies between 0 and 1, as expected. Why Other Options Are Wrong: 0.33, 0.44, 0.57, 0.77: do not satisfy e = √(1/2.25). Common Pitfalls: Taking a linear ratio rather than square root; mixing heights with velocities directly. Final Answer: 0.67

Question 13

Atwood machine (two masses over a smooth pulley)  If two bodies of masses M1 and M2 (with M1 > M2) are connected by a light, inextensible string passing over a smooth, frictionless pulley, what is the expression for the tension T in the string?

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Introduction / Context: In classical mechanics, the Atwood machine illustrates motion with two different masses M1 and M2 connected by a light, inextensible string over a smooth pulley. The system demonstrates uniformly accelerated motion, tension distribution in a string, and Newton’s second law in connected-body problems. The quantity of interest here is the string tension T under ideal conditions (no pulley friction, massless string). Given Data / Assumptions: M1 > M2 so the system accelerates with M1 moving downward and M2 moving upward. String is light (massless) and inextensible; pulley is smooth (frictionless) and massless. Acceleration due to gravity is g and is uniform. Tension is uniform throughout the string. Concept / Approach: The connected bodies share the same magnitude of acceleration a due to the inextensible string. Apply Newton’s second law separately to each mass, write equations for forces along the line of motion, and solve simultaneously to eliminate a. The tension T must be less than M1 * g and greater than M2 * g to accelerate the system consistently with M1 > M2. Step-by-Step Solution: For M1 (downward positive): M1 * g − T = M1 * aFor M2 (upward positive): T − M2 * g = M2 * aAdd the two equations to eliminate T: M1 * g − M2 * g = (M1 + M2) * aTherefore a = g * (M1 − M2) / (M1 + M2)Substitute a into M1 * g − T = M1 * a → T = M1 * g − M1 * g * (M1 − M2)/(M1 + M2) = (2 * M1 * M2 * g)/(M1 + M2) Verification / Alternative check: By symmetry, exchanging M1 and M2 should not change T except through the product M1 * M2 and the sum M1 + M2. The expression T = (2 * M1 * M2 * g)/(M1 + M2) satisfies this symmetry and fits limiting cases (e.g., if M1 ≈ M2, T → nearly M * g). Why Other Options Are Wrong: (M1 + M2) * g / 2: Ignores dynamics; tension is not a simple average of weights. (M1 − M2) * g: This is related to net driving force, not tension. (M1 * M2 * g)/(M1 + M2): Misses the factor 2 that arises from combining the two equations. M1 * g: Would be the tension only if the system were static with equal masses. Common Pitfalls: Incorrect sign conventions, assuming T equals the weight of one mass, or forgetting that both masses accelerate together. Also, including pulley inertia or string mass without stating it changes the result. Final Answer: T = (2 * M1 * M2 * g) / (M1 + M2)

Question 14

Statics of rigid bodies: A force applied at a point on the surface of a rigid body may be considered to act at which location without changing the external effect on the body?

Accepted Answer

Introduction / Context: In engineering statics, the “principle of transmissibility” states that a force acting on a rigid body may be shifted anywhere along its line of action without altering the external mechanical effect (resultant force and moment about any point external to the body). This concept is fundamental to simplifying force systems and drawing free-body diagrams. Given Data / Assumptions: The body is rigid (no deformation), so internal stress redistribution is not considered. We are evaluating external effects (resultant forces and moments) on the body. The same force vector (magnitude and direction) is retained. Concept / Approach: The external effect of a force on a rigid body is fully described by the force vector and its moment about a reference point. Moving the force along its own line of action does not change the moment about any reference point outside the body, because the perpendicular distance to the line of action remains the same. Thus, equivalent systems result when the point of application is transmitted along the line of action. Step-by-Step Solution: Identify the force vector F and its line of action (an infinite line in the force direction passing through the original point).Shift F along this line to a more convenient point for analysis (e.g., to an intersection with a coordinate axis or to a joint).Confirm that the moment of F about any point external to the body remains unchanged when moved along its line of action. Verification / Alternative check: A quick check is to compute the moment M = r × F about a reference O using two different position vectors r1 and r2 to two points on the same line of action; their difference is parallel to F, so (r1 − r2) × F = 0, confirming equal moments. Why Other Options Are Wrong: Centre of gravity: Not required for a general force; only weight acts at the centre of gravity equivalently. Periphery or surface-normal points: Arbitrary shifts off the line of action change the induced moment. Only original point: Overly restrictive and contradicts the transmissibility principle. Common Pitfalls: Applying the principle to deformable bodies (where internal effects do change) or shifting a force off its line of action without adding a balancing couple (which would otherwise be needed to retain equivalence). Final Answer: At any point on the line of action of the force

Question 15

Friction fundamentals (dry sliding): What is the correct definition of the angle of friction between two contacting surfaces?

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Introduction / Context: The angle of friction is a geometric way to express the limiting friction condition between two dry, rough surfaces. It provides an equivalent angular measure of the coefficient of friction and is central to wedge, block-on-incline, and screw mechanics analyses. Given Data / Assumptions: Surfaces are dry with Coulomb friction behavior. Impending motion (limiting friction) is considered for defining the angle. Normal reaction acts perpendicular to the contact surface; friction acts tangentially. Concept / Approach: At impending motion, friction reaches its limiting value F_lim = μ * N where μ is the coefficient of friction and N is the normal reaction. The vector sum of N and F_lim is the resultant R at the threshold of motion. The angle of friction φ is defined by tan φ = F_lim / N = μ. Thus φ is the angle between R and N at the limit state. Step-by-Step Solution: Write limiting friction: F_lim = μ * N.Define angle φ s.t. tan φ = F_lim / N = μ.Recognize that φ is the angle between the normal reaction N and the resultant R = N + F_lim at impending motion. Verification / Alternative check: On an inclined plane, impending slip occurs when W * sin θ = μ * W * cos θ giving tan θ = μ, hence θ = φ; the plane angle of repose equals the angle of friction, which confirms the definition. Why Other Options Are Wrong: Ratio F/N is μ, not an angle; φ is the angle whose tangent is μ. “Force of friction when the body is in motion” defines kinetic friction, not the angle. “Magnitude of friction at the instant…” is again a force, not an angle. Angle between friction and surface is always 0 degrees by definition of tangential contact. Common Pitfalls: Confusing μ (a ratio) with φ (an angle), and mixing limiting (static) with kinetic friction. Final Answer: The angle between the normal reaction and the resultant of the normal reaction and the limiting friction

Question 16

One-dimensional kinematics (with a turning point): A particle moves along a straight line with position x(t) = t^2 − 10 t + 30 (x in metres, t in seconds). What is the total distance travelled between t = 0 s and t = 10 s?

Accepted Answer

Introduction / Context: Total distance travelled differs from net displacement when the motion includes reversals (turning points). For polynomial position-time functions, identifying turning instants via velocity analysis is essential to sum absolute path segments correctly. Given Data / Assumptions: Position: x(t) = t^2 − 10 t + 30 (metres). Time interval: t from 0 s to 10 s. Straight-line motion; continuous differentiability allows critical-point analysis using velocity. Concept / Approach: Compute velocity v(t) = dx/dt. Turning points occur where v(t) = 0 (and acceleration indicates direction change). Evaluate positions at the start, at each turning instant within the interval, and at the end. Sum absolute differences to get total distance; net displacement would be the algebraic difference x(10) − x(0). Step-by-Step Solution: v(t) = d/dt (t^2 − 10 t + 30) = 2 t − 10.Set v(t) = 0 → 2 t − 10 = 0 → t = 5 s (turning point within [0, 10]).Positions: x(0) = 0 − 0 + 30 = 30 m.x(5) = 25 − 50 + 30 = 5 m.x(10) = 100 − 100 + 30 = 30 m.Distances: from t=0 to 5 → |30 − 5| = 25 m; from t=5 to 10 → |5 − 30| = 25 m.Total distance = 25 + 25 = 50 m. Verification / Alternative check: Acceleration a(t) = dv/dt = 2 m/s^2 > 0 confirms a single minimum at t = 5 s. Net displacement x(10) − x(0) = 30 − 30 = 0 m, while total distance is 50 m, consistent with a “down then up” path. Why Other Options Are Wrong: 0 m: This is the net displacement, not the total distance. 30 m or 60 m: Do not match the sum of absolute segment lengths determined by the turning point. None of these: Incorrect because 50 m is attainable and correct. Common Pitfalls: Confusing displacement with distance; forgetting to split the time interval at v = 0; sign errors in evaluating absolute differences. Final Answer: 50 m

Question 17

Newton’s Laws of Motion – identify the correct consolidated statement set Which of the following options correctly lists Newton’s three fundamental laws of motion as used in engineering mechanics?

Accepted Answer

Introduction / Context: Newton’s laws of motion form the bedrock of classical mechanics used across civil, mechanical, and aerospace engineering. They define how forces influence motion and how bodies interact. Mastery of these laws supports analysis of structures, machines, and vehicles under loads. Given Data / Assumptions: The laws apply to inertial frames (non-accelerating reference frames). Motion is described using mass, velocity, momentum, and force. Forces are idealized as vectors acting at points or distributed as needed. Concept / Approach: Law 1 (Inertia): A body maintains uniform motion or rest unless a net external force acts. Law 2 (Dynamics): Net force equals time rate of change of momentum; for constant mass, F = m * a. Law 3 (Interaction): Forces between two bodies are equal in magnitude and opposite in direction along the same line of action. Step-by-Step Solution: Identify statement (a) as Law 1 (inertia).Identify statement (b) as Law 2 (momentum–force relation).Identify statement (c) as Law 3 (action–reaction pair).Since all three are correct, choose the option consolidating them. Verification / Alternative check: Applying to a simple block: if no net force, velocity remains constant (Law 1); pushing with constant force yields constant acceleration (Law 2); hand feels an opposite push from the block (Law 3). Why Other Options Are Wrong: (a), (b), and (c) are individually correct but incomplete. (e) omits the crucial interaction principle in (c). Common Pitfalls: Confusing reaction with resistance or friction; forgetting that action–reaction forces act on different bodies; assuming Law 1 requires zero forces rather than zero net force. Final Answer: All of the above

Question 18

Simple pendulum – determine the correct time period expression For a simple pendulum of length l oscillating with small angular displacement under gravity g, what is the correct expression for the period of one complete oscillation?

Accepted Answer

Introduction / Context: The simple pendulum is a foundational vibration system used to approximate clocks, seismic instruments, and sway of slender structures. Its time period depends on geometric length and gravitational acceleration, assuming small-angle motion and a massless, inextensible string. Given Data / Assumptions: Point mass bob, string length l, gravity g. Small angular amplitude so that sin θ ≈ θ (in radians). No air resistance or pivot friction; string mass negligible. Concept / Approach: For small oscillations, the linearized equation reduces to simple harmonic motion with angular frequency ω = sqrt(g / l). The period T equals 2π divided by ω, giving the classic square–root relationship between T, l, and g. Step-by-Step Solution: Set up equation: angular acceleration = −(g / l) * θ.Compare with SHM form: θ¨ + ω^2 * θ = 0, so ω = sqrt(g / l).Compute period: T = 2π / ω = 2π * sqrt(l / g). Verification / Alternative check: Dimensional check: sqrt(l / g) has units of time^(1), ensuring T is in seconds. Longer pendulums have larger periods, consistent with observation. Why Other Options Are Wrong: (b) inverts the ratio. (c) multiplies l and g incorrectly. (d) lacks the square root. (e) incorrect constant factor. Common Pitfalls: Using large angles where the small-angle approximation fails; forgetting that mass cancels and does not affect T. Final Answer: T = 2π * sqrt(l / g)

Question 19

Uniform circular motion – compute string tension for a horizontal whirl A 1 kg stone is tied to a light string of length 1 m and whirled in a horizontal circle at constant angular speed 5 rad/s (idealized, neglecting gravity effects on the motion pl…

Accepted Answer

Introduction / Context: In ideal uniform circular motion, the string provides the inward (centripetal) force needed to keep a mass moving in a circle. For horizontal whirl models in introductory mechanics, the vertical weight is treated as balanced by other means or neglected so that tension equals the required centripetal force. Given Data / Assumptions: Mass m = 1 kg, radius r = 1 m, angular speed ω = 5 rad/s. String is light and inextensible; motion is in a horizontal plane. Air resistance neglected; tension provides centripetal force. Concept / Approach: Centripetal force required is m * r * ω^2. In the horizontal whirl idealization, this equals string tension T. This is a common modeling step for teaching rotational dynamics and determining allowable speeds or required strengths of tethers and fasteners. Step-by-Step Solution: Compute ω^2 = 5^2 = 25.T = m * r * ω^2 = 1 * 1 * 25 = 25 N. Verification / Alternative check: Using linear speed v = r * ω = 1 * 5 = 5 m/s, centripetal force m * v^2 / r = 1 * 25 / 1 = 25 N, confirming the result. Why Other Options Are Wrong: 5 N, 10 N, 15 N, and 20 N underpredict the required centripetal force for the given speed and radius; only 25 N satisfies m * r * ω^2. Common Pitfalls: Multiplying by radius twice (using r^2), or forgetting the square on ω; confusing tension in a conical pendulum (which must also balance weight) with the horizontal whirl idealization used here. Final Answer: 25 N

Question 20

Three-force equilibrium on a rigid body – necessary geometric condition Three non-parallel forces acting on a rigid body keep it in equilibrium. Besides being coplanar, what must be true of their lines of action?

Accepted Answer

Introduction / Context: The three-force rule is central in statics for frames, arches, and linkages. When exactly three forces act on a rigid body in equilibrium, their geometric relation must satisfy both force and moment balance simultaneously. Given Data / Assumptions: Exactly three forces act; the body is rigid. Forces are coplanar and not all parallel. No couple moments applied externally. Concept / Approach: Equilibrium requires ΣF = 0 and ΣM = 0. With three non-parallel forces, if their lines of action are not concurrent, a net moment remains. The only way for both force and moment sums to vanish is for the three lines of action to intersect at a common point (or be parallel with magnitudes balancing, the special case excluded here). Step-by-Step Solution: Assume three non-parallel forces F1, F2, F3 act on a rigid body.If not concurrent, choose the intersection of any two lines of action as the moment center; the third force creates a non-zero moment that cannot be balanced.Therefore, all three must meet at a single point; magnitudes then form a force triangle to satisfy ΣF = 0. Verification / Alternative check: Graphical statics confirms: draw the force polygon closed; concurrency of lines of action follows from the funicular polygon closing to a point pole. Why Other Options Are Wrong: (b) All-parallel is a different special case. (c) Concurrent and parallel together is contradictory unless all forces are the same line. (d) Skew lines violate coplanarity. (e) Magnitude triangle alone does not ensure zero moment unless concurrency holds. Common Pitfalls: Forgetting the moment equilibrium; assuming a force triangle is sufficient without checking concurrency of lines of action. Final Answer: They must be concurrent (intersect at a single point)

Applied Mechanics Questions