Q: Seconds pendulum standard: What is the conventional length of a seconds pendulum (time period 2 s) under standard conditions?

Introduction / Context: A seconds pendulum has a time period of 2 seconds (1 second for a swing in one direction). Its length under standard gravity is a classical benchmark in elementary mechanics and metrology. Given Data / Assumptions: Simple pendulum approximation (point mass, light string) under standard gravity g ≈ 9.81 m/s^2. Small angular oscillations (small-angle approximation). Concept / Approach: The period of a simple pendulum is T = 2 * π * sqrt(L / g). For a seconds pendulum, T = 2 s. Solve for L to find the benchmark length commonly quoted in handbooks. Step-by-Step Solution: Start with T = 2 * π * sqrt(L / g).Rearrange: L = (T / (2 * π))^2 * g.Substitute T = 2 s and g ≈ 9.81 m/s^2: L ≈ (2 / (2π))^2 * 9.81 ≈ (1/π)^2 * 9.81 ≈ 0.994 m.Convert to centimetres: 0.994 m ≈ 99.4 cm. Verification / Alternative check: Using g = 9.80665 m/s^2 (standard gravity) gives L ≈ 0.9936 m, still ≈ 99.4 cm. Slight local variations in g cause minor deviations but the conventional answer remains ~99.4 cm. Why Other Options Are Wrong: 99.0 cm and 100 cm are rough guesses, not the standard derived value. 101–101.10 cm are too long for T = 2 s under standard g. Common Pitfalls: Forgetting to square the ratio T/(2π) when solving for L. Final Answer: 99.4 cm

Q: Definition of mechanical work: In basic engineering mechanics, how is work defined for a constant force acting along its line of action?

Introduction / Context: Work quantifies the energy transfer when a force causes a displacement. This concept underpins power calculations, efficiency assessments, and energy methods in structures and machines. Given Data / Assumptions: Constant force F acting along the direction of motion. Displacement s measured along the line of action. No changes in direction or variable forces (for this definition). Concept / Approach: For a constant, collinear force and displacement, work W is the scalar product of force and displacement: W = F * s. In the general case with an angle θ between force and displacement, W = F * s * cos θ; here θ = 0°, so cos θ = 1. Step-by-Step Solution: Identify force magnitude F and displacement s in the same direction.Compute work using the scalar product: W = F * s.Units check: N * m = joule in SI. Verification / Alternative check: Energy perspective: the change in kinetic energy equals the net work done (work–energy theorem), which reduces to F * s for a constant net force. Why Other Options Are Wrong: Force x velocity equals power, not work. Force x acceleration is not a standard energy quantity and has units inconsistent with energy. “None of these” is false because the correct definition is provided. Common Pitfalls: Ignoring direction: only the component of force along the displacement does work. Final Answer: force x distance

Q: Units of second moment of area (area moment of inertia): What is the correct dimensional unit for the moment of inertia of an area?

Introduction / Context: The area moment of inertia (second moment of area) characterizes an area's resistance to bending about an axis. Correct units are crucial when using beam formulas for deflection, slope, and stress (e.g., E * I relations). Given Data / Assumptions: I = ∫ y^2 dA (or ∫ x^2 dA depending on axis). Area element dA has units of m^2; distance squared has units of m^2. Concept / Approach: Because the definition multiplies an area element by the square of a length, the resulting unit is m^2 * m^2 = m^4. This is distinct from mass moment of inertia (kg·m^2), which involves mass distribution and rotation dynamics. Step-by-Step Solution: Write the definition: I = ∫ y^2 dA (about the x-axis) or I = ∫ x^2 dA (about the y-axis).Assign units: y^2 in m^2; dA in m^2.Multiply: m^2 * m^2 = m^4. Verification / Alternative check: Example: rectangle b by h about its centroidal axis has I = b * h^3 / 12. Units: m * m^3 = m^4, confirming the dimension. Why Other Options Are Wrong: m and m^2 correspond to length and area, not second moment of area. m^3 is volume, again incorrect. “None of these” fails because m^4 is correct. Common Pitfalls: Confusing area moment of inertia (m^4) with mass moment of inertia (kg·m^2). Final Answer: m^4

Q: Elevator dynamics — if the cable tension when a lift moves upward is twice the tension when it moves downward, determine the lift’s acceleration (express your answer as a multiple of g).

Introduction / Context: Elevator (lift) problems are classic applications of Newton’s second law. Cable tension differs depending on whether the car accelerates up or down. Relating two measured tensions allows you to solve for the unknown acceleration without needing the car’s mass explicitly. Given Data / Assumptions: Tension upward case: T_up. Tension downward case: T_down. Given T_up = 2 * T_down. Lift mass = m, weight W = m * g. Same magnitude of acceleration a in both cases (one upward, one downward) over the short comparison interval. Concept / Approach: Apply Newton’s second law along the vertical for each motion state, taking upward as positive. When accelerating upward, tension must exceed weight; when accelerating downward, tension is less than weight. Two linear equations in T_up and T_down eliminate T and isolate a in terms of g. Step-by-Step Solution: Upward acceleration: T_up − W = m * a ⇒ T_up = W + m * a.Downward acceleration: W − T_down = m * a ⇒ T_down = W − m * a.Given relation: T_up = 2 * T_down.Substitute: W + m * a = 2 (W − m * a) ⇒ W + m * a = 2W − 2 m * a.Collect terms: 3 m * a = W ⇒ a = W / (3 m) = (m * g) / (3 m) = g / 3. Verification / Alternative check (if short method exists): Check sign consistency: with a = g/3, T_up = W + m * g/3 = 4W/3 and T_down = W − m * g/3 = 2W/3; indeed T_up = 2 * T_down. Why Other Options Are Wrong: g/6 and g/4 produce T_up/T_down ratios different from 2; g/2 doubles the discrepancy; “none” is unnecessary because g/3 satisfies the given relation exactly. Common Pitfalls (misconceptions, mistakes): Mixing up direction of motion with direction of acceleration; forgetting that downward acceleration reduces cable tension. Final Answer: g/3

Q: Ideal machine with large velocity ratio — an ideal machine has mechanical advantage (MA) = 100. If the load moves 2 m, how far must the effort move?

Introduction / Context: For an ideal (lossless) simple machine, mechanical advantage equals velocity ratio. This means the distance the effort travels is amplified relative to the load movement by the same factor that the effort reduces the necessary force. Understanding this equivalence is central to hoists, jacks, and block-and-tackle systems. Given Data / Assumptions: Mechanical advantage MA = 100 (ideal, so efficiency = 100%). Load displacement s_load = 2 m. Velocity ratio VR = distance moved by effort / distance moved by load. Concept / Approach: In an ideal machine, MA = VR. Therefore the kinematic requirement is that the effort must move VR times the distance of the load. Multiply the given load displacement by MA to obtain the effort displacement. Step-by-Step Solution: Equate MA and VR: MA = VR = 100.Write VR = s_effort / s_load.Solve: s_effort = VR * s_load = 100 * 2 = 200 m. Verification / Alternative check (if short method exists): A power balance for an ideal machine gives Effort * s_effort = Load * s_load; with Load/Effort = MA = 100, s_effort/s_load = 100, consistent. Why Other Options Are Wrong: 20 m reflects MA = 10; 2.5 m and 2 m ignore the high velocity ratio; 0.02 m reverses the relation. Common Pitfalls (misconceptions, mistakes): Confusing MA with efficiency; forgetting that in ideal machines the gain in force is paid for by a proportional loss in distance moved. Final Answer: 200 m

Q: Stopping distance and time under uniform deceleration: A car moving at 20 m/s is brought to rest uniformly within a distance of 40 m. How long does it take to stop?

Introduction / Context: Uniformly accelerated (or decelerated) rectilinear motion is governed by constant-acceleration kinematics. Stopping a vehicle in a given distance from a known initial speed is a standard application of these formulas. Given Data / Assumptions: Initial speed u = 20 m/s. Final speed v = 0 m/s (comes to rest). Stopping distance s = 40 m. Acceleration a is constant (uniform deceleration). Concept / Approach: Use v^2 = u^2 + 2 a s to find the constant acceleration a. Then use v = u + a t to compute the stopping time t. Signs indicate deceleration (a negative value), but magnitudes are sufficient to get time. Step-by-Step Solution: Apply v^2 = u^2 + 2 a s → 0 = 20^2 + 2 * a * 40.Solve for a → 0 = 400 + 80 a → a = −400 / 80 = −5 m/s^2.Use v = u + a t → 0 = 20 + (−5) t → t = 20 / 5 = 4 s. Verification / Alternative check: Average speed during uniform deceleration is (u + v)/2 = 10 m/s. Time = distance / average speed = 40 / 10 = 4 s, confirming the result. Why Other Options Are Wrong: 2 s and 3 s imply much larger decelerations; 5 s and 6 s imply smaller decelerations insufficient to stop within 40 m. Common Pitfalls: Sign errors in a; forgetting that average speed equals mid-value when acceleration is constant; mixing units. Final Answer: 4 s

Question 1

Resultant of two coplanar forces: If two forces P and Q act at an angle θ, which expression corresponds to the square of the resultant magnitude?

Accepted Answer

Introduction / Context: Combining two forces acting at an angle is a staple in engineering mechanics. The magnitude of the resultant can be obtained via the law of cosines applied to the force triangle or via vector components. Some MCQs list the expression for R^2 rather than R, so it is important to recognize both forms. Given Data / Assumptions: Two forces P and Q act at an included angle θ (0° ≤ θ ≤ 180°). Planar (coplanar) force system. Concept / Approach: Construct the triangle of forces. By the law of cosines on the magnitude triangle, the resultant magnitude R satisfies R^2 = P^2 + Q^2 + 2PQ cos θ. Equivalently, from components, R_x = P + Q cos θ and R_y = Q sin θ if P is taken as the reference direction, and then R^2 = R_x^2 + R_y^2 simplifies to the same expression. Step-by-Step Solution: Set up force triangle with sides P, Q, and R opposite angle θ between P and Q.Apply law of cosines: R^2 = P^2 + Q^2 − 2PQ cos(π − θ) = P^2 + Q^2 + 2PQ cos θ.Conclude the correct algebraic form. Verification / Alternative check: Component method corroborates the same result: R^2 = (P + Q cos θ)^2 + (Q sin θ)^2 expands to P^2 + Q^2 + 2PQ cos θ. Why Other Options Are Wrong: Terms with sin θ or tan θ do not arise from the magnitude relation. P + Q + 2PQ cos θ has wrong dimensions (mixes linear and quadratic terms). P^2 + Q^2 − 2PQ cos θ corresponds to the case where angle between P and (−Q) is θ, not the standard included angle here. Common Pitfalls: Forgetting that many textbooks present R = sqrt(P^2 + Q^2 + 2PQ cos θ); this option lists the squared form. Final Answer: P^2 + Q^2 + 2PQ cos θ

Question 2

Impact terminology in dynamics: The complete duration covering both compression (collision) and restitution phases between two bodies is termed what?

Accepted Answer

Introduction / Context: During an impact, two bodies experience a short-duration interaction with a compression phase (approach) followed by a restitution phase (separation). The terminology clarifies time intervals used in defining the coefficient of restitution and analyzing impulse–momentum. Given Data / Assumptions: Two colliding bodies in one-dimensional normal impact. Compression and restitution are consecutive sub-intervals. Concept / Approach: The phrase period of impact is commonly used for the entire contact time from first touch to separation, which includes both collision (compression) and restitution. This period is when impulsive forces act and momentum exchange occurs. Step-by-Step Solution: Identify sub-intervals: compression (velocities converge) then restitution (velocities separate).Define total contact duration as the sum of these phases.Select the standard term: period of impact. Verification / Alternative check: Textbook definitions pair the coefficient of restitution with end states across the period of impact, not only the compression part. Why Other Options Are Wrong: “Time of collision” and “period of collision” are ambiguous and often refer only to compression. “All the above” is not precise because some listed terms are ambiguous. “Contact interval” is descriptive but not the standard term in most syllabi. Common Pitfalls: Equating “collision time” with the entire impact when some sources restrict it to compression; instructors typically prefer “period of impact.” Final Answer: period of impact

Question 3

Cable with uniform horizontal load: A cable carries w = 0.5 tonne per horizontal metre over a 400 m span between level supports with central dip (sag) of 20 m. What is the minimum tension in the cable and where does it occur?

Accepted Answer

Introduction / Context: Cables under uniformly distributed load per horizontal span (e.g., suspension bridges) take a parabolic shape. The key design quantity is the horizontal component of tension, which attains its minimum value at the lowest point of the cable and increases toward the supports due to vertical reactions. Given Data / Assumptions: Uniform load per horizontal length w = 0.5 tonne/m. Span L = 400 m between level supports. Midspan sag (dip) y = 20 m. Self-weight of cable is included in w; effects like temperature and secondary bending neglected. Concept / Approach: For a parabolic cable under uniform horizontal load, the horizontal tension H is given by H = w * L^2 / (8 * y). The minimum cable tension occurs where slope is zero (at the lowest point), and equals T_min = H. Support tensions exceed H due to the vertical component from reactions. Step-by-Step Solution: Use H = w * L^2 / (8 * y).Substitute: H = 0.5 * (400^2) / (8 * 20).Compute intermediate: 400^2 = 160000; 8 * 20 = 160.Calculate H: H = 0.5 * 160000 / 160 = 0.5 * 1000 = 500 tonnes.Conclude T_min = H and it occurs at midspan (the lowest point). Verification / Alternative check: Dimensional sanity: tonnes per metre * metre^2 / metre yields tonnes; trends make sense—greater sag lowers H, smaller sag raises H. Why Other Options Are Wrong: 200-tonne options misapply the formula or confuse units. “500 tonnes at each support” is incorrect; support tensions are larger than H due to vertical components and are not equal to the midspan minimum. Common Pitfalls: Using w as load per cable length (catenary case) instead of per horizontal length; here the given w is per horizontal metre. Final Answer: 500 tonnes at the centre

Question 4

Projectile motion concept: A particle is launched with speed v at an angle α above the horizontal. At which point in its flight is the centripetal (normal) component of acceleration maximum?

Accepted Answer

Introduction / Context: In projectile motion under uniform gravity, total acceleration is constant and equals g downward. However, its decomposition into tangential (along velocity) and normal/centripetal (perpendicular to velocity) components varies along the path. Understanding this split is useful in dynamics and flight trajectories. Given Data / Assumptions: No air resistance; acceleration due to gravity g is constant and vertical. Velocity direction changes along the parabolic path. Concept / Approach: Resolve g into components relative to the instantaneous velocity direction. Let ψ be the flight-path angle (angle of velocity with horizontal). Then tangential acceleration a_t = g * sin(ψ) and normal (centripetal) acceleration a_n = g * cos(ψ). Since |cos(ψ)| ≤ 1, the maximum possible a_n is g, attained when ψ = 0°, i.e., when the velocity is horizontal—at the top of the trajectory. Step-by-Step Solution: Identify the top point where vertical velocity component is zero and velocity is purely horizontal.Compute a_n expression: a_n = g * cos(ψ).At the top, ψ = 0°, so a_n = g (maximum). Verification / Alternative check: An equivalent view uses curvature κ and speed v: a_n = v^2 * κ. At the top, although speed is minimum, the curvature is maximum for a projectile, and the product equals g—consistent with the component resolution result. Why Other Options Are Wrong: At start/impact, ψ = ±α, so a_n = g * cos α < g. “Elsewhere” is incorrect because the maximum is specifically at the peak. “Constant” is wrong; only total acceleration is constant, not its components. Common Pitfalls: Confusing maximum speed with maximum normal acceleration; they occur at different points. Final Answer: at the top of the trajectory

Question 5

Seconds pendulum standard: What is the conventional length of a seconds pendulum (time period 2 s) under standard conditions?

Accepted Answer

Introduction / Context: A seconds pendulum has a time period of 2 seconds (1 second for a swing in one direction). Its length under standard gravity is a classical benchmark in elementary mechanics and metrology. Given Data / Assumptions: Simple pendulum approximation (point mass, light string) under standard gravity g ≈ 9.81 m/s^2. Small angular oscillations (small-angle approximation). Concept / Approach: The period of a simple pendulum is T = 2 * π * sqrt(L / g). For a seconds pendulum, T = 2 s. Solve for L to find the benchmark length commonly quoted in handbooks. Step-by-Step Solution: Start with T = 2 * π * sqrt(L / g).Rearrange: L = (T / (2 * π))^2 * g.Substitute T = 2 s and g ≈ 9.81 m/s^2: L ≈ (2 / (2π))^2 * 9.81 ≈ (1/π)^2 * 9.81 ≈ 0.994 m.Convert to centimetres: 0.994 m ≈ 99.4 cm. Verification / Alternative check: Using g = 9.80665 m/s^2 (standard gravity) gives L ≈ 0.9936 m, still ≈ 99.4 cm. Slight local variations in g cause minor deviations but the conventional answer remains ~99.4 cm. Why Other Options Are Wrong: 99.0 cm and 100 cm are rough guesses, not the standard derived value. 101–101.10 cm are too long for T = 2 s under standard g. Common Pitfalls: Forgetting to square the ratio T/(2π) when solving for L. Final Answer: 99.4 cm

Question 6

Definition of mechanical work: In basic engineering mechanics, how is work defined for a constant force acting along its line of action?

Accepted Answer

Introduction / Context: Work quantifies the energy transfer when a force causes a displacement. This concept underpins power calculations, efficiency assessments, and energy methods in structures and machines. Given Data / Assumptions: Constant force F acting along the direction of motion. Displacement s measured along the line of action. No changes in direction or variable forces (for this definition). Concept / Approach: For a constant, collinear force and displacement, work W is the scalar product of force and displacement: W = F * s. In the general case with an angle θ between force and displacement, W = F * s * cos θ; here θ = 0°, so cos θ = 1. Step-by-Step Solution: Identify force magnitude F and displacement s in the same direction.Compute work using the scalar product: W = F * s.Units check: N * m = joule in SI. Verification / Alternative check: Energy perspective: the change in kinetic energy equals the net work done (work–energy theorem), which reduces to F * s for a constant net force. Why Other Options Are Wrong: Force x velocity equals power, not work. Force x acceleration is not a standard energy quantity and has units inconsistent with energy. “None of these” is false because the correct definition is provided. Common Pitfalls: Ignoring direction: only the component of force along the displacement does work. Final Answer: force x distance

Question 7

Units of second moment of area (area moment of inertia): What is the correct dimensional unit for the moment of inertia of an area?

Accepted Answer

Introduction / Context: The area moment of inertia (second moment of area) characterizes an area's resistance to bending about an axis. Correct units are crucial when using beam formulas for deflection, slope, and stress (e.g., E * I relations). Given Data / Assumptions: I = ∫ y^2 dA (or ∫ x^2 dA depending on axis). Area element dA has units of m^2; distance squared has units of m^2. Concept / Approach: Because the definition multiplies an area element by the square of a length, the resulting unit is m^2 * m^2 = m^4. This is distinct from mass moment of inertia (kg·m^2), which involves mass distribution and rotation dynamics. Step-by-Step Solution: Write the definition: I = ∫ y^2 dA (about the x-axis) or I = ∫ x^2 dA (about the y-axis).Assign units: y^2 in m^2; dA in m^2.Multiply: m^2 * m^2 = m^4. Verification / Alternative check: Example: rectangle b by h about its centroidal axis has I = b * h^3 / 12. Units: m * m^3 = m^4, confirming the dimension. Why Other Options Are Wrong: m and m^2 correspond to length and area, not second moment of area. m^3 is volume, again incorrect. “None of these” fails because m^4 is correct. Common Pitfalls: Confusing area moment of inertia (m^4) with mass moment of inertia (kg·m^2). Final Answer: m^4

Question 8

Orbital mechanics concept check — a satellite continues moving in its orbit around Earth primarily because of which physical cause?

Accepted Answer

Introduction / Context: In orbital mechanics, it is important to distinguish between the real physical forces acting on a satellite and the kinematic quantities used to describe its motion. A satellite stays in orbit because a central force continuously curves its trajectory, producing the required centripetal acceleration. Given Data / Assumptions: The satellite is in a stable, bound Earth orbit. Earth's gravity is the dominant central force; atmospheric drag and other perturbations are neglected for this conceptual question. Descriptions may use either inertial or rotating frames, but forces must be identified correctly. Concept / Approach: A body moving in a circular or elliptical path requires a centripetal acceleration directed toward the focus/center. This acceleration must be provided by a real force. For Earth satellites, that real force is the gravitational attraction between Earth and the satellite. “Centripetal” is not a separate force; it is the required inward acceleration. “Centrifugal” is a pseudo-force that appears only in a rotating (non-inertial) frame and is not the physical cause of orbital motion. Step-by-Step Solution: Recognize the satellite follows a curved path → requires inward (centripetal) acceleration.Identify the real physical interaction providing this: Earth–satellite gravitation.Conclude: gravitational force supplies the centripetal requirement that keeps the satellite in orbit. Verification / Alternative check (if short method exists): Set gravitational force equal to m * v^2 / r for a circular orbit: G * M * m / r^2 = m * v^2 / r → confirms gravity provides the centripetal requirement. Why Other Options Are Wrong: “Centripetal force” describes the role, not the source; “centrifugal force” is frame-dependent and not a real cause; “none” and solar radiation pressure are incorrect for sustained Earth orbits. Common Pitfalls (misconceptions, mistakes): Confusing centripetal acceleration with an independent new force; treating centrifugal force as real in an inertial frame. Final Answer: Gravitational force

Question 9

Elevator dynamics — if the cable tension when a lift moves upward is twice the tension when it moves downward, determine the lift’s acceleration (express your answer as a multiple of g).

Accepted Answer

Introduction / Context: Elevator (lift) problems are classic applications of Newton’s second law. Cable tension differs depending on whether the car accelerates up or down. Relating two measured tensions allows you to solve for the unknown acceleration without needing the car’s mass explicitly. Given Data / Assumptions: Tension upward case: T_up. Tension downward case: T_down. Given T_up = 2 * T_down. Lift mass = m, weight W = m * g. Same magnitude of acceleration a in both cases (one upward, one downward) over the short comparison interval. Concept / Approach: Apply Newton’s second law along the vertical for each motion state, taking upward as positive. When accelerating upward, tension must exceed weight; when accelerating downward, tension is less than weight. Two linear equations in T_up and T_down eliminate T and isolate a in terms of g. Step-by-Step Solution: Upward acceleration: T_up − W = m * a ⇒ T_up = W + m * a.Downward acceleration: W − T_down = m * a ⇒ T_down = W − m * a.Given relation: T_up = 2 * T_down.Substitute: W + m * a = 2 (W − m * a) ⇒ W + m * a = 2W − 2 m * a.Collect terms: 3 m * a = W ⇒ a = W / (3 m) = (m * g) / (3 m) = g / 3. Verification / Alternative check (if short method exists): Check sign consistency: with a = g/3, T_up = W + m * g/3 = 4W/3 and T_down = W − m * g/3 = 2W/3; indeed T_up = 2 * T_down. Why Other Options Are Wrong: g/6 and g/4 produce T_up/T_down ratios different from 2; g/2 doubles the discrepancy; “none” is unnecessary because g/3 satisfies the given relation exactly. Common Pitfalls (misconceptions, mistakes): Mixing up direction of motion with direction of acceleration; forgetting that downward acceleration reduces cable tension. Final Answer: g/3

Question 10

Ideal machine with large velocity ratio — an ideal machine has mechanical advantage (MA) = 100. If the load moves 2 m, how far must the effort move?

Accepted Answer

Introduction / Context: For an ideal (lossless) simple machine, mechanical advantage equals velocity ratio. This means the distance the effort travels is amplified relative to the load movement by the same factor that the effort reduces the necessary force. Understanding this equivalence is central to hoists, jacks, and block-and-tackle systems. Given Data / Assumptions: Mechanical advantage MA = 100 (ideal, so efficiency = 100%). Load displacement s_load = 2 m. Velocity ratio VR = distance moved by effort / distance moved by load. Concept / Approach: In an ideal machine, MA = VR. Therefore the kinematic requirement is that the effort must move VR times the distance of the load. Multiply the given load displacement by MA to obtain the effort displacement. Step-by-Step Solution: Equate MA and VR: MA = VR = 100.Write VR = s_effort / s_load.Solve: s_effort = VR * s_load = 100 * 2 = 200 m. Verification / Alternative check (if short method exists): A power balance for an ideal machine gives Effort * s_effort = Load * s_load; with Load/Effort = MA = 100, s_effort/s_load = 100, consistent. Why Other Options Are Wrong: 20 m reflects MA = 10; 2.5 m and 2 m ignore the high velocity ratio; 0.02 m reverses the relation. Common Pitfalls (misconceptions, mistakes): Confusing MA with efficiency; forgetting that in ideal machines the gain in force is paid for by a proportional loss in distance moved. Final Answer: 200 m

Question 11

Stopping distance and time under uniform deceleration: A car moving at 20 m/s is brought to rest uniformly within a distance of 40 m. How long does it take to stop?

Accepted Answer

Introduction / Context: Uniformly accelerated (or decelerated) rectilinear motion is governed by constant-acceleration kinematics. Stopping a vehicle in a given distance from a known initial speed is a standard application of these formulas. Given Data / Assumptions: Initial speed u = 20 m/s. Final speed v = 0 m/s (comes to rest). Stopping distance s = 40 m. Acceleration a is constant (uniform deceleration). Concept / Approach: Use v^2 = u^2 + 2 a s to find the constant acceleration a. Then use v = u + a t to compute the stopping time t. Signs indicate deceleration (a negative value), but magnitudes are sufficient to get time. Step-by-Step Solution: Apply v^2 = u^2 + 2 a s → 0 = 20^2 + 2 * a * 40.Solve for a → 0 = 400 + 80 a → a = −400 / 80 = −5 m/s^2.Use v = u + a t → 0 = 20 + (−5) t → t = 20 / 5 = 4 s. Verification / Alternative check: Average speed during uniform deceleration is (u + v)/2 = 10 m/s. Time = distance / average speed = 40 / 10 = 4 s, confirming the result. Why Other Options Are Wrong: 2 s and 3 s imply much larger decelerations; 5 s and 6 s imply smaller decelerations insufficient to stop within 40 m. Common Pitfalls: Sign errors in a; forgetting that average speed equals mid-value when acceleration is constant; mixing units. Final Answer: 4 s

Question 12

What determines the effect of a force on a rigid body? The overall mechanical effect of a single force acting on a body depends on which attributes?

Accepted Answer

Introduction / Context: In engineering mechanics, a force is completely defined by its magnitude, direction, and line of action (including point of application). These characteristics determine the translational and rotational effects (resultant force and moment) on a rigid body. Given Data / Assumptions: Rigid body idealization (no deformation). Single force acting; moments arise from its line of action relative to a reference point. Statics principles apply. Concept / Approach: The translational effect is governed by the force vector (magnitude and direction). The rotational effect (moment) is M = r × F, which depends on the position vector r from the reference point to any point on the line of action of F. Two forces of the same magnitude and direction acting at different points can produce different moments and, therefore, different overall effects. Step-by-Step Solution: Identify attributes: magnitude, direction, line of action.Relate to equations: ΣF → translation; ΣM → rotation via position (moment arm).Hence, all three attributes jointly determine the effect. Verification / Alternative check: Varignon’s theorem and the principle of transmissibility (limited to rigid bodies along the same line of action) illustrate how changing position affects moments even if magnitude and direction remain the same. Why Other Options Are Wrong: Choosing only one attribute ignores the others; the full description requires all three. Common Pitfalls: Confusing the principle of transmissibility with independence from position; forgetting that moments depend on the perpendicular distance to the line of action. Final Answer: All the above

Question 13

Resultant of two collinear opposite forces: Two forces P and Q (P > Q) act along the same straight line but in opposite directions. What is the magnitude of their single resultant?

Accepted Answer

Introduction / Context: When two forces act collinearly and oppositely, they combine algebraically along that line. The stronger force partially cancels the weaker, leaving a single resultant acting in the direction of the stronger force. This principle underpins many equilibrium checks in trusses, cables, and straight-line pulls. Given Data / Assumptions: Two forces P and Q on the same line. P > Q; directions are opposite. Rigid-body statics; no other forces. Concept / Approach: The net force along one dimension is the signed sum of components. Take the positive direction as that of P. Then the resultant R = P − Q, positive, acting in the direction of P. If Q were greater, the magnitude would be Q − P acting in Q’s direction. Step-by-Step Solution: Assign sign convention: + along P.Sum forces: R = +P + (−Q) = P − Q.Conclude magnitude is P − Q, direction along P. Verification / Alternative check: Free-body diagram confirms that the larger of the two determines direction, while magnitude is the absolute difference. Why Other Options Are Wrong: (a) P + Q would apply if the forces acted in the same direction; (b) and (c) are dimensionally incorrect for a resultant; (e) Q − P would be negative under our convention, contradicting P > Q. Common Pitfalls: Forgetting to account for direction; reporting a negative number as a magnitude without indicating direction. Final Answer: P − Q

Question 14

Centroid of a semicircular arc (mechanics of solids) For a semicircular arc of radius r, determine the perpendicular distance of the centroid (center of gravity of the arc length) from the diameter along the axis of symmetry. State the result in ter…

Accepted Answer

Introduction / Context: The centroid of line elements such as circular arcs is a staple result in engineering mechanics and is frequently used in structural analysis, machine design, and composite area calculations. For a semicircular arc, the centroid lies on the radius of symmetry midway between the arc ends, at some distance above the diameter. Knowing this exact distance allows quick evaluation of moments and equivalent point loads for distributed cables, rings, or curved bars. Given Data / Assumptions: Semi-circular arc of radius r (thin line element, uniform line density). Centroid sought along the symmetry axis perpendicular to the diameter. Classical planar statics; thickness of the arc is negligible compared with r. Concept / Approach: For line elements, centroid coordinates are obtained by integrating position weighted by infinitesimal length ds, then dividing by total length. For a semicircular arc, symmetry gives x̄ = 0 and the centroid lies on the axis through the circle center. The known closed-form result places the centroid at a distance 2r/π from the circle’s center along the radius of symmetry, equivalently measured from the diameter since the diameter passes through the center. Step-by-Step Solution: Parameterize the arc with angle θ from −90° to +90° (or 0 to π).Infinitesimal length: ds = r dθ.Vertical coordinate of each element: y = r sin θ (with a suitable reference).Centroid ordinate: ȳ = (∫ y ds) / (∫ ds) = (∫ r sin θ (r dθ)) / (r ∫ dθ) = (r ∫ sin θ dθ) / (∫ dθ).Evaluating from −π/2 to +π/2 (or 0 to π) gives ȳ = 2r/π measured from the center along the symmetry radius toward the arc. Verification / Alternative check: The numeric factor 2/π ≈ 0.6366 lies between 0.5 and 1.0, which is sensible: the centroid of the arc must be closer to the center than the arc itself at distance r, yet above the diameter (0). Why Other Options Are Wrong: (b) r/π and (d) 4r/π^2 are too small; (c) πr/4 is too large; (e) r/2 underestimates the known exact value. Common Pitfalls: Confusing the centroid of a semicircular area (located at 4r/(3π)) with that of a semicircular arc (2r/π); mixing reference points (from center vs from diameter) leads to the same magnitude here because the diameter passes through the center. Final Answer: 2r/π

Question 15

Couples in engineering mechanics: identify the correct characteristics Which statements about a pure couple (a free vector system of two equal, opposite, parallel forces separated by a distance) are correct?

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Introduction / Context: Couples are fundamental in statics. They represent a pure rotational effect independent of the choice of origin. Recognizing their properties is key for reducing force systems and analyzing beams, frames, mechanisms, and machine elements subjected to torques. Given Data / Assumptions: A couple consists of two equal, opposite, parallel forces separated by a finite arm. Rigid body mechanics with classical moment definitions. Moments are computed about arbitrary points in the plane/space. Concept / Approach: Because net force is zero, a couple does not translate a body but tends to rotate it. Its moment M = F * d is a free vector (in 2D, a scalar with sign) and is invariant about any reference point. Balance of a couple requires an equal and opposite couple; a single force cannot null a free moment since any single force introduces a resultant force not equal to zero. Step-by-Step Solution: (a) Sum of forces: +F and −F cancel → 0.(b) Moment of a couple is independent of the origin → same about any point.(c) To cancel M, need −M, i.e., an opposite couple.(d) A single force cannot produce zero net force and nonzero moment simultaneously. Verification / Alternative check: Replacing distributed loads by equivalent resultant and couple shows invariance of the couple part under shift of the line of action, confirming (b). Why Other Options Are Wrong: Each individual statement (a)–(d) is correct; therefore the comprehensive correct choice is “All of the above”. Common Pitfalls: Confusing a couple with a torque applied via a force at an offset; forgetting that moving a couple anywhere in the body does not change its effect. Final Answer: All of the above.

Question 16

Seconds pendulum that gains 2 minutes per day A Second’s pendulum (period ≈ 2 seconds) gains 2 minutes in one day. To make it keep correct time, how should its effective length be adjusted?

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Introduction / Context: Pendulum clocks depend on the simple pendulum relation between period and length. Timekeeping errors arise if the period deviates from its calibrated value. Understanding how to correct the period through length adjustment is a classic dynamics application. Given Data / Assumptions: Second’s pendulum nominal period T0 ≈ 2 s. Clock gains 2 minutes per day → it runs fast → period is too short. Small oscillations; air resistance and amplitude effects neglected. Concept / Approach: The ideal pendulum period is T = 2π * √(L/g). Period increases with length. If the clock gains time, the pendulum beats too quickly; therefore, increase the length to slow it down. Changing the bob mass does not change T (for small oscillations), since mass cancels out of the period expression. Step-by-Step Solution: Observed: clock fast by 2 minutes/day → needs a longer period.Use dependence: T ∝ √L → to increase T, increase L.Therefore adjust the effective length downward or upward? Upward, i.e., lengthen the pendulum. Verification / Alternative check: Quantitatively, if the error is small, ΔT/T ≈ (1/2) ΔL/L. A positive ΔT requires positive ΔL, confirming the qualitative reasoning. Why Other Options Are Wrong: (a) Decreasing length makes the period shorter, worsening the gain. (c) and (d) changing mass does not affect period for small swings. (e) is unnecessary because a correct mechanical adjustment exists. Common Pitfalls: Confusing amplitude effects or believing bob weight affects period; ignoring that friction/escapement mainly influences amplitude, not the ideal period. Final Answer: Must be increased

Question 17

Resultant force being zero: motion implications (Newton’s first law) The resultant of all forces acting on a body is zero. Which of the following kinematic states must be true for the body in classical mechanics?

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Introduction / Context: Newton’s first law states that if the net external force on a body is zero, the body either remains at rest or continues to move with uniform velocity in a straight line. Many test questions probe recognition of these allowable states versus those that necessarily require a nonzero resultant force. Given Data / Assumptions: Classical particle mechanics; rotational dynamics not specifically included in the options except implicitly. Resultant force equals zero. No constraints forcing curved motion by other means (e.g., tension providing centripetal force). Concept / Approach: Zero net force implies zero linear acceleration. Two states are compatible: rest and uniform straight-line motion. The listed options do not include constant-velocity straight motion, but “does not move at all” is compatible with the condition and is the only correct option among those provided. Variable velocity, curved paths, or speeding up inherently require nonzero net force (tangential or normal components). Rotation with angular acceleration requires a net moment; for a particle this also implies forces not balancing in effect. Step-by-Step Solution: Net force F_res = 0 → linear acceleration a = 0.A body at rest (v = 0) satisfies a = 0 → admissible.Curved path or variable speed → needs a ≠ 0 → not admissible. Verification / Alternative check: Consider uniform circular motion: even with constant speed, direction changes so a_n = v^2/R ≠ 0, hence a net force is required; this contradicts F_res = 0. Why Other Options Are Wrong: (a), (b), (c), and (e) imply nonzero linear acceleration or change of direction; they cannot occur with zero resultant force. Common Pitfalls: Forgetting that constant velocity straight motion would also be acceptable, but since it is not listed, choosing the compatible state “at rest” is correct. Final Answer: It does not move at all (can be at rest).

Question 18

Suspension bridge cable behavior with temperature rise Identify the incorrect statement about a suspension bridge cable when temperature increases (all else equal).

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Introduction / Context: Suspension bridge main cables behave like flexible strings supported at towers and anchored at ends. Temperature changes alter cable length and geometry. Understanding the qualitative effects of thermal expansion is important for expansion joint design, hanger lengths, and serviceability checks. Given Data / Assumptions: Uniform temperature rise; no traffic or wind changes. Anchorage and tower supports permit geometric accommodation. Cable modeled as a flexible catenary/parabola under uniform load. Concept / Approach: Thermal expansion increases the unstressed length. For fixed support spacing, an increase in length necessarily increases sag (dip) to accommodate the extra length. The corresponding horizontal component of cable tension typically drops as sag increases, for the same uniform load and span, because H ≈ wL²/(8f) for a parabolic approximation; larger f (sag) gives smaller H. Step-by-Step Solution: Temperature ↑ → cable length ↑ (thermal expansion).Support span fixed → to fit extra length, sag f must ↑.From H ≈ wL²/(8f): as f increases, H decreases slightly → consistent with qualitative behavior. Verification / Alternative check: Simple parametric checks with parabolic cable theory confirm that for constant span and load, increased length correlates with increased dip and reduced horizontal tension. Why Other Options Are Wrong: (a) and (b) are true effects; (e) is also generally true. Therefore (c) stating “dip decreases” is incorrect. Common Pitfalls: Confusing the influence of live load (which also increases sag) with temperature; assuming fixed tension regardless of geometry. Final Answer: The dip of the cable decreases.

Question 19

Work classification by angle between force and displacement If the angle between an applied force and the actual direction of motion of a body lies between 90° and 180°, how is the work done by that force classified?

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Introduction / Context: Work is the energy transfer associated with a force acting through a displacement. The sign of work depends on the angle between the force vector and displacement vector, and recognizing the sign is crucial in energy methods and power calculations in mechanics. Given Data / Assumptions: Force magnitude F acts while the body moves a displacement s. Angle θ between F and displacement satisfies 90° < θ ≤ 180°. Classical definition of work applies: W = F s cos θ for constant F and straight motion. Concept / Approach: The cosine of an angle between 90° and 180° is negative. Hence W = F s cos θ becomes negative. Physically, the force acts opposite to the motion’s component, removing energy from the system (e.g., friction opposing motion). Step-by-Step Solution: Start from W = ∫ F · ds. For constant F and straight path, W = F s cos θ.For 90° < θ ≤ 180°, cos θ < 0.Therefore W < 0 → negative work. Verification / Alternative check: Examples: kinetic friction on a sliding block (θ = 180°) always does negative work; a brake force opposing wheel motion is another common case. Why Other Options Are Wrong: (c) Zero work occurs only when θ = 90°. (a) Virtual work is a method, not a sign classification. (b) “Imaginary work” is not a standard term; (e) positive work requires θ < 90°. Common Pitfalls: Forgetting the vector dot product; mixing up power sign conventions with work sign. Final Answer: Negative work

Question 20

Defining simple harmonic motion (SHM) via acceleration A body is said to perform simple harmonic motion (SHM) when which of the following conditions on its acceleration is satisfied?

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Introduction / Context: Simple harmonic motion models many oscillatory systems near equilibrium: mass–spring, small-angle pendulum, and torsional oscillators. The defining feature is the linear restoring action that pulls the system back toward equilibrium. Given Data / Assumptions: Small displacements so linearization is valid. Negligible damping and forcing. Motion measured from equilibrium position. Concept / Approach: The kinematic definition of SHM is that acceleration a is proportional to displacement x and directed toward equilibrium: a = −ω² x, where ω is the natural angular frequency. This sign convention (negative) ensures acceleration opposes the displacement, providing restoring behavior and sinusoidal solutions. Step-by-Step Solution: Assume a linear restoring force F = −k x → mass m gives a = F/m = −(k/m) x.Identify ω² = k/m → a = −ω² x.Solution x(t) = A sin(ω t + φ) exhibits sinusoidal motion with period T = 2π/ω. Verification / Alternative check: Small-angle pendulum: restoring torque ≈ −m g L θ, equation becomes θ̈ + (g/L) θ = 0, matching SHM form with ω = √(g/L). Why Other Options Are Wrong: (a) Direction away from equilibrium is antirestoring. (b) Square of displacement gives strongly nonlinear behavior, not SHM. (d) Inverse dependence is not SHM. (e) is false because (c) is correct. Common Pitfalls: Confusing direction of acceleration; overlooking that proportionality constant must be positive in magnitude with a negative sign in the equation. Final Answer: Acceleration is proportional to the displacement and directed towards the reference (restoring).

Applied Mechanics Questions