Q: Varignon’s theorem (principle of moments) in planar statics: Which statement correctly states Varignon’s theorem for a system of coplanar forces about any chosen point?

Introduction / Context: Varignon’s theorem, or the principle of moments, is a cornerstone of engineering statics. It simplifies the computation of the moment of a resultant by allowing moments of individual forces to be summed algebraically about the same reference point. Given Data / Assumptions: System of coplanar forces acting on a rigid body. Moments taken about an arbitrary point O in the plane. Sign convention for moments (clockwise/anticlockwise) must be consistent. Concept / Approach: Let R be the vector sum (resultant) of forces F₁, F₂, …, F_n. Varignon’s theorem states: Moment_O(R) = Σ Moment_O(F_i), where the sum is algebraic (with sign). This permits distributing moment calculations before vector addition, which is especially useful in complex systems. Step-by-Step Solution: Compute r × R after summing forces → moment of resultant.Alternatively, compute Σ(r × F_i) and add algebraically.Varignon guarantees these two routes give the same scalar moment about O in planar problems. Verification / Alternative check: Expanding R = ΣF_i and using cross-product linearity shows r × R = Σ(r × F_i), which directly proves the theorem. Why Other Options Are Wrong: a, b restrict to two forces and use “arithmetical” rather than algebraic (sign matters). c says “arithmetical sum” but ignores signs; incorrect in general. e confuses forces with moments (different physical dimensions). Common Pitfalls: Dropping signs; mixing moment arms for different forces; forgetting all forces must be referenced to the same point. Final Answer: Algebraic sum of the moments of all forces about any point in their plane equals the moment of their resultant about that point

Q: Force in an inclined light member supporting a weight: A light member is hinged to a vertical wall and supports a weight W at its free end. If the member makes an angle θ with the (upper) wall, find the axial force induced in the member.

Introduction / Context: Light two-force members (like links or rods) carry only axial force. When such a member supports a vertical load at its free end, the component of the axial force along the vertical direction must balance the applied weight. Determining the axial force is a frequent step in truss and frame analysis. Given Data / Assumptions: Member is light (self-weight neglected) and hinged at the wall: a classic two-force member. Weight W acts vertically downward at the free end. Member makes an angle θ with the horizontal (upper wall), hence its vertical component is T sin θ. Static equilibrium; no other external forces. Concept / Approach: For a two-force member, internal force is purely axial with magnitude T. Vertical equilibrium at the free end requires the vertical component of T to equal W. Horizontal components are reacted at the wall hinge. Step-by-Step Solution: Let T be axial force along the member.Vertical equilibrium: T * sin θ = W.Solve for T: T = W / sin θ = W cosec θ. Verification / Alternative check: As θ approaches 90°, sin θ → 1 and T → W (near vertical member). As θ gets small, sin θ decreases and T grows large, matching the mechanical intuition of a near-horizontal tie carrying high axial force. Why Other Options Are Wrong: W sec θ, W tan θ: incorrect trigonometric relations for the required vertical balance. W cos θ, W sin θ: these are components, not the required axial magnitude. Common Pitfalls: Using horizontal angle vs. vertical angle inconsistently; forgetting the member is a two-force member so the internal force is axial only. Final Answer: W cosec θ

Q: Kennedy’s theorem (instantaneous centres): According to Kennedy’s theorem, for three bodies having plane motion relative to one another, the three instantaneous centres corresponding to each pair lie on what geometric locus?

Introduction / Context: Kennedy’s theorem (also called Aronhold–Kennedy) is a key kinematic principle for mechanisms. It states a collinearity property of instantaneous centres (ICs), which greatly simplifies velocity analysis in linkages such as the four-bar, slider-crank, and geared systems. Given Data / Assumptions: Three rigid bodies in plane motion with respect to each other. Instantaneous centre defined for each pair at a given instant. Rigid body assumption; small-time instant where ICs exist. Concept / Approach: If the three bodies are numbered 1, 2, and 3, their pairwise ICs are I₁₂, I₂₃, and I₃₁. Kennedy’s theorem states that these three ICs are collinear at any instant. This enables finding an unknown IC by joining the other two and locating their intersection with mechanism geometry. Step-by-Step Solution: Identify two known ICs from constraints (pin joints, sliding contacts).Draw the straight line through these two points.The third IC must lie on the same straight line by the theorem. Verification / Alternative check: Velocity polygons or relative velocity equations can be used to confirm collinearity since the velocity of a point is perpendicular to the line from the IC to that point, leading to the same straight-line condition. Why Other Options Are Wrong: Point, two lines, triangle, circle: none captures the specific collinearity relation among the three ICs. Common Pitfalls: Confusing fixed vs. moving ICs; misidentifying ICs at slider contacts; assuming ICs form a triangle (they do not—only the links do). Final Answer: A straight line

Q: Units of angular acceleration (single correct): Which of the following is a correct unit for angular acceleration in mechanics? Choose the best answer.

Introduction / Context: Angular acceleration quantifies the time rate of change of angular velocity. Correctly identifying its units is fundamental in dynamics, rotational kinematics, and control of rotating machinery, where dimensional consistency ensures correct formulation of equations of motion. Given Data / Assumptions: Angular velocity has units of angle per time (e.g., rad/s). Angular acceleration is the derivative of angular velocity with respect to time. Any angle unit can be used, but the SI coherent unit is the radian. Concept / Approach: If ω is in rad/s, then α = dω/dt has units (rad/s)/s = rad/s^2. Degree-based units like deg/s^2 can be used in non-coherent contexts, but must be converted to radians for SI equations (1 rad ≈ 57.2958°). Linear acceleration m/s^2 applies to translational motion, not angular quantities. Step-by-Step Solution: Angular velocity ω → units rad/s.Angular acceleration α = dω/dt → units rad/s^2.Hence the correct single best unit is rad/s^2. Verification / Alternative check: Dimensional analysis: angles are dimensionless in SI but carried symbolically; time has dimension T. Therefore α has dimension T^−2, matching rad/s^2. Why Other Options Are Wrong: deg/s: angular velocity unit, not acceleration. rev/s: angular velocity unit. m/s^2: linear acceleration, not angular. N/m: stiffness unit, unrelated to acceleration. Common Pitfalls: Mixing angular and linear measures; forgetting the “per second squared”. Final Answer: radians per second squared (rad/s^2)

Q: Perfectly plastic (completely inelastic) impact of identical balls on a smooth table: Ball A (mass 250 g) moves at 10 m/s and strikes an identical stationary ball B on a smooth horizontal table. If the impact is perfectly plastic (the balls stick to…

Introduction / Context: In a perfectly plastic (completely inelastic) collision, colliding bodies stick together and move with a common velocity after impact. On a smooth horizontal table, external impulses are negligible during the short impact, so horizontal momentum is conserved. Given Data / Assumptions: Masses: m_A = m_B = 0.250 kg (identical). Initial velocities: u_A = 10 m/s, u_B = 0 m/s. Perfectly plastic impact ⇒ common velocity v after impact. No external horizontal impulse; momentum conserved along the line of motion. Concept / Approach: Use conservation of linear momentum for the system of the two balls. For perfectly plastic impact, kinetic energy is not conserved, but that does not affect the determination of the common velocity immediately after collision. Step-by-Step Solution: Total initial momentum = m_A * u_A + m_B * u_B = m * 10 + m * 0 = 10 m.After impact, combined mass = 2 m; common speed = v.Conservation: 10 m = 2 m * v ⇒ v = 5 m/s. Verification / Alternative check: Relative speed of separation is zero in a perfectly plastic collision, consistent with sticking together and moving at 5 m/s. Why Other Options Are Wrong: 0 m/s: would violate momentum conservation. 10 m/s: would imply no momentum transfer to B. 2.5 m/s: corresponds to unequal masses or different conditions. “None of these”: incorrect since 5 m/s is valid. Common Pitfalls: Trying to conserve kinetic energy in inelastic collisions; forgetting identical masses make the arithmetic especially simple. Final Answer: 5 m/s

Question 1

Projectile launched at an angle to the vertical: angle for maximum horizontal range A projectile is fired at angle θ to the vertical (so its angle to the horizontal is 90° − θ). Neglecting air resistance and assuming level ground, for which value of…

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Introduction / Context: The range of a projectile on level ground depends on the sine of twice the launch angle measured from the horizontal. When the angle is given relative to the vertical, a simple trigonometric adjustment is needed to find the maximizing condition. Given Data / Assumptions: No air resistance; same launch and landing elevation. Initial speed fixed. Angle to the vertical is θ, so the angle to the horizontal is α = 90° − θ. Concept / Approach: Range R on level ground is R ∝ sin(2α), where α is measured from the horizontal. Since α = 90° − θ, we can express R in terms of θ and maximize it by using properties of the sine function. Step-by-Step Solution: Let α = 90° − θ.Then R ∝ sin(2α) = sin(2(90° − θ)) = sin(180° − 2θ) = sin(2θ).The sine function attains its maximum value 1 at 2θ = 90°.Therefore, θ = 45° (equivalently, α = 45° from the horizontal). Verification / Alternative check: Symmetry of complementary angles confirms maximum range occurs for a 45° elevation from the horizontal, which corresponds to 45° from the vertical as well. Why Other Options Are Wrong: 0° and 90° give zero range; 30° or 60° yield sin(60°) = sin(120°) = √3/2, which is less than 1. Common Pitfalls: Forgetting to convert between angle-to-vertical and angle-to-horizontal; maximizing sin θ instead of sin 2θ. Final Answer: 45°

Question 2

Impact time ratio and coefficient of restitution: A glass ball is projected along a smooth floor to strike a vertical wall. It returns to the original point of projection in twice the time it took to reach the wall. Assuming no floor friction, deter…

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Introduction / Context: When a smooth ball strikes a rigid wall, only the normal (perpendicular) component of velocity changes according to the coefficient of restitution e. On a smooth horizontal floor, motion is effectively one-dimensional. Time and distance relationships before and after impact provide a clean way to compute e without needing masses or forces. Given Data / Assumptions: Smooth (frictionless) floor, so speed along the floor is uniform between wall and point of projection. Time to reach wall = t₁. Time to return to the start after impact = 2 t₁ (stated). Distance from start to wall = L (same both ways). Pre-impact speed toward wall = u; post-impact speed away from wall = e*u. Concept / Approach: The coefficient of restitution e is defined along the line of impact as the ratio of the speed of separation to the speed of approach. Here that is simply e = (speed after rebound) / (speed before impact). Using kinematics on a frictionless floor gives times as distance over speed. Step-by-Step Solution: Let time to wall be t₁ = L / u.Time to return after impact is t₂ = L / (e * u).Given t₂ = 2 t₁ ⇒ L / (e u) = 2 * (L / u).Cancel L and u: 1 / e = 2 ⇒ e = 0.5. Verification / Alternative check: If e halves the speed, the return trip should take double the time of the outgoing trip. This matches the statement, confirming e = 0.5. Why Other Options Are Wrong: 0.25, 0.33, 0.40, 0.55: These would imply return times of 4, ~3, 2.5, or ~1.82 times t₁, respectively, none of which match the given “twice”. Common Pitfalls: Mistaking distance ratios for velocity ratios, or assuming energy methods are needed. Here, constant horizontal speed segments make the time ratio method simplest. Final Answer: 0.50

Question 3

Routh’s rule for a symmetric body: A solid body is symmetrical about two perpendicular semi-axes. Using Routh’s rule, the moment of inertia about an axis through the centre of gravity is given by m*(a^2 + b^2)/n. What is the value of n?

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Introduction / Context: Routh’s rule (a compact statement of standard results) provides formulas for the second moment of mass (moment of inertia) of ellipsoids and other symmetric bodies. It generalizes familiar results such as I = (2/5) m r^2 for a solid sphere and I = (1/5) m (b^2 + c^2) for a solid ellipsoid about a principal axis. Given Data / Assumptions: Homogeneous, rigid body. Symmetry about two perpendicular semi-axes of lengths a and b. Axis passes through the centre of gravity and is a principal axis. Concept / Approach: For a solid ellipsoid with semi-axes a, b, c, the principal moments are: I_x = (1/5) m (b^2 + c^2), I_y = (1/5) m (a^2 + c^2), I_z = (1/5) m (a^2 + b^2). Routh’s compact form expresses each as m*(sum of squares of the two semi-axes perpendicular to the moment axis) divided by 5. Step-by-Step Solution: By Routh’s rule: I_through_CG = m * (a^2 + b^2) / n.Comparing with the standard ellipsoid result gives n = 5. Verification / Alternative check: For a sphere (a = b = c = r), I = (2/5) m r^2, which corresponds to m*(r^2 + r^2)/5 = (2/5) m r^2. Why Other Options Are Wrong: 2, 3, 4, 6: None match the established principal moment formulas for ellipsoids; only 5 reproduces the canonical results. Common Pitfalls: Confusing diameters with semi-axes; using thin-shell formulas instead of solid-body formulas; forgetting the factor 1/5 for homogeneous solids. Final Answer: 5

Question 4

Why elastic bodies rebound: understanding elastic collision behavior In the collision of two elastic bodies, which explanation best captures what happens during and after impact?

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Introduction / Context: Elastic collision theory explains how bodies interact when they strike each other or a surface and rebound. The microscopic basis is reversible deformation: bodies store strain energy during compression and release it during restitution, producing post-impact separation without permanent deformation (ideally). Given Data / Assumptions: Bodies are assumed elastic (no permanent set). Impact is along a well-defined line of impact. External influences like friction and air resistance are negligible during the very short impact interval. Concept / Approach: During an elastic impact there are two phases: compression and restitution. The normal component of relative velocity reduces to zero at maximum compression (bodies are momentarily at rest relative to each other along the impact line), then reverses as elasticity drives them apart. The coefficient of restitution e quantifies how much of the approach speed is recovered as separation speed. Step-by-Step Solution: Compression phase: bodies deform; relative normal speed drops to zero at maximum compression.Restitution phase: stored elastic energy reconverts to kinetic; bodies separate with speed e times the approach speed.Ideal elasticity: e = 1; real materials: 0 < e < 1. Verification / Alternative check: Impulse–momentum analysis shows the integral of contact force over the impact interval equals the change in momentum; the sign reversal of relative velocity during restitution matches option selections a–c. Why Other Options Are Wrong: Individual statements a, b, and c are each true but incomplete alone; together (option d) they fully describe the process. “None of the above” conflicts with well-established impact mechanics. Common Pitfalls: Assuming bodies never come to rest during impact; ignoring the two-phase nature; equating elasticity with no compression (compression is essential). Final Answer: All of the above

Question 5

Varignon’s theorem (principle of moments) in planar statics: Which statement correctly states Varignon’s theorem for a system of coplanar forces about any chosen point?

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Introduction / Context: Varignon’s theorem, or the principle of moments, is a cornerstone of engineering statics. It simplifies the computation of the moment of a resultant by allowing moments of individual forces to be summed algebraically about the same reference point. Given Data / Assumptions: System of coplanar forces acting on a rigid body. Moments taken about an arbitrary point O in the plane. Sign convention for moments (clockwise/anticlockwise) must be consistent. Concept / Approach: Let R be the vector sum (resultant) of forces F₁, F₂, …, F_n. Varignon’s theorem states: Moment_O(R) = Σ Moment_O(F_i), where the sum is algebraic (with sign). This permits distributing moment calculations before vector addition, which is especially useful in complex systems. Step-by-Step Solution: Compute r × R after summing forces → moment of resultant.Alternatively, compute Σ(r × F_i) and add algebraically.Varignon guarantees these two routes give the same scalar moment about O in planar problems. Verification / Alternative check: Expanding R = ΣF_i and using cross-product linearity shows r × R = Σ(r × F_i), which directly proves the theorem. Why Other Options Are Wrong: a, b restrict to two forces and use “arithmetical” rather than algebraic (sign matters). c says “arithmetical sum” but ignores signs; incorrect in general. e confuses forces with moments (different physical dimensions). Common Pitfalls: Dropping signs; mixing moment arms for different forces; forgetting all forces must be referenced to the same point. Final Answer: Algebraic sum of the moments of all forces about any point in their plane equals the moment of their resultant about that point

Question 6

Force in an inclined light member supporting a weight: A light member is hinged to a vertical wall and supports a weight W at its free end. If the member makes an angle θ with the (upper) wall, find the axial force induced in the member.

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Introduction / Context: Light two-force members (like links or rods) carry only axial force. When such a member supports a vertical load at its free end, the component of the axial force along the vertical direction must balance the applied weight. Determining the axial force is a frequent step in truss and frame analysis. Given Data / Assumptions: Member is light (self-weight neglected) and hinged at the wall: a classic two-force member. Weight W acts vertically downward at the free end. Member makes an angle θ with the horizontal (upper wall), hence its vertical component is T sin θ. Static equilibrium; no other external forces. Concept / Approach: For a two-force member, internal force is purely axial with magnitude T. Vertical equilibrium at the free end requires the vertical component of T to equal W. Horizontal components are reacted at the wall hinge. Step-by-Step Solution: Let T be axial force along the member.Vertical equilibrium: T * sin θ = W.Solve for T: T = W / sin θ = W cosec θ. Verification / Alternative check: As θ approaches 90°, sin θ → 1 and T → W (near vertical member). As θ gets small, sin θ decreases and T grows large, matching the mechanical intuition of a near-horizontal tie carrying high axial force. Why Other Options Are Wrong: W sec θ, W tan θ: incorrect trigonometric relations for the required vertical balance. W cos θ, W sin θ: these are components, not the required axial magnitude. Common Pitfalls: Using horizontal angle vs. vertical angle inconsistently; forgetting the member is a two-force member so the internal force is axial only. Final Answer: W cosec θ

Question 7

Kennedy’s theorem (instantaneous centres): According to Kennedy’s theorem, for three bodies having plane motion relative to one another, the three instantaneous centres corresponding to each pair lie on what geometric locus?

Accepted Answer

Introduction / Context: Kennedy’s theorem (also called Aronhold–Kennedy) is a key kinematic principle for mechanisms. It states a collinearity property of instantaneous centres (ICs), which greatly simplifies velocity analysis in linkages such as the four-bar, slider-crank, and geared systems. Given Data / Assumptions: Three rigid bodies in plane motion with respect to each other. Instantaneous centre defined for each pair at a given instant. Rigid body assumption; small-time instant where ICs exist. Concept / Approach: If the three bodies are numbered 1, 2, and 3, their pairwise ICs are I₁₂, I₂₃, and I₃₁. Kennedy’s theorem states that these three ICs are collinear at any instant. This enables finding an unknown IC by joining the other two and locating their intersection with mechanism geometry. Step-by-Step Solution: Identify two known ICs from constraints (pin joints, sliding contacts).Draw the straight line through these two points.The third IC must lie on the same straight line by the theorem. Verification / Alternative check: Velocity polygons or relative velocity equations can be used to confirm collinearity since the velocity of a point is perpendicular to the line from the IC to that point, leading to the same straight-line condition. Why Other Options Are Wrong: Point, two lines, triangle, circle: none captures the specific collinearity relation among the three ICs. Common Pitfalls: Confusing fixed vs. moving ICs; misidentifying ICs at slider contacts; assuming ICs form a triangle (they do not—only the links do). Final Answer: A straight line

Question 8

Units of angular acceleration (single correct): Which of the following is a correct unit for angular acceleration in mechanics? Choose the best answer.

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Introduction / Context: Angular acceleration quantifies the time rate of change of angular velocity. Correctly identifying its units is fundamental in dynamics, rotational kinematics, and control of rotating machinery, where dimensional consistency ensures correct formulation of equations of motion. Given Data / Assumptions: Angular velocity has units of angle per time (e.g., rad/s). Angular acceleration is the derivative of angular velocity with respect to time. Any angle unit can be used, but the SI coherent unit is the radian. Concept / Approach: If ω is in rad/s, then α = dω/dt has units (rad/s)/s = rad/s^2. Degree-based units like deg/s^2 can be used in non-coherent contexts, but must be converted to radians for SI equations (1 rad ≈ 57.2958°). Linear acceleration m/s^2 applies to translational motion, not angular quantities. Step-by-Step Solution: Angular velocity ω → units rad/s.Angular acceleration α = dω/dt → units rad/s^2.Hence the correct single best unit is rad/s^2. Verification / Alternative check: Dimensional analysis: angles are dimensionless in SI but carried symbolically; time has dimension T. Therefore α has dimension T^−2, matching rad/s^2. Why Other Options Are Wrong: deg/s: angular velocity unit, not acceleration. rev/s: angular velocity unit. m/s^2: linear acceleration, not angular. N/m: stiffness unit, unrelated to acceleration. Common Pitfalls: Mixing angular and linear measures; forgetting the “per second squared”. Final Answer: radians per second squared (rad/s^2)

Question 9

Perfectly plastic (completely inelastic) impact of identical balls on a smooth table: Ball A (mass 250 g) moves at 10 m/s and strikes an identical stationary ball B on a smooth horizontal table. If the impact is perfectly plastic (the balls stick to…

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Introduction / Context: In a perfectly plastic (completely inelastic) collision, colliding bodies stick together and move with a common velocity after impact. On a smooth horizontal table, external impulses are negligible during the short impact, so horizontal momentum is conserved. Given Data / Assumptions: Masses: m_A = m_B = 0.250 kg (identical). Initial velocities: u_A = 10 m/s, u_B = 0 m/s. Perfectly plastic impact ⇒ common velocity v after impact. No external horizontal impulse; momentum conserved along the line of motion. Concept / Approach: Use conservation of linear momentum for the system of the two balls. For perfectly plastic impact, kinetic energy is not conserved, but that does not affect the determination of the common velocity immediately after collision. Step-by-Step Solution: Total initial momentum = m_A * u_A + m_B * u_B = m * 10 + m * 0 = 10 m.After impact, combined mass = 2 m; common speed = v.Conservation: 10 m = 2 m * v ⇒ v = 5 m/s. Verification / Alternative check: Relative speed of separation is zero in a perfectly plastic collision, consistent with sticking together and moving at 5 m/s. Why Other Options Are Wrong: 0 m/s: would violate momentum conservation. 10 m/s: would imply no momentum transfer to B. 2.5 m/s: corresponds to unequal masses or different conditions. “None of these”: incorrect since 5 m/s is valid. Common Pitfalls: Trying to conserve kinetic energy in inelastic collisions; forgetting identical masses make the arithmetic especially simple. Final Answer: 5 m/s

Question 10

Equilibrium of a smooth circular cylinder resting on its curved (convex) surface: A smooth solid cylinder lies on a horizontal surface in contact through its convex curved surface. Classify its equilibrium condition for small disturbances.

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Introduction / Context: Equilibrium of bodies is classified as stable, unstable, or neutral based on the change in potential energy (or center-of-gravity position) when the body is slightly disturbed. Smooth round bodies on flat surfaces are classic illustrations used in engineering mechanics courses. Given Data / Assumptions: Smooth solid circular cylinder on a horizontal plane. Contact is along the curved convex surface; negligible friction effects for equilibrium classification. Small disturbances (infinitesimal rotations/translations). Concept / Approach: If a body, after a small disturbance, tends to return to its original position (potential energy increases with disturbance), equilibrium is stable. If it tends to move further away (potential energy decreases), equilibrium is unstable. If its potential energy does not change to first order (height of centre of gravity remains the same), equilibrium is neutral. Step-by-Step Solution: For a circular cylinder rolling slightly on a flat plane, the centre of gravity describes an arc with radius equal to the cylinder radius while its vertical height remains constant.Because the vertical position of the centre of gravity does not change for an infinitesimal roll, the gravitational potential energy is essentially unchanged.Hence the cylinder neither tends to return (no restoring moment) nor to move away (no overturning moment): the condition is neutral. Verification / Alternative check: The same reasoning explains why a perfect sphere on a horizontal plane is also in neutral equilibrium: any small rotation leaves the centre height unchanged. Why Other Options Are Wrong: Stable/Unstable: would require a rise or drop of the centre of gravity with small rotation, which does not occur here. “Out of equilibrium” and “None of these”: incorrect since the body is in equilibrium (no net force/moment initially) and neutral is listed. Common Pitfalls: Confusing neutral equilibrium with static indifference due to friction; here, neutrality is a geometric/gravitational property, independent of friction. Final Answer: Neutral equilibrium

Question 11

Statics of a sphere on two inclined planes  A smooth sphere rests in equilibrium, touching two smooth planes BA and BC that are inclined at 45° and 60° respectively to the horizontal. Comparing the support reactions on the two planes, which plane ex…

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Introduction / Context: This problem tests 2D statics of a smooth body in contact with two smooth planes at different inclinations. A sphere supported by two planes experiences only normal reactions at the contacts (no friction), along with its weight acting vertically downward. The task is to compare reaction magnitudes on planes set at 45° and 60° to the horizontal. Given Data / Assumptions: Sphere is smooth; planes are smooth (reactions are normal to the planes). Plane BA is inclined at 45° to the horizontal. Plane BC is inclined at 60° to the horizontal. Weight W acts vertically downward through the sphere’s center. Static equilibrium: sum of forces in horizontal and vertical directions equals zero. Concept / Approach: For a plane at angle α to the horizontal, the normal reaction makes an angle α to the vertical. Resolve each reaction into vertical and horizontal components and apply equilibrium. Horizontal components of the two reactions must balance each other; vertical components must add up to W. From the horizontal balance, a ratio between the reactions is obtained, which identifies which reaction is larger. Step-by-Step Solution: Let R_A = reaction on plane BA (45°), R_C = reaction on plane BC (60°).Horizontal equilibrium: R_A * sin 45° = R_C * sin 60°.Hence R_A / R_C = sin 60° / sin 45° = (√3/2) / (√2/2) = √3 / √2 ≈ 1.225.Therefore R_A > R_C.Vertical equilibrium (check): R_A * cos 45° + R_C * cos 60° = W (satisfied with the above ratio). Verification / Alternative check: As the plane becomes flatter (smaller inclination angle), its normal is closer to the vertical, so a larger reaction is needed to supply sufficient horizontal balance. Since 45° < 60°, the 45° plane supplies the larger reaction, consistent with R_A > R_C. Why Other Options Are Wrong: Less on BA than on BC: Contradicts the ratio R_A/R_C = √3/√2. Equal reactions: Only possible for equal inclinations, not here. Zero on BA: The sphere would fall; impossible in equilibrium. None of these: Incorrect because there is a determinate comparison. Common Pitfalls: Confusing the angle the normal makes with horizontal versus vertical, or assuming reactions depend only on vertical components without checking horizontal balance. Final Answer: More on BA than on BC

Question 12

Projectile motion basics (no air resistance)  A ball is thrown upward and follows a parabolic trajectory before returning to the ground. Which kinematic quantity remains constant during the entire flight?

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Introduction / Context: Ideal projectile motion (neglecting air resistance) is a cornerstone of engineering mechanics and physics. The motion decomposes into independent horizontal and vertical components. Understanding which quantities remain constant is essential for solving range, time-of-flight, and maximum-height problems quickly and accurately. Given Data / Assumptions: No air resistance; only gravity acts vertically downward. Constant gravitational acceleration g acts all along the flight. Initial velocity has horizontal and vertical components. Concept / Approach: Gravity acts only in the vertical direction, so it does not influence horizontal motion directly. Therefore, in the absence of air drag, the horizontal acceleration is zero. With zero horizontal acceleration, the horizontal component of velocity remains constant throughout the motion. Vertical velocity changes linearly with time due to constant g. Step-by-Step Solution: Resolve initial velocity u into u_x and u_y.Horizontal motion: a_x = 0 → v_x(t) = constant = u_x.Vertical motion: a_y = −g → v_y(t) = u_y − g * t, which varies with time.Because v_x is constant and v_y varies, the speed v = √(v_x^2 + v_y^2) changes during flight. Verification / Alternative check: At the highest point, the vertical component is momentarily zero while the horizontal component is unchanged since launch (ignoring drag). Both before and after the peak, v_x remains equal to u_x. Why Other Options Are Wrong: Vertical component: Changes linearly with time under gravity. Speed: Varies because v_y changes. Kinetic energy: Changes with speed, hence not constant. None of these: Incorrect since the horizontal component does remain constant. Common Pitfalls: Assuming speed is constant because the trajectory is smooth; forgetting to decompose motion; mixing effects of air resistance (not considered here) with the ideal model. Final Answer: Horizontal component of velocity

Question 13

Friction directions on a leaning ladder  A ladder rests on a smooth horizontal floor and leans against a rough vertical wall. Considering impending slip conditions, what is the correct direction of the frictional force at the upper contact with the …

Accepted Answer

Introduction / Context: Equilibrium of a ladder against a wall is a classic statics problem that develops understanding of friction directions at multiple contacts. Here, the ground is smooth (no friction at floor) and the wall is rough (friction available at wall). Determining the correct direction of the frictional force is essential for writing equilibrium equations correctly. Given Data / Assumptions: Smooth floor → zero friction at the floor; only a vertical normal from the floor. Rough vertical wall → normal is horizontal; friction acts vertically along the wall. Ladder tends to slide down at the top under its weight. Impending motion is considered for friction direction. Concept / Approach: Friction acts to oppose the relative motion at the contact. The top end of the ladder tends to move downward along the wall due to gravity. Therefore, at the wall contact, friction must act upward to oppose this impending downward motion. Step-by-Step Solution: Identify contact forces: at floor → vertical normal only; at wall → horizontal normal + vertical friction.Assume impending slip: top tends to move downward along wall.Set friction at wall upward to oppose downward tendency.Use this direction in equilibrium equations ∑F_x = 0, ∑F_y = 0, and ∑M = 0 if solving for magnitudes. Verification / Alternative check: If you incorrectly assume friction is downward at the wall, the vertical forces would both point downward (weight and wall friction), making vertical equilibrium impossible because the floor provides no vertical friction (smooth). Thus the only self-consistent direction is upward at the wall. Why Other Options Are Wrong: Horizontal directions (towards/away from wall) describe the wall normal, not friction. Downward at wall would assist the slide, not resist it. None of these is unnecessary because one option matches the correct direction. Common Pitfalls: Forgetting the floor is smooth; mixing up normal and friction directions; assuming directions first and fixing later—always start by considering the likely slip direction. Final Answer: Upwards at its upper end (vertical)

Question 14

Why does a satellite continue in its orbit around the Earth?  Identify the fundamental cause that maintains the satellite's circular (or near-circular) motion.

Accepted Answer

Introduction / Context: Orbital mechanics explains how satellites remain in continuous “free fall” around Earth. Understanding which force acts as the required inward pull clarifies many topics in dynamics, from banking of roads to planetary motion. Given Data / Assumptions: Satellite moves in a near-circular orbit around Earth. Neglect air resistance at orbital altitude. Earth’s gravitational field provides an attractive inverse-square central force. Concept / Approach: Uniform circular motion requires a radial inward force equal to m * v^2 / r. In orbital motion, this necessary inward (centripetal) force is supplied by gravity: F_g = G * M_E * m / r^2. The orbital speed adjusts so that the gravitational pull exactly matches the centripetal requirement, keeping the body in orbit rather than falling straight down. Step-by-Step Solution: Set centripetal requirement: m * v^2 / r.Set gravitational attraction: G * M_E * m / r^2.Equate for circular orbit: m * v^2 / r = G * M_E * m / r^2 → v^2 = G * M_E / r.Thus gravity provides exactly the inward pull (centripetal force) to maintain the orbit. Verification / Alternative check: Because the satellite is continually “falling” toward Earth while moving tangentially fast enough, the path curves around Earth. No outward “real” force is needed; the so-called centrifugal force is a fictitious force observed only in a rotating frame. Why Other Options Are Wrong: “Gravitational force” alone is incomplete; the correct physical role is that it acts as the centripetal force. Centrifugal force is an apparent force in a non-inertial frame, not the cause in an inertial frame. Air drag at orbital altitudes is negligible and opposes motion, not sustains it. “None of these” is unnecessary because a precise correct option exists. Common Pitfalls: Confusing centrifugal with centripetal; thinking satellites need continuous thrust (they do not for closed orbits, aside from station-keeping adjustments). Final Answer: Centripetal force provided by Earth’s gravity

Question 15

Ladder equilibrium conditions  A heavy ladder rests on a horizontal floor and against a vertical wall. In which case can the ladder not be in equilibrium under its own weight and contact reactions?

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Introduction / Context: Determining whether a ladder can remain in static equilibrium requires understanding the roles of normal and frictional forces at the two contacts (floor and wall). If both contacts are smooth, the only available contact forces are normals, which may be insufficient to balance all components and moments created by the ladder’s weight. Given Data / Assumptions: Weight W acts at the ladder’s center of gravity. Smooth contact → only normal reaction, no friction. Rough contact → friction available up to limiting value μN. Rigid ladder; planar statics (2D). Concept / Approach: For equilibrium in 2D, we need ∑F_x = 0, ∑F_y = 0, and ∑M = 0. With both contacts smooth, the wall provides a horizontal normal and the floor provides a vertical normal. There is no horizontal force at the floor and no vertical friction at the wall. These two normals cannot, in general, generate the required counter-moment to balance the turning effect of the weight unless the geometry is singular, so equilibrium cannot be sustained. Step-by-Step Solution: If both contacts are smooth: Floor gives N_floor upward; Wall gives N_wall horizontal.Forces cannot provide a balancing couple because there is no frictional component to adjust.The net moment due to W about the contacts cannot be countered; the ladder will slip. Verification / Alternative check: Introduce friction at either contact and you can select friction directions/magnitudes to satisfy all three equilibrium equations. Remove friction at both and you lose one necessary degree of freedom. Why Other Options Are Wrong: Floor smooth, wall rough: Wall friction can stabilize the system. Floor rough, wall smooth: Floor friction can stabilize the system. Both rough: Even more likely to achieve equilibrium. None of these: Incorrect since a specific non-equilibrium case exists. Common Pitfalls: Assuming “more smoothness” always helps; in ladder statics, at least one frictional contact is typically needed for equilibrium. Final Answer: Floor smooth and wall smooth

Question 16

Atwood-type system with unequal masses  Two loads of 50 kg and 75 kg are hung at the ends of a light inextensible rope passing over a smooth, frictionless pulley. What is the tension in the rope (expressed in kilogram-force terms)?

Accepted Answer

Introduction / Context: This is a classic Atwood machine problem with unequal masses. With a massless rope and frictionless pulley, both masses share the same magnitude of acceleration, and the rope tension is the same on both sides. Working in kilogram-force (kgf) units simplifies numbers for many exam problems. Given Data / Assumptions: m1 = 75 kg (heavier), m2 = 50 kg (lighter). Light inextensible rope; smooth, frictionless pulley. Use g consistently; when giving tension in kgf, effectively we quote T/g in kg units. Concept / Approach: Apply Newton’s second law to each mass along the line of motion, taking the heavier mass moving downward and the lighter moving upward. Solve for the common acceleration first, then back-substitute to obtain the tension. Step-by-Step Solution: Acceleration: a = (m1 − m2) * g / (m1 + m2) = (75 − 50) * g / 125 = 25 g / 125 = 0.2 g.Tension from heavier side: m1 * g − T = m1 * a → T = m1 (g − a) = 75 g (1 − 0.2) = 60 g.Therefore T/g = 60, so T = 60 kgf. Verification / Alternative check: From the lighter side: T − m2 * g = m2 * a → T = m2 (g + a) = 50 g (1 + 0.2) = 60 g, confirming the same result. Why Other Options Are Wrong: 50 kgf: Equals the lighter weight but tension must exceed it to accelerate upward. 75 kgf: Equals the heavier weight; not possible since the heavier mass accelerates downward. 25 kgf: Confuses net driving force with tension. 65 kgf: Numerical distractor; does not satisfy both equations of motion. Common Pitfalls: Mixing kg, N, and kgf inconsistently; forgetting that the same tension acts on both sides; sign errors when writing equations. Final Answer: 60 kgf

Question 17

Definition check: What is kinetic friction?  Choose the most accurate description for kinetic (dynamic) friction acting between two solid surfaces.

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Introduction / Context: Friction is commonly categorized into static and kinetic types. Static friction applies up to the onset of sliding; kinetic friction applies once sliding occurs. Accurate definitions help avoid conceptual errors in block-on-incline, belt friction, and brake problems. Given Data / Assumptions: Two dry, rough, solid surfaces in contact. Neglect lubrication, wear, and temperature effects. Classical Coulomb friction model applies. Concept / Approach: Static friction F_s adjusts up to its limiting value F_lim = μ_s * N just before motion begins. Once the bodies are in relative motion, kinetic friction F_k = μ_k * N acts, typically with μ_k < μ_s. The problem asks specifically for the definition of kinetic friction. Step-by-Step Solution: Identify whether motion exists: if sliding occurs, friction is kinetic.Define kinetic friction: F_k = μ_k * N acting opposite the direction of relative motion.Note contrast: “just about to move” corresponds to limiting static friction, not kinetic. Verification / Alternative check: In experiments, the pull required to keep a block sliding at constant speed is less than the maximum pull needed to start motion, reflecting μ_k < μ_s. Why Other Options Are Wrong: “Just about to move”: Defines limiting static friction. Angle definition: That is the angle of friction (an angular representation), not kinetic friction. Ratio of limiting friction to normal reaction: That is the coefficient of static friction μ_s. None of these: Unnecessary; one option is correct. Common Pitfalls: Using μ_s after sliding has begun; forgetting that kinetic friction is usually lower than the peak static value. Final Answer: Friction acting when the body is in actual motion

Question 18

Frequency–period relationship in simple harmonic motion (SHM)  For a particle executing SHM with angular frequency ω and period T, which statement correctly gives the relationship involving the frequency f?

Accepted Answer

Introduction / Context: Simple harmonic motion (SHM) appears in oscillating systems such as springs, pendulums (for small angles), and electrical LC circuits. Understanding how frequency, period, and angular frequency interrelate is fundamental for designing timing systems and analyzing vibrations. Given Data / Assumptions: SHM defined by sinusoidal time dependence. Frequency f (cycles per second), period T (time per cycle), and angular frequency ω (radians per second). Small-angle approximation may apply for pendulums, but the relationships f = 1/T and ω = 2π f always hold. Concept / Approach: By definition, period T is the time for one complete cycle, and frequency f is the number of cycles per unit time. Therefore, f = 1/T. Angular frequency relates to frequency by ω = 2π f, so f = ω / (2π), linking all three quantities. Step-by-Step Solution: Start from definition: f = cycles / time → f = 1 / T.Relate to angular frequency: ω = 2π f → f = ω / (2π).Hence frequency is inversely proportional to period and directly proportional to angular frequency. Verification / Alternative check: Dimensional consistency: [f] = s^-1, [T] = s, [ω] = s^-1. f = 1/T and ω = 2π f satisfy these units exactly. Why Other Options Are Wrong: Direct proportionality to T is incorrect; as T increases, f decreases. Inversely proportional to ω is incorrect; f increases with ω. Independence from T and ω contradicts definitions. Proportional to T^2 is not physically meaningful for SHM. Common Pitfalls: Confusing linear frequency f with angular frequency ω; forgetting the factor 2π; assuming frequency changes with amplitude (it does not for ideal SHM). Final Answer: Frequency is inversely proportional to the period (f ∝ 1/T)

Question 19

Vertical circular motion – at which position is string tension maximum? A stone is whirled in a vertical circle using a light, inextensible string. Considering both weight and centripetal requirements, at which position of the path is the tension in…

Accepted Answer

Introduction / Context: In vertical circular motion, the string must supply the inward (centripetal) force while also countering or being aided by the weight of the body. Understanding how tension varies around the path is essential for safe design of rotating machinery, amusement rides, and cable-driven systems. Given Data / Assumptions: Light, inextensible string and a small stone treated as a particle. Motion is in a vertical plane with speed possibly varying due to gravity. Neglect air resistance and any stiffness effects in the string. Concept / Approach: Tension is not constant in vertical circular motion. At any point, the centripetal requirement is m * v^2 / r toward the center. Weight mg acts downward. The string tension T must combine correctly with weight to provide the needed inward acceleration. Positions of special interest are the bottom and the top of the circle because the weight reverses its contribution relative to the required inward direction. Step-by-Step Solution: At the lowest point: inward direction is upward, so T_bottom − mg = m * v_bottom^2 / r → T_bottom = mg + m * v_bottom^2 / r.At the highest point: inward direction is downward, so T_top + mg = m * v_top^2 / r → T_top = m * v_top^2 / r − mg.Since v_bottom ≥ v_top for the same energy (gravity speeds the stone on the way down), and mg adds at the bottom but subtracts at the top, T_bottom is the largest.At horizontal positions the component of weight toward the center is mg * sinθ and does not exceed mg; tension there is between top and bottom values. Verification / Alternative check: Energy conservation between top and bottom shows v_bottom^2 = v_top^2 + 4 g r. Substituting into the expressions above confirms T_bottom − T_top = 2 mg + 4 m g = 6 m g minus terms depending on speeds, ensuring T_bottom > T_top for feasible motion. Why Other Options Are Wrong: Horizontal (a): tension is intermediate. Highest point (b): tension is least and may even approach zero at minimum-speed conditions. Same everywhere (d): false in vertical motion. 45° (e): not the extremum. Common Pitfalls: Assuming constant speed as in horizontal whirling; forgetting the weight reverses its role at top vs. bottom; mixing up inward (centripetal) direction with tension direction. Final Answer: when the stone is at the lowest position

Question 20

Centroid shift after cutting a circular hole from a circular plate From a circular plate of diameter 6 cm (radius 3 cm), a smaller circular disc is cut out whose diameter equals the radius of the original plate (i.e., 3 cm). By how much does the cen…

Accepted Answer

Introduction / Context: Composite area centroids are frequently needed in plate-cut detailing, counterweights, and lightweighting by cut-outs. When material is removed, the centroid of the remaining shape shifts toward the heavier side opposite the removed area. Here we analyze a classical circular-plate-with-circular-hole configuration. Given Data / Assumptions: Original plate: radius R = 3 cm. Removed disc: diameter = R, so r = R/2 = 1.5 cm. The small disc is cut so that it is internally tangent to the outer boundary (its center lies along a radius at distance R − r from the big center). Uniform thickness and density (area centroid analysis applies). Concept / Approach: For removal, use the negative-area method. Let O be the big-plate center and O′ the hole center on the same radius. The centroid shift x along that radius is obtained from first moments of area about O for the remaining lamina. Step-by-Step Solution: Areas: A1 = π * R^2, A2 = π * r^2.Center spacing: d = R − r = 3 − 1.5 = 1.5 cm.Resultant centroid shift: x = (A2 * d) / (A1 − A2).Compute: A1 = 9π, A2 = 2.25π → A1 − A2 = 6.75π.x = (2.25π * 1.5) / (6.75π) = 3.375 / 6.75 = 0.50 cm. Verification / Alternative check: Symmetry check: the shift must be along the line connecting the two centers and less than d (1.5 cm). The value 0.50 cm satisfies both. Why Other Options Are Wrong: 0.25 cm and 0.75 cm do not satisfy the area-ratio result; 1.00 cm and 1.50 cm exceed or misrepresent the computed centroidal displacement based on actual areas and geometry. Common Pitfalls: Placing the small hole center incorrectly (e.g., at R or at R/2 without the tangent condition); forgetting to subtract the removed area in the denominator; mixing diameters with radii. Final Answer: 0.50 cm

Applied Mechanics Questions