Q: Area properties – rectangular section about centroidal X–X axis\nCompute the second moment of area (moment of inertia) about the X–X axis for a rectangle that is 3 cm wide (b = 3 cm) and 4 cm deep (h = 4 cm), taking X–X through the centroid and para…

Introduction / Context: The second moment of area (area moment of inertia) of a plane section is a geometric property used in bending, deflection, and stability calculations. For standard shapes, compact formulas exist. Here we determine I about the centroidal horizontal axis X–X for a rectangular section. Given Data / Assumptions: Width b = 3 cm (horizontal). Depth/height h = 4 cm (vertical). X–X axis passes through the centroid and is parallel to the base (horizontal axis). Use standard formula I_x = b * h^3 / 12. Concept / Approach: For a rectangle, the centroidal second moment about the axis parallel to its base is I_x = b * h^3 / 12. This comes from integrating y^2 dA over the area. Units are in length^4. Step-by-Step Solution: Write the formula: I_x = b * h^3 / 12.Substitute b = 3 cm, h = 4 cm.Compute h^3 = 4^3 = 64 cm^3.Compute b * h^3 = 3 * 64 = 192 cm^4.Divide by 12: I_x = 192 / 12 = 16 cm^4. Verification / Alternative check: Dimensional check: input lengths in cm yield I in cm^4. Also note symmetry: increasing depth strongly increases I because of the h^3 factor, consistent with 16 cm^4 being larger than values using smaller h. Why Other Options Are Wrong: 9 cm^4 or 12 cm^4: arise from using b^3 h / 12 or arithmetic slips. 20 cm^4: not supported by the formula; likely from averaging or misusing 1/10 instead of 1/12. Common Pitfalls: Confusing axes: I_x uses h^3, whereas I_y about the vertical centroidal axis would be h * b^3 / 12. Always align the cubic dimension with the axis perpendicular to bending. Final Answer: 16 cm^4

Q: Force systems – classification of parallel forces\nThree forces of 100 N, 200 N, and 300 N are parallel but act in opposite directions to one another. What is the correct classification of such a system?

Introduction / Context: Engineering mechanics classifies force systems by coplanarity, concurrency, and whether parallel forces act in the same or opposite directions. Correct classification helps choose the right resultant and equilibrium tools. Given Data / Assumptions: Three forces: 100 N, 200 N, 300 N. Lines of action are parallel. Directions are opposed (not all the same way). All forces act in a single plane. Concept / Approach: Parallel forces acting in the same direction are called like parallel forces. When at least one force is opposite to the others, they are called unlike parallel forces. Concurrency is not applicable because parallel forces cannot meet at a single point unless they are collinear; typically they are non-concurrent, but the standard name for “opposite-direction parallel” is “unlike parallel forces”. Step-by-Step Solution: Identify parallelism: all three are parallel ⇒ “parallel forces”.Check directions: at least one is opposite to the others ⇒ “unlike”.Hence, the correct classification is unlike parallel forces. Verification / Alternative check: Resultant of unlike parallel forces is obtained by vector addition along the common direction with appropriate signs; the line of action is located using the principle of moments—typical treatment for unlike parallel systems. Why Other Options Are Wrong: coplanar concurrent: parallel forces do not pass through a common point. coplanar non-concurrent forces: true but incomplete; does not capture the “opposite directions” feature the question targets. like parallel forces: would require all forces in the same direction. Common Pitfalls: Confusing the broader classification (coplanar non-concurrent) with the specific term “unlike parallel forces”. Exams usually expect the specific term when directionality is emphasized. Final Answer: unlike parallel forces

Q: Dynamics foundations – source of D’Alembert’s principle\nD’Alembert’s principle, which converts a dynamic problem into a static equivalent by introducing an inertial force, fundamentally depends on which Newtonian law?

Introduction / Context: D’Alembert’s principle is widely used to analyze dynamics by recasting equations of motion into equilibrium form. It explicitly invokes an “inertia force” to balance applied forces, allowing the use of static tools like free-body diagrams with ΣF = 0. Given Data / Assumptions: Newton’s second law: ΣF = m a for a particle. Inertial (non-accelerating) reference frame. Mass m is constant and motion is classical (non-relativistic). Concept / Approach: Rearrange Newton’s second law to ΣF − m a = 0 and interpret −m a as a fictitious inertia force acting on the particle. This converts a dynamics equation into a statics-like equilibrium condition, which is the essence of D’Alembert’s principle. Step-by-Step Solution: Start from ΣF = m a.Move m a to the left: ΣF − m a = 0.Define an inertia force F_in = −m a acting on the mass.Write equilibrium: ΣF + F_in = 0, enabling static analysis methods. Verification / Alternative check: For rigid bodies, an analogous step yields ΣF − m a_G = 0 and ΣM_G − I_G α = 0, again mimicking statics by adding suitable inertia forces and moments, still grounded in Newton’s second law. Why Other Options Are Wrong: Incorrect / depends on frame: while inertial frames are required, the underlying basis remains Newton’s second law. Only valid for constant velocity: trivial case a = 0 reduces to statics; D’Alembert’s utility is precisely when a ≠ 0. Common Pitfalls: Treating inertia force as a real applied force; it is a mathematical device to achieve equilibrium form, not a physical interaction force. Final Answer: Correct

Q: Timekeeping – identifying a pendulum by its beat\nA pendulum that makes one beat (one complete oscillation to and fro) per second is commonly called what?

Introduction / Context: Classical clocks used pendulums whose period is tied to their length. One special case—the second’s pendulum—has a period convenient for timekeeping. This question asks for the standard name of a pendulum with one oscillation per second. Given Data / Assumptions: “One beat per second” interpreted as one complete to-and-fro oscillation in 2 seconds for traditional terminology (one beat often half-period; in horology, a second’s pendulum has period T = 2 s, giving one swing per second). Small-angle approximation applies: T ≈ 2π√(L/g). g ≈ 9.81 m/s^2 near Earth’s surface. Concept / Approach: A second’s pendulum is defined to have period T = 2 s. With the small-angle formula T = 2π√(L/g), this sets a standard length used historically to regulate clocks. Step-by-Step Solution: Set T = 2 s for a second’s pendulum.Use T = 2π√(L/g) ⇒ L = g*(T/2π)^2.With g ≈ 9.81 m/s^2 and T = 2 s, L ≈ 0.994 m (≈ 99.4 cm), consistent with standard values.Thus, the name is “second’s pendulum”. Verification / Alternative check: Clocks regulated by such pendulums tick once per second because each half swing lasts approximately 1 s, matching practical timekeeping. Why Other Options Are Wrong: simple pendulum: general idealized model, not a specific period. compound pendulum and torsional pendulum: other physical realizations; period depends on different parameters. Common Pitfalls: Confusing “beat” with “period”. In horology, a beat commonly refers to a half-swing; a second’s pendulum has one beat per second and a full period of two seconds. Final Answer: second's pendulum

Q: Units and dimensions – SI unit of force\nIdentify the correct SI (International System of Units) unit for force used in engineering mechanics.

Introduction / Context: Consistent units are essential in calculations. Force has a specific SI unit tied directly to Newton’s second law, linking mass, acceleration, and force. Given Data / Assumptions: SI base units: kilogram (kg), meter (m), second (s). Force defined via F = m * a. Concept / Approach: In SI, the unit of force is the newton (N), defined as the force required to accelerate a mass of 1 kg at 1 m/s^2, so 1 N = 1 kg·m/s^2. Step-by-Step Solution: Start from F = m * a.Insert SI base units: kg for mass and m/s^2 for acceleration.Thus, the derived unit is kg·m/s^2, named the newton (N). Verification / Alternative check: Compare with other systems: in the cgs system, the unit of force is the dyne where 1 dyne = 1 g·cm/s^2 = 10^-5 N, confirming that dyne is not the SI unit. Why Other Options Are Wrong: dyne: cgs unit, not SI. kilogram: SI base unit of mass, not force. watt: SI unit of power (J/s), not force. Common Pitfalls: Mixing cgs and SI units in the same problem leads to magnitude errors by factors of 10^n. Final Answer: newton

Question 1

Efficiency of a lifting machine — definitions and equivalent forms

The efficiency of a lifting machine is defined as which of the following?

Accepted Answer

Introduction / Context: Machine efficiency quantifies how effectively a device converts input work into useful output. Recognizing equivalent definitions helps connect kinematic and force-based viewpoints (velocity ratio vs. mechanical advantage). Given Data / Assumptions: Efficiency η is a dimensionless ratio lying between 0 and 1 for real machines. Mechanical Advantage (M.A.) = Load / Effort. Velocity Ratio (V.R.) = distance moved by effort / distance moved by load. Concept / Approach: By definition, η = output / input = useful work out / work in. Using M.A. and V.R., it can be shown that η = M.A. / V.R. because Work = Force * Distance, so (Load * load distance) / (Effort * effort distance) = (Load/Effort) / (effort distance/load distance) = M.A. / V.R. Step-by-Step Solution: Start with η = Work_out / Work_in.Work_out = Load * load distance; Work_in = Effort * effort distance.η = (Load * load distance) / (Effort * effort distance).Rewrite in terms of M.A. and V.R.: η = (Load/Effort) / (effort distance/load distance) = M.A. / V.R. Verification / Alternative check: For an ideal machine (η = 1), M.A. = V.R. — a standard consistency check used in exam problems. Why Other Options Are Wrong: Options A, B, and C are all correct forms of the same concept; therefore, the combined choice “All of the above” is correct. “Velocity ratio to mechanical advantage” inverts the correct relation and would give values greater than 1 for ideal machines, which is not the standard convention. Common Pitfalls: Confusing efficiency with M.A. itself; overlooking distances when comparing forces; forgetting that η cannot exceed 1 for real machines. Final Answer: All of the above

Question 2

Energy of a car’s wheels while moving — what forms are present?

Accepted Answer

Introduction / Context: Rolling motion combines translation of the center of mass and rotation about that center. Recognizing both energy contributions is critical for braking, fuel economy, and vehicle dynamics analyses. Given Data / Assumptions: Wheels are rigid and roll without slipping. Vehicle moves with speed v; wheel has angular speed ω with v = ω r. Gravitational potential energy changes are not under consideration on level ground. Concept / Approach: Total kinetic energy of a rolling rigid wheel equals translational kinetic energy of its center of mass plus rotational kinetic energy about its center. Thus, KE_total = (1/2) m v^2 + (1/2) I ω^2. For pure rolling, v and ω are linked through the radius r. Step-by-Step Solution: Write translational KE: K_trans = (1/2) m v^2.Write rotational KE: K_rot = (1/2) I ω^2.Relate rolling: v = ω r; substitute if needed to compare magnitudes.Add contributions: K_total = (1/2) m v^2 + (1/2) I ω^2. Verification / Alternative check: Special case: A thin hoop (I = m r^2) gives K_total = (1/2) m v^2 + (1/2) m r^2 (v^2 / r^2) = m v^2, demonstrating substantial rotational contribution. Why Other Options Are Wrong: Potential energy only: Irrelevant on level motion; energy form is kinetic. Translation only / Rotation only: Neglect one essential component of rolling motion. No energy: Rolling without slipping does not eliminate kinetic energy; it simply avoids sliding friction losses. Common Pitfalls: Forgetting rotational inertia effects; underestimating braking distances by ignoring rotational energy of wheels and drivetrain components. Final Answer: Kinetic energy of translation and rotation both

Question 3

Force system classification — non-coplanar, non-concurrent forces

Which description correctly defines non-coplanar non-concurrent forces?

Accepted Answer

Introduction / Context: Classifying force systems helps determine the appropriate equilibrium conditions and solution methods. Non-coplanar non-concurrent systems are the most general in three-dimensional statics. Given Data / Assumptions: Rigid-body statics in three dimensions. Forces can be characterized by concurrency (meeting at a point) and coplanarity (lying in one plane). Concept / Approach: A non-coplanar non-concurrent force system has no single point of intersection for all lines of action, and those lines of action do not lie in a single plane. Such systems require full 3D equilibrium equations: ΣF_x = ΣF_y = ΣF_z = 0 and ΣM_x = ΣM_y = ΣM_z = 0. Step-by-Step Solution: Check concurrency: “do not meet at a point” → non-concurrent.Check coplanarity: “do not lie in the same plane” → non-coplanar.Thus, the correct definition includes both conditions simultaneously.Conclude that the accurate description is option B. Verification / Alternative check: Examples include forces acting on 3D frameworks with members not confined to a plane. Such systems cannot be reduced to 2D equations. Why Other Options Are Wrong: Meet at a point but not coplanar: That would be a concurrent but non-coplanar system. Not concurrent but coplanar: That describes non-concurrent coplanar forces. Meet at one point and lie in the same plane: Concurrent coplanar, not asked. Common Pitfalls: Confusing concurrency with coplanarity; assuming that non-concurrent implies 3D automatically (non-concurrent can be purely 2D). Final Answer: Forces that do not meet at one point and their lines of action do not lie in the same plane

Question 4

Machine performance — relation between V.R. and M.A. when efficiency is constant

If the efficiency of a lifting machine is kept constant, the velocity ratio (V.R.) is _____ proportional to the mechanical advantage (M.A.).

Accepted Answer

Introduction / Context: Efficiency links force and motion characteristics of machines. Understanding how velocity ratio and mechanical advantage scale under constant efficiency is useful for design trade-offs. Given Data / Assumptions: Efficiency η is constant for the comparisons. Mechanical Advantage M.A. = Load / Effort. Velocity Ratio V.R. = effort distance / load distance. Concept / Approach: The standard relation is η = M.A. / V.R. If η is constant, then M.A. is proportional to V.R., i.e., V.R. ∝ M.A. Therefore, V.R. is directly proportional to M.A. under fixed efficiency. Step-by-Step Solution: Start with η = M.A. / V.R.Rearrange: V.R. = M.A. / η.With η constant, V.R. ∝ M.A.Hence, proportionality is direct, not inverse. Verification / Alternative check: Consider η = 0.5. If M.A. doubles, then V.R. doubles as well (V.R. = 2 * M.A.), confirming direct proportionality. Why Other Options Are Wrong: Inversely proportional: Would require η ∝ M.A. * V.R., which is not the standard relation. Not / exponential / logarithmic: No basis in the machine efficiency equation. Common Pitfalls: Confusing the inverse relationship between M.A. and V.R. only in an ideal machine where η = 1 (in that special case M.A. = V.R., still a direct relation, not inverse). Final Answer: directly

Question 5

Impact mechanics – definition check
In a perfectly or partially elastic collision, the coefficient of restitution is defined as the ratio of the relative speed of separation after impact to the relative speed of approach before impact. Evaluate the…

Accepted Answer

Introduction / Context: The coefficient of restitution, usually denoted by e, quantifies how “bouncy” a collision is. It compares how fast bodies separate after impact with how fast they were approaching before impact. This parameter is central to solving impact problems in engineering mechanics and physics. Given Data / Assumptions: Two bodies collide with finite relative velocity along the line of impact. Coefficient of restitution e is a material/interaction property for the particular impact condition. Analysis is along the common normal at the point of contact (line of impact). Concept / Approach: By definition, e = (relative speed of separation along line of impact) / (relative speed of approach along line of impact). For perfectly elastic impact, e = 1; for perfectly plastic impact, e = 0; for real impacts, 0 < e < 1. This definition holds for both direct (collinear) and oblique impacts when components are resolved along the line of impact. Step-by-Step Solution: Identify pre-impact normal components: u1n and u2n along the line of impact.Identify post-impact normal components: v1n and v2n along the same line.Compute approach speed: |u1n − u2n|.Compute separation speed: |v2n − v1n| (bodies moving apart after impact).Define e = (|v2n − v1n|) / (|u1n − u2n|). This exactly matches the statement. Verification / Alternative check: For a head-on elastic collision of equal masses, velocities exchange and e = 1. For a perfectly plastic collision, bodies stick and separation speed is zero, giving e = 0. Both special cases align with the stated ratio definition. Why Other Options Are Wrong: False: contradicts the formal definition used universally. True only for head-on collisions / equal masses: incorrect; the definition holds generally when resolved along the impact line. Common Pitfalls: Confusing total speed with the component along the line of impact. Tangential components may remain unchanged for smooth impacts and must not be included in e. Final Answer: True

Question 6

Area properties – rectangular section about centroidal X–X axis
Compute the second moment of area (moment of inertia) about the X–X axis for a rectangle that is 3 cm wide (b = 3 cm) and 4 cm deep (h = 4 cm), taking X–X through the centroid and para…

Accepted Answer

Introduction / Context: The second moment of area (area moment of inertia) of a plane section is a geometric property used in bending, deflection, and stability calculations. For standard shapes, compact formulas exist. Here we determine I about the centroidal horizontal axis X–X for a rectangular section. Given Data / Assumptions: Width b = 3 cm (horizontal). Depth/height h = 4 cm (vertical). X–X axis passes through the centroid and is parallel to the base (horizontal axis). Use standard formula I_x = b * h^3 / 12. Concept / Approach: For a rectangle, the centroidal second moment about the axis parallel to its base is I_x = b * h^3 / 12. This comes from integrating y^2 dA over the area. Units are in length^4. Step-by-Step Solution: Write the formula: I_x = b * h^3 / 12.Substitute b = 3 cm, h = 4 cm.Compute h^3 = 4^3 = 64 cm^3.Compute b * h^3 = 3 * 64 = 192 cm^4.Divide by 12: I_x = 192 / 12 = 16 cm^4. Verification / Alternative check: Dimensional check: input lengths in cm yield I in cm^4. Also note symmetry: increasing depth strongly increases I because of the h^3 factor, consistent with 16 cm^4 being larger than values using smaller h. Why Other Options Are Wrong: 9 cm^4 or 12 cm^4: arise from using b^3 h / 12 or arithmetic slips. 20 cm^4: not supported by the formula; likely from averaging or misusing 1/10 instead of 1/12. Common Pitfalls: Confusing axes: I_x uses h^3, whereas I_y about the vertical centroidal axis would be h * b^3 / 12. Always align the cubic dimension with the axis perpendicular to bending. Final Answer: 16 cm^4

Question 7

Force systems – classification of parallel forces
Three forces of 100 N, 200 N, and 300 N are parallel but act in opposite directions to one another. What is the correct classification of such a system?

Accepted Answer

Introduction / Context: Engineering mechanics classifies force systems by coplanarity, concurrency, and whether parallel forces act in the same or opposite directions. Correct classification helps choose the right resultant and equilibrium tools. Given Data / Assumptions: Three forces: 100 N, 200 N, 300 N. Lines of action are parallel. Directions are opposed (not all the same way). All forces act in a single plane. Concept / Approach: Parallel forces acting in the same direction are called like parallel forces. When at least one force is opposite to the others, they are called unlike parallel forces. Concurrency is not applicable because parallel forces cannot meet at a single point unless they are collinear; typically they are non-concurrent, but the standard name for “opposite-direction parallel” is “unlike parallel forces”. Step-by-Step Solution: Identify parallelism: all three are parallel ⇒ “parallel forces”.Check directions: at least one is opposite to the others ⇒ “unlike”.Hence, the correct classification is unlike parallel forces. Verification / Alternative check: Resultant of unlike parallel forces is obtained by vector addition along the common direction with appropriate signs; the line of action is located using the principle of moments—typical treatment for unlike parallel systems. Why Other Options Are Wrong: coplanar concurrent: parallel forces do not pass through a common point. coplanar non-concurrent forces: true but incomplete; does not capture the “opposite directions” feature the question targets. like parallel forces: would require all forces in the same direction. Common Pitfalls: Confusing the broader classification (coplanar non-concurrent) with the specific term “unlike parallel forces”. Exams usually expect the specific term when directionality is emphasized. Final Answer: unlike parallel forces

Question 8

Dynamics foundations – source of D’Alembert’s principle
D’Alembert’s principle, which converts a dynamic problem into a static equivalent by introducing an inertial force, fundamentally depends on which Newtonian law?

Accepted Answer

Introduction / Context: D’Alembert’s principle is widely used to analyze dynamics by recasting equations of motion into equilibrium form. It explicitly invokes an “inertia force” to balance applied forces, allowing the use of static tools like free-body diagrams with ΣF = 0. Given Data / Assumptions: Newton’s second law: ΣF = m a for a particle. Inertial (non-accelerating) reference frame. Mass m is constant and motion is classical (non-relativistic). Concept / Approach: Rearrange Newton’s second law to ΣF − m a = 0 and interpret −m a as a fictitious inertia force acting on the particle. This converts a dynamics equation into a statics-like equilibrium condition, which is the essence of D’Alembert’s principle. Step-by-Step Solution: Start from ΣF = m a.Move m a to the left: ΣF − m a = 0.Define an inertia force F_in = −m a acting on the mass.Write equilibrium: ΣF + F_in = 0, enabling static analysis methods. Verification / Alternative check: For rigid bodies, an analogous step yields ΣF − m a_G = 0 and ΣM_G − I_G α = 0, again mimicking statics by adding suitable inertia forces and moments, still grounded in Newton’s second law. Why Other Options Are Wrong: Incorrect / depends on frame: while inertial frames are required, the underlying basis remains Newton’s second law. Only valid for constant velocity: trivial case a = 0 reduces to statics; D’Alembert’s utility is precisely when a ≠ 0. Common Pitfalls: Treating inertia force as a real applied force; it is a mathematical device to achieve equilibrium form, not a physical interaction force. Final Answer: Correct

Question 9

Timekeeping – identifying a pendulum by its beat
A pendulum that makes one beat (one complete oscillation to and fro) per second is commonly called what?

Accepted Answer

Introduction / Context: Classical clocks used pendulums whose period is tied to their length. One special case—the second’s pendulum—has a period convenient for timekeeping. This question asks for the standard name of a pendulum with one oscillation per second. Given Data / Assumptions: “One beat per second” interpreted as one complete to-and-fro oscillation in 2 seconds for traditional terminology (one beat often half-period; in horology, a second’s pendulum has period T = 2 s, giving one swing per second). Small-angle approximation applies: T ≈ 2π√(L/g). g ≈ 9.81 m/s^2 near Earth’s surface. Concept / Approach: A second’s pendulum is defined to have period T = 2 s. With the small-angle formula T = 2π√(L/g), this sets a standard length used historically to regulate clocks. Step-by-Step Solution: Set T = 2 s for a second’s pendulum.Use T = 2π√(L/g) ⇒ L = g*(T/2π)^2.With g ≈ 9.81 m/s^2 and T = 2 s, L ≈ 0.994 m (≈ 99.4 cm), consistent with standard values.Thus, the name is “second’s pendulum”. Verification / Alternative check: Clocks regulated by such pendulums tick once per second because each half swing lasts approximately 1 s, matching practical timekeeping. Why Other Options Are Wrong: simple pendulum: general idealized model, not a specific period. compound pendulum and torsional pendulum: other physical realizations; period depends on different parameters. Common Pitfalls: Confusing “beat” with “period”. In horology, a beat commonly refers to a half-swing; a second’s pendulum has one beat per second and a full period of two seconds. Final Answer: second's pendulum

Question 10

Units and dimensions – SI unit of force
Identify the correct SI (International System of Units) unit for force used in engineering mechanics.

Accepted Answer

Introduction / Context: Consistent units are essential in calculations. Force has a specific SI unit tied directly to Newton’s second law, linking mass, acceleration, and force. Given Data / Assumptions: SI base units: kilogram (kg), meter (m), second (s). Force defined via F = m * a. Concept / Approach: In SI, the unit of force is the newton (N), defined as the force required to accelerate a mass of 1 kg at 1 m/s^2, so 1 N = 1 kg·m/s^2. Step-by-Step Solution: Start from F = m * a.Insert SI base units: kg for mass and m/s^2 for acceleration.Thus, the derived unit is kg·m/s^2, named the newton (N). Verification / Alternative check: Compare with other systems: in the cgs system, the unit of force is the dyne where 1 dyne = 1 g·cm/s^2 = 10^-5 N, confirming that dyne is not the SI unit. Why Other Options Are Wrong: dyne: cgs unit, not SI. kilogram: SI base unit of mass, not force. watt: SI unit of power (J/s), not force. Common Pitfalls: Mixing cgs and SI units in the same problem leads to magnitude errors by factors of 10^n. Final Answer: newton

Question 11

Kinematics terminology – is negative acceleration the same as retardation?
Assess the statement: “Negative acceleration is called retardation.”

Accepted Answer

Introduction / Context: In introductory mechanics, the term “retardation” is commonly used to describe acceleration that reduces the magnitude of velocity. Sign conventions can be confusing, so this statement needs careful interpretation. Given Data / Assumptions: A one-dimensional problem with a chosen positive direction. Acceleration a can be positive or negative relative to that axis. Velocity may be positive or negative depending on motion direction. Concept / Approach: Retardation (deceleration) is any acceleration that acts opposite to the direction of velocity, thereby reducing speed. When motion is in the positive direction (v > 0) and acceleration is negative (a < 0), the acceleration is retarding. Many elementary texts equate “negative acceleration” with “retardation” under this typical convention. Step-by-Step Discussion: Choose positive x upward or to the right; suppose v > 0.If a < 0, speed decreases ⇒ this is deceleration/retardation.If v < 0 (moving in negative direction) with a > 0, acceleration is still retarding because it opposes velocity—even though its sign is positive. Thus, strictly speaking, “retardation” is about opposing velocity, not merely the sign.However, in many exam contexts, “negative acceleration” is accepted as “retardation” relative to a chosen positive velocity direction. Verification / Alternative check: Consider free fall upward throw with upward positive: a = −g. During ascent, speed reduces; the negative acceleration is retarding. During descent (v negative), the same a = −g increases the speed (not retarding), illustrating the nuance. Why Other Options Are Wrong: False: too strict; although nuance exists, the statement reflects standard elementary usage. Other conditional options introduce unnecessary restrictions. Common Pitfalls: Assuming sign alone defines retardation without checking the direction of velocity. Always compare acceleration direction with velocity direction. Final Answer: True

Question 12

Standard length – second’s pendulum
What is the approximate length of a second’s pendulum at standard gravity (period ≈ 2 s), expressed in centimeters?

Accepted Answer

Introduction / Context: A second’s pendulum is a simple pendulum whose period is about 2 seconds. Its length provides a classic calibration for early clocks and an instructive example of the small-angle pendulum formula. Given Data / Assumptions: Target period T ≈ 2 s. Small-angle approximation holds: T = 2π√(L/g). Standard gravity g ≈ 9.81 m/s^2 at sea level at mid-latitudes. Concept / Approach: Rearrange the period formula to find the length L that yields T = 2 s. Convert meters to centimeters to compare with options. Step-by-Step Solution: Start with T = 2π√(L/g).Solve for L: L = g * (T / (2π))^2.Compute T/(2π) ≈ 2 / 6.283 ≈ 0.3183.Square: ≈ 0.1013. Multiply by g: 9.81 * 0.1013 ≈ 0.993 m.Convert to cm: ≈ 99.3–99.4 cm, matching 99.4 cm. Verification / Alternative check: Local variations in g change L slightly (about ±0.3%), but the standard textbook value remains ≈ 99.39 cm. Why Other Options Are Wrong: 94.9 cm, 100 cm, 101 cm: outside the accepted range for standard gravity with small-angle motion. Common Pitfalls: Using g = 10 m/s^2 for estimates gives L ≈ 1.013 m (≈ 101.3 cm), which is a rough approximation and not the standard precise value. Final Answer: 99.4 cm

Question 13

Machine classification – behavior of a screw jack
A screw jack used for lifting loads typically exhibits which reversibility characteristic under normal design (self-locking) conditions?

Accepted Answer

Introduction / Context: Screw jacks convert rotary effort into linear lifting using a power screw. The interplay of helix angle and friction determines whether the device can back-drive under load. Understanding this classification is important for safety and design. Given Data / Assumptions: Standard square/ACME thread with typical friction conditions. Helix angle α is small; friction angle φ is relatively large. Self-locking is desired for safety (load should not run down when effort is removed). Concept / Approach: A machine is non-reversible (self-locking) if the load cannot drive the machine backward to deliver output as input. For a screw jack, the criterion for self-locking is φ > α (or equivalently μ > tan α). Under this condition, back-driving does not occur, and the screw jack is non-reversible. Step-by-Step Reasoning: Relate friction angle to coefficient: tan φ = μ.Compute helix angle from screw geometry: tan α = lead / (π d_m).If μ > tan α (i.e., φ > α), then the torque from the load cannot overcome friction to rotate the screw backward.Hence, with typical design values, a screw jack is non-reversible (self-locking). Verification / Alternative check: The effort to lower a load normally exceeds zero; if friction is sufficiently high, the load will not descend unless an effort is applied in the lowering direction—hallmark of non-reversibility. Why Other Options Are Wrong: a reversible machine: would back-drive easily, unsafe for lifting. an ideal machine: would imply zero friction and 100% efficiency, not realistic. none of these: unnecessary; the standard term is non-reversible/self-locking. Common Pitfalls: Assuming all screws can back-drive. Many power screws are intentionally designed to be self-locking for holding loads without external braking. Final Answer: a non-reversible machine

Question 14

Basics of units in mechanics
In engineering mechanics and physics, the SI unit named “joule” is used to measure which physical quantity?

Accepted Answer

Introduction / Context: The International System of Units (SI) assigns standard units to physical quantities so engineers and scientists can communicate unambiguously. The “joule” is frequently encountered in mechanics, thermodynamics, and electrical engineering. This question tests whether you can correctly associate the joule with the right physical quantity. Given Data / Assumptions: We use SI units. Force is measured in newton (N), work/energy in joule (J), power in watt (W), and velocity in m/s. Work is defined as force applied through a distance in the direction of motion. Concept / Approach: Work and energy share the same SI unit because work done on a system changes its energy. By definition, 1 joule is the work done when a force of 1 newton acts through a distance of 1 metre: 1 J = 1 N * 1 m. Recognizing this mapping avoids confusing “power,” which measures the rate at which work is done and is expressed in watts (1 W = 1 J/s). Step-by-Step Solution: Identify quantity linked to joule.Recall definition: Work W = Force * displacement (in direction of force).Unit analysis: N * m = joule (J).Therefore, “joule” is the unit of work (and energy). Verification / Alternative check: Energy forms (potential, kinetic, electrical) are all measured in joules, reinforcing that joule is the energy/work unit. Power being J/s confirms its distinct unit, watt. Why Other Options Are Wrong: Force → newton (N), not joule. Power → watt (W), not joule. Velocity → metres per second (m/s), not joule. Common Pitfalls: Confusing energy (J) with power (W). Power is a rate; energy/work is an amount. Final Answer: work

Question 15

Kinetic energy from rest under uniform acceleration
A body of mass m starts from rest and reaches velocity v in time t. What is its translational kinetic energy at speed v?

Accepted Answer

Introduction / Context: Kinetic energy quantifies the energy of motion. It appears in work–energy calculations, impact analysis, and machine design. This item checks recall and interpretation of the kinetic energy expression for translational motion. Given Data / Assumptions: Mass = m (constant). Initial speed = 0 (starts from rest); final speed = v reached in time t. We are asked for kinetic energy at speed v, not for the acceleration or distance. Concept / Approach: The standard formula for translational kinetic energy is KE = 1/2 * m * v^2. It comes from integrating work done by a net force to increase speed from 0 to v. The time t is incidental here; KE depends only on instantaneous mass and speed, not explicitly on time. Step-by-Step Solution: Use work–energy relation: Work done on the particle equals change in kinetic energy.Starting from rest, KE_initial = 0.At speed v, KE_final = 1/2 * m * v^2.Therefore KE = 0.5 * m * v^2. Verification / Alternative check: From dynamics: Let constant acceleration be a. With v^2 = 2 * a * s, the work W = Force * distance = m * a * s = m * (v^2 / (2 * s)) * s = 1/2 * m * v^2, matching the formula. Why Other Options Are Wrong: mv^2: misses the factor 1/2. mgv^2 and 0.5 mgv^2: insert g (acceleration due to gravity) incorrectly; g is irrelevant to KE unless converting between gravitational potential and kinetic energy. Common Pitfalls: Including g by habit from weight-related problems. Confusing power or momentum with kinetic energy. Final Answer: 0.5 mv2

Question 16

Equilibrium of concurrent forces at a point
A set of forces acting at a single point will be in equilibrium if which condition is satisfied?

Accepted Answer

Introduction / Context: Forces that meet at a point are common in pin-jointed structures, cable junctions, and particle mechanics. Recognizing the correct equilibrium condition simplifies problem solving and prevents unnecessary calculations. Given Data / Assumptions: All forces are concurrent at a single point. We assume coplanar conditions unless otherwise stated. Static equilibrium requires no net acceleration. Concept / Approach: For a particle (point) in equilibrium, the vector sum of all external forces must be zero. In 2D, this is equivalent to requiring both ΣF_x = 0 and ΣF_y = 0 simultaneously. Merely having ΣF_y = 0 is insufficient unless ΣF_x = 0 also holds. Angles being “equally inclined” does not guarantee net zero force. Step-by-Step Solution: Write vector condition: ΣF = 0.In components: ΣF_x = 0 and ΣF_y = 0.Therefore the correct single statement among options is: “sum of all the forces is zero”.Hence select option (b). Verification / Alternative check: If ΣF = 0, the point does not accelerate (Newton’s second law: ΣF = m * a); with m > 0, this implies a = 0, confirming equilibrium. Why Other Options Are Wrong: Equally inclined forces may still produce a nonzero resultant. ΣV = 0 alone ignores horizontal components; equilibrium demands both directions be zero. “None of these” is incorrect because (b) is correct. Common Pitfalls: Checking only one direction (often vertical) and forgetting horizontal balance. Final Answer: sum of all the forces is zero

Question 17

Definition check — non-coplanar concurrent forces
Forces that meet at one point but whose lines of action do not lie in the same plane are called non-coplanar concurrent forces. Choose the correct word to complete the definition.

Accepted Answer

Introduction / Context: Classifying force systems (coplanar/non-coplanar and concurrent/non-concurrent) is foundational in statics. It dictates which equilibrium equations are needed and whether moments must be considered about axes in 3D. Given Data / Assumptions: Forces intersect at a common point (concurrent). Question distinguishes whether the lines of action are in one plane. Concept / Approach: “Coplanar” means existing in the same geometric plane; “non-coplanar” means spatially distributed in 3D. When forces meet at a point but extend in multiple planes, they are concurrent yet non-coplanar. Hence the missing phrase is “do not lie” in the same plane. Step-by-Step Solution: Identify concurrency: lines meet at one point.Identify planarity: lines of action are not confined to a single plane.Therefore use the phrase “do not lie”.Select option (b). Verification / Alternative check: Equilibrium of non-coplanar concurrent forces requires ΣF_x = 0, ΣF_y = 0, and ΣF_z = 0 (three component equations). This directly reflects the 3D nature of non-coplanar systems. Why Other Options Are Wrong: “Lie” would imply coplanar concurrent forces, a different category. Common Pitfalls: Assuming “concurrent” automatically means “coplanar.” They are independent descriptors. Final Answer: do not lie

Question 18

Parallel force systems — same direction case
Forces whose lines of action are parallel and point in the same direction are called:

Accepted Answer

Introduction / Context: Parallel force systems occur in beams, trusses under uniform loads, and multiple ropes lifting together. Proper naming clarifies how resultants are formed and where they act. Given Data / Assumptions: Lines of action are parallel. All forces act in the same sense (e.g., all upward). Concept / Approach: Parallel forces with the same sense are called “like parallel forces.” When they are in opposite senses, they are “unlike parallel forces.” “Coplanar concurrent/non-concurrent” describe intersection behavior and planarity, not the same-direction characteristic being tested here. Step-by-Step Solution: Recognize parallelism and identical direction.Recall terminology: like parallel forces.Select option (c). Verification / Alternative check: Resultant of like parallel forces equals the algebraic sum of magnitudes and acts in the same direction; its line of action is found using moments about a reference point, consistent with the definition. Why Other Options Are Wrong: Coplanar concurrent forces intersect at a point, not generally true for parallel forces. Coplanar non-concurrent is broad and does not indicate direction sense. Unlike parallel forces are opposite in sense, not “same direction.” Common Pitfalls: Mixing “parallel” classification with “concurrent” classification; they address different geometric relationships. Final Answer: like parallel forces

Question 19

Moment of inertia of a solid cone about its axis
For a solid circular cone of mass m and base radius r, what is the moment of inertia about its own vertical (symmetry) axis?

Accepted Answer

Introduction / Context: Moments of inertia about symmetry axes are essential in rotational dynamics and vibration analysis. For common solids (disc, cylinder, sphere, cone), standard formulas save derivation time during exams and design work. Given Data / Assumptions: Homogeneous solid cone. Mass = m; base radius = r. Axis of rotation is the cone’s vertical (symmetry) axis through the apex and center of base. Concept / Approach: The cone can be treated as a stack of differential disks of radius varying linearly from 0 at the tip to r at the base. Integrating each disk’s moment of inertia (1/2 * dm * ρ^2) over the height yields the standard result I_axis = (3/10) * m * r^2. Step-by-Step Solution (outline): Let radius vary with height: ρ = (r/h) * x.For a thin disk: dI = (1/2) * ρ^2 * dm.Relate dm to density and geometry; integrate from x = 0 to h.Result simplifies to I = 3/10 * m * r^2. Verification / Alternative check: Compare with solid cylinder I = 1/2 * m * r^2: since a cone concentrates less mass near the rim than a cylinder of the same r, its I must be less than 0.5 * m * r^2. The value 0.3 * m * r^2 satisfies this expectation. Why Other Options Are Wrong: 3mr^2/5, 2mr^2/5, 4mr^2/5 are larger than the cylinder’s 1/2 mr^2, which is inconsistent for a cone. Common Pitfalls: Confusing the cone’s axial I with its I about a centroidal diameter, which has a different value. Final Answer: 3mr2/10

Question 20

Newton’s laws — identify the first law statement
Which of the following correctly states Newton’s first law of motion (law of inertia)?

Accepted Answer

Introduction / Context: Newton’s three laws form the backbone of classical mechanics. The first law defines inertia and establishes that acceleration requires a net external force. Distinguishing the wording of each law prevents conceptual mix-ups in problem solving. Given Data / Assumptions: Idealized point mass. No net external force implies constant velocity (which includes rest). Concept / Approach: First law (inertia): without a net external force, a body’s velocity remains constant in magnitude and direction. Second law: ΣF = m * a (rate of change of momentum proportional to applied force). Third law: action–reaction pairs equal in magnitude and opposite in direction on different bodies. Step-by-Step Solution: Scan options to match the law of inertia.Option (a) states precisely that velocity remains constant unless acted on by an external force.Hence, select option (a) as Newton’s first law. Verification / Alternative check: In the absence of resultant force, from ΣF = m * a, we have a = 0, which means velocity is unchanged—consistent with the first law’s qualitative statement. Why Other Options Are Wrong: (b) describes Newton’s second law. (c) describes Newton’s third law. (d) is false because (a) is correct. Common Pitfalls: Equating “no force” with “body must be at rest.” No net force implies constant velocity, which could be nonzero. Final Answer: every body continues in its state of rest or of uniform motion, in a straight line, unless it is acted upon by some external force

Engineering Mechanics Questions