Statics Quiz: Fbd Particle
20 questions · exam conditions
0:00
Fbd ParticleQuestion 1 of 20

A particle slides along a rough horizontal surface while being pulled by a cable at angle θ above horizontal. The cable tension is T, particle weight is W, and kinetic friction coefficient is μₖ. In drawing the free-body diagram, what is the most common error students make regarding force directions?

Drawing the normal force equal in magnitude to the weight W
Drawing the friction force in the same direction as the cable tension
Drawing the cable tension perpendicular to the surface instead of at angle θ
Drawing the weight force at angle θ to account for the cable orientation
Drawing the normal force at angle θ to be perpendicular to the cable
← Back to quizzes

Statics Quiz

Statics Quiz: Fbd Particle

Practice Fbd Particle in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fbd Particle, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A particle slides along a rough horizontal surface while being pulled by a cable at angle θ above horizontal. The cable tension is T, particle weight is W, and kinetic friction coefficient is μₖ. In drawing the free-body diagram, what is the most common error students make regarding force directions?

  1. Drawing the normal force equal in magnitude to the weight W
  2. Drawing the friction force in the same direction as the cable tension (correct answer)
  3. Drawing the cable tension perpendicular to the surface instead of at angle θ
  4. Drawing the weight force at angle θ to account for the cable orientation
  5. Drawing the normal force at angle θ to be perpendicular to the cable
Explanation: When analyzing forces on a particle being pulled along a rough surface, you must carefully consider how each force responds to the motion and applied loads. The friction force always opposes the direction of motion (or impending motion). Since the particle slides horizontally along the surface, kinetic friction acts horizontally opposite to the velocity direction. This is independent of the cable tension's direction - friction responds to the motion, not to other applied forces. The cable tension has both horizontal and vertical components (Tcos⁡θT\cos\theta and Tsin⁡θT\sin\theta), but friction only opposes the horizontal sliding motion. Looking at the incorrect choices: Choice A reflects a common misconception about normal force magnitude. While the normal force often equals weight for objects on horizontal surfaces, here the cable's vertical component Tsin⁡θT\sin\theta reduces the normal force, making N=W−Tsin⁡θN = W - T\sin\theta. Choice C misrepresents the cable geometry - the tension acts along the cable at angle θ, not perpendicular to the surface. Choice D incorrectly suggests that weight direction changes based on other forces, but weight always acts vertically downward regardless of how the object is pulled. Choice B represents the most common error because students sometimes think friction should oppose the total applied force rather than the motion itself. They might draw friction pointing opposite to the cable tension's direction instead of opposite to the sliding direction. Study tip: Remember that kinetic friction always opposes motion direction, while the normal force adjusts based on all vertical force components. Draw motion direction first, then friction opposing it.

Question 2

When drawing a free-body diagram for a particle that is accelerating up a rough inclined plane due to an applied force, which statement about force representation is most critical?

  1. The applied force should be drawn larger than other forces to indicate it causes acceleration
  2. All forces should be drawn to scale to show their relative magnitudes accurately
  3. The friction force direction depends on whether the particle would slide up or down without the applied force
  4. The normal force should be drawn perpendicular to the incline, and friction should oppose the direction of motion (correct answer)
  5. Weight should be resolved into components before drawing the free-body diagram
Explanation: When analyzing forces on a particle moving on an inclined plane, the fundamental principle is that each force must be drawn in its correct direction based on physical constraints and definitions, regardless of the motion details. The correct approach always requires drawing the normal force perpendicular to the inclined surface and friction opposing the direction of motion. The normal force acts perpendicular because it represents the contact force preventing the particle from penetrating the surface. Friction, by definition, always opposes relative motion between surfaces—since the particle moves up the incline, kinetic friction points down the incline. This makes option D correct. Option A reflects a common misconception about free-body diagrams. Force arrows should represent direction and point of application, not relative magnitudes. Drawing the applied force larger doesn't convey meaningful information about the physics and can be misleading. Option B is impractical and unnecessary. While forces could be drawn to scale, this isn't required for analyzing the problem. Free-body diagrams are primarily qualitative tools for identifying forces and their directions. Option C contains a critical error in thinking. The friction direction depends on the actual motion occurring, not on hypothetical scenarios. Since the problem states the particle is accelerating up the incline, kinetic friction must oppose this upward motion, pointing down the incline. Remember this rule: in free-body diagrams, focus first on getting directions right based on physical constraints (normal forces perpendicular to surfaces) and definitions (friction opposes motion), then worry about magnitudes through equations.

Question 3

A charged particle with charge +2 μC is held in equilibrium by three forces: gravity (0.001 N downward), an electric force from a nearby charge, and a magnetic force due to its motion through a magnetic field. When drawing the free-body diagram, what is the key consideration for the magnetic force?

  1. The magnetic force should be drawn perpendicular to both the velocity vector and magnetic field direction
  2. The magnetic force should be drawn parallel to the magnetic field lines in the region
  3. The magnetic force magnitude must equal the vector sum of gravity and electric force for equilibrium
  4. The magnetic force should be omitted since magnetic forces cannot maintain static equilibrium (correct answer)
  5. The magnetic force should be drawn proportional to the charge magnitude and opposite to the electric force
Explanation: When analyzing forces in statics problems, you must carefully consider whether each force can actually contribute to maintaining equilibrium. The fundamental issue here is that magnetic forces have a unique property that disqualifies them from static equilibrium situations. Magnetic forces on moving charged particles are always perpendicular to the particle's velocity vector (following the right-hand rule for F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}). This perpendicular relationship means the magnetic force does no work on the particle and cannot change its kinetic energy. More critically for statics, if a particle is truly at rest (zero velocity), then F⃗magnetic=qv⃗×B⃗=0\vec{F}_{magnetic} = q\vec{v} \times \vec{B} = 0 since v⃗=0\vec{v} = 0. A stationary charged particle experiences no magnetic force regardless of the magnetic field strength. Option A incorrectly suggests drawing the magnetic force perpendicular to velocity and field, which would be correct for a moving particle but irrelevant here since the particle is in equilibrium (stationary). Option B misunderstands magnetic force direction entirely—magnetic forces are never parallel to field lines for moving charges. Option C applies correct equilibrium logic (forces must sum to zero) but fails to recognize that magnetic forces cannot exist in static situations. The correct answer is D because magnetic forces simply cannot maintain static equilibrium. Only conservative forces like gravity, electric forces, and contact forces can balance each other when an object is at rest. Study tip: Remember that "equilibrium" in statics means zero velocity, and zero velocity means zero magnetic force. If you see magnetic forces mentioned in equilibrium problems, immediately question whether true static equilibrium is possible.

Question 4

A particle rests on a horizontal surface with coefficient of static friction μ = 0.3. A force F is applied at 45° above horizontal. As F gradually increases from zero, at what point does the nature of forces in the free-body diagram change?

  1. When F exceeds the weight of the particle, the normal force becomes zero
  2. When the horizontal component of F exceeds maximum static friction, kinetic friction replaces static friction (correct answer)
  3. When F equals 0.3 times the weight, the friction force direction reverses
  4. When F reaches a critical value, the normal force changes from compression to tension
  5. The forces in the free-body diagram remain the same regardless of F magnitude
Explanation: When analyzing static equilibrium problems involving friction, you need to track how forces evolve as loading conditions change. The key insight is understanding what causes transitions between different equilibrium states. As force F increases from zero, the particle initially remains in static equilibrium. The horizontal component Fx=Fcos⁡(45°)F_x = F\cos(45°) creates a tendency to slide, which static friction fsf_s resists. The vertical component Fy=Fsin⁡(45°)F_y = F\sin(45°) reduces the normal force: N=W−Fsin⁡(45°)N = W - F\sin(45°). Static friction can vary from zero up to its maximum value fs,max=μNf_{s,max} = \mu N. The critical transition occurs when the horizontal component equals maximum static friction: Fcos⁡(45°)=μ(W−Fsin⁡(45°))F\cos(45°) = \mu(W - F\sin(45°)). Beyond this point, static friction cannot provide sufficient resistance, kinetic friction takes over, and the particle begins sliding. This represents a fundamental change in the free-body diagram's force nature. Let's examine why the other options are incorrect: A suggests normal force becomes zero when F exceeds weight, but with F at 45°, only the vertical component Fsin⁡(45°)F\sin(45°) affects normal force. F would need to be much larger than W. C incorrectly focuses on friction direction reversal at an arbitrary force value, missing the actual transition point. D mentions normal force changing to tension, which is impossible on a surface that can only push, not pull. Study tip: In friction problems, always identify what triggers the transition from static to kinetic friction—it's when the applied horizontal force component exceeds μN\mu N.

Question 5

A particle is simultaneously acted upon by its weight (50 N), a magnetic force (30 N northward), an electric force (40 N eastward), and contact forces from touching two surfaces. If the particle is in equilibrium, what can be concluded about the contact forces in the free-body diagram?

  1. The contact forces must sum to 50 N upward to balance weight only
  2. The contact forces must provide a net force of 50 N upward, 30 N southward, and 40 N westward (correct answer)
  3. Each surface provides a normal force perpendicular to its surface and friction forces as needed for equilibrium
  4. The contact forces are indeterminate without knowing the surface orientations and friction coefficients
  5. The two surfaces must provide equal and opposite contact forces to maintain equilibrium
Explanation: When analyzing equilibrium problems with multiple forces, remember that the net force in every direction must equal zero. This means all forces acting on the particle—whether applied forces or contact forces—must balance perfectly. Let's identify what we know: the particle experiences a 50 N weight (downward), 30 N magnetic force (northward), and 40 N electric force (eastward). For equilibrium, the contact forces must provide exactly the opposite of these three forces combined. The contact forces must therefore provide: 50 N upward (to balance weight), 30 N southward (to balance the northward magnetic force), and 40 N westward (to balance the eastward electric force). This is precisely what answer B states. Answer A incorrectly suggests the contact forces only need to balance the weight, ignoring the magnetic and electric forces entirely. Answer C makes a true general statement about how contact forces work (normal forces perpendicular to surfaces, friction as needed), but this doesn't answer what the contact forces must specifically provide for this equilibrium situation. Answer D suggests we can't determine the contact forces without knowing surface details, but equilibrium requirements dictate exactly what net contact force is needed regardless of how the surfaces are oriented—the surfaces must collectively provide the required force components. Study tip: In equilibrium problems, always start with ∑F=0\sum F = 0 in each direction. The unknown forces (like contact forces) must provide exactly what's needed to make each directional sum equal zero.

Question 6

A particle undergoes projectile motion near Earth's surface. At a point during flight where the particle is moving upward and to the right, which statement about the free-body diagram is correct?

  1. Include weight (downward) and air resistance (opposing velocity direction: downward and leftward) (correct answer)
  2. Include weight (downward), air resistance (downward and leftward), and lift force (perpendicular to velocity)
  3. Include weight (downward) only, since air resistance is negligible for particles
  4. Include weight (downward) and drag force (in the direction opposite to displacement from launch point)
  5. Include weight, normal force from air contact, and viscous drag proportional to velocity squared
Explanation: When analyzing projectile motion, you need to identify all forces acting on the particle by considering both gravity and air resistance effects. The key is understanding that air resistance always opposes the direction of motion, not the direction of displacement. Answer A is correct because it includes both essential forces: weight (which always acts downward due to gravity) and air resistance that opposes the velocity direction. Since the particle moves upward and to the right, air resistance acts downward and leftward—exactly opposite to the velocity vector. This represents the complete free-body diagram for realistic projectile motion. Answer B incorrectly adds a lift force perpendicular to velocity. Lift forces occur when objects have asymmetric shapes or spin (like airplane wings or spinning balls), but a simple particle in projectile motion experiences no lift—only drag opposing its motion. Answer C assumes negligible air resistance, which might apply in idealized physics problems but contradicts the realistic scenario implied by the question. When air resistance is negligible, you'd only include weight, but the question setup suggests considering all relevant forces. Answer D confuses drag direction with displacement direction. Drag always opposes velocity (the direction of motion), not displacement from the launch point. These directions can be completely different—for example, at the peak of trajectory, velocity is horizontal while displacement from launch is mostly vertical. Study tip: Always remember that air resistance opposes velocity direction, not displacement. Draw the velocity vector first, then draw air resistance pointing exactly opposite to it.

Question 7

A particle moves in a vertical circle on the end of a string. At the bottom of the circle, the particle has speed v and the string tension is 3mg, where m is the particle mass and g is gravitational acceleration. In the free-body diagram at this position, which forces should be shown?

  1. String tension (3mg upward), weight (mg downward), and centripetal acceleration (2mg upward)
  2. String tension (3mg upward) and weight (mg downward) only (correct answer)
  3. String tension (3mg upward), weight (mg downward), and centrifugal force (2mg downward)
  4. Weight (mg downward) and net centripetal force (2mg upward) only
  5. String tension (3mg toward center), weight (mg downward), and normal force from circular constraint
Explanation: When analyzing circular motion problems, you need to distinguish between forces (which belong in free-body diagrams) and accelerations (which don't). Free-body diagrams show only the actual physical forces acting on an object. At the bottom of the vertical circle, two forces act on the particle: the string tension pulling upward (3mg) and the gravitational weight pulling downward (mg). These are the only real forces present, making option B correct. Let's examine why the other options are wrong: Option A incorrectly includes "centripetal acceleration" as a force in the free-body diagram. Centripetal acceleration is not a force—it's the result of forces. The net upward force (3mg - mg = 2mg) creates the centripetal acceleration, but acceleration itself doesn't appear in free-body diagrams. Option C includes "centrifugal force," which is a fictitious force that only appears when analyzing motion from a rotating reference frame. In the standard inertial reference frame we use for most physics problems, centrifugal force doesn't exist. Additionally, if it were included, it would point outward (upward at the bottom), not downward. Option D omits the individual forces and shows only the "net centripetal force." Free-body diagrams must show the actual forces acting on the object (tension and weight), not their resultant. Study tip: Remember that free-body diagrams show only real, physical forces—never accelerations, fictitious forces, or net forces. When you see circular motion, identify the actual forces first (gravity, tension, normal forces, etc.), then use Newton's second law to relate their net effect to the centripetal acceleration.

Question 8

A particle is sliding down a curved surface under the influence of gravity. At the point of contact with the surface, the normal force acts radially inward with magnitude 50 N, and the tangential friction force has magnitude 8 N. How should these forces be represented in the free-body diagram?

  1. Normal force pointing toward the center of curvature, friction force tangent to the path opposing motion direction (correct answer)
  2. Normal force pointing away from the surface, friction force pointing toward the center of curvature
  3. Both forces should be resolved into horizontal and vertical components before drawing the diagram
  4. Normal force perpendicular to the average surface slope, friction force opposing the component of weight along the surface
  5. Normal force of 50 N pointing radially outward, friction force of 8 N pointing radially inward
Explanation: When analyzing forces on particles moving along curved paths, you must carefully consider how contact forces behave at the point of interaction. The surface exerts two distinct forces: a normal force perpendicular to the surface and a friction force parallel to the surface. For curved motion, the normal force always points toward the center of curvature (radially inward) because it provides the centripetal force component needed to keep the particle following the curved path. Since the problem states this force acts "radially inward," it's pointing toward the center of curvature. The friction force acts tangentially to the path, opposing the direction of motion as the particle slides down. Answer A correctly describes both forces: normal force toward the center of curvature and friction force tangent to the path opposing motion. Answer B incorrectly suggests the normal force points away from the surface, which would be radially outward—this cannot provide the required centripetal force for curved motion. It also wrongly places friction toward the center of curvature. Answer C suggests resolving forces into components before drawing the free-body diagram, but this misses the point. Free-body diagrams should show actual forces first, then resolve them if needed for calculations. Answer D describes forces as if the surface were straight or only slightly curved, using "average surface slope" and treating this like inclined plane motion rather than curved path dynamics. Remember: on curved surfaces, normal forces point toward the center of curvature to provide centripetal acceleration, while friction always opposes the direction of sliding motion tangentially.

Question 9

A particle is constrained to move along a helical path with both horizontal circular motion and vertical displacement. At the instant shown, the particle moves with speed v in a direction making angle α with the horizontal. For the free-body diagram, which forces are required to maintain this motion?

  1. Weight, normal force from the helical track (toward center of circular motion), and tangential friction force (correct answer)
  2. Weight, normal force (perpendicular to velocity), friction force (parallel to velocity), and helical constraint force
  3. Weight and constraint forces in both radial and tangential directions to maintain the helical path
  4. Weight, centripetal force (toward center), and tangential force (along velocity direction)
  5. Weight, normal force from track, friction force, and Coriolis force due to the helical motion
Explanation: When analyzing motion along a curved path, you need to identify the actual physical forces acting on the particle, not fictitious or resultant forces. A particle on a helical track experiences three-dimensional motion combining circular motion in the horizontal plane with vertical displacement. The correct answer is A because it identifies the three real physical forces present. Weight acts vertically downward due to gravity. The normal force from the helical track provides the centripetal acceleration needed for circular motion, pointing toward the center of the circular path. The tangential friction force acts along the helical path to either accelerate or decelerate the particle as needed to maintain the prescribed motion. Option B is incorrect because it double-counts forces. The "normal force perpendicular to velocity" and "helical constraint force" are the same force described differently, and "friction force parallel to velocity" is just the tangential component already covered in option A. Option C is wrong because it uses vague terminology. "Constraint forces in radial and tangential directions" doesn't specify the actual physical forces involved. In statics and dynamics, you must identify specific forces like normal forces, friction, and weight. Option D fails because "centripetal force" isn't a separate physical force—it's the net result of other forces (in this case, the horizontal component of the normal force from the track). Similarly, "tangential force" is too generic and doesn't specify that it's friction providing this component. Study tip: Always draw free-body diagrams showing actual physical forces (weight, normal, friction, tension, etc.), never fictitious forces like "centripetal" or vague "constraint forces."

Question 10

A particle with mass 2 kg rests on a rough inclined plane (angle 35°, μₛ = 0.6). A horizontal force of 15 N is applied to the particle. When constructing the free-body diagram, which approach correctly represents the applied force?

  1. Draw the 15 N force horizontally, then resolve it into components parallel and perpendicular to the incline
  2. Draw the 15 N force as components: 15cos(35°) parallel to incline and 15sin(35°) perpendicular to incline
  3. Draw the 15 N force horizontally without resolving it into components in the free-body diagram (correct answer)
  4. Draw the 15 N force perpendicular to the incline since horizontal forces cannot act on inclined objects
  5. Draw the 15 N force at 35° to horizontal to align with the incline coordinate system
Explanation: When constructing free-body diagrams in statics, the fundamental principle is to represent forces exactly as they are applied to the object, without modification. The free-body diagram is your starting point for analysis—it shows the actual forces acting on the particle before you begin any mathematical manipulation. Answer C is correct because the 15 N force is applied horizontally, so you must draw it horizontally on your free-body diagram. This preserves the true nature of the applied force. After completing the free-body diagram, you would then resolve this horizontal force (along with weight) into components parallel and perpendicular to the incline for your equilibrium equations. Answer A describes the correct overall problem-solving process but incorrectly suggests doing the resolution within the free-body diagram itself. The resolution step comes after the free-body diagram is complete. Answer B makes a critical error by assuming the 15 N force acts at a 35° angle to the incline, when it actually acts horizontally. This leads to incorrect component calculations—the horizontal force's components relative to the incline would involve different trigonometric relationships. Answer D is fundamentally wrong, as horizontal forces certainly can act on objects on inclined planes, and there's no physical reason to redraw a horizontal force as perpendicular. Remember: your free-body diagram must faithfully represent the actual forces before any mathematical manipulation. Draw forces as they're applied, then resolve them into convenient coordinate systems during your analysis phase.

Question 11

A small block rests on an inclined plane that makes a 25° angle with the horizontal. The coefficient of static friction between the block and plane is 0.4. In the free-body diagram for this particle, how many forces should be shown if the block is on the verge of sliding down the plane?

  1. Two forces: weight and normal force only, since friction is internal to the contact
  2. Three forces: weight, normal force, and friction force acting up the incline (correct answer)
  3. Four forces: weight, normal force, friction force, and applied force needed to prevent sliding
  4. Three forces: weight component parallel to incline, normal force, and friction force
  5. Five forces: weight, normal force, friction force, weight component parallel to incline, and weight component perpendicular to incline
Explanation: When analyzing forces on inclined planes, you need to identify all external forces acting on the object. This requires understanding what constitutes a force in a free-body diagram versus components of forces. For a block on the verge of sliding down an incline, three distinct forces act on it. The weight W=mgW = mg acts vertically downward from the block's center of mass. The normal force NN acts perpendicular to the inclined surface, away from the plane. The static friction force fsf_s acts parallel to the surface, pointing up the incline to oppose the impending downward motion. Since the block is on the verge of sliding, friction reaches its maximum value: fs=μsNf_s = \mu_s N. Choice A incorrectly suggests friction is internal to contact. Friction is an external force that must appear in free-body diagrams—it's the contact force parallel to the surface that prevents or opposes motion. Choice C adds an unnecessary applied force. The problem states the block is naturally on the verge of sliding due to gravity and friction alone; no additional external force is mentioned or required. Choice D makes a critical conceptual error by listing "weight component parallel to incline" as a separate force. Weight components (mgsin⁡θmg\sin\theta and mgcos⁡θmg\cos\theta) are mathematical breakdowns of the single weight force for calculation purposes, not separate physical forces. Study tip: In free-body diagrams, always draw the complete force vectors first (weight, normal, friction), then resolve them into components during calculations. Never list force components as separate forces in your initial diagram.

Question 12

A particle is held in equilibrium by four forces. Three of these forces have magnitudes of 10 N, 15 N, and 12 N respectively. When constructing the free-body diagram, what is the most important principle to remember about the fourth force?

  1. Its magnitude must equal the arithmetic sum of the other three forces (37 N)
  2. Its direction must be opposite to the direction of the largest of the other three forces
  3. Its vector sum with the other three forces must equal zero, making its magnitude and direction uniquely determined (correct answer)
  4. Its magnitude must be between the smallest and largest of the other three forces
  5. It must be drawn parallel to one of the coordinate axes to simplify the equilibrium equations
Explanation: When analyzing equilibrium problems, you're dealing with one of the fundamental principles of statics: a particle in equilibrium has zero net force acting on it. This means all forces must balance out perfectly. For a particle held in equilibrium by multiple forces, the vector sum of all forces must equal zero. This is Newton's First Law applied to static situations. Since you know three of the four forces (10 N, 15 N, and 12 N), the fourth force is completely determined by this equilibrium requirement. Its magnitude and direction must be exactly what's needed to make the total vector sum zero. Answer C correctly identifies this principle. Let's examine why the other options miss the mark. Answer A assumes you simply add the magnitudes arithmetically (10 + 15 + 12 = 37 N), but this ignores that forces are vectors with both magnitude and direction. Forces pointing in different directions don't add this way. Answer B incorrectly suggests the fourth force must oppose only the largest force. While the fourth force might need a component opposing the 15 N force, it must actually balance the vector sum of all three forces, not just the largest one. Answer D claims the fourth force magnitude must fall between 10 N and 15 N, but depending on how the three known forces are oriented, the fourth force could be much smaller or larger than this range. Remember this key principle: equilibrium means the vector sum of all forces equals zero. This constraint uniquely determines any unknown force once you know all the others.

Question 13

A particle is constrained to move along a frictionless track. At the position shown, the track curves with radius 2 m, the particle has speed 4 m/s, and experiences a 10 N driving force tangent to the track. Which statement about the free-body diagram is correct?

  1. Include normal force (perpendicular to track), driving force (tangent to track), weight, and centrifugal force (radially outward)
  2. Include normal force (toward center of curvature), driving force (tangent to track), and weight only (correct answer)
  3. Include normal force (away from center), driving force, weight, and centripetal acceleration as a force
  4. Include only the driving force and weight, since normal force is internal to the constraint
  5. Include normal force, driving force, weight, and the constraint force maintaining the particle on the track
Explanation: A free-body diagram shows only external forces: normal force from track (pointing toward center of curvature), driving force (tangent to track), and weight. Centrifugal force and centripetal acceleration are not real forces. Choice A includes fictitious centrifugal force. Choice C includes acceleration as a force. Choice D omits normal force. Choice E adds a redundant constraint force.

Question 14

A particle is held in equilibrium at point P by four cables as shown in the figure. Cables 1, 2, and 3 extend to fixed points, while cable 4 supports a hanging mass M. If cable 1 breaks, what changes must be made to the free-body diagram for the particle?

  1. Remove the tension force from cable 1 and add an unbalanced force to show acceleration
  2. Remove the tension force from cable 1; other forces remain unchanged until new equilibrium (correct answer)
  3. Replace cable 1 tension with a constraint force representing the broken connection
  4. Remove cable 1 tension and proportionally increase the other cable tensions to maintain equilibrium
  5. Remove cable 1 tension and add reaction forces at the fixed connection points
Explanation: When cable 1 breaks, its tension force is removed from the free-body diagram. The other forces (cable tensions, particle weight) remain the same in the FBD because they represent physical forces that still exist. The system will no longer be in equilibrium, but this is determined by analysis, not by modifying the FBD. Choice A adds non-physical forces. Choice C creates fictitious constraint forces. Choice D prematurely analyzes equilibrium. Choice E adds forces not acting on the particle.

Question 15

A particle is suspended by a string and simultaneously pulled by a horizontal spring as shown. The string makes a 20° angle with the vertical when in equilibrium. If the spring force is 15 N and the particle weighs 25 N, which forces must be included in the free-body diagram?

  1. String tension, spring force, weight, and the equilibrant force to balance the system
  2. String tension of 26.6 N, horizontal spring force of 15 N, and weight of 25 N
  3. String tension, spring force, and weight only - magnitudes are determined by analysis, not required for the diagram (correct answer)
  4. Horizontal component of string tension, vertical component of string tension, spring force, and weight
  5. Spring force, weight, and the resultant tension force combining string effects
Explanation: A free-body diagram shows all external forces acting on the particle: string tension (along the string direction), spring force (horizontal), and weight (vertical downward). The diagram identifies forces qualitatively; magnitudes are calculated during analysis. Choice A adds a non-existent equilibrant. Choice B prematurely calculates values. Choice D incorrectly shows force components instead of actual forces. Choice E combines forces incorrectly.

Question 16

Two particles A and B are connected by a light, inextensible rope passing over a massless pulley as illustrated. Particle A (mass 5 kg) rests on a 30° incline with friction coefficient 0.2, while particle B (mass 3 kg) hangs vertically. When drawing the free-body diagram for particle A, which forces should be included?

  1. Weight of A, normal force from incline, friction force, and tension force equal to weight of B
  2. Weight of A, normal force from incline, friction force, rope tension, and weight of particle B
  3. Weight of A, normal force from incline, friction force, and rope tension (magnitude determined by system analysis) (correct answer)
  4. Components of weight parallel and perpendicular to incline, friction force, and rope tension
  5. Weight of A, normal forces from incline and rope, friction force, and constraint force from the rope connection
Explanation: For particle A's FBD: include its weight (acting vertically), normal force from incline (perpendicular to surface), friction force (parallel to surface), and rope tension (along rope direction). The rope tension magnitude is found through analysis, not assumed equal to B's weight. Choice A assumes tension equals B's weight. Choice B incorrectly includes B's weight acting on A. Choice D shows components instead of actual forces. Choice E adds fictitious constraint forces.

Question 17

A particle rests on an inclined plane (angle θ\theta) and is connected to a hanging mass via a string over a pulley. The coefficient of static friction is μs\mu_s. In the free-body diagram for the particle on the incline, which statement correctly identifies all forces that must be included?

  1. Weight component parallel to incline, weight component perpendicular to incline, normal force, friction force (if motion is impending), and string tension (correct answer)
  2. Total weight vector, normal force, friction force (always present), string tension, and pulley reaction force transmitted through the string
  3. Weight vector, normal force, friction force (only if sliding occurs), string tension, and component of hanging mass weight
  4. Weight vector, normal force, static friction force (if required for equilibrium), string tension, and applied force from the pulley system
Explanation: A proper free-body diagram for a particle includes only forces acting directly on that particle. The weight should be shown as components (parallel and perpendicular to the incline) for easier analysis. Normal force acts perpendicular to the surface. Friction force exists when needed for equilibrium or when motion is impending. String tension acts along the string direction. Choice B incorrectly includes pulley reaction and shows weight as total vector, Choice C incorrectly includes hanging mass effects and wrong friction condition, Choice D incorrectly includes pulley forces and uses vague terminology.

Question 18

A particle is subjected to three forces in a plane: a vertical force of 50 N downward, a horizontal force of 30 N to the right, and a third force F at angle ϕ\phi above the horizontal. When drawing the free-body diagram for static equilibrium, which constraint must be satisfied?

  1. Force F must have magnitude 502+302\sqrt{50^2 + 30^2} and be directed at angle ϕ=tan⁡−1(50/30)\phi = \tan^{-1}(50/30) above the negative x-direction
  2. Force F must have horizontal component 30 N leftward and vertical component 50 N upward, regardless of the angle ϕ\phi chosen
  3. Force F must satisfy Fcos⁡ϕ=−30F\cos\phi = -30 N and Fsin⁡ϕ=50F\sin\phi = 50 N, which uniquely determines both F and ϕ\phi (correct answer)
  4. Force F can have any magnitude and angle ϕ\phi as long as its resultant with the other forces has zero net moment about the particle
Explanation: For static equilibrium of a particle, ∑Fx=0\sum F_x = 0 and ∑Fy=0\sum F_y = 0. Taking rightward as positive x and upward as positive y: 30+Fcos⁡ϕ=030 + F\cos\phi = 0 gives Fcos⁡ϕ=−30F\cos\phi = -30 N, and −50+Fsin⁡ϕ=0-50 + F\sin\phi = 0 gives Fsin⁡ϕ=50F\sin\phi = 50 N. These two equations uniquely determine both F and ϕ\phi. Choice A gives the correct magnitude but wrong angle direction, Choice B incorrectly suggests ϕ\phi can be chosen arbitrarily, Choice D incorrectly introduces moment concepts irrelevant to particle equilibrium.

Question 19

A particle is connected to two identical springs and a damper in a 2D configuration. Both springs have the same rest length L0L_0 and spring constant k. One spring connects the particle to a fixed point directly above, the other to a fixed point directly to the right. When drawing the free-body diagram for small oscillations about equilibrium, which approach correctly represents the spring forces?

  1. Each spring force has magnitude k times the distance from particle to attachment point, directed along the spring toward the attachment point
  2. Each spring force has magnitude k times displacement from equilibrium, directed along the respective coordinate axes toward the equilibrium position
  3. Each spring force has components proportional to the displacement components, with the total spring force being kx2+y2k\sqrt{x^2 + y^2} toward the equilibrium position
  4. Each spring force has magnitude k times the extension beyond rest length, directed along the spring toward the attachment point (correct answer)
Explanation: When analyzing spring forces in multi-dimensional systems, you must distinguish between the actual physical behavior of springs and simplified approximations used for small oscillations. Springs always exert forces based on their deformation from their natural rest length, regardless of the coordinate system or equilibrium position. Answer D correctly captures this fundamental principle. Each spring force equals k times the extension beyond rest length L0L_0, directed along the spring toward its attachment point. This is Hooke's law in its pure form: F⃗=−k(ℓ−L0)r^\vec{F} = -k(\ell - L_0)\hat{r}, where ℓ\ell is the current spring length and r^\hat{r} is the unit vector along the spring. Answer A incorrectly suggests the force depends on the total distance to the attachment point, ignoring that springs only respond to deformation beyond their rest length. A spring at its natural length exerts no force, regardless of that length. Answer B represents a common approximation used for small oscillations about equilibrium, where spring forces are linearized and treated as restoring forces proportional to displacement from equilibrium. While useful for analysis, this isn't the actual spring force—it's a mathematical simplification. Answer C incorrectly combines both springs into a single effective force, which doesn't represent how individual springs behave. Each spring acts independently based on its own deformation. Remember: Real springs always follow Hooke's law based on their extension from rest length. The linearized "displacement from equilibrium" approach in Answer B is a useful approximation for small oscillation analysis, but when asked about the actual spring forces, always return to the fundamental physics.

Question 20

A particle slides down a curved frictionless track and is momentarily at rest at point P where the track has radius of curvature R. At this instant, when drawing the free-body diagram, which statement correctly describes the forces acting on the particle?

  1. Weight (mg) acting vertically downward, normal force from track acting toward center of curvature, and centripetal force (mv2/Rmv^2/R) acting toward center of curvature
  2. Weight (mg) acting vertically downward and normal force from track acting toward center of curvature, with no centripetal force since velocity is zero (correct answer)
  3. Weight (mg) acting vertically downward, normal force from track acting perpendicular to track surface away from center of curvature, and tangential component of weight
  4. Weight (mg) acting vertically downward, normal force from track, centripetal acceleration (an=v2/Ra_n = v^2/R), and tangential acceleration component
Explanation: Free-body diagrams show only forces, not accelerations. At the instant described, the particle is momentarily at rest (v = 0), so centripetal acceleration is zero, but normal force still exists to support the particle against the component of weight toward the center of curvature. The forces are weight (mg downward) and normal force (toward center). Choice A incorrectly includes centripetal force as a separate force, Choice C incorrectly shows normal force direction, Choice D incorrectly includes accelerations in a force diagram.