Statics Quiz: Fbd Rigid Body
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Fbd Rigid BodyQuestion 1 of 20

A uniform rod of length LL and weight WW is supported by a cable at one end and rests on a smooth inclined surface at the other end. When constructing the free-body diagram, which forces act on the rod?

Weight WW, cable tension TT along the cable direction, and normal force NN perpendicular to the inclined surface
Weight WW, cable tension TT vertically upward, normal force NN perpendicular to the inclined surface, and friction force along the incline
Weight WW at the center, cable tension TT along the cable, normal force NN vertically upward from the inclined surface
Weight WW, cable tension TT with horizontal and vertical components, and reaction force RR normal to the inclined surface with friction component
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Statics Quiz

Statics Quiz: Fbd Rigid Body

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

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

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Question 1

A uniform rod of length LL and weight WW is supported by a cable at one end and rests on a smooth inclined surface at the other end. When constructing the free-body diagram, which forces act on the rod?

  1. Weight WW, cable tension TT along the cable direction, and normal force NN perpendicular to the inclined surface (correct answer)
  2. Weight WW, cable tension TT vertically upward, normal force NN perpendicular to the inclined surface, and friction force along the incline
  3. Weight WW at the center, cable tension TT along the cable, normal force NN vertically upward from the inclined surface
  4. Weight WW, cable tension TT with horizontal and vertical components, and reaction force RR normal to the inclined surface with friction component
Explanation: The correct FBD includes: weight W acting downward at the center of mass, cable tension T acting along the cable direction, and normal force N perpendicular to the smooth inclined surface. Since the surface is smooth, there is no friction force. The cable tension acts along the cable, not vertically.

Question 2

A rigid body is in contact with a rough horizontal surface at point C and is also supported by a smooth pin at point A. For the free-body diagram of this body, what should be shown at the contact point C?

  1. Only a normal force perpendicular to the surface, since friction is negligible for rigid bodies
  2. Only a friction force parallel to the surface, since normal forces don't affect rigid body motion
  3. Both a normal force perpendicular to the surface and a friction force parallel to the surface (correct answer)
  4. A single reaction force at an angle determined by the coefficient of friction
  5. No forces, since the pin support at A provides all necessary constraint
Explanation: When analyzing forces at contact points in statics, you need to consider what types of forces can develop based on the surface conditions. A "rough" surface means friction can occur, while "smooth" means frictionless. At point C, where the rigid body contacts the rough horizontal surface, two independent force components can develop. The normal force acts perpendicular to the surface (upward) to prevent the body from penetrating the surface. Simultaneously, because the surface is rough, a friction force can act parallel to the surface (horizontally) to resist any tendency for sliding. These are separate, independent forces that must both be included in your free-body diagram. Option A incorrectly suggests friction is negligible for rigid bodies. This confuses the concept—friction depends on surface roughness, not whether the body is rigid or deformable. Option B makes the opposite error, ignoring the normal force entirely. Normal forces are essential in statics problems as they often provide crucial vertical equilibrium. Option D represents a common misconception where students try to combine the normal and friction forces into a single resultant. While you could mathematically combine them later, the proper approach in free-body diagrams is to show them as separate components. The correct answer is C: show both forces as independent components at point C. Study tip: Always decompose contact forces into their perpendicular and parallel components relative to the contact surface. Rough surfaces allow both components; smooth surfaces only allow the perpendicular (normal) component.

Question 3

A rigid L-shaped bracket is fixed to a wall at point O and supports a horizontal force F at its free end. When constructing the free-body diagram for this bracket, what reactions must be shown at the fixed support?

  1. A vertical reaction force and a horizontal reaction force only
  2. A single reaction force in the direction opposite to the applied force F
  3. Two perpendicular reaction forces and a reaction moment about point O (correct answer)
  4. A reaction moment about point O only, since the applied force creates no net force
  5. Three reaction forces: horizontal, vertical, and one parallel to the bracket arm
Explanation: When analyzing fixed supports in statics, you need to understand that a fixed connection can resist forces in any direction AND prevent rotation. This is fundamentally different from other support types like pins or rollers. For this L-shaped bracket fixed at point O, the applied horizontal force F creates two effects: it tries to push the bracket horizontally and it creates a tendency to rotate the bracket about point O. A fixed support must provide whatever reactions are necessary to maintain equilibrium. The correct answer is C because a fixed support provides two perpendicular reaction forces (horizontal and vertical components) plus a reaction moment. Even though the applied force F is only horizontal, the bracket's geometry means this force creates a moment about the fixed point O. The vertical reaction force is needed because the moment equilibrium requires it - without a vertical reaction force at the support, you cannot balance the moment created by the horizontal force F acting at a distance from point O. Answer A is incomplete because it omits the reaction moment, which is essential when forces create rotational effects about the fixed point. Answer B oversimplifies the problem by assuming only one reaction force is needed - fixed supports are much more constrained than this suggests. Answer D incorrectly claims there's no net force, but the horizontal force F definitely exists and must be balanced by a horizontal reaction component. Remember: fixed supports always provide two force components and one moment reaction. When you see "fixed support" on statics problems, immediately think "three reactions: Fx, Fy, and M."

Question 4

A uniform rectangular plate is suspended by three cables attached at different points on the plate. When drawing the free-body diagram for the plate in static equilibrium, how should the cable forces be represented?

  1. As three force vectors, each acting along the direction of its respective cable toward the attachment point
  2. As three force vectors, each acting along the direction of its respective cable away from the attachment point (correct answer)
  3. As vertical forces only, since cables cannot support horizontal loads
  4. As force components: one horizontal and one vertical component for each cable
  5. As single resultant force acting at the center of gravity of the plate
Explanation: When analyzing cable forces in statics problems, you need to understand how tension works in cables and how to properly represent forces on free-body diagrams. Cables can only pull on objects - they cannot push. For any cable attached to an object, the tension force acts away from the attachment point along the cable's direction. Think of it this way: the cable is pulling the plate toward the other end of the cable, so the force vector on the plate points away from the plate along the cable line. This is exactly what option B describes - three force vectors acting along each cable's direction but away from their respective attachment points. Looking at the incorrect options: Option A has the direction backwards - it shows forces acting toward the attachment points, which would represent the plate pushing on the cables rather than cables pulling on the plate. Option C incorrectly assumes cables only provide vertical support; in reality, cables oriented at angles provide both horizontal and vertical force components naturally through their directional pull. Option D unnecessarily breaks down the forces into components - while you might resolve forces into components during calculations, the actual cable force is a single vector along the cable direction. Study tip: Remember that cables, ropes, and chains always pull, never push. When drawing free-body diagrams, always show tension forces pointing away from the object along the cable's line of action. This fundamental principle will help you correctly orient forces in any cable or rope problem.

Question 5

A rigid frame member is connected at one end by a pin joint and at the other end by a roller support. A uniformly distributed load acts along the member's length. In the free-body diagram for this member, what forces should be shown due to its own weight?

  1. No force, since the member's weight is internal to the system
  2. A concentrated force equal to the total weight, acting at the geometric center of the member
  3. A concentrated force equal to the total weight, acting at the center of mass of the member (correct answer)
  4. A distributed force with the same intensity as the applied distributed load
  5. Two concentrated forces, each equal to half the weight, acting at the pin and roller supports
Explanation: When analyzing free-body diagrams in statics, you must account for all external forces acting on a body, including its own weight. The key principle is that weight always acts as a concentrated force at the center of mass (centroid for uniform objects). The member's weight is an external gravitational force that must be included in your free-body diagram. Since the problem doesn't specify non-uniform density, you can assume the member is uniform, meaning its center of mass coincides with its geometric center. However, the correct terminology and concept is that weight acts at the center of mass, making C the most precise answer. Let's examine why the other options are incorrect: A is wrong because weight is never internal to a system - it's always an external force due to gravity that must appear in free-body diagrams. B uses imprecise language by saying "geometric center." While this happens to be correct for uniform members, the fundamental principle is that weight acts at the center of mass, not the geometric center. For non-uniform objects, these points differ. D incorrectly suggests representing weight as a distributed load. Weight is always treated as a concentrated force in free-body diagrams, even though gravity technically acts on every particle of the object. Remember this pattern: in statics problems, always represent an object's weight as a single concentrated force acting downward at the center of mass. For uniform objects, this conveniently occurs at the geometric center, but the underlying principle is center of mass.

Question 6

A rigid beam is supported by a cable at one end and rests against a smooth wall at the other end. When constructing the free-body diagram for this beam, which statement about the wall reaction is correct?

  1. The wall reaction acts perpendicular to the wall surface, since smooth surfaces cannot exert friction forces (correct answer)
  2. The wall reaction acts perpendicular to the beam, since that's the direction of contact
  3. The wall reaction has both normal and tangential components, regardless of surface smoothness
  4. The wall reaction acts at an angle determined by the coefficient of static friction
  5. No wall reaction exists, since the cable provides all necessary support
Explanation: When analyzing contact forces in statics problems, the key principle is that the type of contact determines the direction and components of the reaction force. Surface properties directly dictate what forces can be transmitted. A smooth wall contact can only exert a normal force perpendicular to the wall surface. Since smooth surfaces have no friction capability, they cannot resist tangential (parallel) forces along their surface. The wall reaction must therefore act perpendicular to the wall surface itself, pointing horizontally away from the wall toward the beam. This is exactly what option A states correctly. Option B incorrectly suggests the reaction acts perpendicular to the beam rather than perpendicular to the wall surface. The reaction direction depends on the constraint (the wall), not the geometry of the object being supported. Option C is wrong because it claims both normal and tangential components exist regardless of surface smoothness. This directly contradicts the fundamental principle that smooth surfaces cannot provide friction forces - there can be no tangential component. Option D incorrectly introduces a friction coefficient, which is irrelevant for smooth surfaces. Smooth contacts by definition have no friction, so there's no coefficient to consider, and the reaction acts purely normal to the surface. Remember this pattern: smooth contacts always produce reactions perpendicular to the contact surface, while rough contacts can have both normal and tangential components. Always identify the surface type first, then determine the allowable reaction directions.

Question 7

A rigid bar AB is connected to a fixed support by a pin at A and is connected to a sliding collar at B that moves along a vertical guide rod. When the bar is in equilibrium under an applied horizontal force at its midpoint, what reactions should appear in the free-body diagram at point B?

  1. Both horizontal and vertical reaction components, since the collar constrains motion in both directions
  2. Only a horizontal reaction component, since the collar can slide vertically but not horizontally (correct answer)
  3. Only a vertical reaction component, since the collar prevents vertical motion but allows horizontal sliding
  4. No reaction components, since the collar can slide freely along the guide
  5. A reaction force perpendicular to the bar AB, regardless of the guide rod orientation
Explanation: When analyzing supports and connections in statics problems, you need to carefully consider what motions each constraint allows or prevents. The key is understanding that reaction forces only develop in directions where motion is constrained. A sliding collar on a vertical guide rod is a classic example of a constraint that works in only one direction. The collar can slide freely up and down along the vertical rod, but it cannot move horizontally because the rod blocks that motion. Since reaction forces only appear where movement is restricted, point B can only exert a horizontal reaction force on the bar. Think of it this way: if the bar tries to move horizontally at point B, the collar hits the guide rod and creates a reaction force. But if the bar tries to move vertically at point B, the collar simply slides along the rod with no resistance, so no vertical reaction develops. Looking at the wrong answers: Choice A incorrectly assumes the collar constrains motion in both directions, but sliding collars specifically allow motion parallel to the guide. Choice C has the directions backwards - it's the horizontal motion that's prevented, not the vertical. Choice D misunderstands that while the collar slides freely vertically, it still provides constraint (and therefore reaction) horizontally. Study tip: For any support or connection, ask yourself "What directions of motion does this prevent?" Reaction components only exist in those constrained directions. Pins prevent motion in all directions, rollers prevent motion perpendicular to the rolling surface, and sliding collars prevent motion perpendicular to the guide.

Question 8

A rigid triangular plate is suspended in 3D space by three cables attached to its vertices. Each cable makes a different angle with the horizontal. When drawing the free-body diagram for the plate, how should the cable tensions be represented?

  1. As three force vectors, each with magnitude T and direction along the respective cable (correct answer)
  2. As nine force components: three components (x, y, z) for each of the three cables
  3. As three vertical forces, since cables can only support loads in tension vertically
  4. As three force vectors of equal magnitude, since the plate is in equilibrium
  5. As a single resultant force acting at the centroid of the triangle
Explanation: When analyzing cable-supported structures in statics, you need to understand how tension forces work and how to properly represent them in free-body diagrams. Cables can only pull (never push), and the tension force always acts along the cable's length in the direction away from the point of attachment. The correct approach is A: represent each cable tension as a single force vector with magnitude T acting along the cable's direction. This is fundamentally how tension works - it's a single force that pulls along the cable's axis. The magnitude may be unknown (and different for each cable), but the direction is clearly defined by the cable's geometry. B is unnecessarily complicated. While you could break each tension into x, y, and z components for calculation purposes, this isn't how you should represent the forces conceptually in the free-body diagram. The tension itself is the fundamental force, not its components. C reflects a major misconception. Cables don't only support vertical loads - they support loads along their length, regardless of orientation. A cable at any angle still provides tension along that angled direction. D assumes equal magnitudes, which is incorrect. Even though the plate is in equilibrium, this doesn't mean all tension forces are equal. The equilibrium equations (ΣFx=0\Sigma F_x = 0, ΣFy=0\Sigma F_y = 0, ΣFz=0\Sigma F_z = 0, and moment equilibrium) will determine the actual magnitudes, which are typically different for each cable. Study tip: Always remember that cables provide tension along their length, never perpendicular to it. Draw tension vectors as arrows pointing away from the attachment point, following the cable's direction.

Question 9

A rigid rod is supported by a universal joint at one end and a cable at the other end. The rod lies in a horizontal plane and supports a vertical downward force at its midpoint. In the free-body diagram, what reactions does the universal joint provide?

  1. Three force components: two horizontal and one vertical
  2. Two force components in the horizontal plane only
  3. Three force components and two moment components (correct answer)
  4. Three force components and three moment components
  5. Two moment components about horizontal axes only
Explanation: When analyzing support reactions in 3D statics problems, you need to understand what constraints each type of support provides. A universal joint (also called a ball joint) allows rotation about all three axes but prevents translation in any direction. For this problem, the universal joint provides three force components (one in each coordinate direction: x, y, and z) since it prevents the rod from translating. The vertical component balances the downward load, while the horizontal components handle any horizontal forces or maintain equilibrium. The key insight is recognizing that this is a 3D problem, not 2D. Even though the rod lies horizontally and the load is vertical, the cable creates a 3D force system. The universal joint must provide moment reactions about two axes (typically the horizontal axes) because it's the only support that can resist moments in this configuration. The joint cannot provide a moment about the rod's longitudinal axis since it's designed to allow rotation about that axis. Option A is wrong because it ignores the moment components that the universal joint must provide. Option B incorrectly assumes this is purely a 2D problem and misses both the vertical force component and all moment components. Option D incorrectly suggests the joint provides three moment components, but a universal joint cannot resist rotation about all three axes—it specifically allows rotation about the rod's axis. Study tip: Always identify whether you're dealing with 2D or 3D equilibrium first. In 3D problems, count both force and moment components that each support can provide based on its physical constraints.

Question 10

A rigid body is in contact with a curved surface at point P. The surface is rough, and the normal to the surface at P makes an angle θ with the vertical. In the free-body diagram for the rigid body, how should the contact forces at P be oriented?

  1. Normal force at angle θ from vertical, friction force at angle (θ + 90°) from vertical (correct answer)
  2. Normal force vertical, friction force horizontal, regardless of surface orientation
  3. Both normal and friction forces at angle θ from vertical
  4. Normal force horizontal, friction force at angle θ from vertical
  5. Single contact force at angle φ from normal, where tan φ equals the coefficient of friction
Explanation: When analyzing contact forces on rigid bodies against curved surfaces, you must carefully consider how the surface geometry affects force directions. Contact forces always act according to the local surface properties at the point of contact, not according to global reference frames. The correct approach recognizes that the normal force always acts perpendicular to the surface at the contact point. Since the surface normal makes angle θ with the vertical, the normal force also acts at angle θ from vertical. The friction force, by definition, acts parallel to the surface at the contact point, which means it's perpendicular to the normal force. This places the friction force at angle (θ + 90°) from vertical, making answer A correct. Answer B is wrong because it ignores the surface orientation entirely. Normal and friction forces aren't magically aligned with global vertical and horizontal directions—they respond to local surface geometry. Answer C incorrectly assumes both forces act in the same direction, violating the fundamental principle that normal and friction forces are perpendicular to each other. Answer D arbitrarily assigns horizontal orientation to the normal force, which would only be correct if the surface happened to be vertical at point P. Study tip: Always remember that contact forces are "local" phenomena. The normal force is always perpendicular to the surface at the contact point, and friction is always parallel to that surface (perpendicular to the normal). Sketch the surface tangent and normal at the contact point first, then draw your forces relative to those local directions, not global coordinates.

Question 11

A rigid assembly consists of two bars welded together at a right angle. The assembly is supported by a ball-and-socket joint at one end of the first bar and by two cables attached to the end of the second bar. When drawing the free-body diagram for the entire assembly, what should be shown at the welded junction between the bars?

  1. Internal forces and moments that maintain the connection between the bars
  2. External reaction forces where the weld interfaces with the environment
  3. Nothing, since the junction is internal to the body being analyzed (correct answer)
  4. A moment about the weld axis to represent the rigidity of the connection
  5. Force components that transfer loads between the two bars
Explanation: When analyzing rigid bodies in statics, understanding what to include in your free-body diagram is crucial. The key principle is that free-body diagrams show only external forces and moments acting on the system you're analyzing. Since you're drawing a free-body diagram for the entire assembly (both bars together), the welded junction is an internal connection within your system. Internal forces and moments exist at this junction to maintain the rigid connection, but they don't appear on the free-body diagram because they're internal to the body being analyzed. This is why answer C is correct—you show nothing at the welded junction. Let's examine why the other options are wrong. Answer A incorrectly suggests showing internal forces and moments. While these forces do exist physically, they're internal to your chosen system and thus don't belong on the free-body diagram. Answer B mischaracterizes the weld as interfacing with the environment—but the weld is completely internal since both bars are part of your system. Answer D suggests showing only a moment, but even if you were to show something at the junction (which you shouldn't), you'd need to show all internal forces and moments, not just one component. Remember this distinction: if you're analyzing the entire assembly, connections between parts of that assembly are internal. If you were analyzing just one bar separately, then forces at the weld would be external to that individual bar. Always clearly define your system boundary first—this determines what's internal versus external.

Question 12

A truss member AB is connected to other members at both ends through pin joints. When isolating member AB for analysis, which forces should appear on its free-body diagram?

  1. Only the axial force along the member's centerline, since pins cannot transmit moments
  2. Pin reaction forces at each end, each having both horizontal and vertical components (correct answer)
  3. Axial forces at both ends plus the weight of the member acting at its centroid
  4. Shear and moment reactions at both pin connections to maintain local equilibrium
  5. The resultant of all forces from connected members, applied at the geometric center
Explanation: When analyzing truss members, you're dealing with two-force members connected by pin joints. The key insight is understanding what pin joints can and cannot transmit, and how this affects your free-body diagram. Pin joints allow rotation but prevent translation in any direction. This means they can exert forces in both horizontal and vertical directions at the connection point, but they cannot transmit moments (since rotation is free). When you isolate member AB, you must show all external forces acting on it. Answer B is correct because each pin connection can exert reaction forces with both horizontal and vertical components. These reactions represent how the rest of the truss pushes or pulls on member AB at each end. You don't know the direction or magnitude of these components initially – that's what your equilibrium equations will solve for. Answer A is incorrect because while it's true that pins can't transmit moments, this doesn't mean only axial forces exist. The pin reactions have horizontal and vertical components that may not align with the member's axis. Answer C is wrong because it suggests axial forces are separate from pin reactions. The pin reactions ARE the forces acting on the member – they just happen to be in horizontal/vertical components rather than axial form. Answer D is incorrect because pins cannot transmit moments. Shear and moment reactions occur at fixed connections, not pins. Remember: For any member with pin connections, always start your free-body diagram with horizontal and vertical force components at each pin. You can resolve these into axial and transverse components later if needed.

Question 13

A uniform rigid beam is simply supported (pin at left end, roller at right end) and carries a point load and a distributed load. When constructing the free-body diagram, which of the following correctly represents the distributed load?

  1. As its resultant force acting at the geometric center of the loaded region
  2. As its resultant force acting at the centroid of the load distribution (correct answer)
  3. As the actual distributed load along the beam, without replacing it with a resultant
  4. As two concentrated forces at the ends of the loaded region
  5. As a distributed load with intensity equal to the average load value
Explanation: When analyzing distributed loads in statics, you need to understand how to properly represent them for equilibrium calculations. A distributed load creates a continuous force per unit length along a beam, but for analysis purposes, we replace it with an equivalent concentrated force. The key principle is that this equivalent force must have the same magnitude and create the same moment effect as the original distributed load. The magnitude equals the area under the load distribution curve, and this resultant force acts at the centroid of the load distribution - the point where the distributed load would balance if it were a physical shape. Option B is correct because the centroid of the load distribution is the proper location for the resultant force. This ensures that both the force magnitude and the moment arm are correctly represented in your equilibrium equations. Option A confuses the geometric center (midpoint of the loaded region) with the centroid of the load distribution. These are only the same for uniform loads - for non-uniform loads like triangular distributions, the centroid shifts toward the heavier side. Option C, while technically accurate for visualization, is impractical for calculations. Free-body diagrams are meant to simplify analysis by replacing distributed loads with equivalent concentrated forces. Option D incorrectly splits the load into end forces, which would create entirely different moment effects and lead to wrong results. Remember: always replace distributed loads with their resultant at the centroid of the load shape, not the geometric midpoint of the loaded region. This distinction becomes crucial when dealing with non-uniform load distributions.

Question 14

A rigid ladder leans against a smooth vertical wall and rests on a rough horizontal floor. A person stands on the ladder at a known position. For the free-body diagram of the ladder, what forces act at the wall contact point?

  1. A horizontal force away from the wall and a vertical friction force
  2. A horizontal force toward the wall and a vertical force supporting part of the weight
  3. Only a horizontal force away from the wall, since the wall is smooth (correct answer)
  4. No forces, since the smooth wall cannot support the ladder
  5. A force perpendicular to the ladder at the contact angle
Explanation: When analyzing contact forces in statics problems, the key principle is that smooth surfaces can only exert normal forces perpendicular to the contact surface, while rough surfaces can exert both normal and friction forces. Since the wall is described as smooth, it can only push the ladder away from itself with a horizontal normal force. The wall cannot exert any vertical forces because smooth surfaces have no friction. Think of trying to lean against a frictionless wall — it can only push you horizontally away from it, never upward or downward. The horizontal force points away from the wall (toward the ladder) because the ladder is pushing against the wall, and by Newton's third law, the wall pushes back on the ladder with equal magnitude in the opposite direction. Option A is incorrect because smooth walls cannot exert vertical friction forces — friction requires surface roughness. Option B incorrectly suggests the wall provides vertical support and shows a fundamental misunderstanding of smooth surface behavior. Option D is wrong because smooth walls can still exert normal forces perpendicular to their surface; "smooth" doesn't mean "no contact forces," just "no friction forces." Remember this pattern: smooth surfaces only give normal forces perpendicular to the contact surface, while rough surfaces can provide both normal and friction components. When you see "smooth" in a statics problem, immediately eliminate any answer choices that include friction or forces parallel to that smooth surface.

Question 15

A uniform beam is supported by a pin at point A and a roller at point B. The beam supports a concentrated load P at its midpoint and a uniformly distributed load w along its entire length. When drawing the free-body diagram for this beam, which of the following statements about the reaction forces is correct?

  1. The pin reaction at A has only a vertical component since the roller at B provides horizontal constraint
  2. The pin reaction at A has both horizontal and vertical components, while the roller reaction at B has only a vertical component (correct answer)
  3. Both the pin and roller reactions have horizontal and vertical components to maintain equilibrium
  4. The roller reaction at B has both horizontal and vertical components since it must resist the distributed load
  5. The pin reaction at A has only a horizontal component while the roller provides all vertical support
Explanation: When analyzing support reactions in statics, you need to understand how different support types constrain motion. This determines which force components each support can provide. A pin support can resist motion in any direction, so it provides both horizontal and vertical reaction components. A roller support, however, can only resist motion perpendicular to the rolling surface - typically just vertical motion for a horizontal surface. For this beam, the pin at A must provide both horizontal (AxA_x) and vertical (AyA_y) components because no other support can resist horizontal forces. The roller at B can only provide a vertical component (ByB_y) since it cannot resist horizontal motion. This makes option B correct. Option A is wrong because it misunderstands the pin's role - the pin doesn't lose its horizontal capacity just because a roller is present. In fact, the pin must provide horizontal resistance precisely because the roller cannot. Option C incorrectly assumes rollers can provide horizontal forces. Rollers are specifically designed to allow motion parallel to the surface while preventing perpendicular motion. Option D makes the same error as C, suggesting the roller can resist horizontal forces from the distributed load. The distributed load's horizontal effects (if any) must be resisted by the pin, not the roller. Remember this key principle: identify what each support type can and cannot do before drawing your free-body diagram. Pins resist all directions, rollers resist only perpendicular to the rolling surface, and any unrestrained directions must be handled by other supports.

Question 16

A rigid frame consists of two members connected by a pin joint at point C. When drawing separate free-body diagrams for each member, which principle governs the forces at the pin connection?

  1. The pin forces on each member must be equal in magnitude and direction to maintain connection
  2. The pin forces on each member must be equal in magnitude but opposite in direction per Newton's third law (correct answer)
  3. The pin forces depend on the relative stiffness of each member and may have different magnitudes
  4. The pin connection transmits only moment between members, not forces, since it allows rotation
Explanation: At a pin connection between two members, Newton's third law requires that the force exerted by the pin on member 1 is equal and opposite to the force exerted by the pin on member 2. This ensures force equilibrium at the pin itself and is fundamental to analyzing multi-member structures.

Question 17

A rigid door is supported by two hinges aligned along a vertical axis. The door is subjected to a horizontal force applied at the door handle. When drawing the free-body diagram, how should the hinge reactions be represented?

  1. Each hinge provides three force components and three moment components
  2. Each hinge provides two horizontal force components only, since vertical loads are balanced by both hinges
  3. The upper hinge provides horizontal reactions, the lower hinge provides the vertical reaction
  4. Each hinge provides horizontal force components; vertical reactions are statically indeterminate (correct answer)
  5. Each hinge provides two horizontal force components and one moment about the hinge axis
Explanation: When analyzing hinge-supported structures in statics, you need to understand how constraints affect the forces and whether all reactions can be determined from equilibrium equations alone. For a door with two hinges on a vertical axis, each hinge can resist forces perpendicular to the hinge axis (horizontal forces) but together they share the vertical load. Since both hinges are aligned vertically, you have three equilibrium equations available: two horizontal force equilibrium equations and one moment equilibrium about any point. However, you have four unknown reaction components: horizontal reactions at each hinge (2 unknowns) plus vertical reactions at each hinge (2 more unknowns). The horizontal reactions can be determined because the applied horizontal force and moment create a determinate system for horizontal equilibrium. But the vertical reactions cannot be uniquely determined from statics alone - the total vertical reaction (supporting the door's weight) is known, but how this load splits between the upper and lower hinges depends on factors like hinge stiffness, door deflection, and manufacturing tolerances. Option A is wrong because hinges don't provide moment resistance about the hinge axis - they're pin connections. Option B incorrectly assumes vertical reactions are automatically balanced and ignores that horizontal reactions are still needed. Option C incorrectly suggests you can arbitrarily assign vertical reaction to one hinge - this isn't justified by statics principles. Study tip: Remember that statically indeterminate problems occur when you have more unknown forces than available equilibrium equations. Always count your unknowns versus your equations to identify indeterminacy.

Question 18

Refer to the figure. A rigid plate is supported by three short compression members (struts) that can be assumed to act as two-force members. When drawing the free-body diagram for the plate, how should the forces from these struts be represented?

  1. As three forces, each acting along the centerline of its respective strut toward the plate (correct answer)
  2. As three forces, each acting along the centerline of its respective strut away from the plate
  3. As force components at each connection point, with both normal and tangential components
  4. As three vertical forces only, since struts primarily resist vertical loads
  5. As moments about each connection point, since struts resist rotation of the plate
Explanation: Two-force members (struts in compression) can only exert forces along their centerlines. Since these are compression members supporting the plate, they push on the plate, so the forces act toward the plate along each strut's centerline. Choice B would be correct for tension members. Choice C incorrectly decomposes forces that must act along specific directions. Choice D incorrectly restricts forces to vertical direction. Choice E incorrectly represents strut forces as moments.

Question 19

Refer to the figure. A rigid beam is supported by a pin at A and a link BC that connects the beam to a fixed support at C. When drawing the free-body diagram for the beam only, how should the force from link BC be represented?

  1. As a force acting along the beam from B toward A
  2. As a force acting along link BC from B toward C (correct answer)
  3. As a force acting along link BC from C toward B
  4. As horizontal and vertical force components at point B
  5. As a moment about point B, since the link prevents rotation
Explanation: Link BC is a two-force member that can only exert forces along its centerline. When isolating the beam, we show the effect of the link on the beam. If the link is in tension, it pulls on the beam at point B in the direction from B toward C. If in compression, it pushes in the opposite direction. The direction depends on the loading, but the force acts along the link. Choice A incorrectly shows force along the beam. Choice C shows the force in the wrong direction for typical loading. Choice D unnecessarily decomposes a force with known direction. Choice E incorrectly represents the link force as a moment.

Question 20

Refer to the diagram. A rigid L-bracket is welded to a vertical post at point O. The bracket supports a vertical load P at point A. In the free-body diagram for the bracket alone, what reactions appear at the welded connection?

  1. A vertical reaction force equal to P, since horizontal equilibrium is automatically satisfied
  2. A vertical reaction force and a reaction moment about point O
  3. Two reaction force components and a reaction moment about point O (correct answer)
  4. Three reaction force components, since this is a 3D welded connection
  5. Only a reaction moment, since the applied load P creates no net force on the system
Explanation: A welded connection (fixed support) can resist forces in any direction and moments about any axis in the plane. The vertical load P requires a vertical reaction force for force equilibrium. The load P also creates a moment about point O, requiring a reaction moment. Additionally, the geometry may require a horizontal reaction force component for complete equilibrium. Choice A ignores the moment equilibrium. Choice B omits the horizontal force component. Choice D assumes 3D when the problem appears planar. Choice E incorrectly ignores force equilibrium.