FBD: Rigid Body - Statics

1. A rigid triangular bracket supports a horizontal load $P$ at its apex and is mounted to a wall using two pin connections at its base. When analyzing the free-body diagram of this bracket, which statement about the reaction forces is most accurate?
   Answer: The system has four unknown reaction components but only three equilibrium equations, making it statically indeterminate With two pin supports, there are four unknown reaction components (two at each pin). However, there are only three equilibrium equations available in 2D (ΣFx = 0, ΣFy = 0, ΣM = 0). Since 4 unknowns > 3 equations, the system is statically indeterminate and requires additional analysis methods.
2. 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?
   Answer: The pin reaction at A has both horizontal and vertical components, while the roller reaction at B has only a vertical component 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 ($A_x$) and vertical ($A_y$) components because no other support can resist horizontal forces. The roller at B can only provide a vertical component ($B_y$) 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.
3. 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?
   Answer: Two reaction force components and a reaction moment about point O 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.
4. 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?
   Answer: As its resultant force acting at the centroid of the load distribution 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.
5. 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?
   Answer: Normal force at angle θ from vertical, friction force at angle (θ + 90°) from vertical 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.
6. 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?
   Answer: The pin forces on each member must be equal in magnitude but opposite in direction per Newton's third law 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.
7. 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?
   Answer: As three force vectors, each acting along the direction of its respective cable away from the attachment point 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.
8. 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?
   Answer: As three forces, each acting along the centerline of its respective strut toward the plate 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.
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?
   Answer: Three force components and two moment components 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.
10. 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?
    Answer: Pin reaction forces at each end, each having both horizontal and vertical components 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.
11. 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?
    Answer: Only a horizontal force away from the wall, since the wall is smooth 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.
12. A uniform circular disk of radius $R$ and weight $W$ rests against a vertical wall and on a horizontal floor. Both surfaces are rough with friction coefficients $\mu_w$ and $\mu_f$ respectively. In the free-body diagram, how many force components act on the disk?
    Answer: Five forces: weight, normal forces from both surfaces, and friction forces from both surfaces The disk experiences five forces: weight W (downward), normal force from wall Nw (horizontal), friction force from wall fw (vertical), normal force from floor Nf (vertical), and friction force from floor ff (horizontal). Both surfaces are rough, so friction forces exist at both contact points.
13. 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?
    Answer: Two perpendicular reaction forces and a reaction moment about point O 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."
14. 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?
    Answer: Both a normal force perpendicular to the surface and a friction force parallel to the surface 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.
15. 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?
    Answer: A concentrated force equal to the total weight, acting at the center of mass of the member 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.
16. A uniform beam of weight $W$ is supported by a pin at point A and rests against a smooth wall at point B. The beam makes an angle $\theta$ with the horizontal. When drawing the free-body diagram for this beam, which statement about the reaction forces is correct?
    Answer: The reaction at A has both horizontal and vertical components, while the reaction at B is purely horizontal At pin support A, both horizontal and vertical reaction components exist (Ax and Ay). At the smooth wall B, friction is absent, so the reaction force is purely normal to the wall surface (horizontal only). This is a key distinction between different support types that students often confuse.
17. 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?
    Answer: As a force acting along link BC from B toward C 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.
18. 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?
    Answer: Only a horizontal reaction component, since the collar can slide vertically but not horizontally 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.
19. 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?
    Answer: Nothing, since the junction is internal to the body being analyzed 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.
20. A rigid compound beam consists of two segments connected by an internal hinge. When constructing the free-body diagram for the entire beam system versus individual beam segments, which statement is correct?
    Answer: Individual segment free-body diagrams show internal hinge forces, while the entire system diagram does not When analyzing compound beams with internal hinges, you need to understand how free-body diagrams change depending on what you're isolating. The key insight is that internal forces only appear when you "cut" through connections to isolate parts of a system. When you draw a free-body diagram of the entire compound beam system, the internal hinge forces are truly internal to the system boundary. Since free-body diagrams only show external forces acting on the isolated body, these hinge forces don't appear. You'll see only the external supports, loads, and reactions. However, when you create separate free-body diagrams for individual beam segments, you must "cut" through the hinge connection. This exposes the internal hinge forces, which now become external forces on each segment's diagram. The hinge forces appear as equal and opposite pairs on the two segments (Newton's third law), but they do appear on the individual diagrams. Option A reverses this relationship incorrectly. Option B is wrong because while the hinge forces do appear as equal and opposite pairs on individual segments, they don't appear on both the system and segment diagrams—only on the segment diagrams. Option C incorrectly suggests hinges don't affect force transmission; hinges actually prevent moment transfer while allowing force transmission. Study tip: Remember that internal forces become visible only when you "cut" through connections. Practice identifying what's inside versus outside your system boundary—this skill applies to any structure analysis problem.

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