Word Problems to FBDs - Statics

1. A 2-meter uniform rod weighing 80 N is pinned at one end to a wall and supported by a cable attached to the other end. The rod makes a 40° angle with the horizontal, and the cable makes a 70° angle with the horizontal. When writing the moment equilibrium equation about the pin for the free body diagram, which forces create moments about this point?
   Answer: The weight of the rod and the cable tension only When solving moment equilibrium problems, you need to identify which forces create moments about your chosen reference point. A force only creates a moment if it has a perpendicular distance (moment arm) from the reference point. For this pinned rod system, let's examine each force about the pin point. The weight of the rod (80 N) acts downward at the center of the rod (1 meter from the pin), creating a moment arm of $1 \cos(40°)$ meters horizontally. The cable tension acts at the rod's end (2 meters from the pin) at a 70° angle, also creating a moment about the pin. These two forces have clear moment arms and generate rotational effects about the pin. However, the reaction forces at the pin (both horizontal and vertical components) act directly through the pin point itself. Since their line of action passes through the reference point, their moment arm is zero, so they create no moment about the pin. Answer A incorrectly includes the pin reaction forces, which don't create moments since they act through the reference point. Answer C wrongly excludes the weight while including the vertical pin reaction, missing that weight has a significant moment arm. Answer D incorrectly suggests only weight creates a moment, ignoring the cable tension's clear moment arm from the pin. The correct answer is B: only the weight and cable tension create moments about the pin. Study tip: When writing moment equations, systematically check each force's line of action. If it passes through your reference point, it creates zero moment and can be ignored in your moment equilibrium equation.
2. A triangular plate is subjected to three forces at its vertices. Before drawing the free body diagram, it's determined that the plate is a three-force member in equilibrium. This means that when constructing the FBD and writing equilibrium equations, which condition must be satisfied?
   Answer: The three forces must be concurrent (their lines of action meet at a single point) When analyzing three-force members in equilibrium, you're dealing with a fundamental principle of statics that applies to any rigid body subjected to exactly three forces. The key insight is understanding how these forces must be arranged geometrically to satisfy equilibrium conditions. For a three-force member in equilibrium, the lines of action of all three forces must intersect at a single point - they must be concurrent. This is because if the forces weren't concurrent, there would be a net moment about the intersection point of any two forces, violating rotational equilibrium. When forces are concurrent, the sum of moments about their intersection point is automatically zero, and you only need to ensure the vector sum of forces equals zero for complete equilibrium. Looking at the incorrect options: Choice A assumes equal magnitudes and 120° spacing, which only applies to very specific symmetric loading cases and isn't a general requirement. Choice C describes a special case with parallel and perpendicular forces, but this arrangement doesn't guarantee equilibrium and isn't the general rule for three-force members. Choice D refers to force polygons used in graphical analysis - while forces in equilibrium do form a closed polygon when arranged head-to-tail, this is true for any number of forces in equilibrium, not specifically three-force members. Remember this rule: whenever you identify a three-force member, immediately look for or construct the point where all three force lines intersect. This concurrency condition is your key to solving three-force member problems efficiently and is often tested on statics exams.
3. A compound structure consists of a rigid bar ABC pinned at A, with a cable connecting points B and D. Point C rests on a roller support, and point D is a fixed anchor point above and to the right of the structure. A vertical load P acts downward at point B. The bar ABC is straight and horizontal. To analyze this structure, free body diagrams must be drawn for appropriate components. Which approach correctly identifies the bodies to be isolated?
   Answer: Draw one FBD for the entire bar ABC including the cable force at B, since the bar is rigid and acts as a single body Since ABC is described as a rigid bar, it should be treated as a single body. The cable BD exerts an external force on the bar at point B. This gives a statically determinate system with three equilibrium equations for the bar and known cable tension direction. Choice A unnecessarily separates the rigid bar into segments. Choice C incorrectly treats the cable force as both external and internal. Choice D treats joints as separate bodies, which is incorrect for rigid body analysis.
4. A 50-N picture frame hangs on a wall from a single nail. The frame is supported by a wire that makes a 140° angle at the top of the frame (70° on each side from vertical). When drawing the free body diagram for the picture frame, how should the wire tension forces be represented?
   Answer: As two tension forces of equal magnitude, each making 70° with the vertical When analyzing forces in static equilibrium problems involving cables or wires, you must carefully identify how the constraint is physically connected to the object. The key insight is that a single continuous wire creates tension forces at each point where it contacts the frame. In this problem, the wire forms a V-shape over the top of the frame, contacting it at two separate points. At each contact point, the wire exerts a tension force along its direction. Since the wire makes a 140° angle at the top (70° on each side from vertical), you have two tension forces: one pulling at 70° from vertical on the left side, and another pulling at 70° from vertical on the right side. These forces have equal magnitude because it's the same continuous wire under uniform tension. Choice A incorrectly treats this as a single force, ignoring the fact that the wire contacts the frame at two distinct points. Choice C makes an angle error—if each side is 70° from vertical, that's 90° - 70° = 20° from horizontal, but the problem states the wire makes 70° with vertical, not horizontal. Choice D completely misrepresents the geometry by suggesting a single 140° force, which doesn't correspond to any physical reality of how the wire contacts the frame. For equilibrium, the vertical components of these two tension forces must sum to 50 N: $2T\cos(70°) = 50\text{ N}$. Remember: always draw tension forces along the direction of each rope or wire segment at every contact point with your object.
5. A ladder of weight $W$ leans against a smooth vertical wall at angle $\theta$ with the horizontal ground. The ground provides friction with coefficient $\mu$. A person of weight $P$ stands at distance $d$ from the bottom of the ladder, which has total length $L$. For the free body diagram of the ladder-person system, which statement correctly identifies all external forces?
   Answer: Normal force from ground, friction force from ground, normal force from wall, weight $W$ at ladder center, and weight $P$ at distance $d$ from ladder bottom For the ladder-person system FBD, external forces include the three reaction forces from supports and the two separate weights acting at their respective locations. Choice A incorrectly combines the weights. Choice B incorrectly includes internal forces between person and ladder (internal to the system). Choice D incorrectly represents weights as moments rather than forces and adds a non-existent contact moment from the smooth wall.
6. A 100-N block rests on a 30° inclined plane. The coefficient of static friction between the block and plane is 0.4. When drawing the free body diagram for this block, which statement correctly describes how the friction force should be represented?
   Answer: As a force of unknown magnitude directed up the incline, with magnitude to be determined from equilibrium When analyzing static friction problems, remember that static friction is a reactive force—it adjusts its magnitude and direction to prevent motion, up to its maximum limit of $\mu_s N$. The correct approach is B: represent friction as a force of unknown magnitude directed up the incline, then determine its actual value from equilibrium conditions. Since the block is stationary, you'll resolve forces parallel to the incline: the component of weight down the incline ($100\sin(30°) = 50$ N) must be balanced by static friction up the incline. Therefore, friction equals exactly 50 N—less than its maximum possible value of $\mu_s N = 0.4 × 100\cos(30°) ≈ 34.6$ N... wait, this reveals the block would actually slide since 50 N > 34.6 N, but the principle remains: determine friction from equilibrium first. A incorrectly assumes friction always acts at its maximum value $\mu_s N$. Static friction only reaches this maximum when the object is on the verge of sliding—most of the time it's less. C misapplies the cosine function. The $\cos(30°)$ already appears in calculating the normal force $N$, not as an additional factor in the friction formula. D incorrectly splits friction into components. Friction acts as a single force parallel to the contact surface—it has no component perpendicular to the incline. Study tip: In static problems, always assume friction adjusts to maintain equilibrium, then check if this required friction exceeds $\mu_s N$. If it does, the object slides and you'll need kinetic friction instead.
7. A wheel of radius 0.5 m and weight 200 N encounters a curb of height 0.2 m. A horizontal force P is applied at the center of the wheel to just roll it over the curb. When drawing the free body diagram at the instant the wheel is about to go over the curb, which forces should be included and where do they act?
   Answer: Weight at center, applied force P at center, normal force at contact point with curb edge only When analyzing a wheel rolling over a curb, you need to identify the critical moment: when the wheel is just about to go over, meaning it's balanced at the tipping point where it only contacts the curb edge. At this instant, the wheel has lost contact with the ground and is pivoting about the curb edge. The forces acting are: the weight (200 N) acting downward at the center of mass, the applied horizontal force P at the center, and a normal force from the curb acting upward at the contact point (curb edge). Since the wheel is momentarily balanced and not sliding at the curb contact, there's no friction force needed at that point. Answer B correctly identifies these three forces: weight at center, applied force P at center, and normal force at the curb contact only. Answer A incorrectly includes ground contact forces. When the wheel is about to go over the curb, it has already lifted off the ground, so no normal or friction forces exist at the ground contact point. Answer C includes normal forces at both ground and curb contacts, but again, the wheel has lost ground contact at the critical tipping moment. Answer D adds friction at the curb contact, but at the instant of tipping, the wheel isn't sliding relative to the curb edge, so friction isn't required for equilibrium. Study tip: For rolling problems, always identify the exact instant being analyzed. "About to go over" means the wheel has lifted off the ground and contacts only the obstacle edge - this eliminates any forces from the original contact surface.
8. When the worker applies a 800-N horizontal force and the block remains stationary, which free body diagram correctly represents the forces acting on the block?
   Answer: Weight 1960 N downward, normal force 1960 N upward, applied force 800 N rightward, friction force 800 N leftward When analyzing forces on a stationary object, you need to apply the fundamental principle of static equilibrium: all forces must balance perfectly in each direction. For this concrete block in equilibrium, let's examine each force systematically. The weight is $W = mg = 200 \text{ kg} \times 9.8 \text{ m/s}^2 = 1960 \text{ N}$ downward. Since there's no vertical acceleration, the normal force must equal the weight: 1960 N upward. The applied force is 800 N rightward as given. Here's the key insight: since the block remains stationary despite the applied force, static friction must act to prevent motion. For equilibrium, the friction force must exactly balance the applied force, so it equals 800 N leftward. You don't need to calculate maximum possible friction here—you only need the actual friction required for equilibrium. Option A correctly shows this balance: 1960 N up/down forces cancel, and 800 N left/right forces cancel. Option B incorrectly uses 1176 N for friction (which happens to equal the maximum static friction $\mu_s N = 0.6 \times 1960$), but actual friction only equals what's needed to maintain equilibrium. Option C uses 784 N for friction ($\mu_k N = 0.4 \times 1960$), incorrectly applying kinetic friction when the block isn't moving. Option D makes two errors: wrong normal force and claiming no friction exists on a stationary object. Remember: static friction adjusts to whatever value is needed (up to its maximum) to maintain equilibrium. Always check that forces balance in both directions for stationary objects.
9. A pulley system consists of a fixed pulley at the top and a movable pulley supporting a 200 N weight. A rope passes over the fixed pulley, under the movable pulley, and back over the fixed pulley to a 100 N counterweight. The rope is assumed massless and the pulleys are frictionless. When drawing the free body diagram for the movable pulley alone, which forces act on it?
   Answer: Two upward tension forces of 100 N each from the rope segments, downward force of 200 N from the suspended weight, and its own weight if not negligible The movable pulley has rope segments on both sides, each carrying the same tension (100 N) due to the massless rope assumption. These provide two upward forces of 100 N each. The suspended 200 N weight acts downward, along with the pulley's own weight if not negligible. Choice B shows only one rope tension. Choice C incorrectly includes a reaction from the fixed pulley (no direct connection). Choice D incorrectly combines the rope tensions into a resultant.
10. A boom crane consists of a uniform 500-N boom hinged at the base and supported by a cable attached to its top end. The boom makes a 60° angle with the horizontal, and the cable makes a 30° angle with the horizontal. A 1000-N load hangs from the end of the boom. When setting up the equilibrium equations from the free body diagram of the boom, which equation correctly represents the sum of forces in the horizontal direction?
    Answer: $T \cos(30°) - R_x = 0$ When analyzing equilibrium problems with angled forces, you need to carefully decompose each force into its horizontal and vertical components based on the angle each force makes with the horizontal. For the horizontal force equilibrium, you have two forces acting on the boom: the tension force T in the cable (making 30° with horizontal) and the horizontal reaction force Rx at the hinge. The cable tension has a horizontal component of $T \cos(30°)$ pointing to the right (away from the boom), while the hinge reaction Rx points to the left to maintain equilibrium. Setting the sum of horizontal forces equal to zero gives: $T \cos(30°) - R_x = 0$. Looking at the wrong answers: Option B ($T \cos(30°) + R_x = 0$) incorrectly assumes both forces point in the same direction, which would mean no equilibrium force to balance the cable's horizontal pull. Option C ($T \sin(30°) - R_x = 0$) uses sine instead of cosine, confusing the horizontal component with the vertical one—remember that cosine gives you the adjacent side (horizontal) when measuring from the horizontal axis. Option D ($T \cos(60°) - R_x = 0$) uses the boom's angle instead of the cable's angle, mixing up which structural element you're analyzing. The key strategy: always identify which angle belongs to which force, then use cosine for horizontal components and sine for vertical components when measuring angles from the horizontal. Draw clear force diagrams and label each angle carefully to avoid these common mix-ups.
11. A 150-lb person stands on a ladder leaning against a frictionless wall. The ladder makes a 70° angle with the horizontal floor, which has sufficient friction to prevent slipping. For the free body diagram of the ladder-person system, which forces must be included to properly analyze the equilibrium?
    Answer: Weight of system, normal force from floor, friction force from floor, and normal force from wall only When analyzing equilibrium problems involving contact forces, you need to identify all external forces acting on your system while considering the specific conditions at each contact point. For this ladder-person system, start by identifying the contact points: the floor (which has friction) and the wall (which is frictionless). At the floor, two force components can exist: a normal force perpendicular to the surface and a friction force parallel to the surface. Since the floor "has sufficient friction to prevent slipping," the friction force is present and necessary for equilibrium. At the frictionless wall, only a normal force can exist—friction forces cannot develop on frictionless surfaces. The weight of the system acts downward at the center of gravity. Choice B correctly includes all necessary forces: the system's weight, normal force from floor, friction force from floor, and normal force from wall. Choice A incorrectly adds a friction force from the wall, but the problem explicitly states the wall is frictionless—this force cannot exist. Choice C makes a fundamental error by claiming friction forces are "internal." Friction from the floor is an external force acting on the ladder-person system and is essential for preventing the ladder from sliding. Choice D mentions "two normal forces from floor," which misunderstands how contact forces work. At any single contact surface, there's only one normal force (perpendicular to the surface), though it can be combined with a friction force (parallel to the surface). Remember: always read contact conditions carefully—frictionless surfaces can only provide normal forces, while surfaces with friction can provide both normal and friction forces.
12. A uniform 200 kg beam of length 8 m is supported by a pin at its left end and a roller at a point 6 m from the left end. A 500 N downward force is applied at the right end of the beam. When constructing the free body diagram for this beam, which forces and moments must be included for a complete static analysis?
    Answer: Vertical reaction at pin, horizontal reaction at pin, vertical reaction at roller, applied 500 N force, and weight of beam acting at the geometric center A pin support provides both horizontal and vertical reactions, a roller provides only vertical reaction, and the uniform beam's weight acts at its center. Choice B incorrectly includes a moment reaction at the pin (pins don't resist moments) and shows distributed weight instead of concentrated. Choice C incorrectly includes horizontal reaction at the roller. Choice D oversimplifies by combining reactions and incorrectly places the reaction moment.
13. A uniform 20-kg bar is suspended horizontally by two vertical cables attached at points 1 m and 3 m from the left end of the 4-m bar. An additional 50-N downward force is applied at the right end. When setting up the equilibrium equations from the free body diagram, what is the correct sum of moments equation about the left end of the bar?
    Answer: $-T_1(1) - T_2(3) + 196(2) + 50(4) = 0$ When analyzing moment equilibrium problems, you need to carefully consider both the magnitude and direction of each moment about your chosen reference point. The key is establishing a consistent sign convention and applying it systematically. Let's set up the moment equation about the left end of the bar. Using the standard convention where counterclockwise moments are positive and clockwise moments are negative: The tension forces $T_1$ and $T_2$ act upward, creating clockwise moments about the left end, so they contribute $-T_1(1)$ and $-T_2(3)$ respectively. The weight of the bar ($20 \times 9.8 = 196$ N) acts downward at the center (2 m from left end), creating a counterclockwise moment of $+196(2)$. The 50-N force at the right end also acts downward, creating a counterclockwise moment of $+50(4)$. This gives us: $-T_1(1) - T_2(3) + 196(2) + 50(4) = 0$, which matches answer C. Answer A incorrectly treats the tension moments as positive (counterclockwise). Answer B makes the same sign error with tensions and also incorrectly assigns positive signs to the downward forces. Answer D has mixed sign errors, treating $T_2$ as negative while $T_1$ is positive, and reversing the signs on the applied forces. Study tip: Always establish your sign convention first (counterclockwise positive is standard), then systematically apply it. Upward forces create clockwise moments when acting to the right of your reference point, while downward forces create counterclockwise moments. Draw the moment arrows on your diagram to visualize the directions clearly.
14. A compound beam structure consists of beam AB simply supported at A and B, with beam BC cantilevered from B to C. A 500-N downward force is applied at point C. When analyzing beam BC separately, which forces and moments should be included in its free body diagram at the connection point B?
    Answer: Horizontal force, vertical force, and bending moment to represent the full constraint When analyzing compound beam structures, you must carefully consider how forces and moments transfer between connected segments. The key insight is that when beam BC is cantilevered from point B, the connection must provide complete constraint against all possible movements and rotations. At point B, beam BC would naturally want to translate horizontally and vertically, as well as rotate under the 500-N load at C. Since BC is rigidly connected to AB (forming a compound beam), the connection at B must resist all these tendencies. This requires three internal actions: a horizontal force (to prevent horizontal translation), a vertical force (to resist the downward load), and a bending moment (to prevent rotation of BC about point B). Option A is incorrect because it ignores the horizontal force component and the moment needed for equilibrium. The 500-N downward force at C creates both vertical force and moment reactions at B. Option B represents a pin connection assumption, which would allow rotation at B - but compound beams have rigid connections that transfer moments. Option C misses the horizontal force component, which is necessary even though the applied load is purely vertical (the geometry and constraints create horizontal reactions). The complete free body diagram of segment BC must show all three components at B: horizontal force, vertical force, and bending moment. Study tip: For compound beam problems, always remember that rigid connections between beam segments transfer three things: horizontal force, vertical force, and moment. Don't let the direction of applied loads fool you into thinking certain reaction components don't exist.
15. A cantilever beam of length $3L$ is fixed at its left end and free at its right end. The beam supports a uniformly distributed load of intensity $w$ per unit length over the middle third of its span (from distance $L$ to distance $2L$ from the fixed end). When drawing the free body diagram for the entire beam, how should the distributed load and fixed support reactions be represented?
    Answer: Equivalent concentrated load $wL$ at distance $1.5L$ from fixed end, vertical reaction at fixed end, horizontal reaction at fixed end, and moment reaction at fixed end (magnitudes to be determined) The distributed load can be replaced by its resultant $wL$ acting at the centroid of the loaded region (1.5L from the fixed end). The fixed support provides three reaction components: vertical, horizontal, and moment, with magnitudes determined from equilibrium. Choice A retains the distributed load unnecessarily and assumes specific reaction values. Choice B omits the horizontal reaction and assumes the moment value. Choice D incorrectly extends the distributed load over the entire beam length.
16. A horizontal beam is supported by a pin at point A and a roller at point B. The beam carries a uniformly distributed load of 50 lb/ft over its entire 8-foot length and a concentrated load of 200 lb at a point 3 feet from the left end. When drawing the free body diagram for this beam, which of the following best describes the reaction components that should be included?
    Answer: One horizontal and two vertical reaction components: horizontal and vertical at the pin, vertical only at the roller When analyzing support reactions in statics, you must understand how different types of supports constrain motion. A pin support prevents translation in both horizontal and vertical directions but allows rotation, while a roller support prevents translation in only one direction (typically vertical) and allows both rotation and horizontal movement. For this beam, the pin at point A can resist forces in both horizontal and vertical directions, so it provides two reaction components: $A_x$ (horizontal) and $A_y$ (vertical). The roller at point B can only resist vertical forces, providing one reaction component: $B_y$ (vertical). This gives you three total reaction components to solve for using the three equilibrium equations available for a 2D static system. Option A is incorrect because it assumes the roller provides both horizontal and vertical reactions. Rollers cannot resist horizontal forces - that's their defining characteristic. Option C misunderstands how distributed loads work; while the 50 lb/ft load is spread across the beam, you still represent the roller reaction as a single vertical force. The distributed load affects the magnitude of reactions, not the number of reaction components. Option D ignores the pin's ability to resist horizontal forces. Even though all applied loads are vertical, the pin must still be capable of providing horizontal reaction if needed for equilibrium. Remember this key principle: the number of reaction components depends solely on the support types, not on the applied loads. Always identify your support types first, then determine how many directions each support can resist forces.
17. A 150 N sign is suspended from a horizontal beam by two cables. The left cable makes a 60° angle with the horizontal beam, while the right cable makes a 45° angle with the horizontal beam. Both cables are in tension and the system is in static equilibrium. When drawing the free body diagram for the junction point where both cables meet the sign, which statement correctly describes the force analysis required?
    Answer: Three forces act at the junction: two tension forces directed along the cables away from their attachment points on the beam, and the weight of the sign acting vertically downward At the junction point, three forces act: two tension forces in the cables (directed away from the beam attachments toward the junction) and the weight. The tension forces pull away from the attachment points, not toward them. Choice A incorrectly shows tensions toward attachment points. Choice C confuses the FBD with component analysis. Choice D incorrectly combines the cable tensions into a resultant.
18. A sign weighing 80 N is suspended from a horizontal beam by two cables. The left cable makes a 60° angle with the horizontal, and the right cable makes a 45° angle with the horizontal. Both cables attach to the same point on the sign. When setting up the free body diagram for the sign, which equilibrium equation correctly represents the horizontal force balance?
    Answer: $T_1 \cos(60°) - T_2 \cos(45°) = 0$ When analyzing suspended objects in statics, you need to carefully track force directions and apply equilibrium conditions. For this sign suspended by two cables at different angles, the key is understanding how tension forces resolve into horizontal and vertical components. Start by drawing a free body diagram of the sign. The weight (80 N) acts downward, while tensions $T_1$ and $T_2$ act along their respective cables. Since the left cable angles upward at 60° from horizontal, its horizontal component is $T_1 \cos(60°)$ pointing leftward (negative direction). The right cable angles upward at 45°, so its horizontal component is $T_2 \cos(45°)$ pointing rightward (positive direction). For horizontal equilibrium, the sum of all horizontal forces must equal zero. This gives us: $-T_1 \cos(60°) + T_2 \cos(45°) = 0$, which rearranges to $T_1 \cos(60°) - T_2 \cos(45°) = 0$. This matches answer B. Answer A incorrectly adds both horizontal components, ignoring that they point in opposite directions. Answer C uses sine functions instead of cosine, confusing horizontal and vertical components—sine gives you the vertical parts of the tensions. Answer D shows the correct relationship but as an equation rather than the equilibrium condition that equals zero. Remember this pattern: for equilibrium problems, always establish a coordinate system first, resolve forces into components using the correct trigonometric functions (cosine for horizontal, sine for vertical), and set the sum of forces in each direction equal to zero.
19. A rigid frame consists of two members AB and BC connected by a pin at B. Member AB is horizontal with a pin support at A, and member BC is inclined at 30° below horizontal with a roller support at C. A 1000 N vertical load acts downward at point B. When drawing separate free body diagrams for each member using the standard convention where internal forces are drawn away from the joint on each member, what forces act at the internal pin connection B?
    Answer: Member AB has horizontal and vertical forces at B pointing away from joint B; member BC has equal and opposite horizontal and vertical forces at B pointing toward joint B Using the standard convention where internal forces are drawn away from the joint on each member, Newton's third law requires that the forces be equal in magnitude but opposite in direction. This means if member AB shows forces pointing away from joint B, then member BC must show equal and opposite forces pointing toward joint B. Choice B violates Newton's third law. Choice C reverses the assumed convention. Choice D incorrectly assumes forces align with member directions.
20. A uniform 8-foot plank weighing 40 lb is used as a seesaw with the fulcrum placed 3 feet from the left end. A 60-lb child sits on the left end while a 45-lb child sits on the right end. The system is not in equilibrium. When drawing the free body diagram for the plank, how should the reaction at the fulcrum be represented?
    Answer: As an upward force to be determined from vertical force equilibrium When analyzing free body diagrams in statics problems, you must distinguish between what forces exist versus what forces you know the magnitude of. The fulcrum reaction is a real force that must be included in your diagram, but its magnitude isn't predetermined—it depends on achieving equilibrium. The correct approach is B: represent the fulcrum reaction as an upward force to be determined from vertical force equilibrium. In statics, reaction forces adjust to whatever magnitude is needed to satisfy equilibrium conditions. You'll find this force by applying $\sum F_y = 0$, which gives you: $R - 40 - 60 - 45 = 0$, so $R = 145 \text{ lb}$. A assumes you already know the reaction equals 145 lb, but this puts the cart before the horse. While this happens to be the correct magnitude, you shouldn't assume it when drawing the free body diagram—you determine it through equilibrium analysis. C incorrectly suggests the reaction only balances the plank's weight. This ignores that the children's weights are transmitted through the plank to the fulcrum. The fulcrum supports the entire system. D misunderstands what "unbalanced system" means. The rotational imbalance (different moments about the fulcrum) doesn't create horizontal forces on the fulcrum. The fulcrum is a pin support that only provides vertical reaction. Study tip: Always draw reaction forces as unknowns in your free body diagrams, then solve for their magnitudes using equilibrium equations. Don't assume their values based on intuition—let the math determine them.

Start this quiz