Statics Quiz: Word Problems To Fbds

20 questions · exam conditions

0:00

Word Problems To FbdsQuestion 1 of 20

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?

Two horizontal and two vertical reaction components, one at each support

One horizontal and two vertical reaction components: horizontal and vertical at the pin, vertical only at the roller

Three vertical reaction components: one at the pin and two at the roller to account for the distributed load

Two vertical reaction components only, since the loads are all vertical

One horizontal and one vertical reaction component at the pin only, since the roller provides no constraint

← Back to quizzes

Statics Quiz

Statics Quiz: Word Problems To Fbds

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

What this quiz covers

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

How to use this quiz

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

All questions

Question 1

Two horizontal and two vertical reaction components, one at each support
One horizontal and two vertical reaction components: horizontal and vertical at the pin, vertical only at the roller (correct answer)
Three vertical reaction components: one at the pin and two at the roller to account for the distributed load
Two vertical reaction components only, since the loads are all vertical
One horizontal and one vertical reaction component at the pin only, since the roller provides no constraint

Explanation: 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.

Question 2

Two identical 50-N blocks are connected by a light rope passing over a frictionless pulley. One block rests on a horizontal surface with coefficient of kinetic friction 0.3, while the other hangs vertically. The system is sliding with the hanging block descending. When drawing separate free body diagrams for each block, how should the rope tension be represented?

The same magnitude T for both blocks, directed along the rope away from each block (correct answer)
Different magnitudes T₁ and T₂ for each block, since they have different accelerations
The same magnitude T for both blocks, but directed toward the pulley for both
Tension T for the hanging block only, since the horizontal block experiences friction instead of tension
Different magnitudes because the rope changes direction at the pulley, creating different forces

Explanation: When analyzing pulley systems in statics, the key principle is that a massless, inextensible rope has the same tension throughout its entire length. This fundamental concept applies regardless of what forces act on the objects the rope connects. In this problem, both blocks are connected by the same continuous rope passing over a frictionless pulley. Since the rope is light (massless) and the pulley is frictionless, the tension must be uniform throughout the rope. Each block experiences this same tension magnitude T, and the tension always pulls away from each block along the direction of the rope. For the hanging block, tension acts upward; for the horizontal block, tension acts horizontally toward the pulley. Option A correctly identifies that both blocks experience the same tension magnitude T directed away from each block along the rope. Option B is wrong because different accelerations don't create different rope tensions - the rope constraint ensures both blocks have the same acceleration magnitude, and tension remains constant throughout the rope regardless. Option C incorrectly suggests tension direction - tension always pulls on an object, so it acts away from each block, not toward the pulley from both blocks. Option D misunderstands force analysis - the horizontal block experiences both friction (from the surface) and tension (from the rope) simultaneously; these are separate forces acting on the same object. Remember this key rule: in problems with massless ropes and frictionless pulleys, tension magnitude is always the same throughout the rope, regardless of the different forces (like friction) acting on connected objects.

Question 3

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?

Weight of system, normal force from floor, friction force from floor, normal force from wall, and friction force from wall
Weight of system, normal force from floor, friction force from floor, and normal force from wall only (correct answer)
Weight of system, normal force from floor, and normal force from wall only, since friction forces are internal
Weight of system, two normal forces from floor (vertical and horizontal), and normal force from wall
Weight of system, normal force from floor, friction force from floor, normal force from wall, and moment reaction from wall

Explanation: 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.

Question 4

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?

$T_1(1) + T_2(3) - 196(2) - 50(4) = 0$
$T_1(1) + T_2(3) + 196(2) + 50(4) = 0$
$-T_1(1) - T_2(3) + 196(2) + 50(4) = 0$ (correct answer)
$T_1(1) - T_2(3) - 196(2) + 50(4) = 0$
$-T_1(1) + T_2(3) - 196(2) - 50(4) = 0$

Explanation: 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.

Question 5

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?

All three forces must be equal in magnitude and separated by 120° angles
The three forces must be concurrent (their lines of action meet at a single point) (correct answer)
Two forces must be parallel and the third must be perpendicular to them
The forces must form a closed triangle when arranged head-to-tail
The algebraic sum of moments about any point must equal the sum of the force magnitudes

Explanation: 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.

Question 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?

As a force of magnitude $\mu_s N = 0.4N$ directed up the incline, where N is the normal force
As a force of unknown magnitude directed up the incline, with magnitude to be determined from equilibrium (correct answer)
As a force of magnitude $\mu_s N \cos(30°)$ directed parallel to the incline
As two components: one parallel and one perpendicular to the incline, each of magnitude $0.4N$
As a force that should not be included until it's determined whether the block is sliding

Explanation: 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.

Question 7

A cantilever beam is fixed at the wall and extends 3 meters horizontally. It supports a triangular distributed load that varies linearly from 0 at the free end to 60 N/m at the wall. When replacing this distributed load with an equivalent point load for the free body diagram, what should be the magnitude and location of this equivalent load?

90 N located 1.0 m from the wall (correct answer)
90 N located 1.5 m from the wall
180 N located 1.0 m from the wall
180 N located 2.0 m from the wall
270 N located 1.0 m from the wall

Explanation: When you encounter distributed loads on beams, you need to replace them with equivalent point loads to simplify your analysis. This requires finding both the total force (area under the load diagram) and where that force acts (centroid of the load diagram). For a triangular distributed load, the total force equals the area of the triangle. With a base of 3 meters and height of 60 N/m, the area is

\frac{1}{2} \times 3 \text{ m} \times 60 \text{ N/m} = 90 \text{ N}

. The equivalent force acts at the centroid of the triangle. For any triangle, the centroid is located at one-third the base length from the side with maximum height. Since the maximum load is at the wall, the centroid is

\frac{1}{3} \times 3 \text{ m} = 1.0 \text{ m}

from the wall. Looking at the wrong answers: Choice B correctly calculates the 90 N force but incorrectly places it at 1.5 m from the wall—this would be the centroid if the load were uniformly distributed, not triangular. Choice C doubles the force to 180 N, which would happen if you calculated the area as base times height instead of using the triangle formula. Choice D combines both errors: wrong force magnitude and incorrect centroid location. Remember that for triangular loads, the centroid is always at one-third the base from the maximum load end, and the equivalent force is half the maximum load times the base length. These formulas are essential for beam analysis problems.

Question 8

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?

The weight of the rod, the cable tension, and the reaction forces at the pin
The weight of the rod and the cable tension only (correct answer)
The cable tension and the vertical reaction at the pin only
The weight of the rod only, since it acts at the center of mass
All forces must be included to satisfy moment equilibrium

Explanation: 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.

Question 9

A construction worker needs to move a heavy 200-kg concrete block. The block sits on a rough horizontal surface with coefficient of static friction 0.6 and coefficient of kinetic friction 0.4. The worker applies a horizontal pushing force.

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?

Weight 1960 N downward, normal force 1960 N upward, applied force 800 N rightward, friction force 800 N leftward (correct answer)
Weight 1960 N downward, normal force 1960 N upward, applied force 800 N rightward, friction force 1176 N leftward
Weight 1960 N downward, normal force 1960 N upward, applied force 800 N rightward, friction force 784 N leftward
Weight 1960 N downward, normal force 800 N upward, applied force 800 N rightward, no friction force since block is stationary
Weight 1960 N downward, normal force 2760 N upward, applied force 800 N rightward, friction force 800 N leftward

Explanation: 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.

Question 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?

$T \cos(30°) - R_x = 0$ (correct answer)
$T \cos(30°) + R_x = 0$
$T \sin(30°) - R_x = 0$
$T \cos(60°) - R_x = 0$
$T \cos(30°) - R_x - 1000 = 0$

Explanation: 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.

Question 11

A uniform L-shaped bracket is mounted to a wall with two bolts arranged vertically. The bracket supports a 150-N downward load at its free end, which is 0.8 m horizontally from the wall. The bolts are spaced 0.3 m apart vertically. When drawing the free body diagram for this bracket, how should the bolt reactions be represented?

As vertical forces only at each bolt location, since the load is vertical
As horizontal and vertical force components at each bolt location (correct answer)
As a single resultant force at the center point between the two bolts
As vertical forces at the bolt locations plus a horizontal force at the upper bolt only
As force components at the bolt locations plus a couple moment to resist the applied moment

Explanation: When analyzing bolted connections in statics, you must consider all possible forces that the connection can resist. Bolts can transfer forces in multiple directions, and the applied loads will generally create reactions in more than one direction. For this L-shaped bracket, the 150-N downward load creates both vertical forces and a moment about the bolt connection. The horizontal distance of 0.8 m means the load generates a moment of

150 \times 0.8 = 120

N·m that tries to overturn the bracket. This overturning moment must be resisted by the bolts, which can only happen if they develop both horizontal and vertical reaction forces. The horizontal components create a force couple (tension in one bolt, compression in the other) to resist the overturning moment, while the vertical components balance the applied load. Answer choice A is incorrect because representing only vertical forces ignores the overturning moment that must be equilibrated. Choice C fails because replacing the actual bolt locations with a single resultant point doesn't capture how the bolts actually resist the applied moment through their spacing. Choice D is wrong because both bolts must develop horizontal forces to create the necessary force couple—you can't resist an overturning moment with a horizontal force at just one location. The correct answer is B: both bolts must have horizontal and vertical force components to maintain equilibrium. Study tip: For bolted connections, always check if applied loads create moments about the connection. If they do, the bolts must develop forces perpendicular to their spacing to resist that moment.

Question 12

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?

As a single upward force equal to 50 N at the point where the wire contacts the frame
As two tension forces of equal magnitude, each making 70° with the vertical (correct answer)
As two tension forces of equal magnitude, each making 20° with the horizontal
As a single tension force making 140° with the horizontal
As two horizontal force components and one vertical component representing the wire support

Explanation: 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.

Question 13

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?

Weight at center, applied force P at center, normal force at contact point with ground, friction at contact point with ground
Weight at center, applied force P at center, normal force at contact point with curb edge only (correct answer)
Weight at center, applied force P at center, normal forces at both ground contact and curb contact points
Weight at center, applied force P at center, normal force at curb contact, friction at curb contact
Weight at center, applied force P at center, contact forces at ground and curb with both normal and friction components

Explanation: 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.

Question 14

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?

As an upward force equal to the total weight of the system (145 lb)
As an upward force to be determined from vertical force equilibrium (correct answer)
As an upward force of 40 lb since only the plank contacts the fulcrum directly
As both vertical and horizontal reaction components since the system is unbalanced
As an upward force and a moment reaction to account for the unbalanced moments

Explanation: 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.

Question 15

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?

$T_1 \cos(60°) + T_2 \cos(45°) = 0$
$T_1 \cos(60°) - T_2 \cos(45°) = 0$ (correct answer)
$T_1 \sin(60°) - T_2 \sin(45°) = 0$
$T_1 \cos(60°) = T_2 \cos(45°)$
$T_1 \sin(60°) + T_2 \sin(45°) = 80$

Explanation: 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.

Question 16

A rigid frame consists of two members connected at a joint. Member AB is horizontal and member BC is inclined. When analyzing this structure, the connection at point B is modeled as a pin joint. In the free body diagrams of the individual members, how should the forces at joint B be represented?

As the same force components on both members, with the same directions
As equal magnitude force components on both members, but opposite in direction (correct answer)
As different magnitude forces since the members have different orientations
As a moment on one member and force components on the other
As force components only on the member that was cut first during analysis

Explanation: When analyzing structures with pin joints in statics, you're applying Newton's third law at the connection points. A pin joint can transmit forces in any direction but cannot transmit moments, making the force analysis straightforward once you understand the action-reaction principle. At joint B, the two members exert forces on each other. According to Newton's third law, these forces must be equal in magnitude but opposite in direction. If member AB exerts a force on member BC at point B, then member BC exerts an equal and opposite force on member AB at the same point. This is true for both the horizontal and vertical force components at the joint. Answer B correctly captures this fundamental principle - the force components have equal magnitudes but point in opposite directions on each member's free body diagram. Answer A violates Newton's third law by suggesting forces act in the same direction, which would mean the joint is pushing on both members simultaneously without any reaction force. Answer C incorrectly assumes that member orientation affects force magnitudes at the joint - while the geometry affects how forces are distributed through each member, the forces the members exert on each other at the connection point are still action-reaction pairs. Answer D misunderstands pin joint behavior; pins cannot transmit moments, only forces. Remember this pattern: at any pin connection in your free body diagrams, the force components on adjacent members are always equal and opposite. This applies regardless of member orientations or loading conditions - Newton's third law is non-negotiable in statics problems.

Question 17

A 10-kg box sits on a 30° inclined conveyor belt that is moving upward at constant velocity. The coefficient of kinetic friction between the box and belt is 0.2. If the box moves with the belt (no relative motion), what forces should appear on the free body diagram of the box?

Weight, normal force, kinetic friction down the incline, and belt tension force
Weight, normal force, and static friction up the incline only (correct answer)
Weight, normal force, kinetic friction up the incline, and air resistance
Weight, normal force, static friction up the incline, and centrifugal force
Weight, normal force perpendicular to incline, and component of weight parallel to incline only

Explanation: When analyzing forces on objects moving with inclined surfaces, you must carefully distinguish between static and kinetic friction based on relative motion between the surfaces. Since the box moves with the belt at constant velocity, there's no relative motion between the box and belt surface. This means any friction present must be static friction, not kinetic friction. The box is in equilibrium (zero acceleration), so all forces must balance. The forces acting on the box are: weight (

mg

) acting vertically downward, normal force perpendicular to the inclined surface, and static friction acting parallel to the surface. Since the component of weight down the incline would cause the box to slide without friction, static friction must act up the incline to maintain equilibrium. No other forces are needed for this equilibrium condition. Option A incorrectly includes kinetic friction and introduces an unnecessary "belt tension force." Kinetic friction only exists when surfaces slide relative to each other, which isn't happening here. The belt's tension is an internal force within the belt system, not a force acting on the box. Option C makes the same kinetic friction error as A and adds air resistance, which isn't significant enough to include in typical statics problems unless specifically mentioned. Option D correctly identifies static friction but incorrectly includes centrifugal force. Centrifugal force only appears in problems involving circular motion, not straight-line inclined motion. Remember: friction type depends on relative motion between surfaces. No sliding means static friction; sliding means kinetic friction. Always check what's actually moving relative to what.

Question 18

A uniform circular disk of weight W and radius R rolls without slipping down a rough inclined plane of angle θ. When drawing the free body diagram for this rolling disk, which forces should be included and how should they be positioned?

Weight W at center, normal force N at contact point, friction f at contact point, all acting at their actual points of application (correct answer)
Weight W at center, normal force N at center, friction f at center, all forces moved to the center of mass
Weight W at center, normal force N perpendicular to incline through center, friction f parallel to incline through center
Weight W at center, contact force C at contact point with components N and f
Weight W at center, normal force N at contact point, friction f at contact point, but friction shown in direction opposing motion

Explanation: When analyzing rolling motion problems in statics, you need to carefully consider where forces actually act on the object, as this affects both equilibrium equations and the constraint relationships for rolling without slipping. For a disk rolling down an incline, three forces act on the system: the weight W acts at the center of mass (the geometric center), the normal force N acts perpendicular to the incline at the contact point, and the friction force f acts parallel to the incline at the contact point. These forces must be drawn at their actual points of application to properly analyze the rolling motion, making option A correct. Option B incorrectly moves all forces to the center of mass. While you can sometimes translate forces in pure translational problems, rolling motion involves rotation, so you must preserve the moment arms created by forces acting away from the center of mass. Moving N and f to the center would eliminate their rotational effects. Option C places the normal and friction forces at the center rather than the contact point. This misrepresents where these contact forces actually occur and would give incorrect moment calculations about any point other than the center. Option D treats the contact forces as a single resultant force C. While mathematically possible, this approach obscures the individual normal and friction components needed for rolling analysis, where you typically need to consider the no-slip constraint

a = \alpha R

. Study tip: Always draw contact forces (normal, friction) at the actual contact point in rolling problems, and remember that the point of application matters for rotational motion analysis.

Question 19

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?

Vertical reaction force only, since beam AB provides only vertical support
Horizontal and vertical reaction forces, but no moment, since B is a pin connection
Vertical reaction force and bending moment, since BC is fixed to AB at B
Horizontal force, vertical force, and bending moment to represent the full constraint (correct answer)
Only a bending moment, since the cantilever requires moment support at the fixed end

Explanation: 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.

Question 20

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?

Normal force from ground, friction force from ground, normal force from wall, weight $W + P$ acting at the combined center of mass of the system
Normal force from ground, friction force from ground, normal force from wall, weight $W$ at ladder center, weight $P$ at person's location, and internal contact forces between person and ladder
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 (correct answer)
Normal force from ground, friction force from ground, normal force from wall, moment due to weight $W$ , moment due to weight $P$ , and contact moment from wall

Explanation: 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.