Statics and Dynamics Quiz: Free Body Diagrams

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Free Body DiagramsQuestion 1 of 9

A rigid frame consists of a horizontal bar pinned to a wall at point A and supported by a two-force link (a slender rod pinned at both ends) connecting point B on the bar to point C on the wall. Point B is to the right of A, and point C is directly above A, making the two-force link oriented diagonally. A vertical load $F$ is applied at the free end D of the bar, which extends beyond B.

When drawing the FBD of the horizontal bar alone (isolated from the wall and the two-force link), which of the following correctly describes the force exerted by the two-force link on the bar at point B?

A force perpendicular to the bar's horizontal axis at point B, because the link is oriented diagonally and only its component perpendicular to the bar contributes to moment equilibrium about A, making the parallel component irrelevant on this FBD.

Two perpendicular force components

B_x

and

B_y

at point B, because the link is pinned at B and a pin connection always introduces two unknown force components regardless of the nature of the connected member.

A single vertical force at B equal in magnitude to the applied load

F

, because the link must carry all of the vertical equilibrium load and connects vertically from B to C directly above A.

A single force acting along the line connecting B and C (i.e., along the axis of the link), which may be tension or compression, because a two-force member can only exert a force along its own axis and this is the contact force the link exerts on the bar at B.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Free Body Diagrams

Practice Free Body Diagrams in Statics and Dynamics 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 Free Body Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

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

All questions

Question 1

A force perpendicular to the bar's horizontal axis at point B, because the link is oriented diagonally and only its component perpendicular to the bar contributes to moment equilibrium about A, making the parallel component irrelevant on this FBD.
Two perpendicular force components $B_x$ and $B_y$ at point B, because the link is pinned at B and a pin connection always introduces two unknown force components regardless of the nature of the connected member.
A single vertical force at B equal in magnitude to the applied load $F$ , because the link must carry all of the vertical equilibrium load and connects vertically from B to C directly above A.
A single force acting along the line connecting B and C (i.e., along the axis of the link), which may be tension or compression, because a two-force member can only exert a force along its own axis and this is the contact force the link exerts on the bar at B. (correct answer)

Explanation: Whenever you isolate a member in a free body diagram, you must ask: what kind of member is exerting the force, and what does that constrain? Here, the key concept is the two-force member — a member loaded only at two points, with no other forces or couples acting on it. This is the central idea being tested. A two-force member can only be in equilibrium if the two forces it carries are equal, opposite, and collinear — meaning they must act along the line connecting its two pins. The slender rod connecting B and C is pinned at both ends with no other loads applied, so it is a classic two-force member. When you draw the FBD of the bar alone, the rod exerts a single force on point B directed along the line from B to C. It may be tension (pulling B toward C) or compression (pushing B away from C), but its direction is fixed by the link's geometry. This makes D correct. A is wrong because it invents a rule about perpendicular components and moment equilibrium — no such rule applies here. The force is along B-to-C, full stop, regardless of moment calculations. B describes what you'd use for a general pin connecting two non-two-force members. If you don't yet know the member is a two-force member,

B_x

and

B_y

seems safe — but recognizing the two-force member lets you replace both unknowns with one, along a known direction. C is wrong because the link is not vertical; it runs diagonally from B to C, so the force it carries is diagonal, not purely vertical. Study tip: Before assigning unknowns at any pin, always check whether the connected member qualifies as a two-force member. If it does, you collapse two unknowns into one with a known direction — a powerful simplification on any statics problem.

Question 2

A pin-connected truss has members AB, BC, and AC forming a triangle. Joint A is pinned to a wall (providing $A_x$ and $A_y$ ), joint C rests on a roller on a horizontal surface (providing a vertical reaction $C_y$ ), and joint B has an external load $\vec{P}$ applied at an angle. A student draws the FBD of member AB alone (not the entire truss).

Which of the following correctly identifies what must appear on the FBD of member AB in isolation?

The external pin reactions $A_x$ and $A_y$ at joint A, the applied load $\vec{P}$ at joint B, and the internal axial force in member BC at joint B directed along BC — but not the force in member AC, because AC connects joints A and C and does not attach to joint B.
The external pin reactions $A_x$ and $A_y$ at joint A, the roller reaction $C_y$ at joint C, the applied load $\vec{P}$ at joint B, and forces in all other members — because isolating one member requires showing every force the truss exerts on it, including reactions at non-adjacent joints.
The external pin reactions $A_x$ and $A_y$ at joint A, the applied load $\vec{P}$ at joint B, the internal force in member BC at joint B directed along BC, and the internal force in member AC at joint A directed along AC — because both BC and AC connect to the joints that bound member AB. (correct answer)
The applied load $\vec{P}$ at joint B and single axial resultants at joints A and B only — with no separate pin reaction components, because member AB behaves as a two-force member and two-force members carry only axial forces along their length.

Explanation: When isolating a single member in a truss problem, your job is to identify every force acting directly on that member at each of its joints. Member AB is bounded by joint A and joint B — so you must account for every connection at those two joints specifically. At joint A, the external wall pin supplies reactions

A_x

and

A_y

. But joint A is also where member AC attaches to member AB. When you cut AB free from the rest of the truss, member AC pulls or pushes on joint A with an internal axial force directed along AC. Similarly, at joint B, member BC connects to AB, so it exerts an internal axial force directed along BC — plus the external load

\vec{P}

is applied there. That gives you the complete picture:

A_x

A_y

, the force from AC at joint A, the force from BC at joint B, and

\vec{P}

at joint B. This is exactly what C describes. A is wrong because it ignores the force member AC exerts on joint A. AC does attach to joint A — that's a direct contact point on member AB, so the interaction force must appear on the FBD. B is wrong because it includes

C_y

, the roller reaction at joint C. Joint C does not touch member AB at all, so that reaction is irrelevant to AB's free body. D exploits a tempting misconception: member AB itself is a two-force member, but you're not drawing AB's internal load — you're drawing all external forces acting on it at its joints, which include pin components and other member forces. Study tip: Always ask "which joints bound the member I'm isolating, and what touches the truss at each of those joints?" Every connection at those joints — external reactions and adjacent member forces — belongs on the FBD.

Question 3

Two blocks, Block 1 (mass $m_1$ ) and Block 2 (mass $m_2$ ), are stacked vertically. Block 1 rests on a rough horizontal surface, and Block 2 sits on top of Block 1. A horizontal force $P$ is applied to Block 2 only. The system is in static equilibrium. The coefficient of static friction between Block 1 and the floor is $\mu_1$ , and between Block 1 and Block 2 is $\mu_2$ .

A student draws an FBD of Block 1 alone. Which of the following correctly lists every distinct external force that must appear on that FBD?

Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting opposite to $P$ , normal force $N_2 = m_2 g$ downward from Block 2, and friction force from Block 2 acting on Block 1 in the direction of $P$ — by Newton's third law, since friction on Block 2 from Block 1 opposes $P$ , friction on Block 1 from Block 2 acts in the direction of $P$ . (correct answer)
Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting horizontally, and normal force $N_2$ downward from Block 2 — but no friction from Block 2, because friction only acts at the floor interface where slipping is imminent.
Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting horizontally, normal force from Block 2 equal to $(m_1+m_2)g$ downward, and friction force from Block 2 acting in the same direction as $P$ .
Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting horizontally, normal force $N_2$ downward from Block 2, the applied force $P$ acting horizontally, and friction from Block 2 acting horizontally on Block 1 — because $P$ is transmitted through contact and must appear on every FBD in the system.

Explanation: When drawing a Free Body Diagram, your golden rule is this: isolate the body and show every force exerted on it by external agents — nothing more, nothing less. For Block 1, the external agents are: the floor below, Block 2 above, and gravity. Answer A correctly captures all of these. Block 1's own weight

m_1 g

acts downward. The floor exerts an upward normal force

N_1

and a horizontal friction force opposing

P

(since the whole system would slide in the direction of

P

without floor friction). Block 2 pushes down on Block 1 with

N_2 = m_2 g

(just Block 2's weight, since Block 2 is in vertical equilibrium). Critically, Block 2 also exerts a friction force on Block 1 — and by Newton's Third Law, since Block 1's friction on Block 2 opposes

P

, Block 2's friction on Block 1 must point in the direction of

P

. A is correct. Answer B incorrectly omits friction from Block 2. Friction acts at any interface where surfaces are in contact and a tangential force exists — not only where slipping is "imminent." Block 2 is being pushed by

P

, so it absolutely exerts friction on Block 1. Answer C uses

N_2 = (m_1 + m_2)g

, which is the floor's normal force — not the contact force Block 2 exerts on Block 1. That's a classic mix-up between

N_1

and

N_2

. Answer D includes

P

on Block 1's FBD. Force

P

is applied directly to Block 2 only; it never directly touches Block 1 and cannot appear on Block 1's FBD. Study tip: Always ask, "What objects physically touch this body?" — those contacts generate the forces. Newton's Third Law pairs live at interfaces, so every contact surface produces two forces: one on each body.

Question 4

A rigid body is subjected to a system of coplanar forces. After careful analysis, an engineer concludes that the FBD shows the body is in equilibrium. A colleague then points out that the engineer forgot to include a couple (pure moment) of magnitude $M_0$ that is applied to the body.

If the couple $M_0$ is now added to the FBD, which of the following statements about the revised equilibrium conditions is most accurate?

Adding the couple $M_0$ will change all three equilibrium equations — $\sum F_x = 0$ , $\sum F_y = 0$ , and $\sum M = 0$ — because a couple has both force components and a moment, and all three are nonzero for a general couple.
Adding the couple $M_0$ affects only the moment equilibrium equation $\sum M = 0$ and not the force equilibrium equations $\sum F_x = 0$ or $\sum F_y = 0$ , because a couple consists of two equal and opposite forces whose net force resultant is zero, contributing only a net moment. (correct answer)
Adding the couple $M_0$ affects $\sum F_x = 0$ and $\sum M = 0$ but not $\sum F_y = 0$ , because the moment arm of the couple creates a horizontal force component but no vertical component in the standard 2D reference frame.
Adding the couple $M_0$ affects only $\sum F_y = 0$ and $\sum M = 0$ because couples, by convention in 2D statics, are always represented as vertical force pairs, contributing a vertical net force and a net moment.

Explanation: Whenever you see a question involving couples (pure moments) on a statics exam, the critical concept to recall is the fundamental definition: a couple consists of two forces that are equal in magnitude, opposite in direction, and separated by a perpendicular distance. Because those two forces are equal and opposite, they cancel each other out vectorially — their net force resultant is exactly zero. Mathematically,

\sum F_x = 0

and

\sum F_y = 0

are unaffected because the couple contributes nothing to either force component. What the couple does produce is a net moment,

M_0 = F \cdot d

, which is the whole point of a pure moment. This means only the moment equation

\sum M = 0

is disrupted, making B the correct answer. Since the original body was in equilibrium but the moment equation now has an unbalanced

M_0

, the revised system is no longer in equilibrium unless something else changes. Choice A is wrong because it incorrectly treats a couple as though it has nonzero force components — it does not. A couple has zero net force by definition, so

\sum F_x

and

\sum F_y

remain unchanged. Choice C is wrong for a similar reason: it fabricates a horizontal force component arising from the couple's moment arm, which is a misunderstanding of how couples are resolved. Choice D is wrong because it invents a convention that couples are "always vertical force pairs" — couples have no net force in any direction, vertical or otherwise. Your study tip: memorize the phrase "a couple has zero net force but nonzero net moment." On exam questions involving couples, immediately eliminate any answer that claims force equations are affected.

Question 5

A student draws an FBD of Block 1 alone. Which of the following correctly lists every distinct external force that must appear on that FBD?

Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting opposite to $P$ , normal force $N_2 = m_2 g$ downward from Block 2, and friction force from Block 2 acting on Block 1 in the direction of $P$ — by Newton's third law, since friction on Block 2 from Block 1 opposes $P$ , friction on Block 1 from Block 2 acts in the direction of $P$ . (correct answer)
Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting horizontally, and normal force $N_2$ downward from Block 2 — but no friction from Block 2, because friction only acts at the floor interface where slipping is imminent.
Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting horizontally, normal force from Block 2 equal to $(m_1+m_2)g$ downward, and friction force from Block 2 acting in the same direction as $P$ .
Weight $m_1 g$ downward, normal force $N_1$ upward from the floor, friction force from the floor acting horizontally, normal force $N_2$ downward from Block 2, the applied force $P$ acting horizontally, and friction from Block 2 acting horizontally on Block 1 — because $P$ is transmitted through contact and must appear on every FBD in the system.

Explanation: When drawing a free body diagram (FBD) of a single object, your job is to isolate that object and identify every force exerted on it by external agents — nothing more, nothing less. Newton's Third Law is the key tool here: for every contact interface, forces come in action-reaction pairs, so you must carefully track which direction each force points on which body. For Block 1's FBD, there are two contact interfaces: Block 1 with the floor, and Block 1 with Block 2. The floor exerts an upward normal force

N_1

and a friction force opposing the tendency of Block 1 to slide (opposite to

P

). Block 2 presses down on Block 1 with normal force

N_2 = m_2 g

. Critically, since Block 2 tends to slide in the direction of

P

relative to Block 1, friction on Block 2 from Block 1 opposes

P

— meaning by Newton's Third Law, friction on Block 1 from Block 2 acts in the direction of $P$ . Add Block 1's own weight

m_1 g

downward, and you have a complete, balanced FBD. This is exactly what A describes. B is wrong because it omits friction from Block 2 entirely — friction acts at every interface where surfaces interact, not just where slipping is "most imminent." C incorrectly states the normal force from Block 2 equals

(m_1 + m_2)g

; Block 2 only pushes down with its own weight,

m_2 g

. D incorrectly includes

P

on Block 1's FBD —

P

is applied directly to Block 2 only, not to Block 1. Study tip: Before drawing any FBD, physically ask: "What objects touch this body?" Each contact generates a normal force and potentially a friction force — account for both at every interface, and apply Newton's Third Law carefully to get directions right.

Question 6

A simply supported beam AB of length $L$ and negligible weight has a pin at A and a roller at B. The roller at B is oriented so that it sits on a surface inclined at 30° from the horizontal. A vertical load $Q$ is applied at the midpoint of the beam.

On the FBD of the beam, which description of the reaction at B is correct?

The reaction at B is a single force directed vertically upward, because rollers always provide a vertical reaction regardless of the surface on which they rest, and only the pin at A can provide a horizontal reaction component.
The reaction at B is a single force directed perpendicular to the inclined surface — i.e., at 60° from horizontal (or 30° from vertical) — because a roller can only exert a force normal to the surface on which it rolls, and this direction is perpendicular to the 30°-inclined surface. (correct answer)
The reaction at B is two force components: one perpendicular and one parallel to the inclined surface, because the roller contacts the inclined surface and must resist both normal and tangential components to maintain the beam in equilibrium.
The reaction at B is a single force directed at 30° from horizontal, parallel to the inclined surface, because the roller is constrained to move along the inclined surface and therefore can only push in the direction of motion — along, not perpendicular to, the surface.

Explanation: When analyzing reactions at supports, your first job is to identify what motion each support prevents — that determines the direction of its reaction force. A roller support is the simplest of all: it prevents translation in exactly one direction — perpendicular to the surface it rests on — and allows free movement along that surface. Because it cannot develop friction or tangential resistance, it produces a single force normal to the contact surface. When that surface is inclined at 30° from horizontal, the normal to it points 60° from horizontal (or equivalently, 30° from vertical). So the reaction at B is a single force in that perpendicular direction, confirming that B is correct. Choice A contains a common and dangerous misconception — that rollers always react vertically. This is only true when the roller rests on a horizontal surface. The reaction direction depends on the surface orientation, not some fixed property of rollers in general. Choice C would be appropriate for a pin or fixed support, which can resist forces in any direction. A frictionless roller, however, cannot develop a tangential (parallel) component; if it could, it wouldn't be a roller — it would behave like a pin. Choice D confuses the direction of permitted motion with the direction of the reaction force. These are exactly opposite: a roller moves freely along the surface precisely because it cannot push in that direction — it can only push perpendicular to it. Study tip: For any roller, ask yourself: "What direction is the surface?" Then rotate 90° — that's your reaction. Rollers react normal to their surface, always.

Question 7

A uniform slender rod of mass $m$ and length $L$ is hinged (pinned) to a ceiling at its upper end A. A horizontal force $H$ is applied at the rod's lower end, causing it to hang in static equilibrium at angle $\theta$ from the vertical. A student is constructing the FBD of the rod.

The student claims: 'Since this is a three-force body, I can represent the two pin reaction components at A as a single resultant force whose line of action must pass through the intersection of the weight's line of action and the horizontal force's line of action.' Which of the following best evaluates this claim?

The claim is correct in identifying a three-force body, but incorrect in the concurrency requirement: for a three-force body in equilibrium, the three forces must be parallel — not concurrent — and the pin reaction must therefore be horizontal to remain parallel to $H$ .
The claim is incorrect because the three-force body theorem requires all three forces to be applied at the same point; since the weight acts at the midpoint and $H$ acts at the lower end, the forces are not concurrent at the outset, so the theorem cannot be used for this rod.
The claim is incorrect because the ceiling pin is a fixed support that resists rotation as well as translation, so it provides three reaction components ( $A_x$ , $A_y$ , and a couple $M_A$ ) and cannot be represented as a single resultant force regardless of loading.
The claim is correct: the rod is a three-force body (weight at midpoint, horizontal force at lower end, pin reaction at A), and for equilibrium the three force lines of action must be concurrent. The pin's two components may be combined into one resultant passing through the intersection of the weight's and $H$ 's lines of action — this also determines the pin reaction's direction without additional calculation. (correct answer)

Explanation: Whenever you see a body with exactly three forces acting on it, the three-force body theorem should come to mind: for a rigid body in equilibrium under exactly three forces, those forces must either all be parallel or all pass through a single common point (concurrent). This question tests whether you understand both what qualifies a body as a three-force body and how the concurrency condition is applied. The rod here has exactly three forces: the weight

mg

acting downward at the midpoint, the horizontal force

H

at the lower end, and the pin reaction at A. Because the weight's line of action and

H

's line of action are not parallel, they intersect at some point P. For equilibrium to hold, the pin reaction at A must also pass through P — this is the concurrency requirement. Crucially, you can legitimately combine the pin's two scalar components (

A_x

and

A_y

) into a single resultant vector, and that resultant's direction is determined entirely by the geometry of P relative to A. No moment equations are needed to find its direction. That makes D correct. A is wrong because it confuses the two sub-cases of the theorem: three-force bodies in equilibrium require concurrency (or all forces parallel), not exclusively parallelism. B is wrong because "concurrent" in the theorem refers to the lines of action intersecting — not the points of application. Forces acting at different points can still satisfy the theorem. C is wrong because a simple pin (hinge) provides only two reaction components (

A_x

A_y

), not a couple; a fixed (clamped) support would add a moment reaction, but a pin does not. Study tip: Always distinguish between a pin support (two force components, no moment) and a fixed support (two force components plus a couple). Mixing these up is one of the most common FBD errors on statics exams.

Question 8

A rigid beam of length $2a$ is supported by three vertical wires: one at each end (A and C) and one at the midpoint (B). The beam is horizontal and carries a concentrated load $Q$ at a distance $a/2$ from end A. Each wire has the same cross-sectional area and is made of the same material, but wire B is shorter in length than wires A and C.

A student draws the FBD of the beam and writes the equilibrium equations, obtaining: $T_A + T_B + T_C = Q$ , $T_C(2a) - T_B(a) - Q(a/2) = 0$ (moments about A). The student then concludes that the system is statically determinate and solves for all three tensions. Which of the following best identifies the flaw in the student's FBD analysis?

The student's moment equation is written incorrectly: the moment of $T_B$ should have a negative sign because B is between A and C, so all wire tensions create counterclockwise moments about A, and the sign convention is violated in the written equation.
The student's FBD and equilibrium equations are both correct as written, and the system is indeed statically determinate with 2 equations and 3 unknowns because one equation is redundant — summing vertical forces and taking moments gives 2 independent equations for 3 unknowns, which is uniquely solvable if additional boundary conditions on wire lengths are applied.
The FBD is correct in identifying the three unknown tensions and the applied load, but the system is statically indeterminate to the first degree: there are 3 unknown tensions and only 2 independent equilibrium equations for a 2D beam in vertical loading, so the problem cannot be solved from equilibrium alone and requires a compatibility (deformation) equation incorporating the different wire lengths. (correct answer)
The FBD incorrectly omits the horizontal reactions at the wire attachment points; each wire, when carrying tension, also exerts a horizontal component on the beam due to the wires' inclination from vertical, and these horizontal forces must be included as additional unknowns before equilibrium can be applied.

Explanation: Whenever you see a beam supported by multiple redundant supports (more unknown reactions than available equilibrium equations), your first instinct should be to count unknowns versus equations before attempting a solution. For a 2D beam loaded purely in the vertical direction, you have exactly two independent equilibrium equations: sum of vertical forces (

\sum F_y = 0

) and sum of moments about any point (

\sum M = 0

). Here, the beam has three unknown tensions —

T_A

T_B

, and

T_C

— giving you 3 unknowns and only 2 equations. That's one more unknown than equations, making the system statically indeterminate to the first degree. The extra equation must come from a compatibility condition: the deformation of each wire must be geometrically consistent with the beam remaining rigid. Because wire B is shorter than A and C, it stretches differently under load, and that difference in stiffness is precisely what the compatibility equation captures. Answer C correctly identifies this flaw. Answer A is wrong because the moment equation as written is actually sign-consistent:

T_B

at the midpoint creates a clockwise moment about A (opposing

T_C

), so a negative sign is appropriate — the student's equation is not the error here. Answer B is wrong because 2 independent equations for 3 unknowns is not uniquely solvable from equilibrium alone. Calling one equation "redundant" is backwards — it's an unknown that's redundant, not an equation. Answer D is wrong because the problem states the wires are vertical, meaning they carry no horizontal component by definition. Study tip: Always count unknowns versus equilibrium equations first. If unknowns exceed equations, you need a compatibility (deformation) condition — this is the hallmark of statically indeterminate problems.

Question 9

A rigid L-shaped bracket is welded to a wall at point A (a fixed support) and has a cable attached at the outer corner B, running at 30° above horizontal to an anchor on the same wall above A. A concentrated load $F$ hangs vertically from the midpoint M of the horizontal arm. The horizontal arm has length $a$ and the vertical arm has length $b$ .

For the FBD of the L-shaped bracket as a single rigid body isolated from the wall, which statement correctly characterizes the reaction at the fixed wall support A?

The fixed support at A provides exactly two reaction components: a horizontal force $A_x$ and a vertical force $A_y$ , because the cable already supplies the moment resistance needed for equilibrium, making a reaction couple at A unnecessary.
The fixed support at A provides three reaction components: a horizontal force $A_x$ , a vertical force $A_y$ , and a reaction couple (moment) $M_A$ , because a fixed (cantilever) wall connection constrains translation in two directions and prevents rotation independently of what other loads are present. (correct answer)
The fixed support at A provides three reaction components: a horizontal force $A_x$ , a vertical force $A_y$ , and a reaction couple $M_A$ , but $M_A$ can be set to zero on the FBD as a simplification when the cable provides a non-zero moment about A, since any non-zero cable moment proves rotational equilibrium is already satisfied.
The fixed support at A provides three reaction components: a horizontal force $A_x$ , a vertical force $A_y$ , and a reaction couple $M_A$ , but only $A_x$ and the cable tension appear on the FBD because the applied load $F$ and the couple $M_A$ are internal to the bracket-wall system and cancel each other.

Explanation: Whenever you see a question about support reactions, your first instinct should be to recall what each support type physically constrains — this is independent of what other forces or moments happen to be acting on the body. A fixed (cantilever) wall support prevents three things simultaneously: translation in the horizontal direction, translation in the vertical direction, and rotation. Because it prevents all three, it must supply three independent reaction components:

A_x

A_y

, and a reaction couple

M_A

. This is a structural fact about the support itself — it doesn't change based on what loads or other connections exist on the body. That reasoning confirms B as correct. A is tempting because students often think, "if the cable already creates a moment about A, then A doesn't need to supply one." This is flawed reasoning. The cable's moment is an applied moment that must be balanced, not one that eliminates the need for rotational resistance from the support. A fixed wall always independently constrains rotation. C contains a subtle but serious error: it suggests you can set

M_A = 0

on the FBD because the cable provides a nonzero moment. This confuses the result of applying equilibrium equations with what belongs on the FBD. You draw all possible reactions first, then solve — you never pre-eliminate unknowns by assuming equilibrium is "already satisfied." D incorrectly claims

F

and

M_A

are internal to the bracket-wall system and can be omitted. Once you isolate the bracket from the wall,

F

is clearly an external applied load and must appear on the FBD. Study tip: Always determine support reactions from the support type alone, not from the other loads present. Draw every reaction a support is capable of providing, then let the equilibrium equations tell you their values.