CELE Engineering Mechanics — Equilibrium of Particles and Rigid BodiesRevision Notes
Condensed revision notes for Equilibrium of Particles and Rigid Bodies, built for the final weeks before the CELE 2026. These are the distilled key points you need when there is no time left for full study notes — just the concepts, formulas, and traps Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Engineering Mechanics subtest is marked as "Core" in the official pattern, and Equilibrium of Particles and Rigid Bodies appears in position 2nd of 8 in the CELE Engineering Mechanics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Equilibrium of Particles and Rigid Bodies - Revision Notes
Equilibrium is the cornerstone of structural analysis. Every beam reaction, truss member force, and foundation design in Philippine civil engineering practice begins with one fundamental principle: the net force and net moment on a body are zero. For the PRC Civil Engineer Licensure Examination (CE Board), mastery of Free-Body Diagrams (FBDs), the three equilibrium equations, Lami's Theorem, and support reactions is non-negotiable. This chapter covers 2D particle and rigid-body equilibrium — topics that appear in virtually every board exam cycle under Engineering Mechanics and Structural Engineering subjects.
Sections
Formulas
Example
For a pin-supported beam with an inclined 100 kN load at 30°: horizontal component = 100 cos30° = 86.6 kN must be balanced by the pin's horizontal reaction Hx = 86.6 kN.
Formula
ΣFx = 0
Variables
Fx = horizontal component of each force (positive → right)
Application
Ensures no net horizontal translation of the body
Example
ΣFy = RA + RB − 50 − 60 = 0 means RA + RB = 110 kN for a beam with those two loads.
Formula
ΣFy = 0
Variables
Fy = vertical component of each force (positive ↑ up)
Application
Ensures no net vertical translation of the body
Example
Taking moments about support A on a 6 m beam: RB(6) = 50(2) + 60(4.5), so RB = 61.67 kN — solved in one step without needing RA first.
Formula
ΣM_O = 0
Variables
M_O = moment of each force about point O; M = F × d (perpendicular distance)
Application
Ensures no net rotation about any chosen point O. Choose O at an unknown force to eliminate it.
Exam Tips
- ALWAYS draw the FBD first — partial credit can be earned from a correct FBD even if the arithmetic is wrong.
- Take ΣM = 0 about the pin of a pin-roller beam to find the roller reaction in ONE equation with ONE unknown.
- For beams, use the alternate moment equation ΣM = 0 about the roller to find the pin vertical reaction — then check with ΣFy = 0.
- For a 2D problem: count unknowns before writing equations. More than 3 unknowns → indeterminate.
Key Points
- A body is in STATIC EQUILIBRIUM when it has zero acceleration (translational and rotational).
- For a 2D coplanar system, equilibrium requires THREE conditions: ΣFx = 0, ΣFy = 0, ΣM = 0.
- For a 3D system, six equations apply: ΣFx = ΣFy = ΣFz = 0 and ΣMx = ΣMy = ΣMz = 0.
- Newton's First Law is the physical basis: a body at rest stays at rest unless acted upon by an unbalanced force.
- A statically determinate structure has exactly as many unknowns as available equilibrium equations.
- A statically indeterminate structure has MORE unknowns than equations — additional compatibility equations are needed.
Definitions
Term
Static Equilibrium
Definition
A state in which the resultant of all forces and moments acting on a body is zero, resulting in zero acceleration.
Importance
Fundamental to all structural analysis; every support reaction calculation relies on this.
Term
Statically Determinate
Definition
A structure where all unknown reactions can be found using only the equilibrium equations (unknowns ≤ equations).
Importance
Most board exam problems involve statically determinate beams and frames.
Term
Statically Indeterminate
Definition
A structure with more unknowns than available equilibrium equations; requires compatibility and constitutive relations.
Importance
Degree of indeterminacy = unknowns − equations; appears in advanced structural analysis problems.
Term
Concurrent Forces
Definition
Forces whose lines of action all pass through a single common point.
Importance
For concurrent forces, ΣM = 0 is automatically satisfied — only ΣFx = 0 and ΣFy = 0 are needed.
Section Title
1. The Concept of Equilibrium
Common Mistakes
- Confusing 'equilibrium' with 'forces are equal' — equilibrium means vector sum is zero, not that individual forces are equal.
- Applying only two equations (ΣFx, ΣFy) to a rigid body and ignoring ΣM = 0, leading to an incomplete solution.
- Choosing a moment center that does NOT eliminate any unknown, making the algebra unnecessarily complex.
- Forgetting to include the body's own weight (self-weight) as a downward force at its centroid when the problem specifies it.
Formulas
Example
UDL of 20 kN/m over 3 m: R = 20 × 3 = 60 kN acting at 1.5 m from the start of the UDL.
Formula
R_UDL = w × L
Variables
w = intensity of UDL in kN/m; L = length over which UDL acts in m; R_UDL = resultant force in kN
Application
Replace a UDL with its resultant for moment calculations. The resultant acts at the centroid of the load diagram.
Example
Triangular load from 0 to 30 kN/m over 4 m: R = 0.5(30)(4) = 60 kN at 4/3 = 1.33 m from the 30 kN/m end.
Formula
R_triangular = (1/2) × w_max × L
Variables
w_max = maximum intensity of triangular load in kN/m; L = length in m
Application
Triangular (linearly varying) load resultant acts at L/3 from the larger end.
Exam Tips
- A correct support table: Roller = 1 unknown; Pin = 2 unknowns; Fixed = 3 unknowns. Memorize this.
- For UDL problems, always compute the resultant and its location BEFORE writing any equilibrium equation.
- For triangular loads: resultant = area of the triangle; location = 1/3 from the larger-intensity end.
- Label all forces with correct directions in the FBD — an assumed incorrect direction will give a negative answer, which is acceptable (just reverse the assumed direction).
Key Points
- An FBD is a sketch of the isolated body with ALL external forces and reactions shown as vectors.
- Remove supports and replace each with its proper reaction forces/moments.
- Include: applied loads, distributed loads (replace with resultant), self-weight at centroid, and support reactions.
- Roller support: 1 unknown reaction — perpendicular to the rolling surface.
- Pin (hinge) support: 2 unknown reactions — one horizontal, one vertical (or a resultant at unknown angle).
- Fixed (cantilever) support: 3 unknown reactions — horizontal force, vertical force, and a moment.
- Internal cable/rope: tension only, along the cable direction toward the anchor.
- Smooth surface contact: normal reaction only, perpendicular to the surface at the contact point.
Definitions
Term
Free-Body Diagram (FBD)
Definition
A diagram of a body completely isolated from its environment, with all external forces, moments, and reactions represented as vectors.
Importance
The single most important step in any equilibrium problem. A wrong FBD guarantees a wrong answer.
Term
Roller Support
Definition
A support that prevents translation perpendicular to the rolling surface but allows translation parallel to it and rotation. Provides ONE reaction.
Importance
Commonly appears as the right-end support on simply supported beams in board exam problems.
Term
Pin (Hinge) Support
Definition
A support that prevents translation in all directions but allows free rotation. Provides TWO reactions (Rx and Ry).
Importance
Left-end support of simply supported beams; also found at frame joints.
Term
Fixed Support
Definition
A support that prevents both translation and rotation. Provides THREE reactions (Rx, Ry, and M).
Importance
Used in cantilever beams and fixed frames; gives three unknowns at that one support.
Section Title
2. The Free-Body Diagram (FBD)
Common Mistakes
- Drawing the reaction at a roller in the WRONG direction — always perpendicular to the surface, never parallel.
- Forgetting the moment reaction at a fixed support — fixed ends always have a reaction moment.
- Not replacing distributed loads with their resultant before applying moment equations.
- Placing the UDL resultant at the wrong location — it acts at the centroid of the load diagram (midpoint for UDL, L/3 from larger end for triangular).
- Omitting the horizontal reaction at a pin when inclined or horizontal external forces are present.
Formulas
Example
A 600 N weight hangs from two cords: cord 1 at 60° from horizontal (left), cord 2 at 30° from horizontal (right). The three forces are T₁, T₂, and W=600 N. The angle between T₂ and W = 180°−30° = 150°, between T₁ and W = 180°−60° = 120°, between T₁ and T₂ = 90°. Then: T₁/sin150° = T₂/sin120° = 600/sin90°. So T₁ = 600 sin150° = 300 N; T₂ = 600 sin120° = 519.6 N.
Formula
F₁/sin α₁ = F₂/sin α₂ = F₃/sin α₃
Variables
F₁, F₂, F₃ = magnitudes of three concurrent forces; α₁ = angle between F₂ and F₃; α₂ = angle between F₁ and F₃; α₃ = angle between F₁ and F₂
Application
Directly solves for unknown force magnitudes when exactly three concurrent forces are in equilibrium — faster than the component method.
Example
For the same 600 N problem: −T₁cos60° + T₂cos30° = 0 → T₂ = T₁(0.5/0.866) = 0.5774T₁. Then ΣFy=0: T₁sin60° + T₂sin30° = 600 → 0.866T₁ + 0.2887T₁ = 600 → T₁ = 519.6 N, T₂ = 300 N.
Formula
ΣFx = 0: −T₁cosθ₁ + T₂cosθ₂ = 0
Variables
T₁, T₂ = cord tensions; θ₁, θ₂ = angles cords make with horizontal
Application
Component method for particle in equilibrium under two cord tensions and a vertical weight.
Exam Tips
- For three-force particle problems, use Lami's Theorem first — it is faster than setting up and solving two simultaneous equations.
- To find Lami's angles: draw the three forces head-to-tail forming a triangle; the Lami angle for each force = 180° minus the interior angle of the force triangle at that vertex.
- Alternatively: Lami angle α₁ = 360° − (angle of F₂ from +x) + (angle of F₃ from +x), measured going around. It is easier to just identify the angle BETWEEN the other two vectors visually.
- Always verify using ΣFx = 0 and ΣFy = 0 after applying Lami — takes 30 seconds and catches errors.
Key Points
- A PARTICLE is modeled as a point — it has mass but no size, so moments are irrelevant.
- Particle equilibrium uses only ΣFx = 0 and ΣFy = 0.
- The most common particle equilibrium problem: a joint/knot holding cables or rods at various angles with a hanging weight.
- For exactly THREE concurrent, coplanar forces in equilibrium, Lami's Theorem provides a direct, elegant solution.
- Lami's angle: each angle αᵢ is the angle BETWEEN THE OTHER TWO forces — NOT the angle each force makes with any axis.
- Lami's Theorem is derived from the Sine Rule applied to the force triangle.
- When four or more concurrent forces act, use the component method (ΣFx = 0, ΣFy = 0) — Lami does not apply.
Definitions
Term
Particle
Definition
An idealized body with mass concentrated at a single point; size and shape are ignored. All forces act at that point (concurrent by definition).
Importance
Simplifies analysis to ΣFx = 0 and ΣFy = 0 only; used for joint analysis in trusses and cable systems.
Term
Lami's Theorem
Definition
For three concurrent coplanar forces in equilibrium, each force is proportional to the sine of the angle between the other two forces.
Importance
Time-saving shortcut in board exams — reduces a two-equation system to direct proportion when only three concurrent forces act.
Term
Concurrent Forces
Definition
Forces whose lines of action intersect at a common point.
Importance
Concurrent systems have no net moment about the point of concurrency — ΣM is automatically zero.
Section Title
3. Particle Equilibrium and Lami's Theorem
Common Mistakes
- Using the wrong angle in Lami's Theorem — the angle must be BETWEEN the other two forces (i.e., the angle OPPOSITE to the force in the sine), not the angle a force makes with a reference axis.
- Applying Lami's Theorem when four or more forces are present — it is strictly for THREE concurrent forces only.
- Sign errors in the component method — define positive directions clearly and be consistent throughout.
- Forgetting that the weight W always acts vertically downward and must be included as one of the concurrent forces.
Formulas
Example
A truss diagonal member inclined at 45° carries force F. Its horizontal component = F cos45° = 0.707F and vertical component = F sin45° = 0.707F.
Formula
Force in two-force member = F along the member axis
Variables
F = axial force (tension positive, compression negative); member axis = line connecting the two pin joints
Application
Direction of force in a truss member is always along its axis — the angle of the member determines the components.
Exam Tips
- In any truss problem, EVERY member is a two-force member — immediately write force directions along member axes.
- When you see a member with pins at both ends and no loads between — it's a two-force member. State this explicitly in your solution.
- Three-force member shortcut: locate the intersection of any two known force lines — the third force must also pass through that point. This gives the direction of the unknown reaction.
- On board exams, the phrase 'smooth pin at both ends, no other loads' is the signal for a two-force member.
Key Points
- A TWO-FORCE MEMBER is a structural element loaded at exactly two points with no other applied forces or moments.
- In a two-force member, the resultant force at each point must be equal, opposite, and collinear — the force acts ALONG the line joining the two load points.
- Two-force members are in pure tension (T) or pure compression (C) — no bending.
- All members of a simple truss are two-force members (pin-jointed, loaded only at joints).
- A THREE-FORCE MEMBER is in equilibrium under exactly three forces. For equilibrium, the three forces must be CONCURRENT (meet at a single point) or all PARALLEL.
- The three-force member principle is a powerful tool: if two force directions are known, the third must pass through their intersection point.
- Frames and machines often contain three-force members — identifying them shortens the solution.
Definitions
Term
Two-Force Member
Definition
A structural element acted upon by forces at only two points (no moments or other loads). The internal force is purely axial (along the member's axis).
Importance
Recognizing two-force members in trusses and frames instantly tells you the direction of the internal force, reducing unknowns.
Term
Three-Force Member
Definition
A structural element in equilibrium under exactly three forces; equilibrium requires these forces to be concurrent or parallel.
Importance
Using the concurrency requirement allows you to determine an unknown reaction direction, often making a problem solvable with one equation instead of two.
Term
Axial Force
Definition
A force acting along the longitudinal axis of a member, producing pure tension or pure compression.
Importance
Axial forces are the only internal forces in two-force members — no shear, no bending.
Section Title
4. Two-Force and Three-Force Members
Common Mistakes
- Assuming a member is a two-force member when it actually has an applied mid-span load — it must have loads at ONLY the two end points.
- Drawing the force in a two-force member perpendicular to the member axis — it must be ALONG the axis.
- For three-force members: forgetting that the forces must be concurrent — drawing them incorrectly non-concurrent yields an impossible equilibrium state.
- Misidentifying a two-force member in a frame where one of the 'pins' actually has another member framing in (which applies a third force at that point).
Formulas
Example
Beam AB, L = 6 m. Point load 50 kN at x = 2 m from A; UDL 20 kN/m over x = 3 to 6 m (resultant 60 kN at x = 4.5 m). ΣM_A = 0: R_B(6) = 50(2) + 60(4.5) = 370 → R_B = 61.67 kN. Then R_A = 50 + 60 − 61.67 = 48.33 kN.
Formula
ΣM_A = 0: R_B × L = Σ(F_i × d_i)
Variables
R_B = roller reaction; L = span length; F_i = each external load; d_i = perpendicular distance from A to the line of action of F_i
Application
Finds the roller reaction of a simply supported beam in one equation.
Example
Cantilever 3 m long: 20 kN at free end + UDL 5 kN/m over full length. V = 20 + 5(3) = 35 kN. M = 20(3) + 5(3)(1.5) = 60 + 22.5 = 82.5 kN·m (CW reaction moment at fixed end).
Formula
M_fixed = Σ(F_i × d_i) + Σ(w_i × L_i × arm_i)
Variables
M_fixed = reaction moment at the fixed support; F_i = concentrated loads; d_i = distances from fixed end; w_i = UDL intensities; L_i = UDL lengths
Application
Computes the fixed-end moment reaction for a cantilever beam.
Example
Pin-roller beam: r = 2 + 1 = 3 reactions → DSI = 3 − 3 = 0 (determinate). Fixed-end cantilever: r = 3 reactions → DSI = 0 (determinate). Two pins: r = 4 → DSI = 1 (indeterminate to the 1st degree).
Formula
Degree of Static Indeterminacy (DSI) = r − 3
Variables
r = total number of reaction force components; 3 = available equilibrium equations for 2D
Application
Determines if the structure is determinate (DSI = 0), indeterminate (DSI > 0), or a mechanism (DSI < 0).
Exam Tips
- The CE Board frequently tests simply supported beams with combined point loads and UDLs — practice this specific configuration until it is automatic.
- For cantilever beams: V at fixed end = total vertical load; H at fixed end = total horizontal load; M at fixed end = sum of moments of all loads about the fixed support.
- If a beam has a hinge (internal hinge) at mid-span, it provides one additional condition: ΣM = 0 about the hinge for EITHER half of the beam separately.
- Always verify your reactions: R_A + R_B must equal the total vertical load for a simply supported beam (a 5-second sanity check).
- For board exam time management: if the problem gives three supports, check if it is determinate before attempting — indeterminate structures need different methods.
Key Points
- The standard procedure: (1) Draw FBD, (2) Identify knowns and unknowns, (3) Choose optimal moment center, (4) Write ΣM = 0 to find one reaction, (5) Write ΣFy = 0, (6) Write ΣFx = 0.
- For a simply supported beam (pin at A, roller at B): take ΣM_A = 0 to find R_B directly, then ΣFy = 0 for R_A.
- For a cantilever (fixed at A, free at B): all three reactions (H, V, M) are at A; sum forces and take moments about A to find M_A.
- Moment sign convention: counterclockwise (CCW) positive is most common — be consistent throughout the problem.
- Overhanging beams: treat like simply supported beams; the overhang portion creates negative (hogging) moments over the support.
- When a beam has an inclined load: RESOLVE it into horizontal and vertical components BEFORE applying equilibrium equations.
- For frames: treat the entire frame as a single FBD first to find external reactions, then analyze internal members separately.
Definitions
Term
Moment Center
Definition
The chosen point about which moments are summed in ΣM = 0. Any point in space can be used.
Importance
Choosing a moment center at the location of an unknown force eliminates that unknown from the moment equation, enabling direct solution of the remaining unknown.
Term
Reaction
Definition
A force or moment exerted by a support on the body in response to the applied loads; determined by the type of support.
Importance
Reactions are the primary unknowns in most structural analysis problems.
Term
Overhang
Definition
A portion of a beam that extends beyond its support; it is cantilevered over air and causes hogging moments at the support.
Importance
Overhanging beams are common in board exams; the extended portion's load still contributes to the moment equation.
Section Title
5. Support Reactions of Rigid Bodies — Board Exam Procedure
Common Mistakes
- Taking moments about a point that is NOT at a support when a support reaction is the unknown — this creates unnecessary simultaneous equations.
- Using the wrong sign for moments — pick CCW = positive and STICK with it throughout the entire problem.
- Forgetting to include horizontal equilibrium (ΣFx = 0) when an inclined load is applied — leads to an incorrect pin reaction.
- For overhanging beams, treating the reaction at the interior support as if it cannot be upward — reactions can be upward or downward depending on load position.
- Placing the fixed-end moment in the wrong rotational direction — the fixed-end moment must OPPOSE the rotation tendency caused by the loads.
Connections
- Equilibrium of rigid bodies is the direct precursor to Shear Force and Bending Moment Diagram construction — you CANNOT draw an SFD/BMD without first finding the correct support reactions.
- Particle equilibrium (concurrent forces) is the basis of the METHOD OF JOINTS in truss analysis — each joint is a particle with concurrent member forces.
- Two-force member identification is essential in the METHOD OF SECTIONS for trusses and in frame analysis to reduce unknowns.
- The three equilibrium equations are used repeatedly in ANALYSIS OF FRAMES AND MACHINES — separating a frame into members and applying equilibrium to each part.
- Support reaction calculations directly feed into SOIL BEARING PRESSURE design (foundation engineering) — the column load transferred to a footing is the reaction from the beam above.
- Lami's Theorem connects to the LAW OF SINES in trigonometry — a direct application in mechanics.
- Degree of Static Indeterminacy (DSI) calculated here determines whether FORCE METHOD or DISPLACEMENT METHOD of structural analysis is required in advanced courses.
- Fixed-end moments calculated for cantilevers appear again in the MOMENT DISTRIBUTION METHOD and SLOPE-DEFLECTION METHOD for indeterminate beams.
- The principle of moments (ΣM = 0) is the mechanical basis of LEVER ARM calculations used in structural steel and reinforced concrete design (e.g., moment arm 'jd' in ACI 318-19 beam design).
- RA 544 (Civil Engineering Law of the Philippines) mandates that licensed civil engineers ensure structural adequacy — equilibrium analysis is the first step in demonstrating that any structure can safely carry its design loads.
Exam Strategy
For the CE Board Exam, equilibrium problems consistently appear in the morning session under Engineering Mechanics and in the afternoon session under Structural Engineering. Allocate no more than 4 minutes per problem. The winning strategy is: (1) Identify the system type immediately — particle (concurrent forces only) or rigid body (non-concurrent). (2) Draw the FBD in 30 seconds — do not skip this. (3) For particles with 3 forces, apply Lami's Theorem directly (saves 1–2 minutes). (4) For beams, always take ΣM about the pin support first to isolate the roller reaction in one equation. (5) For cantilevers, the fixed-end quantities are: V = total load, H = total horizontal load, M = total moment of all loads about the fixed end. (6) Check every answer using the remaining equilibrium equation (e.g., after finding both reactions via ΣM and ΣFy, verify with ΣM about the OTHER support — should equal zero). (7) Watch for trick questions: a roller on an inclined surface has a reaction PERPENDICULAR to that inclined surface, not necessarily vertical. (8) For the Structural Engineering board subject, connect these reactions directly to SFD/BMD problems — they are always linked. Prioritize this chapter in your review — it is foundational to at least 5–8 board exam questions spread across two subjects.
Quick Review Questions
A simply supported beam AB has a span of 8 m. A 40 kN concentrated load acts at 3 m from A, and a UDL of 15 kN/m acts over the entire span. Determine the reaction at B.
UDL resultant = 15 × 8 = 120 kN acting at midspan (x = 4 m from A). ΣM_A = 0: R_B(8) = 40(3) + 120(4) = 120 + 480 = 600 kN·m → R_B = 600/8 = 75 kN. Check: R_A = 40 + 120 − 75 = 85 kN. Wait — take moments about A again carefully: R_B(8) = 40(3) + 120(4) = 120 + 480 = 600; R_B = 75 kN. Then R_A = 160 − 75 = 85 kN. R_B = 75 kN.
State Lami's Theorem and identify the key condition for its application.
The theorem is derived from the Sine Rule applied to the closed force triangle. It applies ONLY when three — and exactly three — concurrent forces keep a body (or particle) in equilibrium. For four or more concurrent forces, revert to the component method.
What is the degree of static indeterminacy of a beam fixed at both ends (both ends fixed) with no internal hinges?
Each fixed support provides 3 reactions (Hx, Vy, M). Total reactions r = 3 + 3 = 6. Available equations = 3. DSI = 6 − 3 = 3. This beam requires methods like the three-moment equation or direct stiffness to solve.
A two-force member is inclined at 30° to the horizontal and carries a compressive force of 200 kN. What are the horizontal and vertical components of this force?
For a two-force member, force acts along the member axis. Fₓ = 200 cos30° = 200(0.866) = 173.2 kN. Fy = 200 sin30° = 200(0.5) = 100 kN. Since it is compressive, the force pushes toward the joint (into the member).
A cantilever beam 4 m long is fixed at A (left end) and free at B (right end). It carries a UDL of 10 kN/m over its full length and a 30 kN downward point load at B. Find the vertical reaction and fixed-end moment at A.
ΣFy = 0: V_A = 30 + 10(4) = 30 + 40 = 70 kN (upward). ΣM_A = 0 (taking moments of all loads about A, CCW positive): M_A = 30(4) + 10(4)(2) = 120 + 80 = 200 kN·m. The fixed-end moment acts counterclockwise (opposing the clockwise load moments).
In a three-force member problem, two of the three forces are known to intersect at point P. What does equilibrium require about the third force?
For a three-force member in equilibrium, the three forces must be concurrent — all three lines of action must meet at a single point. This requirement determines the direction of the unknown third force, which is a powerful shortcut in solving frame reactions.
A 500 N block hangs from a pin connected by two wires: Wire 1 is horizontal (to a wall on the left) and Wire 2 makes 60° with the horizontal (upward to the right). Using Lami's Theorem, find the tension in Wire 1 and Wire 2.
Three forces: T₁ (horizontal, pointing left), T₂ (at 60° above horizontal, pointing upper-right), W = 500 N (downward). Identify Lami angles: α for W = angle between T₁ and T₂ = 180° − 60° = 120°. α for T₁ = angle between T₂ and W: T₂ is at 60° from horizontal (= 60° from +x axis), W is at 270° from +x. Angle between them = 270° − 60° = 210°? Use the rule: angle between two forces = 360° minus the acute/obtuse angle between them, going around. More simply: draw the force triangle. T₁/sin(angle between T₂ and W) = T₂/sin(angle between T₁ and W) = W/sin(angle between T₁ and T₂). Angle between T₁ and T₂ = 180° − 60° = 120°. Angle between T₂ and W = 90° + 60° = 150°. Angle between T₁ and W = 90°. Check: 120° + 150° + 90° = 360° ✔. T₁/sin150° = T₂/sin90° = 500/sin120°. T₁ = 500(sin150°)/sin120° = 500(0.5)/0.866 = 288.7 N. T₂ = 500(sin90°)/sin120° = 500(1)/0.866 = 577.4 N.
What are the three support reactions provided by a fixed (cantilever) support in a 2D problem?
A fixed support prevents all motion: horizontal translation → H; vertical translation → V; rotation → M. This is why a cantilever beam (fixed at one end, free at the other) is statically determinate (3 unknowns = 3 equilibrium equations).
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