How unit 4 is examined
This unit covers coplanar force systems, free body and Bow's notation, plane trusses and friction; trusses, friction and force systems carry almost all the marks.
Graphical and Analytical Treatment of Concurrent and non-concurrent Co-planar forces
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Definition. Coplanar forces lie in one plane. <mark>Concurrent coplanar forces have lines of action that meet at one common point; non-concurrent coplanar forces have lines of action in the same plane that do not pass through a single point.</mark>
Key points.
- Concurrent forces (for example, cables meeting at a ring) can only translate the body, so their equilibrium needs $\sum F_x = 0$ and $\sum F_y = 0$.
- Non-concurrent forces can translate and rotate the body, so equilibrium needs $\sum F_x = 0$, $\sum F_y = 0$ and $\sum M = 0$.
- Graphically, concurrent forces are added by the polygon law: draw the forces head to tail to scale, and the closing side is the resultant; for equilibrium the polygon closes.
- Analytically, every force is resolved into $F_x = F\cos\theta$ and $F_y = F\sin\theta$, and the components are added algebraically.
- Non-concurrent forces are handled graphically with the funicular (link) polygon, or analytically by adding moments about a point.
Steps (resultant of concurrent coplanar forces).
- Resolve every force along X and Y: $F_x = F\cos\theta$, $F_y = F\sin\theta$, with correct signs.
- Find $\sum F_x$ and $\sum F_y$ as algebraic sums.
- Magnitude: $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$.
- Direction: $\theta = \tan^{-1}\left|\dfrac{\sum F_y}{\sum F_x}\right|$ with X-axis, placed in the quadrant given by the signs of $\sum F_x$ and $\sum F_y$.
Lami's theorem. If three concurrent coplanar forces keep a body in equilibrium, each force is proportional to the sine of the angle between the other two: $$\frac{P}{\sin\alpha} = \frac{Q}{\sin\beta} = \frac{R}{\sin\gamma}$$ Derivation. Three forces in equilibrium form a closed triangle with sides parallel to the forces. The exterior angle of the triangle opposite a side equals the angle between the other two forces, and $\sin(180^\circ - x) = \sin x$. The law of sines on the triangle then gives the result above.
Example (Nov 2022). A 100 N weight hangs from a string that makes $80^\circ$ with the horizontal, and a horizontal pull P holds it. The angle between T and W is $170^\circ$, between T and P is $100^\circ$, and between P and W is $90^\circ$. $$\frac{P}{\sin 170^\circ} = \frac{100}{\sin 100^\circ} = \frac{T}{\sin 90^\circ}$$ $P = 100 \times 0.1736/0.9848 = 17.63$ N and $T = 100/0.9848 = 101.54$ N. Answer: P = 17.63 N, T = 101.54 N.
Answer frame. Definition: open with the coplanar definition, sketch a ring with three cables (concurrent) and a beam with parallel or skew forces (non-concurrent), then give equilibrium conditions of each and close with one example of each. Resultant: write "Resultant is found by resolution" and give steps 1-4 with $R$ and $\theta$. Lami: state the theorem, draw the force triangle, derive by the sine rule, then solve with the angles marked on the junction.
Pitfall: Using the angle between a force and the horizontal in Lami's theorem; the angles must be those between each pair of forces.
Asked: [4 marks] (Jun 2022, Jun 2023) Explain concurrent and non-concurrent coplanar forces (in detail, with sketches) Asked: [7 marks] (Nov 2022) State and derive Lami's theorem and solve the given problem (W = 100 N, string at 80 degrees) using it Asked: [7 marks] (Jun 2025) Write the steps for finding the resultant of a concurrent coplanar force system
Free body diagram, Force diagram and Bow's notation
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Definition. A free body diagram (FBD) is the body isolated from its surroundings with every external force and reaction drawn on it. The space diagram is the line sketch of the structure with its loads, and the force diagram is the vector polygon drawn to scale from it.
Key points.
- In Bow's notation each space between the external forces, and each space between members, is labelled with a capital letter, and a force is named by the two letters on its sides (for example, AB).
- The external forces are lettered in clockwise order around the joint or structure, and the same order is followed in the force diagram.
- Each force is drawn in the force diagram as a line whose ends are the small letters of its two spaces, so a member force is read directly as the line between two letters.
- Joint by joint the force diagram is built into one closed figure, which gives every member force graphically, and tension or compression is read from the direction of the arrow at the joint.
Asked: [4 marks] (Jun 2022) Explain Force Diagram and Bow's notation
Application of Equilibrium Concepts: Analysis of plane Trusses: Method of joints, Method of Sections
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Definition. A plane truss is a framework of straight members joined at their ends to form triangles, loaded only at the joints. ==In the method of joints each joint is isolated and $\sum F_x = 0$, $\sum F_y = 0$ are applied to the concurrent forces there; in the method of sections the truss is cut through at most three unknown members and $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$ are applied to one part.==
Assumptions (method of joints).
- All members are joined at their ends by smooth frictionless pins or hinges.
- Loads and reactions act only at the joints.
- Each member is straight, and its centroidal axis coincides with the line joining the joint centres.
- The self-weight of members is neglected or lumped at the joints, so each member is a two-force member carrying only axial tension or compression.
Key points.
- Check first that the truss is a perfect frame, $m = 2j - 3$.
- Find the support reactions from overall equilibrium before any joint is cut.
- In the method of joints, start at a joint with only two unknown forces and move on joint by joint.
- Assume every member force to be tension, that is, pulling away from the joint; a negative answer means compression.
- In the method of sections, pass the cut through not more than three members, and take moments about the point where two of the cut members meet, to find the third directly.
- Sections are the quick method for a single internal member; joints are used when all members are needed.
Comparison.
| Basis | Method of joints | Method of sections |
|---|---|---|
| Body isolated | One joint | A part of the truss |
| Force system | Concurrent | Non-concurrent |
| Equations | $\sum F_x = 0$, $\sum F_y = 0$ | $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$ |
| Unknowns per step | At most 2 | At most 3 |
| Best for | All member forces | A few particular members |
| Drawback | Errors carry from joint to joint | Cut needs care to avoid more than 3 unknowns |
Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-01" viewBox="0 0 553 295" width="553" height="295" role="img" aria-label="Truss of Jun 2022 and Dec 2024: A hinge, F roller, 50 kN at B, 30 kN at D; AC = CE = 4 m, EF = 3 m, BC = 5 m, DE = 4 m"><style>#dsfig-u4-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-01 .t{fill:#16181D;font-weight:500}#dsfig-u4-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-01 .dot{fill:#16181D}#dsfig-u4-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-01 .ah{fill:#454C5A}#dsfig-u4-01 .ah.hi{fill:#2340B8}#dsfig-u4-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-01 .e{stroke:#B1B7C3}html.dark #dsfig-u4-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-01 .t{fill:#E6E8ED}html.dark #dsfig-u4-01 .t.inv{fill:#0F1115}html.dark #dsfig-u4-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-01 .dot{fill:#E6E8ED}html.dark #dsfig-u4-01 .ann{fill:#8FA3FF}html.dark #dsfig-u4-01 .lbl{fill:#858D9C}html.dark #dsfig-u4-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-01 .ah{fill:#B1B7C3}html.dark #dsfig-u4-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah10" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh10" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M51.9,240.2 L200.1,54.8"/><path class="e" d="M59,255 L193,255"/><path class="e" d="M212,59 L212,236"/><path class="e" d="M230.4,44.6 L365.6,78.4"/><path class="e" d="M225.4,241.6 L370.6,96.4"/><path class="e" d="M231,255 L365,255"/><path class="e" d="M384,102 L384,236"/><path class="e" d="M395.4,98.2 L501.6,239.8"/><path class="e" d="M403,255 L494,255"/><circle class="n" cx="40" cy="255" r="18"/><text class="t" x="40" y="255" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="212" cy="40" r="18"/><text class="t" x="212" y="40" dy=".35em" text-anchor="middle">B</text><circle class="n" cx="212" cy="255" r="18"/><text class="t" x="212" y="255" dy=".35em" text-anchor="middle">C</text><circle class="n" cx="384" cy="83" r="18"/><text class="t" x="384" y="83" dy=".35em" text-anchor="middle">D</text><circle class="n" cx="384" cy="255" r="18"/><text class="t" x="384" y="255" dy=".35em" text-anchor="middle">E</text><circle class="n" cx="513" cy="255" r="18"/><text class="t" x="513" y="255" dy=".35em" text-anchor="middle">F</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Truss of Jun 2022 and Dec 2024: A hinge, F roller, 50 kN at B, 30 kN at D; AC = CE = 4 m, EF = 3 m, BC = 5 m, DE = 4 m</figcaption></figure>
Example (method of joints, Jun 2022, Dec 2024). Reactions: $\sum M_A = 0$: $F_y \times 11 = 50 \times 4 + 30 \times 8 = 440$, so $F_y = 40$ kN, $A_y = 40$ kN, $A_x = 0$. Lengths: $AB = \sqrt{41}$, $BD = \sqrt{17}$, $CD = 4\sqrt2$, $DF = 5$ m.
| Joint | Working | Result |
|---|---|---|
| F | $DF \times 4/5 = 40$ up; $EF = DF \times 3/5$ | DF = 50 C; EF = 30 T |
| E | $CE = EF$; no vertical load | CE = 30 T; DE = 0 |
| D | $\sum F_x, \sum F_y$ with 30 kN load | BD = 32.98 C; CD = 2.83 T |
| A | $AB \times 5/\sqrt{41} = 40$; $AC = AB \times 4/\sqrt{41}$ | AB = 51.23 C; AC = 32 T |
| C | $\sum F_y$: $BC + 2 = 0$ | BC = 2 C |
Check at B: $\sum F_x = 32 - 32 = 0$ and $\sum F_y = 40 + 2 + 8 - 50 = 0$. Answer: AB 51.23 C, AC 32 T, BC 2 C, BD 32.98 C, CD 2.83 T, CE 30 T, DE 0, DF 50 C, EF 30 T.
Example (method of sections, same truss). Cut through BD, CD and CE and take the left part. Moments about D: $40 \times 8 - 50 \times 4 - CE \times 4 = 0$, so CE = 30 kN tension, which agrees with the joint method.
Answer frame. Definition: open with the definition of a plane truss, draw the truss with reactions, define both methods with the example above, and close with the comparison table. Numerical: write given data, reactions, then a joint table starting from the joint with two unknowns, and box the result table with T/C. Assumptions: write the four assumptions as separate lines.
Pitfall: Forgetting to state tension or compression for each member.
Asked: [10 marks] (Jun 2022, Dec 2024) Calculate the force in the members shown in the figure using the method of joints (also: forces in BC, CE, EF; all members) Asked: [6 marks] (Nov 2022) Write the difference between method of joints and method of sections Asked: [7 marks] (Jun 2023) Define the method of joints and method of sections to analyse a plane truss with a suitable example Asked: [4 marks] (Dec 2024) Write the assumptions used in the method of joints
Frictional force in equilibrium problems
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Definition. Friction is the force that opposes the motion, or tendency to move, between two surfaces in contact. ==Limiting friction is $F = \mu N$, where $\mu = \tan\phi$ is the coefficient of friction and $\phi$ is the angle of friction.==
Key points.
- Friction acts along the contact surface, opposite to the direction of impending motion.
- The normal reaction $N$ is found first from equilibrium perpendicular to the surface; on an incline of angle $\theta$, $N = W\cos\theta$.
- On an incline with the body about to move up under a force $P$ parallel to the plane, friction acts down the plane: $P = W(\sin\theta + \mu\cos\theta) = W\dfrac{\sin(\theta + \phi)}{\cos\phi}$.
- When the body is about to slide down, friction acts up the plane, so the holding force is $F_{\min} = W(\sin\theta - \mu\cos\theta)$.
- Draw a separate FBD for each body of a connected system, and use the same cord tension for both.
- For a pull P at angle $\alpha$ above the horizontal, $N = W - P\sin\alpha$, and P is least when $\tan\alpha = \mu$.
Example 1 (Nov 2022). $M = 10$ kg, $\theta = 45^\circ$, $\mu = 0.5$, $W = 98.1$ N. $F_{\min} = W(\sin 45^\circ - 0.5\cos 45^\circ) = 98.1 \times 0.3536 = 34.68$ N $F_{\max} = W(\sin 45^\circ + 0.5\cos 45^\circ) = 98.1 \times 1.0607 = 104.05$ N Answer: F(min) = 34.68 N, F(max) = 104.05 N.
Example 2 (Jun 2025). With $\mu = \tan\phi$: $\dfrac{180}{210} = \dfrac{\sin(15^\circ + \phi)}{\sin(20^\circ + \phi)}$, which gives $\tan\phi = 0.214$, $\phi = 12.08^\circ$. Then $W = 180\cos\phi/\sin(27.08^\circ) = 386.7$ N. Answer: W = 386.7 N, $\mu = 0.214$.
Example 3 (Jun 2022, Dec 2024). 10 kN block on $30^\circ$ incline moving up; 30 kN block moving left, pulled by P at angle $\alpha$. Incline block: $N_2 = 10\cos 30^\circ = 8.66$; $T = 10\sin 30^\circ + 0.2 \times 8.66 = 6.73$ kN. Horizontal block: $P\cos\alpha = T + 0.2(30 - P\sin\alpha)$, so $P = \dfrac{12.73}{\cos\alpha + 0.2\sin\alpha}$. Least P when $\tan\alpha = 0.2$, $\alpha = 11.31^\circ$: $P = 12.73/\sqrt{1.04}$. Answer: P(least) = 12.48 kN at $\alpha = 11.31^\circ$.
Answer frame. Open with "Friction opposes impending motion; limiting friction is $\mu N$"; draw the FBD with W, N, friction, and the applied force, marking the direction of impending motion; write the two equilibrium equations, substitute, and box the answer with units. For the two-force question, give both limiting cases.
Asked: [6 marks] (Jun 2022, Dec 2024) Find the least value of P for impending motion of the block system (Fig. P-511) to the left, with $\mu = 0.20$ under each block Asked: [7 marks] (Nov 2022) 10 kg block on a $45^\circ$ incline, $\mu_s = 0.5$: find the minimum and maximum force F to prevent slipping Asked: [7 marks] (Jun 2025) 180 N moves a body up a $15^\circ$ incline and 210 N up a $20^\circ$ incline; find the weight and coefficient of friction
Last-minute revision
- Concurrent: lines of action meet at one point; non-concurrent: same plane, no common point.
- Concurrent equilibrium: $\sum F_x = 0$, $\sum F_y = 0$; non-concurrent adds $\sum M = 0$.
- $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$, $\theta = \tan^{-1}|\sum F_y/\sum F_x|$.
- Lami: $P/\sin\alpha = Q/\sin\beta = R/\sin\gamma$, with angles between the other two forces.
- Bow's notation: label spaces clockwise with capitals; the force diagram is the vector polygon.
- Perfect truss: $m = 2j - 3$; joints: two unknowns; sections: three unknowns.
- Assume tension; a negative sign means compression.
- Friction $F = \mu N$, $\mu = \tan\phi$; on an incline $N = W\cos\theta$.
- Up the plane: $P = W(\sin\theta + \mu\cos\theta)$; down: $W(\sin\theta - \mu\cos\theta)$.
- Answers: 34.68 N and 104.05 N; W = 386.7 N, $\mu$ = 0.214; P = 12.48 kN.
Memory hooks
- Concurrent = "meet at a point", non-concurrent = "miss each other".
- Joints = point (two equations); Sections = beam (three equations, take moments).
- Bow: go clockwise, capitals in spaces.
- Friction rule: "Up needs plus, down needs minus" ($\sin\theta \pm \mu\cos\theta$).
- Least P for a pull: tan alpha = mu.
Coverage checklist
- Graphical and Analytical Treatment of Concurrent and non- concurrent Co- planner forces: Jun 2022/Jun 2023 concurrent terms, Nov 2022 Lami, Jun 2025 resultant steps.
- free Diagram, Force Diagram and Bow's notations: Jun 2022 force diagram and Bow's notation.
- Application of Equilibrium Concepts: Analysis of plane Trusses: Method of joints, Method of Sections: Jun 2022/Dec 2024 joints numerical, Nov 2022 difference, Jun 2023 define both, Dec 2024 assumptions.
- Frictional force in equilibrium problems: Jun 2022/Dec 2024 blocks, Nov 2022 incline 45 degrees, Jun 2025 180 N and 210 N.