Equilibrium of a Particle Under Coplanar Forces

A particle is in equilibrium under three coplanar forces: a horizontal force of $10$ N, a force $F$ N at $25°$ above the horizontal, and a force $T$ N at $55°$ below the horizontal (structured exactly like this lesson's main example, different numbers). What is $F$, to 3 s.f.?

• $8.32$ N

  Correct. Vertical equilibrium gives $F\sin25° = T\sin55° \Rightarrow F = T(\sin55°/\sin25°)$. Substituting into the horizontal equation $10 = F\cos25° + T\cos55°$ and solving simultaneously gives $T \approx 4.29$ N and $F \approx 8.32$ N.

• B$11.0$ N — from $10 = F\cos25°$

  This drops T's term from the horizontal resolution entirely — exactly the documented trap this lesson is built around, transferred to new numbers: T genuinely has a horizontal component here too ($T\cos55°$), and an equation that leaves it out is describing a different, incomplete system, not a simplified version of the real one.

• C$4.29$ N

  This is the correct value of T, not F — both unknowns were found correctly, but the final answer reports the wrong one of the two. Worth double-checking which letter a question is actually asking for once both values are in hand.

• D$9.06$ N — from $F = 10\cos25°$

  This applies the "resolve along F, T drops out" shortcut from this lesson's main example — but that shortcut only works when the two unknown forces are mutually perpendicular, which needs their angles to the horizontal to sum to 90°. Here $25°+55°=80°$, not $90°$, so T does NOT have a zero component along F's direction, and this equation silently omits a nonzero T term rather than correctly eliminating it.

Traps tested: Incomplete resolution missing a term · Correct values found but wrong unknown reported · Elimination shortcut misapplied where the 90 degree condition fails

In this lesson's main worked example (5 N, F at 30°, T at 60°), resolving along F's own direction eliminates T completely, because F and T act at exactly 90° to each other. In the system from MCQ 1 (10 N, F at 25°, T at 55°), does resolving along F's own direction eliminate T in the same way?

• No — F and T are 80° apart here (25° + 55°), not 90°, so T's component along F's direction is $T\cos80°$, which is nonzero; resolving along F still leaves an equation containing both unknowns

  Correct. The shortcut depends on one specific geometric fact — the two unknown forces being mutually perpendicular — which itself depends on their angles to the horizontal summing to exactly 90°. $25°+55°=80°\ne90°$, so that condition simply isn't met here, and no amount of resolving in a clever direction changes that; the simultaneous-equations method (or Lami's theorem, correctly applied) is what's actually needed.

• BYes — the elimination trick works for any two forces at different angles to the horizontal, not just ones at 90° to each other

  This overgeneralises a special case into a universal rule. The trick works because resolving along one force's direction is simultaneously resolving PERPENDICULAR to the other — and that coincidence only happens at exactly 90° between the two forces, not at any other angle.

• CNo — the shortcut never actually works, even in the original 30°/60° example; that example's clean numbers were a coincidence of rounding, not a real elimination

  The elimination in the 30°/60° example is exact, not a rounding artefact — it was checked two independent ways (component projection and Lami's theorem) in this lesson's own worked material, and both agree on F = 2.5√3 N precisely. The shortcut is genuinely available there; it's specifically THIS new system (80° between the forces) where it isn't.

• DYes, but only if you resolve along T's direction instead of F's

  Swapping which force you resolve along doesn't fix the underlying issue — the condition that makes the trick work is the ANGLE BETWEEN the two unknown forces, which is 80° regardless of which of the two you choose to resolve along. Neither direction eliminates the other force here.

Traps tested: Elimination shortcut misapplied where the 90 degree condition fails · Genuine shortcut dismissed as a coincidence

A light rigid rod connects a particle to a fixed point, and the rod is found to be under thrust rather than tension. What does this tell you about how the rod is acting on the particle?

• The rod is pushing the particle away from the fixed point, along the rod's own line — unlike a string, which can only pull, a rigid rod can also push, and "thrust" is the name for that pushing case

  Correct. Tension and thrust describe the same physical mechanism — a straight structural member exerting a force along its own length — split by direction: pulling is tension, pushing is thrust. A string can never supply thrust (it would simply go slack instead), which is exactly why 5.2 lists tension and thrust together as a pair specific to rods, not strings.

• BThe rod is under so much compression that it's about to break

  Thrust describes a DIRECTION of force (pushing, along the rod's line), not a magnitude close to some failure limit — a rod can be safely under thrust at any size of force the equilibrium condition happens to require, exactly as a rod under tension can. Nothing about the word "thrust" implies anything is close to breaking.

• CThrust means the rod is under tension, just measured in the opposite direction

  Tension and thrust aren't the same force with a sign flipped — they're genuinely different physical actions (pulling vs. pushing) that only a rigid rod, not a string, is capable of switching between. Reporting one as a signed version of the other loses the actual distinction the two words exist to make.

• DA rod can only ever be in thrust, never in tension — tension is a word reserved for strings

  A rigid rod can supply either — pulling (tension) or pushing (thrust) — depending on what the rest of the system requires for equilibrium; it isn't restricted to one or the other. It's strings that are restricted (to tension only, since they go slack rather than push), not rods.

Traps tested: Thrust mistaken for a statement about structural failure · Thrust and tension conflated as signed versions of the same thing · Rod incorrectly restricted to thrust only

Using Lami's theorem on the same three-force system as MCQ 1 (10 N horizontal; F N at 25°; T N at 55°): the angle opposite the 10 N force (between F's and T's own lines) is 80°; opposite F is 125°; opposite T is 155°. What is T, to 3 s.f.?

• $T = \dfrac{10\sin155°}{\sin80°} \approx 4.29$ N

  Correct, and matching MCQ 1's simultaneous-equations answer for T exactly — the same number reached by two structurally different methods is exactly the cross-check this lesson's method-comparison block demonstrates on the main example.

• B$T = \dfrac{10\sin125°}{\sin80°} \approx 8.32$ N

  This uses 125°, the angle that belongs to F's own line in Lami's theorem, in the equation for T — the two values are individually correct FOR THE FORCE THEY'RE PAIRED WITH, but this answer has swapped which angle goes with which force. (8.32 N is in fact the correct value of F, not T.)

• C$T = \dfrac{10\sin80°}{\sin155°} \approx 23.3$ N

  This inverts Lami's ratio — dividing by $\sin155°$ instead of by $\sin80°$, and multiplying by $\sin80°$ instead of $\sin155°$. Every force in Lami's theorem sits over the sine of ITS OWN opposite angle; the 10 N force's own equation uses $\sin80°$ as its denominator, not as T's numerator.

• D$T = 10\sin25° \approx 4.23$ N

  This is close enough to the correct 4.29 N to look plausible, but it isn't Lami's theorem at all — it uses the given angle (25°) directly rather than the angle Lami's theorem actually needs (155°, opposite T), and drops the $\sin80°$ denominator entirely. A near-miss number is exactly the kind of wrong answer that's easiest to overlook.

Traps tested: Lamis theorem angle assigned to the wrong force · Lamis theorem ratio inverted · Lamis theorem formula abandoned for an unrelated substitution