Moments — Rods and Beams on Two Pivots, "About to Tilt"

A non-uniform rod AB, length 6 m, weight 60 N, rests horizontally on supports at C (1 m from A) and D (5 m from A). The rod's centre of mass is 4 m from A (not the midpoint, since the rod is non-uniform). Find R_D by taking moments about C.

• $R_D = 45\text{ N}$, from $R_D \times 4 = 60 \times 3$

  Correct. Distance from C to D is $5 - 1 = 4\text{ m}$; distance from C to the centre of mass is $4 - 1 = 3\text{ m}$ (using the STATED centre of mass, not the geometric midpoint, since the rod is explicitly non-uniform). $R_D = \frac{180}{4} = 45\text{ N}$.

• B$R_D = 30\text{ N}$, treating the weight as acting at the geometric midpoint, 3 m from A

  The rod is stated to be non-uniform with its centre of mass given as 4 m from A — using the geometric midpoint (3 m) instead ignores information the question gave you specifically because a uniform rod's midpoint shortcut does not apply here.

• C$R_D = 60\text{ N}$, from $R_D \times 4 = 60 \times 4$, using the centre of mass's distance from A instead of from C

  Every distance in a moments equation about C has to be measured FROM C, not from A. The centre of mass is 4 m from A, but only $4 - 1 = 3\text{ m}$ from C — using the wrong reference point for the distance is exactly the error the general marking guidance calls a method error, not an accuracy slip.

• DNot enough information — a non-uniform rod's reactions cannot be found without knowing HOW the mass varies along its length

  The question already supplies the one number a non-uniform rod problem needs: where its centre of mass actually is (4 m from A). Once that's given, the rod's weight is treated as a single force acting there — exactly like a uniform rod's weight at its midpoint — and no further detail about how the mass is distributed is required.

Traps tested: Uniform midpoint assumption applied to a non uniform rod · Distance measured from the wrong reference point · Overclaims uncertainty

A rod rests on supports at C and D. You are told the rod is on the point of tilting about C. Which single fact does that sentence give you?

• $R_D = 0$

  Correct — D is the support losing contact when the rod is about to rotate about C, exactly the pattern confirmed independently in Jan 2023 Q4 and Jun 2024 Q4 (there, framed around a different named pivot, but the same rule: the OTHER support's reaction is zero).

• B$R_C = 0$

  This zeroes the reaction at the named pivot itself — the documented error, not the documented fact. C is the point the rod rotates AROUND, which requires C to still be exerting a (nonzero) upward force.

• CThe rod's weight must be acting exactly at C

  "About to tilt about C" is a statement about the reaction forces at the two supports, not about where the rod's weight acts — the weight's position is fixed by the rod's own mass distribution (its midpoint if uniform) and does not move because of how a load elsewhere on the rod is arranged.

• D$R_C = R_D$

  Nothing about "on the point of tilting" implies the two reactions are equal — in fact the tilting condition typically arises precisely because they have become very UNequal, with one heading to zero while the other carries the full weight.

Traps tested: Reaction at the pivot itself zeroed · Tilting condition confused with weight position · Tilting condition confused with equal reactions

A light rod rests horizontally on two supports, with a single particle of weight 40 N attached to it somewhere between them. What is the total upward force supplied by the two reactions, and why?

• 40 N, split unevenly between the two reactions depending on where the particle is

  Correct. "Light" means the rod itself contributes no weight, so the ONLY downward force is the 40 N particle — resolving vertically, $R_C + R_D = 40$ regardless of where the particle sits. Where exactly it sits changes how that 40 N SPLITS between the two reactions (found from a moments equation), not the total.

• B20 N at each support, since the particle is somewhere between them

  This assumes the particle is exactly midway between the two supports, which the question never states. The total is fixed at 40 N by resolving vertically; the SPLIT between the two supports depends on the particle's exact position and is generally not 20/20 unless that position happens to be symmetric.

• CCannot be found without knowing the rod's own weight

  The rod is stated to be light, which is precisely the piece of information that removes the rod's own weight from the problem entirely — a light rod's weight is not "unknown," it is exactly zero by the definition of the modelling term.

• DMore than 40 N, since both supports are pushing up independently

  Both reactions act upward, but their SUM is fixed by resolving vertically against the only downward force present — the sum cannot exceed the total downward force any more than it can fall short of it, in equilibrium.

Traps tested: Assumes symmetric split without justification · Light rod treated as having unstated weight · Reaction forces summed without equilibrium constraint