Moments & Equilibrium: the Non-Uniform Beam

The question

A 4 m plank rests on two supports. It weighs 250 N, yet the left support pushes up with only 100 N while the right pushes with 150 N. The plank looks symmetrical — so why does one end carry half again as much as the other?

Because the plank is NON-uniform: its centre of gravity is not in the middle. The two supports share the weight, but they share it unequally — and the split tells you exactly where the heavy part hides. Master moments and you can locate that hidden weight without ever cutting the beam open.

The key idea

A rigid body is in equilibrium only when TWO conditions hold simultaneously. (1) Forces balance: the total upward force equals the total downward force, ΣF = 0. (2) Moments balance: about ANY point, the total clockwise moment equals the total anticlockwise moment, Στ = 0. A moment (turning effect) is the force multiplied by the perpendicular distance from the pivot to the force's line of action. Because a force acting through the pivot has zero perpendicular distance, taking moments about a support eliminates that support's reaction from the equation — the standard way to solve for an unknown reaction or an unknown distance.

A beam on two supports is held by two upward reactions, $R_A$ and $R_B$, against its weight $W$ acting down at the centre of gravity. **Step 1 — forces:** $R_A + R_B = W$. The supports split the weight, but they split it according to how close the CoG is to each: the nearer support carries more. **Step 2 — moments:** choose a pivot and set clockwise = anticlockwise. Pivot at A and $R_A$ contributes nothing (its line of action passes through A), so $R_B\times(\text{distance to }B) = W\times(\text{distance to CoG})$ — one equation, one unknown. **Why a NON-uniform beam matters:** if the beam were uniform the CoG would sit at its geometric centre and the maths would already be set; for a non-uniform beam the CoG is hidden, and the unequal reactions are exactly what locate it. **The zero-reaction condition:** loading the beam so that $R_A = 0$ means support A pushes with no force — every newton of upward support now comes from B, and the beam is on the verge of tipping about B. Take moments about B and the load on one side must balance the weight on the other. **Connect it:** this is the seesaw you balanced as a child — heavier child sits closer to the pivot — written as an equation. Same physics scales a child's seesaw, a diving board, and the deck of a bridge.

The formula

• ·moment (N·m) = force F (N) × perpendicular distance d (m) from the pivot to the force's line of action. For equilibrium you write two equations: vertical forces up = forces down (gives one reaction once the other is known)
• ·and clockwise moments = anticlockwise moments about a chosen pivot. Pivot at a support and that support's reaction drops out
• ·leaving a single unknown to solve.

What to remember

• ·Equilibrium needs BOTH conditions: forces balance (up = down, ΣF = 0) and moments balance (clockwise = anticlockwise about any point, Στ = 0).
• ·Moment = force × perpendicular distance; a force acting at the pivot has zero moment, so taking moments about a support eliminates that support's reaction.
• ·A non-uniform beam's centre of gravity is NOT at its midpoint — the unequal reactions locate it; driving one reaction to zero means the other support carries the entire load.