Moment of force

Push a heavy door right at the hinge and it barely budges; push the same door near the outer edge and it swings open with almost no effort. The force you applied did not change — what changed is how far from the hinge you applied it. That turning effect of a force about a fixed point or axis is called the moment of a force, or torque, and it is the idea behind door handles placed far from hinges, long spanners for stiff nuts, and the delicate balance of a see-saw. This article builds the formula for moment, the principle of moments that makes a balanced beam predictable, and the special case of a couple — then works two examples so the arithmetic becomes automatic before your exam.

The core insight

The moment of a force equals force times the perpendicular distance from the pivot to its line of action, and a body stays in rotational balance when total clockwise moment equals total anticlockwise moment about any point.

The core idea

Defining the moment. The moment (or torque) of a force about a point is the product of the force and the perpendicular distance from that point to the line along which the force acts:

τ=F×d\tau = F \times d

where: τ\tau is the moment (newton-metre, N m), FF is the applied force (newton, N), and dd is the perpendicular distance from the pivot to the force's line of action (metre, m) — never the slant length of the rod or handle unless the force happens to act exactly perpendicular to it.

Sense of turning. A moment is not just a size; it also has a rotational sense. A clockwise moment tends to rotate the body clockwise about the pivot; an anticlockwise moment tends to rotate it the other way. The two senses can cancel, which is exactly what happens when a beam balances.

Principle of moments. A body is in rotational equilibrium when, about any chosen point, the sum of clockwise moments equals the sum of anticlockwise moments:

F1d1=F2d2F_1 d_1 = F_2 d_2

where: F1,d1F_1, d_1 are the force and perpendicular distance producing one sense of turning, and F2,d2F_2, d_2 produce the opposite sense. This single rule governs a beam balance, a see-saw resting on its pivot, and a metre rule loaded with weights on either side of a support.

Full equilibrium. A rigid body under several coplanar forces is in complete equilibrium only when two conditions hold together: translational equilibrium, ΣF=0\Sigma F = 0 (the vector sum of all forces is zero, so it does not accelerate sideways), and rotational equilibrium, Στ=0\Sigma \tau = 0 (clockwise moments balance anticlockwise moments about any point, so it does not spin up). A body can satisfy one condition and fail the other — a pair of equal, opposite, offset forces has zero net force yet a non-zero net moment, which is precisely what a couple is.

Figure
A metre rule pivoted at its centre with a 6 newton weight 30 centimetres to the left producing an anticlockwise moment and a 4 newton weight 45 centimetres to the right producing an equal clockwise moment
Principle of moments — the rule balances pivot (50 cm mark) 6 N 30 cm left 4 N 45 cm right anticlockwise moment (6 times 30) equals clockwise moment (4 times 45), both 180 N cm
Derivation ABalancing a loaded metre rule with the principle of moments

The goal is to find, from scratch, exactly where a known weight must sit on a pivoted metre rule so that it balances a second known weight placed on the other side, applying only the moment formula and the equal-and-opposite-sense rule for rotational equilibrium, with no assumptions beyond a rigid, uniform rule pivoted at its 50 cm mark and weights hung vertically. Step 1. Identify what produces each moment: a hanging weight is a vertical force FF, and the perpendicular distance dd from the pivot is simply the horizontal distance along the rule from the pivot to the point where the weight hangs, because the weight acts straight down while the rule is horizontal, making that horizontal distance already perpendicular to the force. Step 2. Assign a rotational sense to each weight by inspection: a weight on the left side of the pivot pulls that side down, which is an anticlockwise turn about the pivot, while a weight on the right side pulls that side down in a clockwise turn — the two senses oppose each other, which is what allows balance. Step 3. Write the principle of moments, F1d1=F2d2F_1 d_1 = F_2 d_2, equating the anticlockwise moment on one side to the clockwise moment on the other, and solve algebraically for the one unknown quantity, whether that unknown is a distance or a force. Step 4. Convert the found distance back into an actual scale-mark reading by adding to or subtracting from the pivot's mark, since dd is a distance from the pivot, not a position on the printed scale.

Worked example

A uniform metre rule is pivoted at its 50 cm mark. A 6 N weight hangs at the 20 cm mark, producing an anticlockwise moment. A second student wants the same 4 N weight from the earlier example to balance it, but this time asks where a 4 N weight must go so the rule balances, and then adds a third check: what happens if a spanner of arm 0.24 m is used with two 8 N forces to tighten a nut instead — does that spanner even need a pivot?

Step 1. Find the moment that must be balanced. The 6 N weight sits at 20 cm, a distance d1=50−20=30d_1 = 50 - 20 = 30 cm from the pivot, giving an anticlockwise moment of F1d1=6×30=180F_1 d_1 = 6 \times 30 = 180 N cm.

Step 2. Apply the principle of moments to the unknown side. Let the 4 N weight sit at distance dd cm to the right of the pivot, giving a clockwise moment F2d=4×dF_2 d = 4 \times d. Setting the moments equal:

6×30=4×d⇒d=1804=45 cm6 \times 30 = 4 \times d \quad \Rightarrow \quad d = \frac{180}{4} = 45\ \text{cm}

Step 3. Convert to a scale reading. Since the pivot is at 50 cm and the weight sits 45 cm to its right, the mark is 50+45=9550 + 45 = 95 cm. The 4 N weight balances the rule at the 95 cm mark.

Step 4. Now check the spanner case, where the same idea of "force times perpendicular distance" appears in a different setting. Two 8 N forces act at opposite ends of a 0.24 m handle, equal in magnitude, opposite in direction, and not along the same line — this is a couple, not a single pivoted weight. Its moment is F×d=8×0.24=1.92F \times d = 8 \times 0.24 = 1.92 N m. A couple needs no external pivot at all: because the two forces are equal and opposite, the net force is zero, so the spanner does not fly off sideways — it only spins, turning the nut with a moment of 1.92 N m.

Common traps

  • Using the slant length of a rod or handle instead of the true perpendicular distance from the pivot to the force's line of action — the two are equal only when the force happens to act perpendicular to the rod.
  • Balancing moments by matching sizes alone and forgetting to check that the two forces have opposite rotational sense; two forces on the same side of a pivot cannot balance each other no matter how their sizes are chosen.
  • Treating a couple as one big resultant force — a couple has zero net force but a non-zero net moment, and it can never be replaced by a single equivalent force.
  • Confusing moment (τ=F×d\tau = F \times d, unit N m, a turning effect) with work (W=F×dW = F \times d in special cases, unit joule, energy transferred) — the formulas look identical but the physical quantities and units are never interchangeable.

Ask the tutor: unsure how to spot the perpendicular distance when a force acts at an angle to the rod, or want to practise balancing a metre rule loaded with three or more weights — ask the AI tutor to walk through it with you.

Why is it the perpendicular distance and not just the distance along the rod that matters

Only the component of distance that is perpendicular to the force's line of action contributes to turning. If a force pulls along a direction that is slanted relative to the rod, part of that pull tries to stretch or compress the rod through the pivot rather than turn it, and that part produces no rotation at all. Measuring straight from the pivot to the line the force acts along, at a right angle, isolates exactly the turning part of the force, which is why τ=F×d\tau = F \times d always uses the perpendicular distance.

Can a body have zero net force but still start rotating

Yes, and this is exactly what a couple demonstrates. Two equal, opposite, parallel forces that do not share the same line of action add up to zero net force, so the body does not accelerate in any straight-line direction, yet each force still produces a moment about any point you choose, and those two moments add up rather than cancel because both forces try to turn the body the same way. The result is pure rotation with no sideways push, which is exactly how a screwdriver or a spanner turns a screw or nut without sliding it anywhere.

If the principle of moments balances a beam, why does the choice of pivot point not matter

The principle of moments as stated for a rule already resting on a fixed pivot only needs moments taken about that pivot, since the pivot itself supplies whatever extra force is needed to keep the rule from translating. For full equilibrium of a body with no fixed support, both conditions matter together: zero net force and zero net moment about any one point. Once those two conditions hold simultaneously, it turns out the net moment is automatically zero about every other point too — but at Class 10 level you only need to apply the principle of moments about the given pivot, not prove that deeper result.

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