Moments, Torque & Equilibrium Calculator

Moments and Torque Guide

What's the key thing to understand about Moment of a Force?

Moment = Force × perpendicular distance from pivot. M = F × d. Units: N·m (newton-metres). The perpendicular distance is crucial — the line of action of the force must be perpendicular to the distance measured. If the force is at angle θ to the lever arm: M = F × d × sin θ. This is equivalent to torque: τ = r × F (cross product). Example: 50N force at 2.0m from pivot: M = 50 × 2 = 100 N·m. Clockwise convention: clockwise = positive moment (or negative, depending on convention — be consistent).

What do I need to know about Principle of Moments?

For a body in rotational equilibrium: sum of clockwise moments = sum of anticlockwise moments. Σ(F × d)_clockwise = Σ(F × d)_anticlockwise. Example: seesaw with 50N at 2m left and 80N at 1.5m right of pivot. Clockwise (right): 80 × 1.5 = 120 N·m. Anticlockwise (left): 50 × 2 = 100 N·m. Not in equilibrium — 20 N·m net clockwise. To balance: move the 80N force: 100/80 = 1.25m from pivot. Applications: levers, beams, bridges, human joints (the forearm is a class 3 lever — force between fulcrum and

What's the key thing to understand about Couples and Pure Torque?

A couple: two equal and opposite forces separated by a perpendicular distance d. Torque of couple = F × d (no net force, only rotation). Examples: turning a steering wheel (two hands applying opposite forces). Using a spanner — the net torque equals F × width of spanner. A couple produces pure rotation with no translation — it is the same regardless of where you take moments. Unlike a single moment, the torque of a couple is the same about any point. This is why the torque output of an engine, a couple produced within the crankshaft, can be usefully quoted as a single figure without needing to specify exactly where on the engine it's measured.

What do I need to know about Centre of Mass?

Centre of mass position: x_cm = Σ(m_i × x_i) / Σm_i. Sum of (mass × distance from reference) divided by total mass. Example: uniform beam 4m long (mass 20kg, CM at 2m from left end). With 50kg person at 1m from left: x_cm = (20×2 + 50×1) / (20+50) = (40+50)/70 = 1.286m from left end. For uniform shapes: CM is at the geometric centroid. Composite objects: combine individual CMs weighted by mass. The object balances when supported at the centre of mass — tipping occurs if the CM moves outside the