A pivoted object at rest: turning effects balance, and forces balance. Asked every year, and not obviously part of "F = ma".
The two conditions for equilibrium
Torque τ = F × d d is the PERPENDICULAR distance from the pivot to the line of the force Rotational balance Στ(clockwise) = Στ(anticlockwise) Taken about any chosen pivot point Translational balance ΣF(up) = ΣF(down) The object is not accelerating in any direction either Exam-style A uniform bench, 621 N and 3.21 m long, on two supports A and B Find the support forces, taking torques about A. Every weight force contributes a clockwise or anticlockwise torque about A, equal to its size × its perpendicular distance from A. Set the sum of clockwise torques about A equal to the sum of anticlockwise torques about A, and solve for the unknown force at B. Once F_B is known, use ΣF(up) = ΣF(down) to find F_A — the two support forces must carry all the weight between them. ✓ Two equations, two unknowns: torques about one support finds the other support's force, then the force balance finds the first. ✕
d is the perpendicular distance, not the slope distance
τ = F × d only works when d is measured perpendicular to the line of the force — usually the horizontal distance from the pivot for a vertical weight force. Using the distance along a diagonal support wire, rather than straight across, is the most common way this calculation goes wrong. ✓
Choosing a pivot that removes an unknown
When a beam or bench is supported at two points and both support forces are unknown, take torques about ONE of the support points. That support force then has zero perpendicular distance from the pivot (it acts AT the pivot), so its torque is automatically zero and it disappears from the equation — leaving one equation with only one unknown to solve. This is a strategic choice, not a different physical setup: taking torques about the other support gives the same final answer via a different equation.