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A 200 N ladder leans against a smooth wall at 60° to the floor. Three forces act — its weight, the wall's push, and whatever the ground supplies — yet the ladder hangs perfectly still. If you slowly lower the foot so the angle shrinks, at some critical angle the ladder suddenly screams across the floor and you hit the deck. What changed? Nothing was added; the ladder just couldn't grip any more. How do you predict the exact angle where the grip runs out?
A 200 N ladder leans against a smooth wall at 60° to the floor. Three forces act — its weight, the wall's push, and whatever the ground supplies — yet the ladder hangs perfectly still. If you slowly lower the foot so the angle shrinks, at some critical angle the ladder suddenly screams across the floor and you hit the deck. What changed? Nothing was added; the ladder just couldn't grip any more. How do you predict the exact angle where the grip runs out?
Standing still is not 'no forces' — it's forces and TURNING effects that all cancel. A rigid body needs THREE conditions at once: horizontal forces balance, vertical forces balance, and torques balance. The smooth wall can only push horizontally, so the floor's friction is the ONLY thing fighting the slide. Lower the angle and the wall pushes harder, demanding more friction — until the floor can't deliver and the ladder goes. Master the three conditions and you can size that friction, find when a ladder slips, and read off the reactions on a loaded bridge.
A rigid body in equilibrium obeys THREE conditions simultaneously: forces balance horizontally (ΣF_x = 0), forces balance vertically (ΣF_y = 0), and turning effects balance (Στ = 0 about ANY point). A single particle only needs the force conditions; an extended body can also rotate, so torque must cancel too. The power move is choosing the pivot wisely: take moments about a point where an unknown force acts, and that force's torque vanishes (zero moment arm), eliminating it from the equation. For a ladder against a SMOOTH wall, the wall supplies only a horizontal normal force, so the floor alone provides vertical support and the friction that stops the slide. For a beam on two supports, taking moments about one support removes its reaction and lets you solve directly for the other.
The one genuinely new idea here is that **'at rest' means two things at once for an extended body: no net force AND no net turning effect.** A point mass can't spin, so suffices; a ladder or beam can rotate, so we add about any pivot — and crucially the pivot is OURS to choose. **Why choosing the pivot is a superpower:** a force has zero moment about any point on its own line of action, so if we take moments about the foot of the ladder, both the ground normal and the friction contribute nothing — one equation, one unknown (). Picking the pivot to delete the forces you don't want is the single most useful habit in statics. **Smooth vs rough contacts:** a smooth (frictionless) surface can only push perpendicular to itself, so a vertical smooth wall gives a purely horizontal ; a rough floor gives both a normal and a friction up to a maximum . The ladder is safe only while the friction it NEEDS, , stays under what the floor can supply: , i.e. . **Why a lower angle slips:** and as the ladder flattens, so the friction demand climbs without bound while the available is fixed — eventually demand wins and the ladder shoots out. **Beams:** the same trick handles bridges and shelves — take moments about one support to expose the reaction at the other. **Connect it:** moments balanced the see-saw of earlier work; here we simply enforce force balance on BOTH axes at the same time.
Two killers. First, thinking a flatter ladder is safer: in fact a SMALLER angle needs MORE friction, because N_w = (W/2)/tanθ blows up as tanθ → 0 — flatten the ladder and the friction demand soars. Second, treating a smooth wall as if it pushes UP (a vertical force): a frictionless wall can only push perpendicular to itself, i.e. purely HORIZONTAL, so the floor must supply ALL the vertical support and ALL the friction. Also remember to enforce all three conditions (ΣF_x, ΣF_y, Στ), not just moments, and to take moments about a point that deletes an unknown (the foot, or one beam support).