Ladders, Hinges & Friction

List the contacts before writing a single equation, because each one decides how many unknowns enter. A smooth surface pushes perpendicular to itself: one unknown, the normal force. A rough surface adds a friction component along itself: two unknowns. A cable pulls along its own length: one unknown that must come out positive. A hinge or pin stops the body translating but not turning, so its force can point anywhere; carry it as two components, Rₓ and R_y. Against these, a planar rigid body offers exactly three independent equations — two force balances and one torque balance. Torque balances about further points are combinations of those three, not new information. So a ladder on a rough floor against a smooth wall (N_f, f, N_w) is determinate, as is a beam held by one hinge and one cable (Rₓ, R_y, T). Roughen the wall too and the ladder carries four unknowns into three equations.

A torque balance holds about every point, so choose the one through which the most unknown forces act: their lever arms are zero and they vanish from that equation. For a ladder the choice is the foot, where N_f and f both act. Take a uniform 4.5 m, 16 kg ladder at θ = 70° against a smooth wall. The wall's horizontal push N_w acts at height L sin θ above the foot; the weight acts at horizontal distance ½L cos θ from it. Balancing, N_w L sin θ = mg (½L cos θ), so N_w = ½mg cot θ = ½ × 157 N × 0.364 = 28.6 N. Horizontal balance then gives f = N_w = 28.6 N toward the wall, and vertical balance gives N_f = mg = 157 N. One torque equation delivered one unknown, and L cancelled: the forces depend on the angle and the weight, not on the length of the ladder.

An equilibrium solution is only a candidate until every contact has been checked against what it can physically do. Friction must satisfy |f| ≤ μₛN: with μₛ = 0.30, the floor under that 70° ladder can grip with up to 0.30 × 157 N = 47.1 N, the ladder asks for 28.6 N, and so it stands using 61% of the available friction. A normal force must come out positive, because a surface can push but not pull, and a tension must come out positive, because a cable can pull but not push. A negative friction is harmless: the assumed direction was backwards and the magnitude still stands. A negative normal force or tension is not: that contact has let go, the model you solved is the wrong one, and you must solve again without it. Report f/(μₛN), not just f, because that ratio says how close the structure is to its limit.

Safety questions ask where equilibrium only just survives: the shallowest angle, the highest rung, the heaviest load. There, and only there, write f = μₛN, and let the quantity you are hunting take friction's place as the unknown. The friction direction must now be right in advance: it opposes the slip that would happen without it, so at the foot of a ladder it points toward the wall. Put a climber of mass M a distance d up the ladder, and torques about the foot give f = (½m + M d/L) g cot θ, while N_f = (m + M)g. Demand grows with d; supply does not. A 6.0 m, 22 kg ladder at 68° on a floor with μₛ = 0.30 lets a 70 kg climber reach only d = L[μₛ(m + M) tan θ − ½m]/M = 4.91 m. Stand it at 76°, one out for every four up, and the same formula gives 8.5 m, beyond the top rung.

Roughen the wall as well and the ladder carries four unknowns: N_f, f_f, N_w and f_w. Short of slipping, three equations cannot share the load between the two frictions, and the rigid-body model has nothing more to say. At impending slip it does, because a rigid ladder cannot slide at its foot without its top sliding down the wall: both contacts reach their limits together, with f_f = μ_f N_f toward the wall and f_w = μ_w N_w upward. Those two relations close the system. For a uniform ladder alone, horizontal balance gives N_w = μ_f N_f, vertical balance gives N_f(1 + μ_wμ_f) = mg, and torques about the foot then give tan θₘᵢₙ = (1 − μ_wμ_f)/(2μ_f). With μ_f = 0.35 and μ_w = 0.30, tan θₘᵢₙ = 0.895/0.70 and θₘᵢₙ = 52.0°, against 55.0° with a smooth wall. The wall's grip enters only as the product μ_wμ_f, so it buys three degrees; the floor does nearly all the work.

A hinge is a contact that never slips and never lets go, so it supplies whatever force equilibrium asks for, in any direction. Carry Rₓ and R_y with a guessed sense and let the signs report the truth. For a horizontal beam hinged at a wall and held by a cable at angle α above it, torques about the hinge give T sin α, and vertical balance gives R_y = (total weight) − T sin α. With the cable fixed at the free end, T sin α is always less than the total weight, so R_y is positive. Attach the cable nearer the wall and hang a heavy load at the tip, and the cable can end up lifting more than the whole weight to hold the torque: R_y turns negative and the hinge pulls the beam down. The hinge force lies along a member only for a light two-force member loaded at its two pins alone. A door on two hinges carries four unknowns: the torque balance fixes the horizontal pair, but only the sum of the vertical components is determined, and the split depends on how the hinges were fitted.