A painter in Hyderabad leans an aluminium ladder against the wall of a new flat. On the rough cement of the balcony it stands firmly. Inside, on the polished marble floor, the same ladder at the same angle creeps outward at the foot and slides down the wall. The painter props it much steeper, and it holds. Even then, he knows not to climb right to the top.
The forces on a ladder
Take a uniform ladder of length \(L\) and weight \(W\), leaning at an angle \(\theta\) to the floor. The wall is smooth, so it can push only at right angles to itself, horizontally. Call that push \(N_w\). The floor is rough. It pushes up with a normal force \(N\), and it can supply friction \(f\) along the floor. The weight acts at the middle of the ladder.
Which way does friction act? Without it, the wall's push would drive the foot away from the wall. So friction at the foot points towards the wall. Now write the conditions of equilibrium. Horizontally, \(f = N_w\). Vertically, \(N = W\). For torques, the foot is the best point, since both unknown floor forces act there.
\[ N_w\,L\sin\theta = W\,\tfrac{L}{2}\cos\theta \]
Why. torques about the foot: the wall's push has moment arm \(L\sin\theta\), the height of the top; the weight has moment arm \(\tfrac{L}{2}\cos\theta\).
\[ N_w = \tfrac12 W\cot\theta \]
Why. divide both sides by \(L\sin\theta\).
\[ f = N_w = \tfrac12 W\cot\theta \]
Why. the horizontal forces balance.
\[ \tfrac12 W\cot\theta \le \mu_s N = \mu_s W \]
Why. static friction cannot exceed its limit \(\mu_s N\), and \(N = W\).
\[ \mu_s \ge \tfrac12\cot\theta \]
Read it. divide by \(W\); the least coefficient of friction that holds the ladder is \(\tfrac12\cot\theta\).
The result does not contain \(L\) or \(W\). A long heavy ladder and a short light one need the same friction at the same angle. It depends on the angle alone. At \(60°\), \(\mu_{\min} = \tfrac12 \times 0.577 \approx 0.29\). At \(45°\), \(\mu_{\min} = 0.5\). A steep ladder has a small \(\cot\theta\), so it needs little friction. A steep ladder is a safe ladder.
A uniform ladder leans at \(60°\) to a rough floor, against a smooth wall. What is the least coefficient of static friction at the floor that keeps it from slipping?
The least safe angle
Read the result the other way. For a given floor, with coefficient \(\mu_s\), the ladder is safe only if \(\cot\theta \le 2\mu_s\), that is \(\tan\theta \ge 1/(2\mu_s)\). At the least angle \(\theta_{\min}\), the friction needed is exactly the limiting friction. Below it, the foot slides away from the wall and the top slides down.
Start the sim at \(\theta = 60°\) and \(\mu = 0.30\). The friction needed is \(100/\tan 60° \approx 57.7\,\text{N}\). The floor can give \(0.30 \times 200 = 60\,\text{N}\), so the ladder stands, with little to spare. The least angle for this floor is \(\tan^{-1}(1/0.6) \approx 59.0°\). Lower the angle to \(59°\) and the friction needed, \(60.1\,\text{N}\), just passes the limit: press play and the ladder slides.
A painter's ladder stands on a floor with \(\mu_s = 0.25\), against a smooth wall. What is the least angle with the floor at which it can stand?
A person on the ladder
A person climbing the ladder adds weight, and adds it at a point that moves. The higher the climber, the larger the moment arm of their weight about the foot. So the wall must push harder, and the friction needed grows. The floor's normal force grows too, but only by the climber's weight, whatever the height. So a ladder that holds a painter on its lowest rung may slip when he nears the top.
A uniform ladder \(5\,\text{m}\) long and of mass \(10\,\text{kg}\) leans against a smooth wall, with \(\sin\theta = 0.8\) and \(\cos\theta = 0.6\). The floor has \(\mu_s = 0.5\). A painter of mass \(50\,\text{kg}\) climbs it. Take \(g = 10\,\text{m/s}^2\). How far along the ladder can he climb before it slips?
The ladder weighs \(100\,\text{N}\) at its middle and the painter \(500\,\text{N}\) at a distance \(x\) along it from the foot. The floor's normal force is \(N = 600\,\text{N}\), so friction can reach \(0.5 \times 600 = 300\,\text{N}\).
Torques about the foot: \(N_w \times 5 \times 0.8 = 100 \times 2.5 \times 0.6 + 500 \times x \times 0.6\). Friction needed: \(f = N_w\).
\(4N_w = 150 + 300x\), so \(N_w = 37.5 + 75x\). The ladder holds while \(37.5 + 75x \le 300\), that is \(x \le 3.5\,\text{m}\).
The painter can climb \(3.5\,\text{m}\) along the ladder, seven-tenths of its length. Any higher, and the foot slides.
A rough wall
A rough wall can also push along itself. As the top slides down, friction at the wall acts upward on the ladder. Let \(\mu_1\) be the coefficient at the floor and \(\mu_2\) at the wall. At the point of slipping, both frictions are at their limits: \(f_1 = \mu_1 N\) at the floor and \(f_2 = \mu_2 N_w\) at the wall. The wall's friction carries part of the weight, so the floor presses less, and a smaller angle becomes safe.
A uniform ladder of weight \(W\) leans with its foot on a floor of \(\mu_1 = 0.5\) and its top on a wall of \(\mu_2 = 0.5\). Find the least angle with the floor at which it can stand. Compare it with a smooth wall.
At the point of slipping: floor friction \(\mu_1 N\) towards the wall, wall push \(N_w\), wall friction \(\mu_2 N_w\) upward, and the weight at the middle.
Horizontal: \(N_w = \mu_1 N\). Vertical: \(N + \mu_2 N_w = W\). Torques about the foot: \(N_w L\sin\theta + \mu_2 N_w L\cos\theta = W\tfrac{L}{2}\cos\theta\).
The first two give \(N = W/(1 + \mu_1\mu_2)\). Then the torque equation becomes \(\tan\theta_{\min} = (1 - \mu_1\mu_2)/(2\mu_1)\). With both coefficients \(0.5\), \(\tan\theta_{\min} = 0.75/1 = 0.75\), so \(\theta_{\min} \approx 36.9°\).
The rough wall lets the ladder stand down to \(36.9°\). With a smooth wall and the same floor, \(\tan\theta_{\min} = 1\) and the least angle is \(45°\).
A ladder that stands just safely against a smooth wall is moved to a rough wall, with the same floor and the same angle. What changes?
The reaction of a hinge
A hinge holds the end of a beam or a door at one point but lets it turn. Unlike a smooth wall, a hinge can push or pull in any direction in the plane. So its force is written as two unknown components, \(H_x\) and \(H_y\). Never guess its direction. Take torques about the hinge, so that both components have no torque, and find the other unknown first.
The beam of Fig 6.1 is \(2\,\text{m}\) long and weighs \(100\,\text{N}\). The string runs from its far end to the wall at \(30°\) to the beam, and a \(100\,\text{N}\) load hangs at the far end. Find the tension \(T\) and the force of the hinge.
On the beam act \(100\,\text{N}\) down at the middle and \(100\,\text{N}\) down at the end. The string pulls with \(T\) towards the wall, and the hinge with \(H_x\) and \(H_y\).
Torques about the hinge: \(T\sin 30° \times 2 = 100 \times 1 + 100 \times 2\). Horizontal: \(H_x = T\cos 30°\). Vertical: \(H_y + T\sin 30° = 200\).
\(T = 300/1 = 300\,\text{N}\). Then \(H_x = 300 \times 0.866 \approx 260\,\text{N}\), away from the wall, and \(H_y = 200 - 150 = 50\,\text{N}\), up.
\(T = 300\,\text{N}\). The hinge pushes with \(\sqrt{260^2 + 50^2} \approx 265\,\text{N}\), at about \(10.9°\) above the beam. It is not along the beam, and not along the string.
In the first state of Fig 6.1, the beam alone feels exactly three forces. The three-force rule then says the hinge force must point at the crossing of the other two lines. The weight's vertical line at \(1\,\text{m}\) meets the string \(1 \times \tan 30° \approx 0.58\,\text{m}\) above the beam. The hinge force of \(100\,\text{N}\) at \(30°\) points exactly there, as the dashed lines show.
A uniform beam \(2\,\text{m}\) long and of weight \(60\,\text{N}\) is hinged to a wall. A string from its far end, at \(30°\) to the beam, holds it level. A \(40\,\text{N}\) load hangs at the far end. What is the tension in the string, in newtons?
The hinge of a level beam held by a string always pushes upward with half the beam's weight, whatever load hangs at the far end. Explain why.
Take torques about the far end of the beam, where the string and the load both act. Their moment arms are zero there. Only the beam's weight, at the middle, and the hinge's vertical component, at the other end, have torques. So \(H_y \times L = W \times L/2\), and \(H_y = W/2\), whatever the load.
Now return to the painter. On the balcony's rough cement, perhaps \(\mu_s = 0.6\), the least angle is \(\tan^{-1}(1/1.2) \approx 40°\). On polished marble, with \(\mu_s\) nearer \(0.2\), it is \(\tan^{-1}(2.5) \approx 68°\). At the same angle the ladder stood outside and slid inside. Propping it steeper made it safe. And the higher he climbs, the more friction he needs, so the top rungs are the risky ones.
Safety guides tell workers to set a ladder so that its foot is one unit out from the wall for every four units of height. That gives \(\tan\theta = 4\) and \(\theta \approx 76°\). By this lesson, the least coefficient of friction is then only \(\tfrac12 \times 0.25 = 0.125\). The margin covers a climber near the top and a floor that is less rough than it looks.
- Drawing friction at the foot pointing away from the wall.The wall's push drives the foot outward, so friction at the foot points towards the wall.
- Putting a friction force on a smooth wall.A smooth wall pushes only at right angles to itself. Friction appears at the wall only when it is rough.
- Guessing the direction of a hinge force.Write it as \(H_x\) and \(H_y\). Take torques about the hinge first, then use the force balance.
- Taking torques about the top of the ladder.The foot carries two unknowns, \(N\) and \(f\). Torques about the foot remove both.