Week 10 · Equilibrium of coplanar forces: ladders on rough ground and a smooth wall, a ladder held by a rope, and a rod about to slip · 3 lessons
1
A ladder on rough ground against a smooth wall: friction at the ground, and how far a person can climb before it slips (with the solution completed)
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 3
GoalFind how far a person can climb a ladder standing on rough ground against a smooth wall, and the least coefficient of friction μ needed.
New words
smooth wall · the wall pushes only at right angles to itself, so its reaction S is horizontallimiting friction · the greatest friction, F = μN; the ladder is about to slip
Explanation
A uniform ladder AB of weight W has its foot A on rough ground and its top B against a smooth wall. Put the wall on the right. Up and to the right are positive for the force sums, and anticlockwise moments are positive. The forces are: the weight W at the middle of the ladder; the weight P of a man at a point of the ladder; the ground reaction N, up, at A; the friction F at A, along the ground towards the wall; and the wall reaction S at B, horizontal and away from the wall (to the left). Vertical forces: N = W + P. Horizontal forces: F = S. Moments about A remove N and F and leave only S. Take a ladder 5 m long whose foot is 3 m from the wall and whose top is 4 m up (a 3-4-5 triangle). If the man is at the fraction f of the length (so s = 5f metres from A), then S × 4 = W × 1.5 + P × 3f, which gives S = 0.75(W/2 + Pf). The ladder stays up while F ≤ μN. The limit F = μN gives the highest position of the man.
Worked examples
A ladder 5 m long, of weight 160 N, stands with its foot 3 m from a smooth wall (the top is 4 m up) on rough ground with μ = 1/2. A man of weight 200 N climbs it. How far along the ladder can he go? (Wall on the right; up, right and anticlockwise are positive.) Step 1: N = 160 + 200 = 360 N. Step 2: limiting friction F = (1/2)(360) = 180 N, so S = 180 N. Step 3: moments about A: 180 × 4 − 160 × 1.5 − 200 × 3f = 0, so 720 = 240 + 600f. Step 4: f = 0.8, so s = 4 m from the foot.
The same ladder and man. What is the least μ that keeps the ladder safe with the man at its middle? (Same conventions.) Moments about A: 4S = 160 × 1.5 + 200 × 1.5 = 540, so S = 135 N. Then F = 135 N and N = 360 N, so μ ≥ 135/360 = 3/8.
Class activity
Lean a long ruler against a book standing on a desk (the book is the wall; you may put a cloth under the ruler's foot) and compare how steep it must be to stay up with and without a coin near its top. Which case slips first?
Practice
1
The 160 N ladder (5 m long, foot 3 m from the wall, top 4 m up) stands on rough ground with no man on it. Find S and the least μ.
4S = 160 × 1.5, so S = 60 N. Then F = 60 N and N = 160 N, so μ ≥ 60/160 = 3/8.
2
The 200 N man stands at the top B of the same 160 N ladder. Find S and the least μ.
4S = 160 × 1.5 + 200 × 3 = 840, so S = 210 N. N = 360 N, so μ ≥ 210/360 = 7/12.
3
A 280 N ladder of the same shape (5 m long, foot 3 m from the wall) stands on ground with μ = 1/2 and carries a 200 N man. How far along the ladder can he climb?
N = 480 N and S = F = 240 N. 240 × 4 = 280 × 1.5 + 600f, so 960 = 420 + 600f and f = 0.9, so s = 4.5 m.
4
Which way does friction at the foot act, and why?
Towards the wall, because the wall pushes the top away from itself and friction must balance that push.
5
Why does a smooth wall give no vertical force?
A smooth surface pushes only at right angles to itself; the wall is vertical, so the push is horizontal.
