Friction in Engineering Mechanics: Ladder and Incline Method

Friction in engineering mechanics is a contact force along the surface that opposes sliding, or the attempt to slide, and it is only as large as equilibrium needs. Its ceiling is limiting friction: the maximum static friction, reached only when the body is about to slip. The moment just before slipping is called impending motion.

Push a heavy almirah gently and it stays put: friction equals your push. Push harder and friction rises to match, until it hits the ceiling and the almirah slides. Three states, three rules (N is the normal reaction):

The laws of friction say that friction acts along the contact surface against sliding, that its maximum value is proportional to the normal reaction, and that this maximum does not depend on the contact area. Kinetic friction is usually smaller than the maximum static friction (OpenStax University Physics, section 6.2).

Use this test. If the question says "about to slip", "limiting", "minimum" or "maximum", write F = μN. If the body is "at rest" or "in equilibrium", find F from equilibrium, then check F ≤ μN. If F comes out bigger than μN, the body does not stay at rest.

The angle of friction φ is the angle whose tangent is the coefficient of friction: tan φ = μ. The angle of repose is the steepest incline on which a block rests without sliding, and it equals φ, so the weight of the block never enters.

Take a block of weight W on a slope of angle θ, just about to slip:

Along the slope:        F = W sinθ
Perpendicular to slope: N = W cosθ
Limiting friction:      F = μN  →  W sinθ = μ W cosθ  →  tanθ = μ

W cancels. A 1 kg block and a 100 kg block on the same two surfaces start sliding at the same angle.

Check first whether the block slides at all: compare the friction needed to hold it, W sinθ, with the friction available, μW cosθ. Only then fix the direction of friction, write the equilibrium equations, and use F = μN if the block is about to slip.

We take g = 9.81 m/s². Every number below is our own working.

Worked example A: does the block slide?

A 12 kg block rests on a rough 25° incline with μ = 0.5. Find the friction force.

• W = 12 × 9.81 = 117.7 N. Friction needed: W sin25° = 49.8 N.
• N = W cos25° = 106.7 N, so friction available: 0.5 × 106.7 = 53.3 N.

Needed is less than available, so the block stays, and the friction is 49.8 N up the slope. Cross-check: tan25° = 0.466 is less than μ = 0.5. Writing F = μN here gives 53.3 N, the wrong answer.

Worked example B: how much force holds it?

An 8 kg block sits on a 35° incline with μ = 0.5. A force P parallel to the slope, pointing up the slope, acts on it. Find the minimum P that stops the block sliding down and the maximum P before it moves up.

• W = 8 × 9.81 = 78.5 N. W sin35° = 45.0 N. N = W cos35° = 64.3 N. μN = 32.1 N.
• tan35° = 0.700 is more than μ = 0.5, so the block slides down on its own and needs P.
• Minimum P: about to slip down, so friction acts up the slope. P + μN = W sinθ, so P = 45.01 − 32.14 = 12.9 N.
• Maximum P: about to slip up, so friction acts down the slope. P = W sinθ + μN = 45.01 + 32.14 = 77.2 N.

Any P from 12.9 N to 77.2 N holds the block; friction is zero at P = 45.0 N.

Solve a ladder friction problem in four moves: decide which surfaces are rough, draw the free body diagram, take moments about the foot, then check F ≤ μN (or set F = μN if the ladder is about to slip).

Notation: ladder length L, angle θ measured from the horizontal, foot A on the floor, top B on the wall. N_f and F_f are the floor's normal reaction and friction; N_w and F_w are the wall's.

Pick the case first: which surfaces are rough?

Read the question for "smooth" (no friction) and "rough" (friction available). Then count unknowns against your three equations.

Pehle surfaces, phir equations. Rough floor with smooth wall is the standard question, so we use it below.

Write the free body diagram and the three equations

Draw the diagram with our free body diagram routine first: the ladder's weight W_l at the midpoint, a person W_p at distance d from the foot, and the end reactions.

 wall
  |
  |B ──► N_w   (smooth wall: no friction)
  | \
  |  \
  |   ● W_p ↓   (person, d from A)
  |    \
  |     \
  |      ● W_l ↓  (at the midpoint)
  |       \
  |        \
  |_________\A_______ floor
   θ at A,  ↑ N_f     ◄── F_f  (towards the wall)

ΣFy = 0:         N_f = W_l + W_p
ΣFx = 0:         F_f = N_w
ΣM about A = 0:  N_w · L sinθ = W_l · (L/2) cosθ + W_p · d cosθ

The third equation gives N_w. Then F_f = N_w, and you compare F_f with μN_f.

Why take moments about the foot?

Take moments about the foot A because the floor's two forces, N_f and F_f, both pass through A and drop out. That leaves one unknown, N_w, so a single equation gives it directly.

Check F ≤ μN and read the climb distance

The ladder holds only if F_f = N_w is at most μN_f. Setting F_f = μN_f gives d_max, the farthest the person can climb. Whether the person's weight appears depends on the ladder.

• Weightless ladder: d_max = μL tanθ. The person's weight cancels, because W_p divides out of N_w ≤ μN_f.
• Weighted ladder: d_max = [μ(W_l + W_p) L tanθ − W_l L/2] / W_p. The answer depends on the person's weight.
• Unloaded ladder (nobody on it), minimum μ: μ ≥ ½ cotθ, equivalent to tanθ ≥ 1/(2μ). The ladder's own weight drops out.

θ here is measured from the horizontal. If the question gives the angle α with the wall, use θ = 90° − α, and tan and cot swap.

Inputs: L = 5 m, θ = 60°, floor μ = 0.4, smooth wall, ladder weight 200 N.

Worked row for 600 N: μ(W_l + W_p) = 0.4 × 800 = 320 N; 320 × 5 × tan60° = 2,771 N·m; subtract W_l L/2 = 200 × 2.5 = 500 N·m; 2,271 / 600 = 3.79 m. A heavier person climbs a little less far; weight cancels only when the ladder weighs nothing.

When the wall is rough as well, the ladder has four unknowns, N_f, F_f, N_w and F_w, but only three equilibrium equations. You can solve it only when the ladder is about to slip, because friction is then limiting at both ends together and gives two more equations.

For a uniform ladder alone, with floor coefficient μ_f and wall coefficient μ_w, wall friction acts upwards and carries part of the weight, so N_f = W − F_w. With F_f = μ_f N_f and F_w = μ_w N_w, N_w = μ_f W / (1 + μ_f μ_w). Moments about the foot then give:

tanθ = (1 − μ_f μ_w) / (2 μ_f)

Inputs: floor μ = 0.5, wall μ = 0.3, uniform ladder, no person. Then tanθ = 0.85 / 1.0, so the minimum angle is 40.4°. With a smooth wall, tanθ = 1, so it is 45.0°. For equal coefficients the formula reduces to (1 − μ²) / (2μ), as in this ladder-friction derivation, a useful cross-check.

Count the forces before you reach for a three-force shortcut (Lami's theorem, which applies only to three concurrent forces). Once friction appears, a body usually has more than three forces; our Lami's theorem page explains when three forces qualify.

Three mistakes come up again and again in friction numericals: drawing friction in the wrong direction, writing F = μN when the body is not about to slip, and taking moments about a poor point.

Use g = 9.81 m/s². Try each before reading the answer.

1. Incline. A 10 kg block rests on a rough 20° incline with μ = 0.4. Does it slide? Find the friction force.

Answer: W = 98.1 N. Needed: 98.1 sin20° = 33.6 N. Available: 0.4 × 98.1 cos20° = 36.9 N. The block stays, with friction 33.6 N up the slope.

2. Ladder with a person. A uniform 6 m ladder weighing 120 N rests at 55° to a rough floor (μ = 0.3) against a smooth wall. A 600 N person climbs it. How far along the ladder can the person go before it slips?

Answer: N_f = 720 N, so F_f max = 0.3 × 720 = 216 N = N_w. Moments about the foot: 216 × 6 sin55° = 120 × 3 cos55° + 600 d cos55°, so d = (216 × 6 × tan55° − 360) / 600 = 2.48 m.

3. Both surfaces rough. A uniform ladder rests at 50° to the floor against a wall with μ = 0.3. Find the floor μ at the verge of slipping.

Answer: tanθ = (1 − μ_f μ_w) / (2μ_f), so μ_f = 1 / (2 tanθ + μ_w) = 1 / (2 × 1.192 + 0.3) = 0.373. With a smooth wall it would be 0.42.

Revise friction in this order: the three states, the slope check, then the ladder cases. The order for the whole subject is in our Engineering Mechanics study plan, and this page's routine is a friction-specific version of the method in how to solve physics numericals.

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