Friction
Friction is the force that resists sliding between two surfaces in contact. It acts along the contact surface, opposite to the direction of motion or of impending motion. Without it, a brick could not rest on a sloping roof, a retaining wall would slide on its base and a vehicle could not brake. With too much of it, piles are hard to drive and sliding formwork sticks. Engineers therefore need to estimate it and decide when to rely on it.
The classical model, attributed to Coulomb, says the limiting friction force is proportional to the normal force pressing the surfaces together. The constant is the coefficient of static friction, μ. Below the limit, friction is only as large as needed to hold the body still; it is a reaction, not a fixed value. Once sliding starts, the resisting force drops a little to the kinetic value. The angle of friction φ is defined by tan φ = μ, and a block on a slope just begins to slide when the slope angle equals φ. This is the same idea, in a different setting, as the angle of internal friction of a sand.
Typical problems are a block on an incline pulled by a rope, a ladder resting against a wall, wedges used to lift heavy formwork, and belt or rope friction round a bollard. The method never changes: draw the free-body diagram, assume the direction of impending motion, put F = μN at each sliding surface, and apply equilibrium. If more than one surface could slip, check each possibility.
In civil practice friction governs the sliding check of gravity walls and dams, the skin friction on piles, the grip of rebar in concrete, the design of bridge bearings with sliding plates, and the tension lost along a prestressing tendon as it rubs the duct. Friction coefficients come from tests or code-specified values, and designers use a safety margin because surfaces can be wetter, smoother or dirtier on site than assumed.
Limiting friction: F = μ NAngle of friction: tan φ = μBlock on incline at rest without help if θ ≤ φRope round a post (capstan): T₁ = T₂ e^(μθ), θ in radians
Fixing the direction of friction before deciding which way the body is about to move.
Problem. A 2 kN crate rests on a plank sloping at 20°. The coefficient of static friction is 0.45. Does it slide, and what pull parallel to the plank is needed to start it moving up?
- Normal force N = 2 cos 20° = 1.879 kN. Down-slope weight component = 2 sin 20° = 0.684 kN.
- Maximum friction = 0.45 × 1.879 = 0.846 kN, which exceeds 0.684 kN, so the crate stays put.
- To move it up, the pull must overcome both the down-slope component and friction (now acting down-slope): P = 0.684 + 0.846 = 1.53 kN.
Answer. It does not slide; about 1.53 kN is needed to start it up the plank
Quick quiz
1Is friction always equal to μN?
No. μN is the maximum; below that, friction equals whatever is needed for equilibrium.
2What is the angle of friction for μ = 0.6?
tan⁻¹ 0.6 ≈ 31°.
3Name two civil checks governed by friction.
Sliding of retaining walls or dams on their base; skin friction on piles. Others: bearings, prestress losses.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- F. P. Beer, E. R. Johnston, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, latest edition.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, latest edition.
- S. Timoshenko, D. H. Young, Engineering Mechanics, McGraw Hill, 4th edition, 1956.
Free and official sources
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