Static Friction

NCEES FE Civil spec: Statics, G. Static friction. FE Reference Handbook: Statics chapter, friction and belt friction.

Static friction is the resisting force that develops between two surfaces in contact before they slide. It is not a fixed value: it grows to whatever is needed for equilibrium, up to a maximum of μ times the normal force. FE Civil problems ask whether a body slips, what force starts motion, or whether it tips before it slides.

What the NCEES spec covers

Statics section G covers dry (Coulomb) friction: blocks on flat and inclined surfaces, impending motion, tipping versus sliding, wedges and occasionally belt friction. Kinetic friction during motion belongs to Dynamics.

Key concepts

Friction is passive. Solve equilibrium first, then compare the required friction with the available maximum $μ_{s} N$ . If required exceeds available, the body slips.
Impending motion means friction is at its maximum and acts opposite the direction the body is about to move.
Angle of friction. A block on an incline with no other loads is on the verge of slipping when $tan θ = μ_{s}$ .
Tipping vs sliding. For a tall block pushed sideways, find the force to slide and the force to tip about the front edge. The smaller one governs.

Core formulas

F \leq μ_{s} N F_{m a x} = μ_{s} N tan ϕ_{s} = μ_{s}

For a block of weight W on an incline at angle θ with a force P parallel to the slope:

P_{hold} = W (sin θ - μ_{s} cos θ) P_{push up} = W (sin θ + μ_{s} cos θ)

Belt (capstan) friction, with $β$ the wrap angle in radians:

F_{1} = F_{2} e^{μ β} (F_{1} > F_{2})

Where it is in the FE Reference Handbook

The Statics chapter has short sections on friction and belt friction. The handbook gives the governing inequality, not the incline formulas above, so practice deriving them from an FBD.

Worked example

A 50 kg crate rests on a 25° incline with $μ_{s} = 0.35$ . (a) Will it slide on its own? (b) What force P parallel to the incline is needed to hold it? (c) What P would start it moving up the slope?

W = 50 (9.81) = 490.5 N W sin 2 5^{\circ} = 207.3 N N = W cos 2 5^{\circ} = 444.5 N

(a) Available friction is $0.35 (444.5) = 155.6 N$ , less than the 207.3 N pulling it down the slope. Equivalently, $tan 2 5^{\circ} = 0.466 > 0.35$ . It slides.

(b) To hold it, friction acts up the slope and helps P:

P = 207.3 - 155.6 = 51.7 N

(c) To push it up, friction acts down the slope and opposes P:

P = 207.3 + 155.6 = 362.9 N

Any P between 51.7 N and 362.9 N keeps the crate at rest.

Common mistakes

Setting friction equal to $μ_{s} N$ when the body is not at impending motion.
Using W instead of the normal force, which changes on an incline or when another force has a vertical component.
Pointing friction the wrong way. It opposes the impending motion, not necessarily the applied force.
Using the kinetic coefficient for a static problem.
Forgetting to check tipping on tall or narrow blocks.

Another worked case is in the static friction example.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A 50 kg block rests on a horizontal surface with coefficient of static friction μ_s = 0.30. What horizontal force is required to start it moving? Use g = 9.81 m/s². See the worked solution or practice the Statics & Equilibrium deck
The coefficient of static friction is μ_s = 0.40 and the normal force is 500 N. What is the maximum static friction force? See the worked solution or practice the Statics & Equilibrium deck
A block slides down a 30° incline with a coefficient of kinetic friction of 0.20. Its acceleration down the slope is most nearly: See the worked solution or practice the Dynamics & Seismic deck
A 20 kg crate on a level floor is pulled by a horizontal 100 N force. The coefficient of kinetic friction is 0.25. The crate's acceleration is most nearly: See the worked solution or practice the Dynamics & Seismic deck

Related Calculators and Examples

Frequently Asked Questions

Is static friction always equal to mu times N?

No. Static friction is only as large as needed for equilibrium. It reaches its maximum value of mu_s times N only when motion is impending.

At what incline angle does a block start to slide?

With no other forces, a block is on the verge of sliding when tan(theta) equals the coefficient of static friction, so theta = arctan(mu_s).

How do I check tipping versus sliding?

Find the applied force that causes sliding (friction at its maximum) and the force that causes tipping (moment about the leading bottom edge equal to zero with the normal force at that edge). The smaller force governs.

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