Statics · Static and kinetic friction
Condition: The theoretical fundamentals distinguish static from sliding friction, develop the condition of sticking with the limiting static friction force, the angle of friction together with the friction wedge, and how to handle several contact points.
Open theoretical foundations →First draw every system on paper first: weight, normal force, static friction force, external forces. The sense of direction of may be assumed arbitrarily — the sign corrects it.
Exercise 1 · Concept check
A body rests on a rough surface. Which statement is correct?
The static friction force is a reaction force — it adjusts to whatever equilibrium demands and can be calculated from it for statically determinate systems. Only its upper limit is given by given.
The sliding friction force on the other hand is an applied force: as soon as the body slips, the equation applies, and its magnitude is fixed independently of equilibrium.
Assign to each statement the term it describes.
limiting static friction force
condition of sticking
Coulomb's law of friction
angle of friction
The fine distinction lies between — a quantity — and the condition of sticking — a statement. Coulomb's law of friction is the corresponding equation for the moving case.
The angle of friction with expresses the same condition geometrically: it spans the friction wedge in which the resultant of which: and must lie.
Build the free-body diagram of the crate resting at rest on the ramp. Careful: the palette offers more forces than belong in it.
Three forces: vertically downwards, normal to the ramp and along the ramp.
The offered sliding friction force belongs not in it: it only occurs once the body slips. As long as the crate is at rest, the static friction force acts — and that is not an applied but a reaction force.
For the sense of direction of you may choose freely; here it is assumed up the slope, because otherwise the crate would slip downwards. If a negative value came out, it would simply point down the slope.
Conclusion: one inequality, two limiting cases
Static friction problems differ from everything before them in the inequality sign. It turns a calculation with one answer into one with a range — and forces you to consider both limiting cases separately.
Exercise 2 · Block under a horizontal force
A block of weight rests on a plane inclined by ; the coefficient of static friction is . It is pressed against the plane by a horizontal Force force. Within what limits must lie for the block to stick?
As soon as acts, its normal component is added: . This is exactly where this exercise differs from the slope-parallel rope force of the theory page — there of independent.
Static friction force from the balance along the plane, counted positive up the slope.
Allowed symbols: G, F, alpha
Notation help: sin(alpha), cos(alpha), brackets allowed.
Along the plane the downslope force acts downwards and the slope-parallel component of the horizontal force, , upwards:
A negative result only means that points the other way — the sense of direction could after all be assumed arbitrarily.
The block threatens to slip downwards, so and :
Now the block threatens to slip upwards, so :
Both limits can be written strikingly briefly with the angle of friction — that is no coincidence but precisely the geometric meaning of the friction wedge.
Resolve the horizontal force into the directions along and normal to the plane — it has a component in both balances.
Write the condition of sticking as and resolve the absolute value into the two cases and up.
After rearranging, on both sides — collect the terms on one side before dividing.
Conclusion: the force changes the normal force too
Because presses against the plane, and with it the limiting static friction force grows as well. That is why the upper limit here is markedly higher than a slope-parallel force would allow — the block “helps itself” the harder one presses.
That is the same effect that makes wedges and screws work.
Exercise 3 · Ladder with self-weight
A ladder of length leans at against a smooth wall; the ground is rough with . The ladder itself weighs — its weight acts at the middle. A person of stands at of the ladder length.
Assign: which balance yields which quantity directly?
The moment balance about contains neither nor — both act there and have no lever arm. It therefore yields alone.
Vertically only against the two weights, horizontally only against the wall force — the smooth wall has no tangential component after all.
The ladder length appears in every term and cancels out.
Required coefficient of static friction at the ground, symbolically.
Allowed symbols: GL, Q, s, alpha
Notation help: cos(alpha)/sin(alpha) the cot; fraction with /.
At the ground, and . The condition of sticking thereby becomes:
That lies below — the ladder holds, but without much reserve.
In the limiting case the inequality becomes an equation:
For that is along the ladder. Without self-weight, would give a different value — the self-weight works in favour of stability here, because it acts lower down than the person.
Put the reference point of the moment balance at the base — two of the three unknown forces act there and drop out.
Both weights act vertically; their lever arms about the base point are and .
Conclusion: the self-weight helps
The ladder itself acts at the middle, the person higher up. Because its lever arm is smaller, the self-weight raises the ground normal force more than it drives the wall force — which makes the ladder safer, not less safe.
That is why the massless calculation from chapter 07 is the conservative estimate: it lies on the safe side.
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| Exercise | Status |
|---|---|
| Exercise 1: Concept check | Open |
| Exercise 2: Block under a horizontal force | Open |
| Exercise 3: Ladder with self-weight | Open |