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7.1 · Distinguish static, limiting, and kinetic friction

Learn to distinguish static, limiting, and kinetic friction through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Friction

Identify the contact condition, choose the correct friction model, and check the force balance

Friction is a contact force directed along the surface where two bodies touch. It opposes relative sliding, or the tendency for relative sliding. The central question is whether the surfaces are sliding relative to each other. If they are not, friction is static; if they are, friction is kinetic. Static friction can take different magnitudes, depending on what is needed to prevent sliding. Only at the threshold of slipping does it reach its maximum, called limiting friction. This lesson uses the simple Coulomb friction model and focuses on distinguishing these three descriptions.

What you will learn

  • Distinguish static friction from kinetic friction by whether the contacting surfaces slide relative to one another.
  • Explain how static friction adjusts up to a maximum value.
  • Identify limiting friction as static friction at the threshold of slipping.
  • Apply the appropriate friction rule in simple planar contact problems and check the result.

1. Contact forces and friction direction

For each example, the system is the block. We represent it as a particle when only force balance is needed; this means we track the forces on the block without examining its rotation. Choose the positive xx-direction along the surface and the positive yy-direction perpendicular to it. On a horizontal surface, right and up are positive.
The normal force NN is perpendicular to the contact surface. Friction is parallel to it. To determine friction's direction, consider which way the surfaces would tend to slide relative to one another without friction. Friction on the block opposes that relative sliding or tendency. If you initially assume a direction, a negative signed result means the actual direction is opposite your assumption.
Draw the free-body diagram for the chosen system: include the block's weight, the normal force, friction, and any applied forces. For a particle in equilibrium, use force balance along each chosen axis. The examples do not require a rigid-body moment calculation; if rotation of a body were part of the question, a particle model would not be enough.
∑Fx=0,∑Fy=0\sum F_x=0,\qquad \sum F_y=0
  • Normal force acts perpendicular to the contact; friction acts parallel to it.
  • Friction opposes relative sliding or the tendency for relative sliding.
  • Use a free-body diagram to keep the forces on the chosen system clear.

2. Static friction and its limiting value

Static friction applies when the contacting surfaces have no relative sliding. In an equilibrium problem, force balance determines the amount of static friction required. It does not automatically take its maximum value. For example, a small horizontal push may need only a small opposing friction force to keep a block at rest.
In the Coulomb model, the coefficient of static friction μs\mu_s sets the maximum available static friction for a given normal force. The magnitude of the required static friction must be no greater than this limit. If equilibrium would require more friction than the limit, the assumed no-slip condition cannot be maintained under this model.
Limiting friction is the special case at the threshold of slipping. The surfaces have not begun to slide relative to each other, but static friction has reached its maximum magnitude. Thus, use the static-friction inequality for ordinary no-slip contact, and use equality when the problem states that slipping is impending.
∣Fs∣≤μsN,∣Fs,lim∣=μsN|F_s|\leq\mu_sN,\qquad |F_{s,\mathrm{lim}}|=\mu_sN
  • Static friction adjusts to the force needed for no relative sliding, within its allowed range.
  • Limiting friction is still static friction; it occurs at the threshold of slipping.
  • The normal force affects the maximum friction in this model.

3. Kinetic friction and selecting the rule

Kinetic friction applies when the contacting surfaces are sliding relative to one another. In the Coulomb model, its magnitude is calculated using the kinetic-friction coefficient μk\mu_k and the normal force. Its direction is opposite the relative sliding direction. Unlike static friction, kinetic friction is not chosen to provide whatever force is required for horizontal equilibrium.
Start by identifying the contact condition. If there is no relative sliding, use static friction and check whether the required value is within the limit. If the block is at the threshold of slipping, use the limiting equality. If the surfaces are sliding, use the kinetic-friction rule. The coefficients alone do not tell you which condition applies; the stated or established contact condition does.
Keep the force balance consistent with the question. A kinetic-friction force can be found from its rule even if other horizontal forces are not given. Do not claim horizontal equilibrium unless the forces and stated conditions support it. In all cases, friction is a force, measured in newtons, while the coefficients are dimensionless.
∣Fk∣=μkN|F_k|=\mu_kN
  • Kinetic friction acts during relative sliding and opposes that sliding.
  • Use the kinetic coefficient only for sliding contact; use the static coefficient for static or limiting contact.
  • Apply equilibrium only where equilibrium is part of the stated model.

Worked example

Static friction below its limit

A 50 N block rests on a horizontal surface. A horizontal 20 N force pushes it right. The coefficient of static friction is 0.50. Determine whether it can remain at rest and find the friction force if it does.
  1. Identify the contact condition
    The block is at rest relative to the surface, so use static friction. Its assumed direction is left because the applied force tends to slide the block right.
  2. Balance vertical forces
    There are no other vertical forces. Vertical equilibrium makes the normal force equal to the 50 N weight.
    ∑Fy=N−50 N=0⇒N=50 N\sum F_y=N-50\,\mathrm{N}=0\quad\Rightarrow\quad N=50\,\mathrm{N}
  3. Find the required friction and test the limit
    Horizontal equilibrium requires friction to balance the applied 20 N force. The static-friction limit is the coefficient times the normal force.
    Fs=20 N,μsN=0.50(50 N)=25 NF_s=20\,\mathrm{N},\qquad \mu_sN=0.50(50\,\mathrm{N})=25\,\mathrm{N}
  4. Check equilibrium and feasibility
    The required 20 N is less than the 25 N limit, so static contact is possible. Both force sums are zero. The particle model checks force balance; it does not evaluate rotation or a moment balance.
    ∑Fx=20−20=0,∑Fy=50−50=0\sum F_x=20-20=0,\qquad \sum F_y=50-50=0
Answer: The block can remain at rest. Static friction is 20 N left, below its 25 N maximum.
Check: The actual friction is the value required for equilibrium, 20 N—not the maximum available value, 25 N.

Worked example

Limiting friction at impending slip

A 60 N block rests on a horizontal surface. The coefficient of static friction is 0.40. A horizontal force is increased until the block is just at the threshold of slipping right. Find the applied force and describe the friction force.
  1. Apply the limiting condition
    The phrase “just at the threshold of slipping” means static friction is at its limiting value. Use the equality for limiting static friction.
    Fs,lim=μsNF_{s,\mathrm{lim}}=\mu_sN
  2. Find the normal force
    The applied force is horizontal, so vertical equilibrium makes the normal force equal to the 60 N weight.
    ∑Fy=N−60 N=0⇒N=60 N\sum F_y=N-60\,\mathrm{N}=0\quad\Rightarrow\quad N=60\,\mathrm{N}
  3. Calculate the applied force
    At the threshold, horizontal equilibrium requires the applied force to balance limiting friction. Substitute the normal force and coefficient.
    P=Fs,lim=0.40(60 N)=24 NP=F_{s,\mathrm{lim}}=0.40(60\,\mathrm{N})=24\,\mathrm{N}
  4. Verify force balance
    The 24 N applied force and 24 N limiting friction cancel horizontally. The normal force and weight cancel vertically. This particle model verifies force balance; no rotation is part of the example.
    ∑Fx=24−24=0,∑Fy=60−60=0\sum F_x=24-24=0,\qquad \sum F_y=60-60=0
Answer: The applied force at the threshold is 24 N right. Limiting static friction is 24 N left.
Check: The surfaces have not begun sliding at the stated threshold, so this is limiting static friction, not kinetic friction.

Worked example

Kinetic friction during sliding

A block is known to be sliding right on a horizontal surface. Its normal force is 80 N and its coefficient of kinetic friction is 0.25. Find the kinetic friction force. No horizontal equilibrium condition is specified.
  1. Select kinetic friction
    The surfaces are sliding relative to one another, so use the kinetic-friction model. Friction acts opposite the stated sliding direction, which is left here.
    ∣Fk∣=μkN|F_k|=\mu_kN
  2. Calculate the magnitude
    Substitute the dimensionless coefficient and the given normal force. The friction magnitude has units of newtons.
    ∣Fk∣=0.25(80 N)=20 N|F_k|=0.25(80\,\mathrm{N})=20\,\mathrm{N}
  3. Check what can be concluded
    The friction force is 20 N left. No horizontal balance can be verified because no other horizontal forces are specified, and the problem does not state horizontal equilibrium. If the block has no other vertical forces, its weight is 80 N and the vertical forces balance. A particle model does not assess moments.
    Fk,x=−20 N,∑Fy=80−80=0F_{k,x}=-20\,\mathrm{N},\qquad \sum F_y=80-80=0
Answer: Kinetic friction has magnitude 20 N and acts left, opposite the block's rightward sliding.
Check: The sliding condition selects the kinetic coefficient. The static-friction limit is not the applicable rule.

Common mistakes and how to avoid them

Setting static friction equal to its maximum in every no-slip problem.
Correction: Use force balance to find the amount required, then check that it is no greater than the limit. Equality applies at impending slip.
Calling friction at impending slip kinetic friction.
Correction: At the threshold, relative sliding has not started. Friction is limiting static friction.
Choosing friction direction from the applied force on the block alone.
Correction: Consider the relative sliding or tendency of the two contacting surfaces. Friction opposes that relative direction.
Using static friction when the surfaces are sliding.
Correction: For relative sliding, use the kinetic-friction rule and direct friction opposite the sliding.

Lesson summary

  • Static friction applies when the surfaces do not slide relative to each other and adjusts to the amount needed, up to a limit.
  • Limiting friction is static friction at the threshold of slipping, with magnitude equal to the static-friction limit.
  • Kinetic friction applies during relative sliding, has magnitude given by the kinetic coefficient times the normal force, and opposes sliding.
  • Identify the contact condition before selecting the friction rule; then use force balance only when equilibrium is specified.

Check your understanding

Question 1

A block remains at rest, and equilibrium requires 12 N of friction. The maximum static friction is 18 N. What is the actual static-friction magnitude?
  1. 6 N
  2. 12 N
  3. 18 N
  4. It cannot be determined without the kinetic coefficient.
Show answer and explanation
12 N
Static friction supplies the 12 N required for equilibrium. Since this is below the 18 N limit, the no-slip condition is possible.

Question 2

At impending slip, which description is correct?
  1. Kinetic friction has reached its maximum.
  2. Static friction is at its limiting value.
  3. Friction is zero because sliding has not started.
  4. Static friction must be less than half its limiting value.
Show answer and explanation
Static friction is at its limiting value.
Impending slip is the threshold condition: the surfaces have not begun sliding, and static friction is at its limit.

Question 3

A surface is sliding left relative to a block. Which way does kinetic friction on the block act?
  1. Left
  2. Right
  3. Perpendicular to the surface
  4. Its direction cannot be related to the sliding direction.
Show answer and explanation
Left
The surface's motion relative to the block is leftward, so the block's motion relative to the surface is rightward. Kinetic friction on the block opposes that relative sliding and therefore acts left.

Key terms

Normal force
The contact-force component perpendicular to the contacting surface.
Static friction
Friction when the contacting surfaces have no relative sliding; its magnitude adjusts up to a maximum.
Limiting friction
Static friction at the threshold of slipping, with magnitude equal to the maximum static friction in the model.
Kinetic friction
Friction when contacting surfaces slide relative to one another; in the Coulomb model, its magnitude is the kinetic coefficient times the normal force.
Coefficient of friction
A dimensionless model parameter used to relate the friction limit or kinetic-friction magnitude to the normal force.

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