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Theory/Dynamics

Dynamics · Chapter 03

Friction

Friction as a contact force: its microscopic origin, the static and kinetic regimes and the empirical laws they obey, the angle of friction and the cone it sweeps, the angle of repose, and the recommended order for solving friction problems.

24 min read · 2 topics

01

Friction, Where It Comes From

Introduction to Friction

Friction is a force that opposes the relative motion or tendency of relative motion between two surfaces in contact. It arises from the electromagnetic interactions between the atoms and molecules of the surfaces. On a microscopic level, surfaces are not perfectly smooth; they have irregularities like hills and valleys. When two surfaces are in contact, these irregularities can interlock. Adhesions can form between the atoms of the two surfaces.

Two surfaces, magnified at the contact.
Two surfaces, magnified at the contact.

When one surface moves or attempts to move over another, these interlocked points must be broken, and the adhesions must be overcome. This resistance to motion is what we experience as friction. There are two main mechanisms that contribute to friction: adhesion (the attractive forces between the surfaces) and ploughing (one surface deforming the other).

Types of Friction: Static

Friction is classified by what the surfaces are doing relative to each other. Static friction acts when there is no sliding. It is self-adjusting: it takes exactly the value needed to prevent motion, growing as the applied force grows, up to a maximum called the limiting friction.

fs≤μsNf_s \le \mu_s Nfs​≤μs​N

Static friction matches the applied force.
Static friction matches the applied force.

At rest on a rough slope.
At rest on a rough slope.

Both figures make the same point. In each, the friction equals the force trying to move the block, the applied force in one, mgsin⁡θmg\sin\thetamgsinθ in the other. μsN\mu_s Nμs​N is the largest value static friction could reach, not the value it has.

Types of Friction: Kinetic and Rolling

Kinetic Friction (fkf_kfk​)

Once the surfaces are sliding, friction settles to a roughly constant value, usually less than the limiting static friction. Always directed against the relative velocity, not against the applied force.

fk=μkNf_k = \mu_k Nfk​=μk​N

The drop from μsN\mu_s Nμs​N to μkN\mu_k Nμk​N at the moment of breaking away explains why a heavy box jerks forward as it starts to slide: the applied force suddenly exceeds the friction opposing it.

Rolling Friction

A wheel or ball rolling on a surface meets a much weaker resistance, caused by the two bodies deforming rather than by contact junctions being sheared apart. This is why it is so much easier to roll a load than to drag it.

Rolling meets far less resistance than dragging.
Rolling meets far less resistance than dragging.

Laws of Friction

The behavior of friction is summarized by the following empirical laws:

Friction against applied force.
Friction against applied force.

  1. Friction is proportional to the normal force. The maximum static friction and the kinetic friction are directly proportional to the normal force between the two surfaces.

    fs,max=μsNf_{s,max} = \mu_s Nfs,max​=μs​N

    fk=μkNf_k = \mu_k Nfk​=μk​N

  2. Friction is independent of the apparent area of contact. As long as the normal force is the same, the friction force does not depend on the area of the surfaces in contact.
  3. Kinetic friction is independent of the relative velocity of the surfaces. The kinetic friction force is approximately constant and does not change with the speed of the object (for low to moderate speeds).
  4. The coefficient of kinetic friction is generally less than the coefficient of static friction. It takes more force to start an object moving than to keep it moving. (μk<μs\mu_k < \mu_sμk​<μs​)

Reading the friction graph

A graph of friction force against applied force shows the two regimes as two distinct branches.

Concept Check: The limiting static friction is 30 N. What is the friction when the applied force is 12 N? And when it is 44 N? Think about the answers first, then study the figure.

Friction against applied force.
Friction against applied force.

The 45° branch is the self-adjusting part: friction equals the applied force point for point, and the block does not move. At the top of it the block breaks away. Friction falls to its kinetic value, which is why a heavy box jerks forward the moment it starts to slide.

The figure below runs the same history in time. The applied force is ramped up steadily, and the two bars show friction answering it.

Press Play. Static friction rises with the applied force to its 30 N limit, then drops to the kinetic 22 N as the block breaks away.
Press Play. Static friction rises with the applied force to its 30 N limit, then drops to the kinetic 22 N as the block breaks away.

The force of friction is a...

It is easier to roll a barrel than to pull it along the road. This statement is correct because:

A 5 kg block sits on a rough floor with μs = 0.4 and μk = 0.3. Consequently, g = 10 gives a limiting static friction of 20 N. You push horizontally with 12 N and the block does not move. What is the friction force?

You slide a brick across a bench, first on its large face and then on its narrow edge, pressing with the same weight. How does the friction force compare?

A block is sliding to the left across a rough floor while you push it to the right, decelerating it. Which way does kinetic friction act on the block?

02

Friction, Angles, Slopes and Problems

Angle of Friction

The angle of friction is the angle that the resultant of the limiting friction and the normal force makes with the normal force. It is denoted by λ\lambdaλ.

tan⁡(λ)=fs,maxN=μsNN=μs\tan(\lambda) = \frac{f_{s,max}}{N} = \frac{\mu_s N}{N} = \mu_stan(λ)=Nfs,max​​=Nμs​N​=μs​

So, the coefficient of static friction is equal to the tangent of the angle of friction.

The angle of friction, λ.
The angle of friction, λ.

Cone of Friction

The cone of friction is the cone generated by revolving the resultant of the normal force and the limiting friction force about the normal force. If the resultant of the applied forces is within the cone of friction, the object will remain in equilibrium.

The cone of friction.
The cone of friction.

Angle of Repose

The angle of repose is the maximum angle of an inclined plane at which a body can remain at rest, held by friction. Beyond this angle, the body will start to slide down. It is denoted by α\alphaα.

At the angle of repose, the component of gravity pulling the object down the incline is equal to the maximum static friction.

mgsin⁡(α)=fs,max=μsN=μsmgcos⁡(α)mg \sin(\alpha) = f_{s,max} = \mu_s N = \mu_s mg \cos(\alpha)mgsin(α)=fs,max​=μs​N=μs​mgcos(α)

tan⁡(α)=μs\tan(\alpha) = \mu_stan(α)=μs​

Thus, the angle of repose is equal to the angle of friction.

Below and at the angle of repose.
Below and at the angle of repose.

Reading the Angle of Repose Figure

Concept Check: Does the angle at which the block starts to slide depend on how heavy the block is? Think about the answer first, then study the figure.

Below and at the angle of repose.
Below and at the angle of repose.

Study the two arrows in the figure. As the slope steepens, mgsin⁡θmg\sin\thetamgsinθ grows and friction grows with it, matching it exactly, the block stays at rest. At α\alphaα friction reaches μsmgcos⁡θ\mu_s mg\cos\thetaμs​mgcosθ and cannot grow any further. The block starts to slide.

Both sides of mgsin⁡α=μsmgcos⁡αmg\sin\alpha = \mu_s mg\cos\alphamgsinα=μs​mgcosα carry the same mgmgmg, so it cancels: the angle of repose is the same for a matchbox and a piano on the same pair of surfaces.

Solving Friction Problems

Here is a general strategy for solving problems that involve friction:

  1. Draw a Free-Body Diagram (FBD): Draw a clear FBD for each object in the system. Show all forces acting on the object, including gravity, normal forces, applied forces, and friction.
  2. Choose a Coordinate System: Choose a convenient coordinate system. It is often helpful to align one axis with the direction of motion or the inclined surface.
  3. Apply Newton's Second Law: Apply Newton's Second Law (ΣF=ma\Sigma F = maΣF=ma) to each object, resolving the forces into their components along the chosen axes.
  4. Analyze the Friction Force:
    • If the object is at rest, use the condition of static equilibrium (ΣF=0\Sigma F = 0ΣF=0). The static friction force will be equal and opposite to the net applied force, up to its maximum value, fs,max=μsNf_{s,max} = \mu_s Nfs,max​=μs​N.
    • If the object is on the verge of moving, the static friction is at its maximum: fs=fs,max=μsNf_s = f_{s,max} = \mu_s Nfs​=fs,max​=μs​N.
    • If the object is moving, use the kinetic friction force, fk=μkNf_k = \mu_k Nfk​=μk​N. The direction of kinetic friction is always opposite to the direction of velocity.
  5. Solve the Equations: Solve the resulting system of equations for the unknown quantities.

The order to solve a friction problem in.
The order to solve a friction problem in.

Example Problem

A 10 kg block is pulled across a horizontal surface by a force of 50 N at an angle of 30° above the horizontal. If the coefficient of kinetic friction is 0.2, what is the acceleration of the block?

This example uses g=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2; the rest of the chapter rounds to 10. Work it with 10 and you should get a=2.83 m/s2a = 2.83\ \text{m/s}^2a=2.83 m/s2 instead of 2.87, the method is what matters, not the third digit.

Solution:

  1. Draw the FBD: The forces acting on the block are gravity (mg), the normal force (N), the applied force (P). Kinetic friction (f_k).

A 10 kg block dragged by a 50 N pull at 30°.
A 10 kg block dragged by a 50 N pull at 30°.

  • Choose Coordinates: Let the x-axis be horizontal and the y-axis be vertical.
  • Apply Newton's Laws:
    • Y-direction: Since there is no vertical acceleration, the net force in the y-direction is zero.

      ΣFy=N+Psin⁡(30°)−mg=0\Sigma F_y = N + P \sin(30°) - mg = 0ΣFy​=N+Psin(30°)−mg=0

      N=mg−Psin⁡(30°)=(10)(9.8)−50(0.5)=98−25=73 NN = mg - P \sin(30°) = (10)(9.8) - 50(0.5) = 98 - 25 = 73 \, \text{N}N=mg−Psin(30°)=(10)(9.8)−50(0.5)=98−25=73N

    • X-direction: The net force in the x-direction causes the acceleration.

      ΣFx=Pcos⁡(30°)−fk=ma\Sigma F_x = P \cos(30°) - f_k = maΣFx​=Pcos(30°)−fk​=ma

  • Calculate Friction: The kinetic friction is fk=μkN=(0.2)(73)=14.6 Nf_k = \mu_k N = (0.2)(73) = 14.6 \, \text{N}fk​=μk​N=(0.2)(73)=14.6N.
  • Solve for Acceleration:

    50cos⁡(30°)−14.6=10a50 \cos(30°) - 14.6 = 10a50cos(30°)−14.6=10a

    50(0.866)−14.6=10a50(0.866) - 14.6 = 10a50(0.866)−14.6=10a

    43.3−14.6=10a43.3 - 14.6 = 10a43.3−14.6=10a

    28.7=10a28.7 = 10a28.7=10a

    a=2.87 m/s2a = 2.87 \, \text{m/s}²a=2.87m/s2

  • Advantages and Disadvantages of Friction

    Friction is often called a 'necessary evil' because it has both useful and harmful effects.

    Advantages of Friction

    • Walking: Friction between our shoes and the ground allows us to walk without slipping.

    Walking is friction pushing you forward.
    Walking is friction pushing you forward.

  • Writing: Friction between a pen or pencil and paper is necessary for writing.
  • Braking: Brakes in vehicles use friction to slow down and stop.
  • Holding objects: Friction allows us to grip and hold objects.
  • Lighting a match: Friction between the matchstick and the striking surface generates heat to ignite the match.
  • Disadvantages of Friction

    • Wear and Tear: Friction causes moving parts of machinery to wear out over time.
    • Energy Loss: Friction opposes motion. Consequently, energy is wasted in overcoming it, usually as heat. This reduces the efficiency of machines.
    • Heat Generation: Friction produces heat, which can damage machinery or cause unwanted effects.
    • Reduces Speed: Friction slows down moving objects.

    Friction doing useful work.
    Friction doing useful work.

    A block is at rest on a rough inclined plane. The frictional force acting on it is...

    Sand poured onto a heap always settles into a cone of the same slope, however much you pour. What sets that slope?

    In the worked example the 50 N pull is angled 30° above the horizontal. If the same 50 N were angled 30° below the horizontal instead, what would happen to the acceleration?

    A crate has μs = 1.0 with the floor. You push it with a force angled downwards at 60° below the horizontal, as hard as you like. Will it ever slide?

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