Friction

Friction is a contact force that acts tangent to the interface between two bodies. In elementary mechanics, friction is usually modeled by empirical relations involving the Normal force. The model is simple, but using it correctly requires more care than merely writing f = μN.

The most important distinction is between static and kinetic friction. Static friction acts when the contacting surfaces do not slide relative to one another. Its magnitude adjusts to whatever value is required by the equations of motion, up to a limiting value. Kinetic friction acts when the surfaces slide relative to one another and is commonly approximated by a nearly constant magnitude proportional to the normal force.

This article develops the standard dry friction model, explains its physical meaning and limitations, and shows how it enters Newton’s second law problems.

1 Friction is a contact interaction

At a rough interface, the total contact force can be decomposed into a component normal to the surface and a component tangent to the surface. The normal component is the normal force N. The tangential component is the friction force f.

For a horizontal block pulled to the right, a friction force may act to the left:

Friction does not have a universally fixed direction. The correct statement is:

(1)

This is more precise than saying that friction “opposes motion.” Friction can accelerate an object. For example, a moving conveyor belt can exert friction on a box and speed the box up.

2 Static friction

Suppose a block rests on a rough horizontal surface and a horizontal applied force F is increased gradually from zero. If the block remains at rest, Newton’s second law in the horizontal direction requires

(2)

Thus static friction is not automatically equal to μ_sN. Instead, it adapts to the force needed to prevent relative slipping, subject to the inequality

(3)

The largest possible static friction magnitude is

(4)

The equality applies only at the threshold of impending motion.

If the equations require a static friction magnitude greater than μ_sN, the assumed no-slip state is physically impossible. The surfaces must begin to slide, and the kinetic friction model then becomes appropriate.

3 Kinetic friction

For two surfaces sliding relative to one another, the elementary Coulomb model uses

(5)

The kinetic friction force points opposite the relative tangential velocity of the surfaces at the contact.

For many common dry surface pairs,

(6)

although the coefficients are empirical properties of the contacting materials and surface condition rather than universal constants.

The idealized transition is often represented schematically as follows:

4 Why the coefficient of friction is dimensionless

Both friction and normal force have units of force. Therefore

(7)

is dimensionless.

The coefficient of friction summarizes complicated microscopic interactions into one empirical number. Real friction can depend on surface contamination, temperature, speed, wear, deformation, lubrication, and many other effects. The elementary model is most useful when those effects can be neglected over the range of interest.

5 The normal force must be found first

Because the friction model contains N, one should normally determine the normal force before calculating friction.

For a block on a horizontal surface with no other vertical forces and no vertical acceleration,

(8)

But this is not a universal relation. If an applied force has a vertical component, the normal force changes.

For an applied force F directed upward at angle α above the horizontal, vertical force balance gives

(9)

so

(10)

The upward component of the pull reduces the normal force and therefore reduces the available friction.

If instead the force is directed downward into the surface, its vertical component increases N and increases the friction magnitude.

6 A reliable static friction procedure

Static friction problems are safest when solved in two stages.

First, assume no slipping and use Newton’s second law to determine the friction force required by that assumption. Call the result f_req.

Second, compare that required magnitude with the maximum available static friction:

(11)

If the inequality is satisfied, the static solution is possible and

(12)

If

(13)

the static solution is impossible and the body begins to slide.

This procedure prevents one of the most common errors in mechanics: setting

(14)

in every static problem.

7 Friction on an inclined plane

For a block on a plane inclined at angle 𝜃, choose axes parallel and perpendicular to the plane. If no other forces have components normal to the plane,

(15)

The component of gravity down the plane is

(16)

If the block is at rest, static friction must oppose the tendency to slide downhill. The required magnitude is

(17)

The largest available static friction is

(18)

Therefore static equilibrium is possible only when

(19)

Canceling mg cos 𝜃 gives

(20)

At the limiting angle 𝜃_s at which slipping is about to begin,

(21)

This angle is often called the angle of repose for a simple block-surface model.

8 Sliding down an incline

If the block slides downhill, kinetic friction acts uphill with magnitude

(22)

Taking downhill as positive,

(23)

Therefore

(24)

and

(25)

The mass cancels. In this idealized model, blocks of different mass but the same friction coefficient have the same acceleration on the same incline.

9 Friction direction must come from the tendency to slip

Consider a block resting on an incline. Gravity tends to move the block downhill, so static friction points uphill.

But suppose an external force pulls the block strongly uphill while the block remains at rest. The tendency to slip may then be uphill, so static friction points downhill.

Thus the direction of static friction cannot be determined solely from the direction of gravity or from the current velocity of the center of mass. One must determine which way the contacting surfaces would tend to slide relative to each other if friction were absent.

10 Friction in connected body systems

Friction often appears together with strings and Pulleys. The basic procedure remains the same:

1. identify whether the surfaces are static or sliding;
2. find the normal force;
3. determine the friction model and direction;
4. write the string constraint if bodies are connected;
5. apply Newton’s second law to each body or to a useful combined system.

For a block m₁ sliding on a horizontal table with kinetic friction and connected over an ideal pulley to a hanging mass m₂, the kinetic friction magnitude is

(26)

If m₂ moves downward and m₁ moves toward the pulley, the system equation is

(27)

Therefore

(28)

The direction assumed in the derivation must still be checked against the sign of the result.

11 Common misconceptions

1. Static friction is not always μ_sN. It satisfies |f_s|≤ μ_sN and only reaches equality at impending slip.
2. Friction is not always opposite the object’s velocity. It opposes relative sliding or the tendency to slide at the contact.
3. The normal force is not always mg. Angled applied forces, curved motion, elevators, and other effects can change N.
4. A friction coefficient is not a force. It is a dimensionless empirical ratio used in the model.
5. Friction need not always slow an object. A moving surface can exert friction that accelerates another body.
6. The elementary model is not exact. Real friction is more complicated than the constant-coefficient Coulomb model.

12 Worked example 1: static friction adjusts to the applied force

A 10.0 kg block rests on a horizontal surface with

(29)

First, a horizontal force of 30.0 N is applied. Determine whether the block moves and find the friction force.

The normal force is

(30)

The maximum static friction is

(31)

To remain at rest, the required friction is only 30.0 N. Since

(32)

the static solution is valid:

(33)

Now increase the applied force to 50.0 N. Static friction cannot supply 50.0 N because its maximum is only 39.2 N. The block slides, so

(34)

Newton’s second law gives

(35)

so

(36)

13 Worked example 2: block on a rough incline

An 8.00 kg block is placed on a 25.0^∘ incline. The coefficients are

(37)

Determine whether the block remains at rest. If it slides, find its acceleration.

The static criterion is

(38)

Here

(39)

Static friction is insufficient, so the block slides downhill.

The acceleration is

(40)

Therefore

(41)

Notice that the mass did not enter the final acceleration.

14 Worked example 3: pulling upward at an angle

A 20.0 kg crate slides on a horizontal floor with μ_k = 0.25. It is pulled by a force of 80.0 N at 30.0^∘ above the horizontal. Find the normal force, kinetic friction, and horizontal acceleration.

Vertical acceleration is zero, so

(42)

Thus

(43)

which gives

(44)

The kinetic friction is

(45)

The horizontal component of the applied force is

(46)

Therefore

(47)

so

(48)

15 Worked example 4: connected bodies with friction

A 6.00 kg block rests on a horizontal table and is connected by a massless inextensible string over an ideal pulley to a hanging 3.00 kg mass. For the table contact,

(49)

Determine whether the system moves. If it does, find the acceleration and Tension.

The largest static friction on the table block is

(50)

If the system were at rest, the hanging mass would require

(51)

The table block would then require 29.4 N of static friction, which exceeds the available 14.7 N. Therefore the system moves.

Once sliding begins,

(52)

For the two-body system,

(53)

Hence

(54)

so

(55)

For the table block,

(56)

so

(57)

16 Practice problems

1. A 12.0 kg block rests on a horizontal floor with μ_s = 0.50. A horizontal force of 20.0 N is applied. Find the actual static friction force.
2. For the block in Problem 1, what is the largest horizontal applied force that can be exerted without causing sliding?
3. A 5.00 kg block slides across a horizontal surface with μ_k = 0.20. A horizontal force of 18.0 N acts in the direction of motion. Find the acceleration.
4. A 15.0 kg crate is pulled by a 60.0 N force at 25.0^∘ above horizontal on a surface with μ_k = 0.30. Find N, f_k, and the horizontal acceleration.
5. The same 15.0 kg crate is instead pushed by a 60.0 N force at 25.0^∘ below horizontal. Find N and f_k. Explain why the friction differs from Problem 4.
6. A block rests on an incline with μ_s = 0.40. What is the largest incline angle for which the block can remain at rest in the simple dry friction model?
7. A block slides down a 30.0^∘ incline with μ _k = 0.20. Find its acceleration.
8. A block is held at rest on an incline by an external force directed uphill. Explain why static friction can point downhill if the external force is sufficiently large.
9. A 7.00 kg block on a horizontal table is connected over an ideal pulley to a hanging 2.00 kg mass. If μ_s = 0.35, determine whether the system can remain at rest.
10. A block just begins to slide when an incline reaches 18.0^∘. Estimate μ _s.
11. A box initially at rest is placed on a conveyor belt moving to the right. Assuming slipping initially occurs, which direction does kinetic friction on the box point? Does friction initially speed up or slow down the box?
12. Two blocks of different mass slide down the same incline with the same μ_k. According to the elementary kinetic friction model, do they have different accelerations? Explain from the equations.

17 Answer check

1. The required friction is 20.0 N, and the maximum available is (0.50)(12.0)(9.81) = 58.9 N. Therefore

   (58)

2. (59)

3. (60)

   (61)

4. (62)

   (63)

   (64)

5. (65)

   (66)

   The downward force component increases the normal force, so the kinetic friction magnitude increases.

6. (67)

7. (68)

8. If the applied uphill force would make the block tend to slip uphill in the absence of friction, static friction must act downhill to oppose that tendency.
9. The hanging Weight is

   (69)

   The maximum static friction is

   (70)

   Therefore the system can remain at rest.

10. (71)

11. The belt slips to the right relative to the box, so kinetic friction on the box points to the right. Friction initially speeds the box up.
12. No. Since

    (72)

    the mass cancels in the elementary model.

18 Summary

The elementary dry friction model is built around two different regimes. Static friction satisfies

(73)

while kinetic friction during sliding is modeled as

(74)

Static friction should normally be solved as an unknown force and checked against its maximum allowed magnitude. The normal force must be determined from the perpendicular dynamics rather than assumed to equal mg. Friction direction is determined by relative slipping or the tendency to slip at the contact.

For a simple block on an incline, the threshold for static equilibrium is

(75)

and a block sliding downhill has acceleration

(76)

The next article, M02-08, develops inclined-plane dynamics more broadly, including combinations of friction, tension, and externally applied forces.

References