# Static friction: coefficient, example, exercise

The **static friction** is the force that arises between two surfaces wherein one surface does not slide relative to one another. It is of great importance, as it allows us to advance when walking, as it is the force present between the floor and the sole of the shoes.

It is also the static friction that appears between the asphalt and the car’s tires. If this force is not present, it is impossible for the car to start moving, as happens in a car trying to start on a frozen surface: the wheels slide, but the car does not advance.

Static friction depends on the roughness of the surfaces in contact and also on the type of material from which they are made. Therefore, the wheels and sports shoes are made of rubber, to increase friction with the asphalt.

In the static friction model, the characteristics of materials and the degree of roughness between surfaces are summarized in a number called the *coefficient of static friction* , which is determined experimentally.

**Static friction coefficient**

The figure above shows a book that is supported on a table with an inclination of 15.7º.

If the surfaces of the book and table were very smooth and polished, the book could not be kept at rest. But since they are not, a force appears tangent to the surfaces in contact called a *static friction* force .

If the tilt angle is large enough, there will not be enough *static friction force* to balance the book and it will start to slide.

In this case, there is also friction between the book and the table, but that would be a *dynamic frictional **force* , also called *kinetic friction* .

There is a boundary between static friction and dynamic friction, which occurs when static friction reaches its maximum value.

Consider in Figure 2 the force diagram of a book of mass m that remains at rest on a plane of inclination α.

The book is kept at rest because the static friction force F balances the system.

If the tilt angle increases a little, the contact surfaces should provide more frictional force, but the amount of static friction the contact surfaces can provide has a maximum limit F _{max} , ie:

F ≤ F _{max} .

The maximum static friction force will depend on the materials and the degree of roughness of the surfaces in contact, as well as the firmness of the grip.

The coefficient of static friction μ _{e} is a positive number that depends on the characteristics of the surfaces in contact. The normal force **N** that the plane exerts on the block is responsible for the degree of tension between the block surface and the plane. This is how they determine the maximum frictional force surfaces provide when there is no slippage:

F _{max} = μ _{and} C

In summary, the static friction force follows the following model:

F ≤ μ _{and} N

**Example: determination of static friction coefficient**

The static friction coefficient is a dimensionless number that is experimentally determined for each surface pair.

We consider the block at rest in Figure 2. The following forces act on it:

– Friction force: **F**

– The weight of the mass block m: **mg**

– Normal strength: **N**

As the block is stationary and has no acceleration, according to Newton’s second law, the net force – a sum vector – is zero:

**F** + **N** + **mg** =

It is considered a fixed XY coordinate system with the X axis along the inclined plane and the Y axis perpendicular to it, as shown in Figure 2.

The forces must be separated according to their Cartesian components, giving rise to the following system of equations:

**-Component X** : -F + mg Sen (α) = 0

**-Component Y** : N – mg Cos (α) = 0

From the first equation, the static friction force value is eliminated:

F = mg Sen (α)

And from the second the value of the normal force:

N = mg Cos (α)

The static friction force F obeys the following model:

F ≤ μ _{and} N

Substituting the values obtained previously in the inequality, we are left with:

mg Sen (α) ≤ μ _{and} mg Cos (α)

Considering that for α values between 0º and 90º, the sine and cosine functions are both positive and that the quotient between sine and cosine is the tangent, we are left with:

Tan (α) ≤ μ _{e}

The equality holds for a particular value of α called the critical angle and which we denote by α * , that is:

μ _{e} = Tan (α * )

The critical angle is determined experimentally by gradually increasing the slope to the exact angle at which the block starts to slide, which is the critical angle α * .

In the book of figure 1, this angle was determined experimentally, resulting in 24°. Therefore, the static friction coefficient is:

μ _{e} = Tan (24º) = 0.45.

It is a positive number between 0 and infinity. If μ _{e} = 0, the surfaces are perfectly smooth. If μ _{and} → ∞ the surfaces are perfectly joined or welded together.

The coefficient of friction value is usually between 0 and 10.

**Exercise**

In sprint or dragster racing, accelerations of up to 4g are achieved during the start, which are achieved exactly when the tires do not slide relative to the pavement.

This is because the static friction coefficient is always greater than the dynamic friction coefficient.

Assuming the total weight of the vehicle plus the driver is 600 kg and the rear wheels support 80% of the weight, determine the static friction force during start-up of 4g and the coefficient of static friction between the tires and the pavement.

**Solution**

According to Newton’s second law, the net force is equal to the vehicle’s total mass by the acceleration it acquires.

As the vehicle is in vertical balance, the normal and the weight cancel each other out, leaving as a resultant force the friction force F that the pavement exerts in the contact area of the traction wheels, leaving:

F = m (4g) = 600 kg (4 x 9.8 m / s ^{2} ) = 23520 N = 2400 kg-f

In other words, the pulling force is 2.4 tons.

The friction force that the wheel exerts on the floor recedes, but its reaction, which is equal and opposite, acts on the rim and advances. This is the force that drives the vehicle.

Obviously, all this force is produced by the engine trying to push the tread backwards through the wheel, but the wheel and tread are coupled by the friction force.

To determine the coefficient of static friction, we use the fact that the F obtained is the maximum possible friction, since we are at the limit of maximum acceleration, therefore:

F = μ _{and} N = μe (0.8 mg)

The fact that the rear-wheel drive wheels support 0.8 times the weight was taken into account. Clearing the friction coefficient, we obtain:

μ _{e} = F / (0.8 mg) = 23520 N / (0.8 x 600 kg x 9.8 m / s ^ 2) = 5.

Conclusion: μ _{e} = 5.