Each of us is familiar with the manifestation of friction. Indeed, any movement in everyday life, whether it is a person walking or moving a vehicle, is impossible without the participation of this force. In physics, it is customary to study three types of friction forces. In this article, we will consider one of them, we will understand what constitutes friction of rest.
Horizontal bar
Before proceeding to the answers to the questions of what is the static friction force and what is it equal to, we consider a simple case with a bar that lies on a horizontal surface.
Let us analyze what forces act on the bar. The first is the weight of the item itself. Let us designate it with the letter P. It is directed vertically downward. Secondly, this is the reaction of the N. support. It is directed vertically upward. Newton’s second law for the case under consideration is written as follows:
m * a = P - N.
The minus sign here reflects the opposite directions of the weight vectors and the support reactions. Since the bar is at rest, the value of a is equal to zero. The latter means that:
P - N = 0 =>
P = N.
The reaction of the support balances the weight of the body and is equal to it in absolute value.
Acting external force on a block on a horizontal surface
Now, to the situation described above, we add one more acting force. Suppose a person began to push a block along a horizontal surface. We denote this force by the letter F. We can notice an amazing situation: if the force F is small, then in spite of its action, the bar continues to rest on the surface. The body weight and the reaction of the support are directed perpendicular to the surface, so their horizontal projections are equal to zero. In other words, the forces P and N cannot provide any opposition to the value of F. In this case, why does the bar remain at rest and not move?
Obviously, there must be a force that is directed against force F. This force is the friction of rest. It is directed against F along a horizontal surface. It acts in the contact area of the lower edge of the bar and the surface. Denote it by the symbol F t . Newton's law for horizontal projection is written as:
F = F t .
Thus, the modulus of the static friction force is always equal to the absolute value of the external forces acting along the horizontal surface.
Beginning of bar movement
To write down the formula of rest friction, we continue the experiment begun in the previous paragraphs of the article. We will increase the absolute value of the external force F. The bar will still remain at rest for some time, but the moment will come when it will begin to move. At this moment, the force of rest friction will acquire a maximum value.
To find this maximum value, take another exactly the same bar as the first one and put it on top. The contact area of the bar with the surface has not changed, but its weight has doubled. It was experimentally established that the force F of separation of the bar from the surface also doubled. This fact allowed us to write the following formula of rest friction:
F t = μ s * P.
That is, the maximum value of the friction force is proportional to the body weight P, where the parameter µ s acts as the proportionality coefficient. The value µ s is called the coefficient of rest friction.
Since the body weight in the experiment is equal to the support reaction force N, the formula for F t can be rewritten as follows:
F t = µ s * N.
Unlike the previous one, this expression can always be used, even when the body is on an inclined plane. The module of the rest friction force is directly proportional to the reaction force of the support with which the surface acts on the body.
Physical Causes of the Force Ft
The question of why static friction appears is complex and requires consideration of contact between bodies at the microscopic and atomic levels.
In the general case, two physical reasons for the appearance of the force F t can be called:
- The mechanical interaction between peaks and troughs.
- Physico-chemical interaction between atoms and molecules of bodies.
No matter how smooth any surface is, it has irregularities and heterogeneities. Roughly, these heterogeneities can be represented in the form of microscopic peaks and troughs. When the peak of one body falls into the hollow of another body, then mechanical adhesion occurs between these bodies. A huge number of microscopic couplings is one of the reasons for the appearance of static friction.
The second reason is the physicochemical interaction between the molecules or atoms that make up the body. It is known that when two neutral atoms approach each other, some electrochemical interactions can occur between them, for example, dipole-dipole or van der Waals interactions. At the beginning of the movement, the bar is forced to overcome these interactions in order to break away from the surface.
Ft Power Features
It was already noted above what the maximum rest friction force is equal to, and also its direction of action is indicated. Here we list other characteristics of the quantity F t .
Rest friction does not depend on the contact area. It is determined solely by the reaction of the support. The larger the contact area, the less deformation of the microscopic peaks and troughs, but the greater their number. This intuitive fact explains why the maximum value of F t does not change if the bar is turned over to a face with a smaller area.
Rest friction and sliding friction are of the same nature, are described by the same formulas, but the second is always less than the first. Sliding friction appears when the block begins to move along the surface.
Force F t in most cases is an unknown quantity. The formula given above for it corresponds to the maximum value of F t at the moment the bar starts to move. To more clearly understand this fact, below is a graph of the dependence of the force F t on the external influence F.
It is seen that with increasing F, the static friction increases linearly, reaches a maximum, and then decreases when the body begins to move. It is already impossible to talk about the force F t during movement, since it is replaced by sliding friction.
Finally, the last important feature of the force F t is that it does not depend on the speed of movement (at relatively high speeds, F t decreases).
Coefficient of friction µs
Since the value µ s appears in the formula for the friction force modulus, a few words should be said about it.
The coefficient of friction µ s is a unique characteristic of two surfaces. It does not depend on body weight, it is determined experimentally. For example, for a tree-tree pair, it varies from 0.25 to 0.5, depending on the type of tree and the quality of the surface treatment of the rubbing bodies. For a waxed wooden surface on wet snow, µ s = 0.14, and for human joints this coefficient takes on very low values (≈0.01).
No matter what the value of µ s for the considered pair of materials, the similar coefficient of sliding friction µ k will always be less. For example, when a tree glides along a tree, it is 0.2, and for human joints it does not exceed 0.003.
Next, we consider the solution of two physical problems in which we apply the acquired knowledge.
Slant bar: Ft force calculation
The first task is quite simple. Suppose that a block of wood lies on a wooden surface. Its mass is 1.5 kg. The surface is inclined at an angle of 15 o to the horizon. It is necessary to determine the force of rest friction, if it is known that the bar does not move.
The catch of this problem is that many people begin to calculate the reaction of the support, and then, using the reference data for the friction coefficient µ s , they use the above formula to determine the maximum value of F t . However, in this case, F t is not maximum. Its module is equal only to external force, which tends to move the bar from a place down the plane. This force is equal to:
F = m * g * sin (α).
Then the friction force F t will be equal to F. Substituting the data into equality, we get the answer: rest friction force on an inclined plane F t = 3.81 Newton.
Slant bar: calculating the maximum angle of inclination
Now we will solve this problem: a wooden block is on a wooden inclined plane. Assuming the coefficient of friction equal to 0.4, it is necessary to find the maximum angle of inclination α of the plane to the horizon at which the block begins to slide.
Sliding will begin when the projection of the body weight onto the plane becomes equal to the maximum rest friction force. We write the corresponding condition:
F = F t =>
m * g * sin (α) = μ s * m * g * cos (α) =>
tg (α) = µ s =>
α = arctan (μ s ).
Substituting the value µ s = 0.4 in the last equation, we obtain α = 21.8 o .