Each of us is familiar with the manifestation of the force 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 figure out what static friction is.
Bar on a horizontal surface
Before proceeding to answer the questions, what is the static friction force and what is it equal to, let's consider a simple case with a bar that lies on a horizontal surface.
Let's analyze what forces act on the bar. The first is the weight of the item itself. Let's denote it with the letter P. It is directed vertically down. Secondly, this is the reaction of the support N. It is directed vertically upwards. Newton's second law for the case under consideration will be written in the following form:
ma=P - N.
The minus sign here reflects the opposite directions of the weight and support reaction vectors. Since the block is at rest, the value of a is 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.
External force acting on a bar on a horizontal surface
Now let's add one more acting force to the situation described above. Let's assume that a person starts pushing a block along a horizontal surface. Let us denote this force by the letter F. One can notice an amazing situation: if the force F is small, then despite its action, the bar continues to rest on the surface. The weight of the body 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 oppose F in any way. In that case, why does the bar remain at rest and not move?
Obviously, there must be a force that is directed against the force F. This force is the static friction. It is directed against F along a horizontal surface. It acts in the area of contact between the lower edge of the bar and the surface. Let's denote it with the symbol Ft. Newton's law for horizontal projection will be written as:
F=Ft.
Thus, the modulus of the static friction force is always equal to the absolute value of the external forces acting along the horizontal surface.
Start of bar movement
To write down the formula for static friction, let's continue the experiment started 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 there will come a moment when it starts to move. At this point, the static friction force will reach its maximum value.
To find this maximum value, take another bar exactly the same 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 found that the force F of the bar detachment from the surface also doubled. This fact made it possible to write the following formula for static friction:
Ft=µsP.
That is, the maximum value of the friction force turns out to be proportional to the weight of the body P, where the parameter µs acts as a proportionality coefficient. The value µs is called the static friction coefficient.
Since the body weight in the experiment is equal to the support reaction force N, the formula for Ft can be rewritten as follows:
Ft=µsN.
Unlike the previous one, this expression can always be used, even when the body is on an inclined plane. The modulus of the static friction force is directly proportional to the support reaction force with which the surface acts on the body.
Physical causes of force Ft
The question of why static friction occurs is complex and requires consideration of contact between bodies at the microscopic and atomic level.
In general, there are two physical causes of forceFt:
- 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 inhomogeneities. Roughly, these inhomogeneities can be represented as microscopic peaks and troughs. When the peak of one body falls into the cavity of another body, mechanical coupling 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 physical and chemical 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 moment of the beginning of the movement, the bar is forced to overcome these interactions in order to break away from the surface.
Features of Ft strength
It has already been noted above what the maximum static friction force is equal to, and also its direction of action is indicated. Here we list other characteristics of the quantity Ft.
Resting friction does not depend on the contact area. It is determined solely by the reaction of the support. The larger the contact area, the smaller the deformation of microscopic peaks and troughs, but the greater their number. This intuitive fact explains why the maximum Ftt will not change if the bar is flipped to the edge with the smallerarea.
Resting friction and sliding friction are of the same nature, described by the same formulas, but the second is always less than the first. Sliding friction occurs when the block begins to move along the surface.
Force Ft is an unknown quantity in most cases. The formula given above for it corresponds to the maximum value of Ft at the moment the bar starts moving. To understand this fact more clearly, below is a graph of the dependence of the force Ft on the external influence F.
It can be seen that with increasing F, the static friction increases linearly, reaches a maximum, and then decreases when the body begins to move. During the movement, it is no longer possible to talk about the force Ft, since it is replaced by sliding friction.
Finally, the last important feature of the Ft strength is that it does not depend on the speed of movement (at relatively high speeds, Ft decreases).
Coefficient of friction µs
Since the formula for the friction modulus contains the value µs, a few words should be said about it.
The coefficient of friction µs is a unique characteristic of the 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 rubbing bodies. For waxed wood surfaces onwet snow µs=0.14, and for human joints this coefficient takes very low values (≈0.01).
Whatever the value of µs for the pair of materials under consideration, the analogous coefficient of sliding friction µk will always be smaller. For example, when sliding a tree on a tree, it is equal to 0.2, and for human joints it does not exceed 0.003.
Next, we will consider the solution of two physical problems in which we can apply the acquired knowledge.
Bar on an inclined surface: force calculation Ft
The first task is quite simple. Suppose there is a block of wood on a wooden surface. Its mass is 1.5 kg. The surface is inclined at an angle of 15o to the horizon. It is necessary to determine the static friction force if it is known that the bar is not moving.
The catch with this problem is that many people start by calculating the reaction of the support, and then using the reference data for the coefficient of friction µs, use the above formula to determine the maximum value of F t. However, in this case, Ft is not the maximum. Its modulus is equal only to the external force, which tends to move the bar from its place down the plane. This force is:
F=mgsin(α).
Then the friction force Ft will be equal to F. Substituting the data into equality, we get the answer: the static friction force on an inclined plane Ft=3.81 newtons.
Bar on an inclined surface: calculationmaximum tilt angle
Now let's solve the following 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 bar will begin to slide.
Sliding will start when the projection of the body weight on the plane becomes equal to the maximum static friction force. Let's write the corresponding condition:
F=Ft=>
mgsin(α)=µsmgcos(α)=>
tg(α)=µs=>
α=arctg(µs).
Substituting the value µs=0, 4 into the last equation, we get α=21, 8o.
