When one surface slides over another or has a tendency to, a force parallel to the interface acts on the two surfaces which resists the sliding. This force is called friction.
This force is due to the roughness of the surfaces, the tiny bumps and valleys in a surface that prevent one surface from smoothly sliding over the other and inter molecular attractive forces between the surfaces.
These are forces between the molecules of one body and those of another, so friction is ultimately electromagnetic in nature. The normal reaction is also the result of all the electromagnetic forces between the molecules at the surface of one body with those of another. In fact, if one body exerts a force F on another , the component perpendicular to the surface of contact is called the normal reaction and the component parallel to the surfaces is called friction.
?Friction opposes motion?
You have heard that many times. You have to be careful how you understand that. Lets say you are standing still, and then you start to walk. You have accelerated in the forward direction.
What is the force that accelerated you forwards?
If you are tempted to say, ? My foot!? , keep in mind that we need an external force to accelerate an object. An object cannot act on itself.
Your feet can push on the rest of your body and accelerate it, but if I considered your entire body as one system, then we need an external force to accelerate its center of mass.
There is only one external force in the horizontal direction: friction. hence, it must be acting in the forward direction. How can that be?
Lets consider another example. A car accelerates forward. What is the force that provided this acceleration? If you are tempted to say, ?Why, the engine!?, imagine the same car out in space, drifting along. You hit the accelerator, the wheels spin, you go nowhere. Your engine is not much good. Put the same car on a frictionless surface ( a lake of ice ), the wheels spin, you go nowhere. Put the car on a road with plenty of friction, then watch it go.
Friction is the external force that accelerates the car forward. It must then point in the forward direction. How can that be?
To understand these two examples, you must look not at the objects as a whole, but at the two surfaces where the friction acts:
When you walk, you place your foot on the ground and push backwards. The foot tends to slide backward. The ground exerts an opposing frictional force, in the forward direction. Friction opposes the relative sliding of two surfaces in contact.
In the case of the wheel too, you can see that the bottom surface of the wheel tends to slide backwards along the road when we accelerate. The road resists this with friction, which acts forwards on the wheel. Of course, the reaction to this is the backward acting frictional force on the road.
There is much more to the acceleration of a car. What happens when we hit the brakes? The car is now decelerating. Which means that the frictional force on the wheels is now backwards. However, the wheels are still spinning forwards, since the car is still moving forwards. I?m afraid you will have to wait till the chapter on Rigid Body Dynamics for the full explanation. In the meantime, ponder over it.
We need an external force to accelerate. We use Newton ?s 3rd law, cleverly pushing the floor back so it pushes us forward. A flat floor can only push us forward if there is friction.
Moral of this story: Friction opposes the relative sliding of two surfaces against one another, it does not oppose motion.
Place a book on a table. What is the force of friction that is acting on the book now?
Answer: Zero. If a force were acting on the book, it would be accelerating in the direction of the force.
Now gently (gently!) push the book in one direction. It does not move. What is the force of friction now?
Answer: The frictional force must be equal to and opposite to the force exerted by you.
Now increase the force you are exerting a little. The book stays put. what can you say about the frictional force now?
Answer: Since the book does not have any acceleration, the frictional force is equal to the applied force, pointing in the opposite direction. Friction has adjusted itself to equal the external force.
As you keep increasing the force, so does friction, up until a point, when the book suddenly moves. So we conclude that friction is able to adjust itself up to a maximum value.
What are the factors that affect this maximum value? Lets call the maximum value fmax. This is the ceiling for friction under the given circumstances. Does fmax depend on the area of contact? Does it depend on the weight of the book? Does it depend on the quality of the surfaces (how polished or rough they are)?
Attach a spring scale to a block and pull gently. The reading on the scale tells you what force you are applying. Gently increase this force, and note its value when the block starts moving. Repeat the experiment several times to get an average value.
Repeat the experiment with the block lying on another side, with a different area of contact. We find the force required to start it moving is independent of the area of contact, as long as the surfaces in contact are similar. That is, fmax is independent of area of contact.
Place another block of the same mass on top of this one. Repeat the experiment. You find that the force is approximately twice as large. Try out various weights on the book. You will find that fmax is proportional to the force of contact between the two surfaces (the normal reaction).
fmax a N
The proportionality constant, denoted ms, is a number that is dependent on the surfaces. It is called the coefficient of static friction.
fmax = ms N
The smoother the two surfaces, the smaller is ms .
Remember that fmax is the maximum value that friction can have given the two surfaces and N, the force with which they are pressed together. The friction itself can have any value upto fmax. It will adjust itself to prevent any relative sliding of the surfaces. Hence,
f £ ms N
Summary:
Given two surfaces in contact, there is a number called the coefficient of static friction, which will enable us to calculate the maximum frictional force, or the ceiling for friction. The harder the two surfaces are pressed together, larger is N, hence larger is the maximum frictional force.
Examples
1) Hold a book against a wall by applying a force perpendicular to the wall
How is it that a perpendicular force is able to balance the weight of the book?
Answer: It is the frictional force between the wall and the book that holds the book up. By pressing against it, we are increasing the normal reaction force R between book and wall:
This increases the maximum possible friction between wall and book. If it is higher than the weight of the book, then the frictional force becomes equal to the weight of the book, thereby balancing it.
Lets say that ms = 0.5 between the wall and the book. Lets say that the book weighs 100 N. [ N stands for newtons here, not normal reaction]. We need a frictional force of 100 N to hold it up. Which means that fmax has to be at least 100 N.
fmax = 0.5 R, so R has to be at least 200 N. So F has to be at least 200 N, since F = R. Hence, we need to apply at least 200 n of force against the book.
What if we apply more than 200 N ? f max will be more than 100 N. But the frictional force f will remain at 100 N, balancing the book.
2) You are a rock climber, climbing a ?chimney?, which is a gap between two stone walls, with no hand holds or grips, but plenty of friction.
One method of negotiating the climb is to place yourself as shown in the pic above, and push against the walls with your feet and back so that you do not slide down.
What you are doing is increasing the normal force between your back and the wall ( and that between your feet and the other wall ) so that the maximum friction value is raised. If it becomes more than your weight, then friction adjusts itself to equal your weight, so that you are in equilibrium.
3) You are standing in a moving bus, holding on to the cylindrical hand railing on top. The driver suddenly brakes. To prevent yourself from falling forwards, you grip the railing harder. Why?
To prevent yourself from falling forwards, you need an external backward force exerted on you. The friction between your palm and the railing is that backward force. It acts since your hand tends to slide along the railing. if you grip hard, you are making the normal reaction large, hence raising the ceiling for friction. Now friction can adjust itself to the needed value.
4) You are holding a tube. Someone is trying to pull it, so that it slides through your grip. To prevent that, you tighten your grip. By increasing a force that is perpendicular to the tube (the normal reaction), you are able to increase a force that is parallel to it (the friction).
5) Perform this simple experiment at home: take a smooth plank and place a smooth block or a book on it. Now slowly ( very very slowly) raise one end, keeping the other end fixed. Hold a protractor against the block so you can measure the angle at which it starts to slide. If the angle is q, what is the coefficient of static friction ms ?
KINETIC FRICTION
It takes less force to start a body moving ( when there is friction) than to keep it moving. Once the two surfaces are sliding against each other what you have is a frictional force that is simply proportional to N.
F = mk N.
Note that we do not have an inequality here. Things are simple now. No more self adjusting force. Also, for a given pair of surfaces, the coefficient of kinetic friction, denoted by mk, is smaller than ms.
eg. What force is required to keep a 3 kg book moving at a constant velocity of 4 m/s on a table? The coefficient of kinetic friction is given as 0.2.
The normal reaction is N = mg = 3 x 9.8 = 29.4 Newtons. The force of friction is , 29.4 x 0.2 = 5.88 Newtons. This is the force you need to apply to the book so that it moves at a constant velocity, no matter what the velocity is.
Put a book or a block on a table and push gently until it starts to move. You will find that once Fmax= ms N is reached, it starts to move, but with a jerk. Once it starts to move, the force of friction suddenly drops ( mk < ms ) and the force you are applying is more than the frictional force, so it is now accelerating. You cannot smoothly take it from rest to uniform motion.
Pull a plug out from a socket by applying a steadily increasing force. Once it starts to move, it will come out fast.
Can the coeff of kinetic friction be larger than the coeff of static friction for any pair of surfaces?
Example 1.
What is the force of friction if a 20 kg force making an angle of 300 with the vertical as shown keeps the 2 kg block moving with a constant velocity? What is the coefficient of kinetic friction?
Ans.
The force diagram is shown below.
In the y-direction:
N + 20 cos (30) - mg = 0 [since the acceleration in this direction is zero]
In the x-direction:
20 sin(30) - f = 0 [ since the acceleration in the x-direction is also zero]
From this, f = 10 newtons.
From the first equation, N = 19.6 - 20Ö3 / 2
\mk = f / N = 10 / 2.28 = 4.39
Anything wrong with mk turning out to be larger than 1?
Example 2
A 3 kg block slides down a 600 ramp. The coefficient of kinetic friction is 0.2. What is the acceleration of the block?
Ans.
The force diagram is shown below:
When solving problems involving inclined planes, it is usually better to use a coordinate system where one of the axes is along the incline and the other perpendicular to the incline.
The reason will become clear soon.
In the y - direction: N - mg cos(q) = 0
[ q is the angle that the incline makes with the horizontal]
In the x-direction: mg sin(q) - f = ma
In our example, q = 600, mg = 3x9.8 = 29.4 newtons, so N = 14.7 newtons.
This gives f = mk N = 2.94 newtons.
a = (25.5 - 2.94 ) / 3 = 7.52 m / s2.
Lets do the same problem in the usual coordinate system ( x- direction along the horizontal ) shown:
In this coordinate system, the acceleration of the block ( along the incline ) has two non zero components: a cos (q) in the x - direction and - a sin(q) in the y-direction. Similarly, N as well as f now have two non-zero components, as opposed to the previous coordinate system, where only one of their two components was non-zero. The equations are:
x-direction: N sin (q) - f cos (q) = m a cos(q)
N (Ö3 / 2) - 0.2 N (1/2) = 3 a (1/2)
y-direction: N cos (q) - mg + f sin (q) = - ma sin (q)
N (1/2) - 29.4 + 0.2 N (Ö3/2) = - 3a (Ö3/2)
we need to solve these two equations for N and a.
You can see that with this coordinate system the equations are not only harder to set up, but are also harder to solve. We get the same answers upon solving them, of course.
An interesting problem:
The top block has a mass of 5 kg and the bottom block has the mass of 10 kg. The coefficients of static and kinetic friction are 0.6 and 0.4 respectively, between any of the pairs of surfaces. A force of 100 N is applied to the bottom block as shown. If the blocks are initially at rest, what are their accelerations immediately after the force is applied?
Ans.
The free body force diagrams:
Please take some time to understand the force diagrams. Make sure you understand where each force is coming from.
The frictional force f1 acting on the top block is to the right because the bottom block tends to drag the top block along with it. The reaction force on the bottom block is to the left.
N1 is the vertical force exerted by the bottom block on the top block ( and the corresponding reaction force).
f2 is the force of friction that the floor exerts on the bottom block.
If you are ever unsure of the direction of the frictional forces between two surfaces, look carefully at the surfaces in contact: friction opposes the relative sliding.
There are three possibilities here: the blocks do not move, the two blocks move as one, and two have different accelerations. Which case is it?
In problems like this one, we have to consider each case separately. One of them will give us sensible answers and the other will give us nonsensical answers.
Case 1: Does the blocks move in the first place? N2 = 150N, ms=0.6, so fmax = 90N. The external force is 100N, so they do move.
Case 2: They move as one.
This assumption implies that the force of friction f1 has not reached fmax.
So we cannot put f1 equal to 0.6 N1.
The two blocks have the same acceleration, a. So treating them as one block,
100 - (0.4) (150) = 15 a. The total mass is 15 kg, the normal reaction force between the ground and the lower block N2 is 150 newtons.
This gives us a = 40 / 15 = 2.67 m / s2.
Every thing looks fine but before we move on to the next problem, we need to check if this answer is meaningful. If the acceleration of the top block is 2.67 m/s2, then the frictional force acting on it must be 5 x 2.67 = 13.33N. We should check if this is greater or less that fmax for the top block.
fmax = 0.6 x 50 = 30N. The frictional force is in fact, less than fmax, hence our answer is consistent. Had it turned out otherwise, we would have to consider Case 3 to see if it gave a consistent answer.
Just for the fun of it, lets examine Case 3: They have different accelerations. This means that there is relative sliding between the blocks, so
f1 = mk N1 = 0.4 x 50 = 20N
This means that the bottom block?s acceleration is given by:
100 ? f1 ? f2 = 100 ? (0.4 x 150) ? 20 = 10 a
a = 2 m/s2.
The top block?s acceleration is given by: f1 = 5a giving a = 4 m/s2. This tells us that upon pulling the bottom block, the top block has a larger acceleration forward than the bottom block! This is not only nonsensical, but more importantly, it is inconsistent with the direction of the friction f1 that we assumed between the blocks.
?Inertia?
The following experiment is often used to explain inertia:
A postcard is placed over a tumbler. A coin is placed over the card. The card is now smartly flicked away horizontally or simply pulled away fast. The coin drops into the tumbler. The teacher then asks the students why the coin dropped into the tumbler and did not move with the card. The students all chorus, ?Inertia!?. The teacher nods, satisfied, and moves on to the next lesson.
Here is another explanation:
When you drag the card out slowly, the coin stays on the card, it does not drop into the glass. Let the coefficient of static friction between the card and the coin be ms. The coin?s mass is m. The maximum value of static friction is therefore, ms mg. When you pull the card, the friction on the coin drags it along with the card.
If the card?s acceleration is small, friction is able to provide the card the necessary acceleration so that it moves with the card. That is, it prevents relative sliding. If the card?s acceleration is large, then friction, having a ceiling, is not able to give the coin the same acceleration. What is this threshold acceleration? Ans. ms g
The card then slides wrt the coin. Now kinetic friction takes over. The acceleration of the coin is mk g. Since the card?s acceleration is more than this, it moves out from under the coin fast.
Let the time taken for the card to clear the glass be t. During t the coin moves ½ mk g t2. If this is less than the distance to the edge of the glass, it falls in. Otherwise, it falls outside the glass. So for the trick to work, t should be small, or in other words, the card should be made to clear the glass fast. Once the card has gone past the coin, the coin is in parabolic free fall.
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