Lecture 7- Friction - Nptel

7/30/2019 Lecture 7- Friction - Nptel

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Lecture 7Friction

Whatever we have studied so far, we have always taken the force applied by one surface on an object tobe normal to the surface. In doing so, we have been making an approximation i.e., we have beenneglecting a very important force viz., the frictional force. In this lecture we look at the frictional force in

various situations.

In this lecture when we talk about friction, we would mean frictional force between two dry surfaces. Thisis known as Coulomb friction. Frictional forces also exist when there is a thin film of liquid between twosurfaces or within a liquid itself. This is known as the viscous force. We will not be talking about suchforces and will focus our attention on Coulomb friction i.e., frictional forces between two dry surfaces only.Frictional force always opposes the motion or tendency of an object to move against another object oragainst a surface. We distinguish between two kinds of frictional forces - static and kinetic - because it isobserved that kinetic frictional force is slightly less than maximum static frictional force.

Let us now perform the following experiment. Put a block on a rough surface and pull it by a force F(seefigure 1). Since the force Fhas a tendency to move the block, the frictional force acts in the oppositedirection and opposes the applied force F. All the forces acting on the block are shown in figure 1. Note

that I have shown the weight and the normal reaction acting at two different points on the block. I leave itfor you to think why should the weight and the normal reaction not act along the same vertical line?

It is observed that the block does not move until the applied force Freaches a maximum value Fmax. Thus

from F = 0 up to F = Fmax, the frictional force adjusts itself so that it is just sufficient to stop the motion. Itwas observed by Coulombs that F maxis proportional to the normal reaction of the surface on the object.You can observe all this while trying to push a table across the room; heavier the table, larger the pushrequired to move it. Thus we can write


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where s is known as the coefficient of static friction. It should be emphasize again that is the maximumpossible value of frictional force, applicable when the object is about to stop, otherwise frictional forcecould be less than, just sufficient to prevent motion. We also note that frictional force is independent ofthe area of contact and depends only on N.

As the applied force Fgoes beyond F, the body starts moving now experience slightly less force

compound to. This force is seem to be when is known as the coefficient of kinetic friction. At low velocitiesit is a constant but decrease slightly at high velocities. A schematic plot of frictional force Fas a functionof the applied force is as shown in figure 2.

Values of frictional coefficients for different materials vary from almost zero (ice on ice) to as large as 0.9(rubber tire on cemented road) always remaining less than 1.

A quick way of estimating the value of static friction is to look at the motion an object on an inclined plane.Its free-body diagram is given in figure 3.


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Since the block has a tendency to slide down, the frictional force points up the inclined plane. As long asthe block is in equilibrium

As is increased, mgsin increases and when it goes past the maximum possible value of friction fmaxtheblock starts sliding down. Thus at the angle at which it slides down we have

Let us now solve a couple of simple standard examples involving static friction/kinetic friction.


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Example 1:A 50kgblock is on an inclined plane of 30. The coefficient of static friction between the blockand the plane is 0.3. We wish to determine the range ofmunder the block will be in equilibrium (seefigure 4).

The very first question that we address first is: why is there a range of m? It is because of the friction. If

friction were absent, there is only one value ofm, , that will balance the 50kgmass. On theother hand, friction can adjust itself according to the kind of motion; it can even change directiondepending upon which way is the 50kgblock slipping. Thus when friction is present, there is range of m,starting from when the 50kgblock has a tendency to slide down the ramp to when mpulls it up the ramp.Let us calculate these values. We first take the case when the 50kgblock is about to slide down the ramp.At that point, the friction will be pointing up with its magnitude at its maximum value. In that case the f reebody diagram of50kgblock is as follows.

A reminder here that a free body diagram is the one where we isolate a body and replace all otherelements in contact with it by the forces they apply on the body. Thus the ramp surface is replaced by its


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normal reaction N, and the frictional force (max) N, and the string attached with mis replaced by thecorresponding tension Tin it. We reiterate that we have taken the direction of up the ramp because wehave assumed the block to be sliding down and we have taken the friction at its maximum possible value.This gives us the smallest possible mass m.

Taking directions along the ramp to be the x-direction and that perpendicular to it to be the y-

direction gives

50g sin30 = T + N

Similarly gives

N= 50g sin30

To get T, we apply the force balance equation to mass mto get

T = mg

Solving these equations with g = 10ms -2 gives

m = 7.68kg.

Now as we start increasing m, the frictional force would become smaller and smaller than its maximumvalue N, eventually changing direction and increasing up to in the Nin the opposite direction. The freebody diagram of50 kgblock will then look like (note that the direction of friction is opposite to that infigure 5) figure 6.

Then (taking the x-axis along the ramp) implies that


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T = 50g sin30 + N

The other two equations remain the same as in the previous case. Solution of these equations gives

m = 42.32kg

Thus we see that due to friction, there is a range of mass mfrom 7.68 kgto 42.32 kgthat can balance the50 kgweight on the ramp; for all the values of m between the values determined above, the frictionalforce will be less than its maximum value. I leave this example by asking you: at what value ofmwill thefrictional force be zero?

We now take certain specific example of friction viz. rotation of a solid cylinder against a dry surface; thisis known as dry thrust waving. We then discuss the case of belt friction and finally the square screwthread and the screw jack. In these discussions we closely follow the book by Shames on EngineeringMechanics.

Example 2: Let us first take the case of a cylinder of radiation Rand mass mkept vertically on a roughsurface. It is to be rotated about the vertical by applying a torque T. We wish to calculate Twhen thecylinder is about to rotate. Or in other words what is value of maximum Tso that the cylinder does notrotate see figure 7). The coefficient of friction between the cylinder and the surface is .

In this example, we will have to consider the torque generated by friction. To do this, let us consider a ringof radius r and thickness 'dr'and see how much frictional force does this experience (see figure 8)?


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If we assume the normal reaction to be evenly distributed, then normal reaction on the ring is

The frictional force on it at radius ris therefore

And the torque due to this friction is therefore

That is the maximum torque that can be tolerated against friction. Of course we have assumed in thisderivation that the weight of the cylinder is evenly distributed. If the weight is concentrated more towardsthe centre, Twould be less and it would be more if the weight is more towards the periphery. The nextexample that we take is that of a block on a ramp again.

Example 3:A block of mass 100 kgis on a ramp of angle 30. We wish to determine the magnitude anddirection of the frictional force for the applied force F = 600N, F = 500Nand F =

100N. This is a problem where we do not know a priori whether the block will bemoving up or down the plane or whether it will have a tendency to move up or down the plane. So whilesolving we have to keep it mind :

1. Whether the block will remain stationary or move.2. Which way does the block have tendency to move or which way does it move?


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That will determine the direction of friction.

To get the answer, we see that if the maximum possible static friction is not able to stop theblock, it will move and in that case the friction will be kinetic. We will check that as we solve the problem.To understand which way will be the friction act, let us first assume that there is no friction and calculatethe corresponding Fofor equilibrium. If applied F is greaterFothan the block will have a tendency to moveup, otherwise it will have a tendency to move down the plane. Free-body diagram of the block looks(friction = 0) as follows.

Taking x-axis along the plane and y-axis perpendicular to it (see figure) we get from the equilibriumconditions

Let us see what happens if we increase Fbeyond 558.6N. In that case the component ofFup the slopewill increase and the block will have a tendency to move up. Let us now answer the second questionwhether at 600N

( Fcos30 - 100gsin 30 )> Max. Friction OR ( Fcos30 -100gsin ) < Max Friction

From

N=100gcos30 +Fsin30


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we get

N=981

=490

=848.68+300=1148.68N

Thus maximum value of frictional force is

On the other hand,

Thus only 29.6Nof frictional force is required to keep the block in equilibrium. This is well below themaximum possible frictional force. So under600N, the block will be in equilibrium and the direction offriction will be down the plane. The free body diagram will look as follows in this case.

Now we consider the case when the horizontal push is changed to 500N.At F = 500N ( Fcos30 - 100gsin30) will be negative so the block will have a tendency to slide down. Letus again calculate the maximum possible frictional force sN and ( 100gsin30 - Fcos30) forF = 500N.

N= 100gcos30- Fsin30


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=848.68+250=1098.68

and

Further

100gsin30-Fcos30=490 - 250 1.732

=57N

which is again well below the maximum frictional force of 219.7N. So the block will remain in equilibriumwith its free-body diagram a follows.

Next we consider the case of the force pushing the block to be 100N. At F= 100N obviously the block asa tendency to slide down since it does so forF= 500N. For 100N case

N = 100gcos30+Fsin30

=848.68+50

=898.68N

and

Further


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This force is larger than the maximum frictional (static) force. Thus the block will start sliding down,

However when the block slides down, the friction will no more be given by sN but by kN. Thus thefrictional force is = .17x 898.68N = 152.8N. Thus the free body diagram of the block will look as givenbelow.

Another way of doing this problem would be to apply the external force Fand calculate the frictional forcerequired to keep the body in equilibrium by assuming a direction for the frictional force. This would giveboth the direction and magnitude of the frictional force required for equilibrium of the block. If this force is

below sN, the body will remain in equilibrium; if it exceeds sN, it will move.

Next we consider friction on a belt or a rope going around an object. For simplicity, we consider the rope

to be going around a pulley and making contact with it over an angle .Let T1 >T2so that the rope hasa tendency to slide to the right and therefore experiences a frictional force to the left.


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Let us now consider a small section of angular width so and see how it gets balanced. Let thenormal to this section be shown by the dotted line.

Normal reaction in this direction balances the components ofTand in the opposite direction sothat.


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The frictional force is balanced by so that

Solving this equation, we get

Thus tension in the rope increases exponentially with the contact angle. Notice that therelationshipT1 =T2does not depend on the radius of the pulley but only on the contact angle. Thus, even ifthe shape of the contact is not perfectly circular, the same derivation applies and the relationship betweenT1 and T2remains the same, depending only on the contact angle.

Because of the relationship , if we take a cylindrical object - say a pencil -and let a string

pass over it, then . If we wrap the string once more, then , and so on. This isnicely demonstrated by balancing a heavy object like a bottle filled with water by a light object like abunch of keys tied on two ends of a string and the string itself going many times over a rough stick. More

you wrap the string, a bigger filled bottle would you be able to balance with the same bunch of keys. Ifyou measure the weights of the objects carefully, you should be able to check the relationship derivedabove and also get the value of the coefficient of friction (see figure 15).


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We unknowingly use the effect discussed above when we wrap a clothes-string over the nail on which it istied many times to make it tight. Similarly boats are tied with the ropes holding them going over a polemany times. We now solve an example where the surface over which the rope passes is not cylindrical.

Example:A 70kgload is being lifted by tying a rope to it and passing the rope over a tree-trunk. Thepersons lifting it have to apply a force of 1800N just when the load starts moving up. What is thecoefficient of static friction between the tree-trunk and the rope if the contact angle between the trunk andthe rope is 120? (see figure 16).


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Since the rope is in contact with the rope over an angle of , the relationship betweenTonthe slide of the persons pulling the rope and Toon the side of the load is (friction is towards the load sincerope is about to move towards the persons pulling it)

Notice that we are not worrying about the shape of the tree-trunk but rather only about the contact angle.

It is given that T = 1800Nand T0=70 x 9.8 = 688N. Therefore we have

Finally, we take the example of a square screw thread and a screw jack. In this case a screw with squarethreads passes through a nut and we wish to consider the action of the nut on the screw. You have seenthis in a jack where a load (say a car) is lifted by rotating a screw. So a load Wis lifted by applying atorque Ton the screw (figure 17).


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The threads on the screw make an angle with the horizontal so that

where pis the pitch and ris the mean radius of the screw. We wish to find the torque T(minimum) that isneeded to lift the load W. When the screw is being lifted up the nut applies a normal reaction and africtional force on the screw shown in figure 16. Keep in mind that the normal reaction and the friction acton the periphery of the screw. Thus they also apply a couple about the vertical axis on the screw. Thefigure also shows the load W. Balancing the force in the vertical direction and the torque about thevertical axis gives

Solving these equations gives

This is the torque needed to lift the load W.


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Let us now say that we raise the load and then release the torque. Will the jack self-lock i.e., would it holdthe load where it is and would not unwind. We want to find the condition for this. When the screw self-locks, it will have a tendency to move down. Thus the free body diagram of the screw in this conditionlooks as follows.

Balancing the vertical force now gives

and if the screw is not to unwind, the torque due to the friction should be large than that due N. Thus forself-locking we require that

Thus is the condition for self-locking. Under this condition the screw will not unwind by itself as torque due

to the frictional force would be sufficient to prevent the unwinding due to the torque arising from N.

Example: Let us now solve a household example of this. Sometimes you see that experimental tables orrefrigerators at home have screw-like contraption at their legs to adjust their heights. Let the radius ofsuch a screw be 2mm. We wish to know what should be minimum number of thread per cm so that thescrew self-locks forS=0.15.


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For self locking we know that

This gives on substitution

If the number threads per cm is Nthen

This gives

Thus there should be a minimum of 6 threads per centimeter in order that the screw self-locks itself.

We conclude this lecture with this example. In this lecture we have learnt about the frictional forces andhow they act under different situations.

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Lecture 7- Friction - Nptel