Engineering Mechanics
Lecture 7 : Friction


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


Example 1: A 50kg block is on an inclined plane of 30°. The coefficient of static friction between the block and the plane is 0.3. We wish to determine the range of m under the block will be in equilibrium (see figure 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 of m , 2, that will balance the 50kg mass. On the other hand, friction can adjust itself according to the kind of motion; it can even change direction depending upon which way is the 50kg block slipping. Thus when friction is present, there is range of m , starting from when the 50kg block has a tendency to slide down the ramp to when m pulls it up the ramp. Let us calculate these values. We first take the case when the 50kg block 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 free body diagram of 50kg block is as follows.


A reminder here that a free body diagram is the one where we isolate a body and replace all other elements in contact with it by the forces they apply on the body. Thus the ramp surface is replaced by its normal reaction N, and the frictional force (max) µN, and the string attached with m is replaced by the corresponding tension T in it. We reiterate that we have taken the direction of up the ramp because we have 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 4 gives

50g sin30 = T + µ N

Similarly 5 gives

N = 50g sin30

To get T, we apply the force balance equation to mass m to 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 maximum value µN , eventually changing direction and increasing up to in the µN in the opposite direction. The free body diagram of 50 kg block will then look like (note that the direction of friction is opposite to that in figure 5) figure 6.


Then 7(taking the x-axis along the ramp) implies that

T = 50g sin30 + µN

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 m from 7.68 kg to 42.32 kg that can balance the 50 kg weight on the ramp; for all the values of m between the values determined above, the frictional force will be less than its maximum value. I leave this example by asking you: at what value of m will the frictional force be zero?

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


Example 2: Let us first take the case of a cylinder of radiation R and mass m kept vertically on a rough surface. It is to be rotated about the vertical by applying a torque T. We wish to calculate T when the cylinder is about to rotate. Or in other words what is value of maximum T so that the cylinder does not rotate see figure 7). The coefficient of friction between the cylinder and the surface is µ.


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


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


The frictional force on it at radius r is 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 this derivation that the weight of the cylinder is evenly distributed. If the weight is concentrated more towards the centre, T would be less and it would be more if the weight is more towards the periphery. The next example that we take is that of a block on a ramp again.


Example 3: A block of mass 100 kg is on a ramp of angle 30°. We wish to determine the magnitude and direction of the frictional force for the applied force F = 600N, F = 500N and F = 100N . 13This is a problem where we do not know a priori whether the block will be moving up or down the plane or whether it will have a tendency to move up or down the plane. So while solving 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?

That will determine the direction of friction.

To get the answer, we see that if the maximum possible static friction 14 is not able to stop the block, 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 calculate the corresponding Fo for equilibrium. If applied F is greater Fo than the block will have a tendency to move up, 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 equilibrium conditions


Let us see what happens if we increase F beyond 558.6N. In that case the component of F up the slope will increase and the block will have a tendency to move up. Let us now answer the second question whether at 600N

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


N=100gcos30° +Fsin30°

we get

N=981 17

=490 18


Thus maximum value of frictional force is


On the other hand,


Thus only 29.6N of frictional force is required to keep the block in equilibrium. This is well below the maximum possible frictional force. So under 600N, the block will be in equilibrium and the direction of friction 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. Let us again calculate the maximum possible frictional force µsN and ( 100gsin30º - Fcos30º ) for F = 500N .

N= 100gcos30° - Fsin30°





100gsin30° -Fcos30° =490 - 250 × 1.732


which is again well below the maximum frictional force of 219.7N. So the block will remain in equilibrium with 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 as a tendency to slide down since it does so for F = 500N. For 100N case

N = 100gcos30° +Fsin30°


                                                       =898.68 N





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 µsN but by µkN. Thus the frictional force is = .17x 898.68N = 152.8N. Thus the free body diagram of the block will look as given below.


Another way of doing this problem would be to apply the external force F and calculate the frictional force required to keep the body in equilibrium by assuming a direction for the frictional force. This would give both the direction and magnitude of the frictional force required for equilibrium of the block. If this force is below µsN, the body will remain in equilibrium; if it exceeds µsN, 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 27.Let T1 >T2 so that the rope has a tendency to slide to the right and therefore experiences a frictional force to the left.


Let us now consider a small section of angular width 29 so and see how it gets balanced. Let the normal to this section be shown by the dotted line.


Normal reaction in this direction balances the components of T and 31 in the opposite direction so that.


The frictional force is balanced by 33 so that


Solving this equation, we get


Thus tension in the rope increases exponentially with the contact angle. Notice that the relationship T1 =T2 does not depend on the radius of the pulley but only on the contact angle. Thus, even if the shape of the contact is not perfectly circular, the same derivation applies and the relationship between T1 and T2 remains the same, depending only on the contact angle.

Because of the relationship 36, if we take a cylindrical object - say a pencil -and let a string pass over it, then 37. If we wrap the string once more, then 38, and so on. This is nicely demonstrated by balancing a heavy object like a bottle filled with water by a light object like a bunch 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. If you measure the weights of the objects carefully, you should be able to check the relationship derived above and also get the value of the coefficient of friction (see figure 15).


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


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


Since the rope is in contact with the rope over an angle of 41, the relationship between T on the slide of the persons pulling the rope and To on the side of the load is (friction is towards the load since rope 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 = 1800N and 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 square threads passes through a nut and we wish to consider the action of the nut on the screw. You have seen this in a jack where a load (say a car) is lifted by rotating a screw. So a load W is lifted by applying a torque T on the screw (figure 17).


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


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


Solving these equations gives


This is the torque needed to lift the load W .

Let us now say that we raise the load and then release the torque. Will the jack self-lock i.e., would it hold the 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 condition looks 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 for self-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 or refrigerators at home have screw-like contraption at their legs to adjust their heights. Let the radius of such a screw be 2mm. We wish to know what should be minimum number of thread per cm so that the screw self-locks for µS =0.15.

For self locking we know that


This gives on substitution


If the number threads per cm is N then


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 and how they act under different situations.