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NORMAL FORCE AND FRICTION
When two solid bodies press against each other contact forces are generated in the plane of contact. These forces are due to the internal stress produced by the deformation of the bodies and they form action – reaction pairs acting on the two bodies. When the net external force pressing one body against the other is not normal to the plane of contact, the force due to internal stress can be resolved into two rectangular components: friction and the normal force (often called normal reaction in Indian textbooks). It is obvious that the line of action of friction does not pass through the centre of mass of the sliding body but the inference about the shift in the line of action of the normal force needs some explanation.
The concept of force as a push or pull is intuitive and is known to everybody. But when a body does not move in spite of a push (or pull) it is not understood intuitively that the ground on which the body rests exerts an equal push (or pull). Only after one understands that force produces acceleration, the inference is drawn about the existence of such a contact force.
Often the contact force that comes into play when a body slides over some other body is resolved into two rectangular components called the normal force and the force of friction. Though effort is made to give some explanation of the nature of friction, no need is felt to explain how the normal force is generated.
As long as the motion is only translational it is not necessary to think about the point of application of the contact forces and all the forces are treated as acting at the centre of mass of the body. But the fact remains that the force of friction being in the plane of contact of the two bodies, it generates a torque which has to be balanced by some other torque if there is no rotational motion. This balancing torque can be only due to the shifting of the line of action of the normal force exerted by the lower body. This shifting of the point of application of the normal force and the friction is not at all intuitively obvious. Hence it is necessary to explain why and how the point of application of these forces gets shifted.
The explanation is not needed to find the magnitude and the direction of the normal force. But so is the case about friction also. The inferences about the contact forces are drawn from the state of motion of the body and the knowledge of the other external forces acting on the body. If a body is at rest when an external force presses it against another body, we infer that the other body exerts a contact force on it and that force is equal and opposite to the external force. When there is also some transverse force acting on the body, we infer that a force of friction opposite to the transverse force comes into play. Since the line of action of the force of friction is confined to the plane of contact the net torque due to the external transverse force and the force of friction cannot become equal to zero. This requires that the line of action of the normal force is shifted by an appropriate amount so that the total torque on the body becomes zero. What makes the line shift is left unexplained in the absence of the discussion of the nature of the normal force. A qualitative explanation of this aspect helps to acquire a better understanding of the contact force called normal force and offers a new perspective to see why the frictional losses are proportional to the normal force.
When a body is at rest on a horizontal plane under the action of weight and the normal force, the two forces have to be equal and opposite acting along the same line passing through the center of mass of the body. When an additional transverse force tending to displace it on the plane is not able to displace it, we infer that the force of static friction acting at the surface of contact of the body with the plane is equal and opposite to the applied force. The inference is based on the first condition for static equilibrium viz,, the sum of the forces acting on the body has to be zero. But the second condition for static equilibrium requires that sum of the torques of the forces also is zero. Since the line of action of the force of static friction does not pass through the centre of mass of the body, the net torque of the applied transverse force and the force of friction cannot be zero. So, we infer that the line of action of the normal force must shift in the direction of the applied transverse force by such amount that the torque due to the normal force about the centre of mass of the body is equal and opposite to the torque due to the applied transverse force and the force of static friction. Since no further calculation is required, the matter is left there without any explanation why the line of action of the normal force gets shifted.
While discussing the motion of a body under the action of forces, when there is no rotational motion, it is possible to treat the body as a point particle where all the forces act. The translational motion of bodies under the applied forces and the friction is generally discussed with this simplification. So, no thought is given to the shifting of the line of action of the normal force, though in the figures it is often shown. Some attempt is made to explain the nature of friction but generally nothing is said about the nature of the normal force. To understand why the line of action of the normal force gets shifted when a transverse force acts on the body, it is essential to understand what produces the normal force.
The weight, or any force which presses the body against the plane, actually exerts a deforming stress on the surface of the plane. The body below the plane gets deformed and the internal stress in the deformed body produces a restoring force. We know from the theory of elasticity that the deformation stops when the restoring force is equal and opposite to the deforming force. This is the mechanism that gives rise to the normal force. If the deforming stress is not equally distributed over the surface, the deformation will not be uniform. Assuming the elastic properties of the material are uniform all over its extent, the deformation will be more where there is greater external stress. The restoring force is the net force obtained by integrating the internal stress over the entire area of contact. So, if the compression of the part of the plane where the resultant of the weight and the transverse force is directed is more, the net restoring force will shift in that direction.
What all this means is that with the transverse force acting on the body, the external normal force no longer produces deforming stress uniformly. The deformation increases in the direction of the transverse force. This becomes obvious in the extreme case where a body is about to get overturned under the action of the applied force and the static friction. As the rear side of the body is being lifted the compression of the bed below it gets reduced but the compression of the front part will increase because the external pressing force exerted by the body acts over the small area of contact.
When a body is sliding or rolling over a horizontal surface (of another body), the normal force is just equal to the weight of the body. For a sliding rectangular body the compressive stress is normal to the plane of contact at every point. But in the case of rolling the plane gets deformed into a curved surface of radius of the rolling body at the area of contact1. When the body is at rest the stress is distributed in such manner that the normal force is the integrated resultant of the internal compressive stress produced by the strain in the lower body. So though for a cylinder or sphere the normal force is radially directed at all parts of the body in contact with the lower body the resultant passes through the centre of mass of the body. For a rolling body the normal force on the front parts is more than that on the rear parts because the motion is actually a continuous collision. As a result though the normal force due to the internal stress is perpendicular at every point of the deformed bed over which the rolling body moves, the resultant force R is not normal to the surface of contact. Its line of action passes above the centre of mass of the rolling body producing a torque which retards the rolling motion .(Had the resultant (of the normal forces) passed through the centre of mass of the rolling body, there would be no torque retarding the angular velocity.) In the case of rolling motion both the point of application and the direction of the net normal force R is shifted in the direction of motion. Both the rolling friction and the normal force N balancing the weight of the rolling body are thus seen to be components of the compressive stress of the deformed plane. No additional hypothesis is needed to explain rolling friction.
Sphere deforming the plane on which it rolls and the point of application of the net force shifted in the direction of motion of the sphere
For a sphere rolling on an ideal plane both the normal force and the rolling friction must be shown to act at a point where the sphere makes no contact with the surface.
Once we see that the normal force is due to deformation of the body, we have a different perspective for understanding why the friction is proportional to the normal force. In compressing the body under it, work is done by the normal force and that work is stored in the form of strain energy. As the body is shifted under the action of the applied transverse force, the strain energy stored in the part of the other body under it gets released and gets converted into heat. The strain energy being proportional to the normal compressing force, the frictional loss is proportional to the normal force. If we postulate that this loss is due to work done against the force of friction, then the force of friction has to be proportional to the normal force. This conclusion which is stated as a law of friction appears to be a logical inference considering the processes involved in the sliding as well as in the rolling motion.
The above approach is not able to explain why the force of static friction arises and why it is self adjusting in both direction and magnitude. The general consensus about static and sliding friction is that they are due to intermolecular attractions between the molecules at the surface of contact. But this conventional theory is not able to give satisfactory explanation for the following observations.
1. There should be no doubt that the normal force is due to repulsive forces (compressive stress) between the molecules. If the external pressing force is normal to the surface of contact there is only normal force and it must be entirely due to repulsive forces between molecules at the interface. But if the direction of external force is changed even by infinitesimal angle the net repulsive force is resolved into friction and normal force and then friction is said to be due to attraction and normal force is due to repulsion between the molecules. When two bodies press against each other without producing any motion the pressing force is equal and opposite to the net contact force exerted by the other body. It appears strange that one component of this force is due to intermolecular attraction whereas the other component is due to intermolecular repulsion.
2. When a body resting on a plane is lifted there is not the slightest evidence that the external force needed to lift the body is larger than the weight of the body for overcoming the attractive forces. If by pressing the body friction is increased due to the attractive forces between molecules, why are the attractive forces not observed while lifting that body?
It is necessary to give a theory of static friction overcoming the above difficulties.
Reference 1. ‘What is Rolling Friction’ by D. A. Desai in Resonance Vol 9 No 12 (pp52-54).
The horse-cart paradox of Newtonian mechanics is discussed in introductory text-books on mechanics in the context of Newton’s laws of motion. The paradox is generally presented in the following way.
If the cart exerts an equal and opposite force on the horse, how is it possible for the horse to pull the cart?
Then the paradox is shown to be resolved by stating that for change in state of motion of a system to be possible an external force is needed, and the force which displaces both the horse and the cart is the push from the ground which is generated as reaction to the backward push on the ground by the horse.
To common sense, it appears that the horse generates internally the force necessary to move him as well as the cart. But this is considered to be a misconception leading to the paradox as above. But the belief that the force which gives impulse to the horse-cart system is generated in the body of the horse persists. The belief that the speed of an automobile is controlled by the power of its engine and not by the static friction between its tyres and road below cannot be discarded as a misconception.
To analyse the problem it is necessary to critically examine how the forward force from the ground is generated by the horse. Then it becomes clear that the paradox arises because of an inference which is arrived at by applying Newton’s Laws of Motion to a system of particles. The inference is stated as if it is a principle in the following way. “A body, which consists of a system of particles, cannot be accelerated unless an external force acts on it.” This is the conclusion arrived at by considering the possible effect of forces generated internally in the system. Though factually the statement is correct it is misinterpreted leading to an inference that the external forces do work and supply kinetic energy while explaining the locomotion of animals and engine-driven vehicles.
Let us examine the arguments leading to the inference. The internal forces are between particles of the body, and so for every force from one particle to another, there must be an equal force from the second particle on the first, and just as in the horse-cart situation, the sum of such forces must give zero resultant. So, the acceleration of the system from its internal forces is zero and if there is any acceleration it must be due to external forces acting on the system. This means no force generated in the body of the horse can accelerate him and the cart. The only external force acting on the system is the contact force between the ground and the system. It is correctly guessed that the contact force between the cart wheels and ground are dissipative. So, the only forces that could give impulse to the system could be between the hooves of the horse and the ground. The force has to be in forward direction and it can only be due to a backward push on ground by the horse. Thus the only possible conclusion that appears to explain the motion is that the forward push from the ground gives the momentum to the horse-cart system.
But does it really resolve the paradox satisfactorily? The points where the external forces act on the horse are where its hooves are firmly pressed against the ground. These points of application of the forces do not get displaced. So, no work is done by the external forces. How can the horse get kinetic energy if the external forces do no work on it?
To wriggle out of this difficulty two types of answers have been given. The first, which is found more often, states that the external force does impart the impulse to the body of the horse and hence does the work increasing its kinetic energy. But this explanation is purely speculative and the impulse cannot be measured experimentally. In cases where elastic potential energy of bodies generate forces to displace bodies in contact, forces do work transferring potential (elastic) energy into kinetic energy of the body set in motion and the point of contacts of such forces do show displacement during the impulses. In the case of motion of the horse, a belief, that the potential energy in the ground is transferred to the body of the horse is without any basis. When the horse stands still, it still presses the ground, but the potential energy stored in ground does not give it upward momentum.
The idea that a static force can do work is itself contradictory and cannot be accepted simply because it is convenient to resolve a paradox.
The second type of answers accepts that the external force is static and so the increase in the kinetic energy of the body must be from internal work in the system. But, here again, we have a paradoxical situation. From Newton’s Laws we get what is known as the work-energy theorem, according to which the increase in the kinetic energy of a body is equal to the work done on it by some force. So, if the motion of the horse is due to the internal energy, we have to accept that forces internal to a body can accelerate it. This is contradicting the assumption that a body cannot be accelerated unless the force acting on it is an external one.
The contradictions arise because the inference regarding the necessity of external force for acceleration of a body is misinterpreted to mean the work is done by the external force. If we accept that the presence of external force is necessary but the work is done by the internal force then it means that the acceleration of the body is under the action of the internal force. This way of resolving the paradox would mean that the increase in the speed of the horse is at the cost of the calories it burns while running and the increase in the speed of the car is dependent on the rate of burning fuel in its engine. This appears agreeing with common sense. To resolve the paradox in a way which does not defy common sense need not be incorrect physics. Let us see how that is possible.
The internal forces in a system add to give zero resultant force. We can express the sum of internal forces in two components: one component, which we represent as FiF, is in the direction of motion of the body and the other, represented as – FiB, in the opposite direction. Then FiF – FiB = 0. When an external force FExt acting in the forward direction on the system is present, the net force acting on the system is given by FiF – FiB + FExt . If M be the mass of the system, then its acceleration is given by a = [FiF – FiB + FExt ]/M.
If we use the condition FiF – FiB = 0, then we get a = FExt /M. The misinterpretation of the equation giving the expression for the net force becomes clear when we note that the external force is static as in the case of the motion of the horse in horse-cart paradox. If we note the empirical condition that FExt is static, we draw an inference that FiB = FExt. If we substitute this condition in the equation giving the acceleration due to the net force on the system, we get a = [FiF – (FiB – FExt )]/M = FiF/M and this tells us that the acceleration is due to the unbalanced internal force. According to this interpretation the acceleration is due to the net force acting on the body as decreed by Newton’s Second Law but the net force is not necessarily external. It is the sum of both external and internal components.
When internal forces are generated in a system by converting some form of energy into mechanical energy, always a pair of equal and opposite forces will appear. (This is demanded by Newton’s Third Law and is true in case of heat engines as well.) When the horse stands still on the ground, its weight (which is the external force exerted by the earth) presses its hooves against the ground producing the contact forces at the interface. These forces are equal and opposite. If the contact force from the ground on the horse is less than its weight, its body begins to sink in the ground, which shows that the contact force produced by the weight can be at the most equal to it in magnitude and never larger. To move forward the horse presses the ground with additional force, which is generated internally. The net contact force can in this case be larger than the weight of the horse. For forward motion, the horse presses the ground backwards, and the forward component of the contact force can never be greater in magnitude than the backward internally generated force which is responsible for its generation. The external forces are either elastic restoring forces or static frictional forces. They are self adjusting in magnitude to the applied forces and have upper limits set by the forces which are responsible for their generation. When the external contact force is less than the internally generated push, hooves of the horse tend to sink in or skid on the ground in the direction of the net force equal to FiB – FExt . This is the case when the tyres of a car on slippery muddy road rotate without accelerating the car. The external force thus generated by the internal push from the body can never give impulse to it and the traditional explanation of the cause of motion of horse-cart system is fundamentally wrong.
The actual motion of the horse is realized from the complex mechanisms of the rigid bones and muscles in his body. But the basic sequence of movements necessary for locomotion appears to be the same for all animals and insects. It consists of alternating pushes and pulls on the body by internal forces. The sequence of the internal forces necessary for locomotion of a caterpillar can illustrate the concepts involved. The sequence is as follows.
In the series of figures shown below the sequence is illustrated.
In Fig (a) the caterpillar is about to start its movements. It makes contacts with the surface on which it crawls with legs below its anterior and posterior segments. The external forces can be at either of these contacts.
In Fig (b) the caterpillar holds firmly the support with the legs under its anterior segment, lifts its posterior segment and pulls it forward by internal forces. The centre of mass of its body is displaced forward because the backward directed internal force (from the pair generated internally) is balanced by the external static force on its legs which hold the support.
Fig.1 Movements of a Caterpillar in Locomotion
In Fig.(c) the caterpillar lowers its posterior segment and with legs under that segment holds the support firmly.
In Fig.(d) the caterpillar frees its anterior, lifts it and stretches its body with the internal forward push. Here the internal backward push is balanced by the external force at the legs under its posterior segment.
The internal forward force which was earlier represented by FiF is dynamic and is represented in the figure by the symbol Fid and the equal and opposite internal backward force by Fis because it is static. The external forward force Fext gets itself adjusted till it just balances the internal force making both of them static.
The displacement S is thus achieved by the internally generated pushes and pulls by the caterpillar and not by the pushes of the supporting surface on the caterpillar’s body.
The directions of the relevant internal forces in the body of a galloping horse and the external static forces acting on its hooves are shown in the figures below.
Fig.2 Horse in Locomotion
When the hooves of front legs of the horse are in contact with the ground, the internal forces tend to pull the front leg and the rest of the body closer. The force on the rest of the body gives impulse to it in the forward direction. But the hooves being held fixed in place by the balancing of the internal force by the external contact force, the legs can only rotate about the fixed points.
Similarly, when the hooves of hind legs are in contact with the ground, the internal forces tend to push apart the rest of the body with respect the hind legs. But the legs only rotate about the fixed points on the ground, while the force Fid imparts an impulse to the rest of the body. As can be seen the forces on the legs in the two situations are in opposite directions but the force on the rest of the body is in the same forward direction giving it the momentum.
The internally generated pushes and pulls may be of different magnitudes, but essentially the sequence remains the same. In the case of animals like kangaroos jumping involves pushes during the time when they are in contact with the ground, pulls on the limbs of the body being executed when they are in air. In the case of vehicles rolling on wheels, the points of contact with the ground are continuously shifted and whether the internal forces are pushes or pulls will depend on which wheels are supplied the power.
In conclusion, the correct way of resolving the horse- cart paradox is to state that the horse generates internal pair of action-reaction forces in its body and gets the backward directed member balanced by obtaining a static force from the ground. If Mh and Mc be the masses of the horse and the cart and the magnitude of the force exerted by the cart on the horse is Fcart, then the acceleration of the horse and the cart is (Fid – Fcart)/Mh = Fcart/Mc.
The external force just cancels fully or partially the internal backward force and does not supply any impulse to the horse-cart system.
The energy equation will show that the internal energy spent is equal to the sum of the kinetic energy of the system and the heat generated in the system and the external body. For example, in the case of a running car, the power of the engine must be equal to the sum of the product of the momentum and acceleration of the car and the rate of generation of heat in the car and the road. This should be in accordance with First Law of Thermodynamics.
It should be noted that for a rigid body, the displacements of both forward and backward directed internal forces will be equal during any displacement of the body and the net work done by the internal forces would be zero. The inference that external force is needed for accelerating a body is true in this case. But bodies capable of locomotion are never rigid. Different segments of the bodies of animals or self propelled vehicles are displaced relative to other segments which are held fixed during their motion and the impulses from the internal forces during these displacements give the momentum and kinetic energy to these bodies. The parts where the external static forces are applied go on interchanging alternately.