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THE AMAZING NORMAL FORCES
Horia I. Petrache Department of Physics, Indiana University Purdue University Indianapolis Indianapolis, IN 46202
November 9, 2012
Abstract
This manuscript is written for students in introductory physics classes to address some of the common difficulties and misconceptions of the normal force, especially the relationship between normal and friction forces. Accordingly, it is intentionally informal and conversational in tone to teach students how to build an intuition to complement mathematical formalism. This is accomplished by beginning with common and everyday experience and then guiding students toward two realizations: ( i ) That real objects are deformable even when deformations are not easily visible, and ( ii ) that the relation between friction and normal forces follows from the action-reaction principle. The traditional formulae under static and kinetic conditions are then analyzed to show that peculiarity of the normal-friction relationship follows readily from observations and knowledge of physics principles.
1. Normal forces: amazing or amusing?
Learning about normal forces can be a life changing event. In introductory physics, we accept and embrace these totally mysterious things. Suddenly, normal forces become a convenient answer to everything: they hold objects on floors, on walls, in elevators and even on ceilings. They lift heavy weights on platforms, let footballs bounce, basketball players jump, and as if this was not enough, they even tell friction what to do. (Ah, the amazing friction forces โ yet another amazing story! [1]) Life before physics becomes inexplicable. This article is about building an intuition about normal forces using the action-reaction law of mechanics and the fact that real objects are deformable. The thinking can start with looking around the room: there are chairs, tables, cabinets, and people, all sitting on the floor. Imagine how many normal forces are all around! Now, if you think about it, the floor must be very, very smart since it knows exactly where and when to apply what normal force. For example, if the teacher moves a chair and takes its place, the floor immediately brings in the correct normal force. Somehow, the floor can tell the teacher from other objects in the room. We conclude that we must find these marvelous people who build such smart floors as they must be the holders of the secret of the normal force. But there is no mystery. Normal forces are deformation forces. They can also be called contact forces. Contact means deformation, and deformation gives rise to contact forces. Normal forces are called as such because they are perpendicular to the surface of contact. In mathematics, normal means perpendicular. We learn that friction forces are also contact forces but they are parallel to the surface rather than perpendicular. Confusion regarding normal and friction forces can arise in at least two occasions. First, we must accept that friction forces are proportional to normal forces. This proportionality appears intuitive based on everyday experience, but is it an experimental result or a fundamental physics law? How to think about the fact that two perpendicular forces are proportional to one another? Second, while for objects in motion we write ๐น๐ = ๐ (^) ๐ ๐น๐ with a definite equal sign, for the static case we write the less
convincing inequality, ๐น๐ โค ๐ (^) ๐ ๐น๐. So maybe friction is not really that proportional to the normal force after all. Or is it? How do I make sure and pass this class?! Perhaps I should just answer B to all questions about friction and normal forces. Fortunately, there is no need to do that. As shown in the following, normal forces and their relationship with friction can be visualized in an intuitive way once we accept that real objects are deformable. We will first discuss objects on flat floors and then move to objects on inclines and see how the action-reaction principle comes to the rescue in each case. We will also see some interesting aspects of action-reaction forces for accelerating objects.
2. Normal forces on floors
Figure 1 shows an "artistic" view of a floor with two objects on it. In such an exaggerated vision, floors are deformable objects (more like bed mattresses) that give in under the weight of
contact forces due to the incline: a normal force F (^) N and a friction force Ff. But why two reaction forces instead of one?!
Figure 2. An object on an incline is shown to experience 3 forces: a gravitational force, a normal force, and a friction force.
According to the action-reaction principle, as the box presses its weight down on the incline, the incline should react with an equal and opposite force. This is shown in Fig. 3.
Figure 3. According to the action-reaction law, when the box acts with a force on the incline, the incline reacts with an equal and opposite force on the box. The reaction force is due to deformation at the contact area exaggeratedly shown here by the hashed area.
Figure 3 shows a pair of action-reaction forces: the box acts with force F on the incline and the incline "responds" with an equal and opposite force on the box as stated by the action-
reaction principle. What is usually not mentioned in textbooks is that we decide to decompose the reaction force into two components for convenience. One component is parallel to the surface of contact and the other is perpendicular to it. We call the former friction force and the second normal force. (The action-reaction principle does not restrict how we might like to decompose forces for our own benefit.) So one other mystery is easily solved: the friction force and the normal force are related to one another precisely because they are the two components of a single reaction force. This decomposition is shown in Fig. 4.
Figure 4. The reaction force experienced by the box due to the incline can be decomposed into two components: one parallel and the other perpendicular to the surface of contact. To complete the free body diagram of the box we need to add the gravitational force acting on the box.
The decomposition into a parallel and a perpendicular component is useful because in general we are interested in how the box moves (or not) along the contact surface. The parallel component (friction) affects the acceleration of the box along the incline, while the perpendicular component (normal force) tells us whether the box remains in contact with the surface or not. (No contact means zero normal force). However, we should not forget that being the two components of a single force, friction and normal force must be related to one another as we learn in class. Let us investigate this in more detail. From Fig. 4, we have: ๐น๐ = ๐น sin ๐ (1) ๐น๐ = ๐น cos ๐ , (2) and assuming that we guessed correctly the sin and the cos , we end up with ๐น๐ = ๐น๐ tan ๐. (3) So for a given tilt angle, the friction force is proportional to the normal force as we are told. This is true as long as the object does not move. However, we should expect that there is a maximum angle ๐๐๐๐ฅ at which the object starts sliding down. The tangent of that maximum angle is called the static friction coefficient, ๐ (^) ๐ . Mathematically, ๐ (^) ๐ = tan ๐๐๐๐ฅ. We then have ๐น๐ = ๐น๐ tan ๐ โค ๐น๐ tan ๐๐๐๐ฅ = ๐ (^) ๐ ๐น๐. (4)
5. Normal forces on objects in motion without friction
We are now in serious trouble. In the static case in Fig. 3, we have a pair of action- reaction forces that are both vertical. The normal component of the force on the box is off the vertical as shown in Fig. 4 but that is OK because friction takes care of the other component making sure that the net reaction force is vertical. But if friction is absent, the net force on the box due to the incline cannot be vertical anymore which means that the force on the incline due to the box should be off the vertical as well in order to obey the action-reaction principle. To see that this is indeed so, let us look at the equations of motion for the general case in which friction is present: ๐๐ ๐ ๐๐๐ โ ๐น๐ = ๐๐ (5) ๐๐ ๐๐๐ ๐ โ ๐น๐ = 0 , (6) which can be written ๐น๐ = ๐๐ ๐ ๐๐๐ โ ๐๐ (7) ๐น๐ = ๐๐ ๐๐๐ ๐. (8) This gives ๐น๐^2 + ๐น๐^2 = (๐๐ ๐ ๐๐๐ โ ๐๐)^2 + (๐๐ ๐๐๐ ๐)^2 โค (๐๐)^2 (9)
and ๐น๐ ๐น๐
โค tan ๐. (10)
These results say that the reaction force of the incline is less than mg and it is not vertical but tilted towards the normal to the surface. In the limit of zero friction, the reaction force is the normal force, as expected. This is illustrated in Fig. 5.
Figure 5. In the absence of friction, the box accelerates down the incline with acceleration ๐ = ๐๐ sin ๐. The action-reaction forces are perpendicular to the surface of contact.
For the non-friction case in Fig. 5, the force on the incline due to the box has to be equal and opposite to the normal force in order for the action-reaction principle to hold. Being tilted away from the vertical, the force on the incline due to the box has both a horizontal and a vertical component given by ๐น๐ฅ = ๐น๐ sin ๐ = ๐๐ cos ๐ sin ๐ (11) ๐น๐ฆ = ๐น๐ cos ๐ = ๐๐ cos ๐ cos ๐. (12) One way to "visualize" the horizontal component acting on the incline is to consider the case where there is no friction between the incline and the floor allowing the incline to move freely. In this case, because of conservation of momentum, the incline moves to the left as the box moves (accelerates) to the right. The horizontal component of the force reduces to zero for ๐ = 0 , as expected. For the vertical component given by Eq. (12), think about the apparent weight of a falling object (like in problems with weights measured in elevators). The box accelerates downwards with acceleration ๐ (^) ๐ฆ = ๐ sin ๐ = ๐๐ sin ๐ sin ๐. Its apparent weight then is ๐(๐ โ ๐ (^) ๐ฆ) =
๐๐(1 โ ๐ ๐๐ 2 ๐) = ๐๐ ๐๐๐ 2 ๐ which is precisely the force Fy in Eq. (12). The box presses down on the incline with a force less than mg because it falls with non-zero acceleration. The vertical component of the forced felt by the incline due to the box is less than mg by a factor of ๐๐๐ 2 ๐ and it becomes equal to mg when ๐ = 0, as expected. For completeness, let us also write the expressions for ๐น๐ฅ and ๐น๐ฆ when friction is present.
We have ๐น๐ฅ = ๐น๐ sin ๐ โ ๐น๐ cos ๐ = ๐๐ (cos ๐ sin ๐ โ ๐ ๐๐๐ 2 ๐) (13)
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