Lecture 6

Normal Force and Apparent Weight

Look at any object resting on a table : nobody will argue that the object is at rest. Nor anyone will argue that the object is feeling the force of gravity. But, given that the object is at rest, there must be another force acting on the body that cancels the effect of gravity. This force is in fact there, and it is provided by the table itself (if we were to examine the problem at the atomic level, we would realize that this force is due to electrostatic repulsion between the electrons in the object and the table). We conclude that the table produces a force balancing the force that the object is exerting on it. And this is exactly what we expect from Newton's third law : if a body A exerts a force (in this case the weight) on an object B, B will exert an equal and opposite force on A. This force is usually called the normal force (in mathematical jargon normal = perpendicular), since it always acts perpendicularly to the surface generating it.

Important example : given an object of weight w, what is the value of the normal force when the object is resting on a non-horizontal surface?

Equally important is the case when the object and the supporting surface are moving with some acceleration. You all have experienced the feeling of being heavier or lighter when an elevator starts or stops (or, even better, if you ride one of those super-scary roller-coasters). Let us examine the problem quantitatively, i.e. let us suppose that we ride an elevator while standing on a bathroom scale. The question is :

what will the reading of the scale be when the elevator is

1.

at rest ?

2.

moving upward with constant acceleration ?

3.

moving down with constant acceleration (less than g) ?

4.

moving down with acceleration g? (the cable has broken !!)

5.

moving up or down with constant velocity ?

Let's see if Newton's laws allow us to verify our intuition. To start, we should realize that the scale reading represents the value of the force that the passenger is exerting onto the scale. And, by Newton's third law, this force is equal and opposite to the force the scale exerts onto the passenger, i.e. what we have called the normal force. If we find, for each situation, the value of the normal force, our problem is solved. According to Newton's second law, the passenger's acceleration is determined by the net force, i.e. the resultant sum of all the forces that he is subjected to. And what are these forces ?

They are of course the weight and the normal force, so that we can write (choosing the positive axis to point up)

F_N - mg = ma

where g is the magnitude of the acceleration of gravity (i.e. g is a positive number). We then have :

F_N = mg + ma = w + ma

Assigning to a all the possible values discussed above we get exactly the results we would expect

Friction

The normal force generated by a surface has the property of not being defined a priori, but to vary according to the force that is applied to the surface (up to a certain limit, after which the surface might collapse). Another type of force that has the same properties is the force of friction. Friction plays a rather fundamental role in everyday's life (and in the evolution of both life and civilization). On the one side you might think that friction is a nuisance: if there was no friction, you could send loads of arbitrary weight to any far destination while expending hardly any effort : just give it a small nudge in the right direction, and the load will keep moving in that direction forever.... But things are not as simple as that, if there was no friction you would not be able to walk... Or if you wanted to give a nudge to a load to send it somewhere, the load would send you in the opposite direction with equal force....

Let us then resign ourselves to a world where friction plays an important part, and let us study its properties.

The first thing one can say is that, when a body is pulled in a certain direction, friction will oppose the force by "pulling" in the opposite direction : friction (a force, therefore a vector) acts in the direction opposite to the motion (or attempted motion).

Increasing the pulling force, the frictional force increases accordingly, until one reaches the "breaking point" . We can then say that friction grows in concert with the force applied, until it reaches a maximum value. For a given object, this maximum value will depend on the properties of the two surfaces in contact. We quantify this by introducing a coefficient of friction, indicated by the greek letter $\mu$: given two bodies pressing on each other, the larger is $\mu$ the larger is the force needed to make them slide across each other. Experience also tells us that, for a given choice of surfaces in contact, friction will increase proportionally to the strength of the force pressing the bodies together. In the common case of a body resting on a surface, the friction is then proportional to the normal force. We can summarize all this by writing :

$F_{friction} \leq \mu F_{normal}$

From this we see that $\mu$ is dimensionless. Pay attention to the fact that this is an equality between vector magnitudes only, and it does not imply equality of direction. In fact friction and normal forces are orthogonal to each other.

Another lesson from experience is that, in general, it takes a bigger effort to put an object into motion than to keep it moving. We account for this effect by refering to static ($\mu_{s}$)and kinetic ($\mu_{k}$) friction coefficients. How can we put all this into use? Example 1 : a 50 kg crate on an inclined plane starts sliding when the angle is 45 degrees. What is the coefficient of (static) friction ?

At the breaking point the component of gravity parallel to the surface is equal and opposite to the friction force. But the friction force is given (in magnitude) by the normal force times the coefficient of friction, and the normal force, inthis case, is the component of gravity perpendicular to the surface. We can then write

$\mu mg \cos 45 = mg\sin 45$, i.e. $\mu = \tan 45$

Example 2 : when at an angle of 60 degrees, the crate moves with an acceleration g/4. What is the coefficient of (kinetic) friction ?.... Try to solve this by yourself, remembering that the acceleration will be given by the net force acting on the body divided its mass, the forces responsible for the motion of the body are only those parallel to the surface, and these are the friction and the component of gravity parallel to the surface

Another "invisible" force : Tension

In a large class of problems, we have to deal with objects that are either moved or restrained by means of a rope, a cable, or the like. For all such cases, it is important to realize that rope itself is at the same time subjected to a force and exerting one (as we should expect anyway, because of Newton's third law).

Example : if you pull a heavy load by means of a rope, you apply a force to the rope (which also means that the rope is applying an equal and opposite force to you). But the force you apply to the rope is transmitted to the load; in other words, the rope applies the same force to the load, therefore the load applies the same force onto the rope.... To summarize, there are 4 forces in this problem but (make sure you remark the distinction) two of them are acting onto the rope, while the other two are applied from the rope to external objects. The force (actually forces, and they are equal and opposite) is what we call the tension. Example 1 : a 50 kg load is hanging from the ceiling, what is the rope's tension ?

Example 2 : I pull the same load horizonatally, with a friction coefficient = 0.5. What is the tension of the rope if the crate moves

1.

with constant velocity v ?

2.

with constant acceleration a ?

You should now be able to answer all these questions by yourself.

Important rule to solve "tension" problems : as long as the rope does not encounter any friction, the tension is transmitted equally to all points of the rope,regardless of how many pulleys and the like it winds around.