If we put a heavy body, for example a brick, on a horizontal plane, it does not accelerate, it is in equilibrium. This implies that the plane, in general the constraint, has developed a force, called normal because it is perpendicular to the plane, which is exactly equal and opposite to the force that the body, the brick, exerts on the plane. The latter may be the weight, as in the example, or not. If we push with our hand on the brick, the magnitude of the normal force is equal to the sum of the weight and our push. Similarly, if we push a wall with a hand, it does not move. The normal force made by the wall is equal and opposite to the push.
The normal force is a contact force. If we raise the brick or take back our hand from the wall, even at very small distances, the force disappears. Contact force is
F
r0 r
0 Fig. 3.9 The force between
two molecules as a function of the distance between their centers
the resultant of the forces between the molecules of the constraint and the molecules of the body. When the two are in contact, molecules on the two surfaces are at distances between their centers equal to molecular diameters. The applied force tends to bring the molecules of the body and of the constraint nearer to each other, namely to reduce their radii. This is opposed by the van der Waals force, which, as we have seen, quickly becomes enormous. For this reason two solid bodies cannot penetrate into each other. We have also seen that the intermolecular force goes quickly to zero at distances larger than the equilibrium position. This explains why the force disappears if we separate the surfaces even by very small distances. Already at a few molecular diameters the surfaces no longer interact.
Contact forces are used in practice when we want to constrain a body to move on a certain trajectory. For example, we have repeatedly used a horizontal plane to force a block to move in that plane; the rails force the train to move on a certain path, etc. The physical systems used for this purpose, the support plane, the rails, etc., are called mechanical constraints, because they constrain the motion. The constraints may inhibit motion on one side only or both, being named respectively unilateral and bilateral. A support plane is unilateral because it does not inhibit a body from rising above its surface. The rails of a train are unilateral but those of a roller coaster are bilateral, the coaster cannot detach from the rail.
Usually, forces produced by the mechanical constraints are not known a priori. They depend on the motion of the body, hence on other forces acting on it. For example the force exerted by the rail on the wheel of a train in a given curve depends on the curvature, but also on the speed of the train and on the mass of the wagon. Indeed, the rail develops a force that is exactly the centripetal force needed to have the wagon moving at that speed on that curvature with its mass. The forces exerted by the constraints are said to be passive, the other ones, which are usually under control, are called active.
We can however, calculate the passive forces if we know the motion of the body and all the active forces acting on it. Let us look at two examples.
l x O y m g T l T (a) (b)
Fig. 3.10 Two different mechanical constraints for the same motion. a simple pendulum, b solid guide
Example E 4.1 We have already studied the pendulum in Sect.2.9. We recall that the simple pendulum is a material body, of mass m, constrained to move on a circular arc of radius l. The easiest way to implement a mechanical constraint is like in Fig.3.10a, with an inextensible wirefixed in Ω that exerts the tension T on the material point. Clearly, the constraint is unilateral, because the wire can fold. We could make it bilateral by using a light bar instead of the wire. In Fig.3.10b the constraint is implemented with a wooden or plastic guide shaped as an arc of a circle of radius l, in which the body can slide. Assuming friction to be negligible, the guide will develop a normal force. We represent it with the same symbol as the tension of the wire, namely T.
In both cases, the second law gives: Tþ mg ¼ ma. We already know the motion and are interested in the constraint force T. We observe that in both cases T is directed always towards the centerΩ. The radial component of the resultant of the forces must be the centripetal one, corresponding to the velocity υ of the body, namely F¼ mt2=l, where υ is the velocity at the considered instant and the minus sign means that the force is towards the center. The radial component of the resultant isT þ mg cos h and we have
T þ mg cos h ¼ mt2=l: ð3:20Þ
Clearly, T is not a constant, rather it depends on the position of the pendulum, which is defined by the angle θ. We could do that using the equation of motion we have found in Sect.2.9. However, it is easier to employ energy conservation. The reason is the term mυ2in the last expression, which is twice the kinetic energy. If the pendulum is abandoned from the initial positionθ0, corresponding to the height y0, the energy conservation equation is mgy0¼ mgy þ mt2=2.
Hence mt2¼ 2mg y0ð yÞ. But, y ¼ l 1 cos hð Þ and y0¼ l 1 cos h0ð Þ, hence y0 y ¼ l cos h cos h0ð Þ, and we can write mt2¼ 2mgl cos h cos h0ð Þ and finally, substituting in Eq. (3.20), T ¼ mg 3 cos h 2 cos h0ð Þ.
Example E 4.2 Consider, in a vertical plane, an inclined guide connected at its lower extreme with a circular guide, as shown in Fig.3.11. We want to study the motion of a material point, a small rigid ball for example, on the circular rail, which
r y x 0 mg N (a) (b)
Fig. 3.11 aThe forces on a ball moving on a vertical circular rail, b motion of the ball in case of detachment
is unilateral, of radius r. We use the incline to launch the ball with a certain initial velocity on that rail. More precisely, we want tofind the minimum initial velocity in order that the ball would travel through the entire circle without detaching from the rail.
Two forces act on the ball, its weight mg and the force of the constraint, which we suppose to be normal, N. The latter is directed as the radius, towards the center. The normal force cannot be directed outwards.
Again, the radial component of the resultant of the forces must be the centripetal force requested by the motion. This component is the sum of N and of the radial component of the weight. The latter is a maximum at the highest point of the guide. To be sure that the ball does not detach, it is then sufficient to verify that in this point. Here, the weight and the constraint normal force are both directed vertically downwards. The condition of non-detachment is then Nþ mg ¼ mt2=r. Solving for the unknown N we have N¼ m tð 2=r gÞ.
The condition of non-detachment is N > 0, hence the termυ2> gr. If the velocity is smaller, the ball detaches following a trajectory as in Fig.3.11b, which gives a sequence of images of the ball in its motion. We can think that in this situation the weight is providing a centripetal force too large for the radius of curvature of the guide, at that velocity. The motion must follow a trajectory with a smaller radius, and the ball detaches.