We're back with first and second laws. Remember that this equation is for the total force acting. In memory of Galileo, here is a ball on a surface. It has weight, which is a vector, which I'll call W, and its direction is down. To hold it up requires an upwards force, which I'll call the normal force N, being supplied by the bench. I call it the normal force because it's at right angles to the bench in the normal direction to the bench. I can also supply this force with my hand, and then I can feel that there's a force. The normal force is one of a Newton pair, and I feel, minus N, the force that the ball exerts on my hand, or on the bench. Now, when I place the ball on the bench, it turns out that the normal force has a magnitude almost exactly equal to its weight. The directions are opposite, so W plus N equals zero. Why they are almost exactly equal is a bit subtle, and you might like to reflect on that before we discuss it next week. This is one example of the situation we call mechanical equilibrium. Total force is zero, so acceleration is zero. When the ball is rolling with constant velocity, the horizontal force is also zero, so that's also mechanical equilibrium. The diagram I've drawn here,a vector representation of all the external forces acting on an object, is called a free body diagram. Let's look at the slider. Here, the upwards force is the lift, L, provided by the compressed air underneath. It's in mechanical equilibrium at rest. At constant velocity, it's also in mechanical equilibrium, but at the ends, there are large horizontal forces and large accelerations. To solve problems in physics, there are several steps and one of the most important is making appropriate approximations. Often it's the most difficult part of solving a problem. This is a simple example, but let's be careful and concentrate. Suppose I accelerate this trolley, mass, one kilogram, by pulling on the string, mass, two grams. What forces are required to give an acceleration of two meters per second per second? First, we can see that it doesn't accelerate upwards or downwards, so by Netwon's second law, the vertical forces add up to zero. Well, that's not very interesting. What horizontal forces are involved? Well, we can identify two Newton pairs. On your left side, the string pulls the trolley to the right, and the trolley pulls the string towards the left. They are a Newton pair, so they have the same magnitude. Call it F Left. On the right, my hand pulls the string to the right, but the string pulls my hand left. These have the same magnitude. Call it F Right. Let's label these using Newton's third law. Now, the only external force acting on the trolley is F Left, so we write F Left is equal to MA, is equal to 1.0 kilograms times 2.0 meters per second per second, equals 2.0 newtons. What about the string? It's also accelerating at A, but two forces act on it. F Left pulls it to the left, and F Right pulls it to the right. So, Newton's second law allows us to calculate the difference between the forces at the left and right end. Because the string is light, this is only 0.004 newtons, so F Left equals F Right to a very good approximation. And I'll call both of these forces the tension in the string. Now, the key to making this approximation is that the string is light. Its mass is much smaller than that of the other object. In physics problems, we often meet light strings. We wouldn't make that approximation if instead of using a string, I used a chain. Strings and chains are also often described as inextensible, meaning that they don't stretch. If I used a rubber band, I couldn't even assume that the acceleration at both ends was the same. So, if the string is light and inextensible, our problem is simple: the force I apply with my hand equals the tension in the string, and the tension in the string accelerates the trolley and the string. But because the string is light, I can write: tension equals mass plus mass of string times A, is approximately equal to mass times A. A further lesson, when I wrote, tension equals mass plus mass of string times A, what happened to the other three force vectors? Let's see that next. There's another important thing about applying Newton's laws, and it concerns internal forces: the forces that one part of a body or system exerts on the other. As we saw before, these add to zero by Newton's third law. So, when we write Newton's second law, we only need to sum the forces external to the object or to the system that we're considering. Sounds easy? Well, let's see how you go here. The man says to the horse, "Gid up. " That's horse language for go. The horse replies, "(Neigh) There's no point! Newton's third law says that the cart will exert a force on me equal and opposite to the force I exert on it. Sum of the forces is equal to zero. So the acceleration will be zero Okay, how would you answer the horse? Get out your secret weapons draw a sketch before you answer. Look at the harness, the thing that links the horse and the cart. We can say that the force that the horse exerts on the harness equals the force that the harness exerts on the cart,but only if the harness is light. However, for the object comprising cart, harness, and horse together, these are both internal forces. If we look just at the cart, the external horizontal force for it is the tension, T, in the harness. That force is internal for the horse plus cart system, but external to the cart. That's what accelerates the cart. Now, if we look at the horse, the harness pulls it backwards. So what makes the horse go forwards? Let's look at the horse's hoofs and the ground. The horse pushes the ground backwards, and the ground pushes the horse forward. Also a Newton pair. So the external force on the horse are F, horse forwards, and T, backwards. The total force of these is positive, so the horse can accelerate forwards. Finally, look at the force that the horse exerts on the ground. This is an external force acting on the Earth. However, the mass of the Earth is so great that the earth doesn't accelerate measurably.
