 In the last couple of videos, we learned about some common everyday forces and how to quantify them. We're now going to slightly increase the complexity and work through some problems where we might have multiple forces acting on an object. For the first problem, let's consider an apple falling from a tree. The first step is always to draw a free body diagram. It's easiest to draw the object as a single point, representing its center of mass. In this case, the only force acting is gravity downwards, equal to mg. We will ignore the drag force for now. We can apply Newton's second law. The net force is equal to the mass times the acceleration. So in this case, the net force is just equal to mg. So mg equals ma. The striking thing about this result is that when there is no air resistance, the acceleration is independent of the mass of the object. This is what you saw in the video with the feather and the hammer being dropped on the moon. In this problem, because of the small distance and the shape of the apple, it's a reasonable assumption to ignore the drag force, but that's not always the case. For the second problem, we'll analyze a car accelerating from a stop position at a traffic light. Again, the first thing we do is draw a free body diagram. Here we have gravity acting downwards and the normal force acting upwards. When you press on the accelerator, this causes the wheels to rotate and the tyre pushes backwards on the road. Because the tyre is not slipping on the road, the part of the tyre that is touching the road is actually stationary for the short time interval that it's in contact with the road. Otherwise it would be freely rotating and you would lose traction. This means that it's static friction rather than kinetic friction that is relevant here. As the tyre pushes back on the road, the static friction force pushes the car forward. This might seem counter-intuitive, but if you imagine a car on ice, the wheels would rotate and spin freely against the ice and the car would not move forward. Let's calculate the maximum acceleration that's possible without spinning the tyres. The maximum acceleration will occur when the frictional force is maximum. Now the largest the frictional force can be is mu s n. This problem is more complicated than the last one, because we have forces parallel to the road and forces perpendicular to the road. What we need to do is to write separate f equals ma equations in both the x and y directions. In the x direction, the only force acting is friction, and so f net in the x direction is equal to mu s n. In the y direction, we have the normal force pointing in the positive y direction and gravity pointing in the negative y direction. If we add these up, we get the net force in the y direction is equal to n minus mg. And because the car is not accelerating upwards or downwards, the acceleration in the y direction is zero. And so f net in the y direction is equal to zero. This allows us to solve for n, which we can then substitute back into our force equation in the x direction. We then apply Newton's second law in the x direction f net equals ma and find that the maximum possible acceleration of the car without spinning the tyres is equal to mu s g. You'll notice that we've solved all of this algebraically without substituting any numbers in. It can be really tempting when you're given numbers in the question to substitute them into the equation right at the start. The problem with this is that it's much harder to check that your answer makes sense. The easiest check to do is making sure that the dimensions on both sides of your final equation match. Here we have acceleration, which has units of meters per second squared, equals to mu s, the coefficient of friction, which has no dimensions, and g, which also has units of meters per second squared. So we're all good to now go ahead and start substituting numbers into the answer. For a standard car on a road, mu s is around 0.9, which gives us an acceleration of 8.8 meters per second squared. This means that every second we can increase the speed of the car by 8.8 meters per second. Let's briefly compare this to the acceleration if you do spin the tyres, also known as burning rubber. If the tyres are moving relative to the road surface, then it's the kinetic, not static, friction force that's pushing the car forward. Basically all the forces are the same, so we can just replace the mu s with mu k. For a standard car, mu k is around 0.7, so this gives us a smaller acceleration, 6.9 meters per second squared. So it's definitely less effective to accelerate when you spin the tyres. For the third problem, we're going to consider a skydiver falling out of an aeroplane. First we draw a free body diagram, we have gravity acting downwards, and the drag force acting upwards. The drag force is a type of friction due to a fluid, in this case air, moving around an object. Unlike friction between solid objects, drag depends on the object's speed. This is something that you would have experienced. When you walk, you don't notice the air moving around you, but if you ride a bike, the air resistance is pretty obvious. For a small, slow-moving object, when the air moves smoothly past it, the drag is proportional to the speed of the object. But for a large, fast-moving object, the movement of the air around the object will be turbulent, and here the drag is proportional to the speed squared. A skydiver is large and fast-moving, so we can express the drag as fd equals cv squared, where v is the speed, and c is a constant, which depends on things like the air density, the shape of the skydiver, and so on. We can apply Newton's second law to this case. So the sum of the forces is equal to ma. The sum of the forces here is mg downwards, minus the drag force upwards. Now we're going to plot a graph of the speed versus time. So to start with, when the skydiver first jumps out of the aeroplane, the skydiver's speed is close to zero, and so the drag force is zero, and the acceleration is just equal to g. You'll remember that on a speed time graph, a constant acceleration is a straight line. As the skydiver accelerates and the speed gets faster, the drag force increases, and the acceleration will decrease. This continues until the point where the drag force is equal to the gravitational force, and now the sum of forces is zero, and the acceleration is zero, and the speed stops increasing. This point is called the terminal velocity. We can calculate the terminal velocity by setting a equal to zero and solving for v.