 Now we're going to look at an example. Let's go with a classic. A pair falling out of a pear tree. So if a 200-gram pair falls out of a 2m pear tree, how strong is the gravitational force that acts on the pear? Firstly, we should draw a diagram of the situation. Now the gravitational force, as we know from our previous video, is equal to mass times the acceleration due to gravity. So the gravitational force, Fg, is equal to mg. Now from our last video, we know that this force is pointing downwards, and we'll model the pear as a point object with the force acting at the centre of mass. So let's plug in our values. We know the mass is 200g, and little g is the acceleration due to gravity, which is 9.8m per second squared. So Fg equals 0.2 times 9.8kgm per second squared, which is equal to 1.96N. Remember that force is measured in Newton's, which is equal to kgm per second squared. Now you may have noticed that the height of the tree doesn't affect our answer. Near the Earth's surface, we approximate the acceleration due to gravity as 9.8m per second squared, regardless of the height of objects above the ground. Force due to gravity also affects objects on the ground. After the pear hits the ground, it is still experiencing the same gravitational force, in this case gravity is cancelled out by the normal force of the Earth's surface that acts to prevent objects sinking into the ground. For our next example, we'll investigate an object on an incline. Say for example an experimentalist going down a slide. The slide is a pretty steep slide with an angle from the ground theta of 35 degrees. Our intrepid experimentalist weighs 60kg. What is the net force on the experimentalist sliding down the slide? Firstly, we'll draw a diagram of the situation. Now the slide is pretty smooth, so we're going to neglect friction to get a rough idea of the net force a person will experience. We'll introduce friction in the next video. So we know that from a stationary start, the experimentalist will slide down the slide. So we can already infer that the net force will point down the slide. But how? Well let's consider the forces at work. On Earth, we have gravity, which is always points directly downwards. But there's also a normal force from the slide pointing perpendicularly to the slide's surface. We know this because the experiment is hopefully not breaking the slide. So how can we work out the net force? Well, let's look at our gravitational force as a vector, which can be broken up into perpendicular components, or at right angles to each other. A common choice of breaking vectors up is using vertical and horizontal axes, although in this example we won't use these axes specifically. So, breaking vectors up essentially means that we can split vectors into two smaller vectors at right angles, which we added together to create the original vector. The net forces on the experimentalist cannot be neatly summed in the same direction, as they're acting at different angles. So let's break up the gravitational force into a component along the ramp, and a component perpendicular to the ramp. Let's use our intuition here. There is no motion or acceleration perpendicular to the ramp. As we've assumed that the experimentalist stays on the slide's surface. So we know that the normal force from the slide cancels out the component of gravity parallel to this normal force, which is perpendicular to the slide's surface. This means that the net force in the experimentalist is entirely the component of gravity perpendicular to the normal force, which is parallel to the slide's surface. To work out what force this is, we can use geometry and trigonometry. As we've already shown, our gravity force vector can be broken into a triangular geometry based on the angle to the ground of the slide. Now, if we look at the location of our force, we can see that the angle to the slide's surface is 35 degrees from the horizontal. Now, the angle of the slide and the angle of the normal force together make up a 90-degree angle. And similarly, the horizontal and vertical components will also make up a 90-degree angle. So we can see that the angle at the top of our triangle is also equal to 35 degrees. Now, if we want to find the side of the triangle that is perpendicular to the normal force, we need to use trigonometry. So if you remember Sokotoa, we label each of the three sides of the triangle, and we find that we want the side of the triangle that is opposite to the angle. The side of the triangle that we already know is the hypotenuse, which is fg. So we have that sine of 35 degrees is equal to our net force divided by our gravitational force. We know that the gravitational force fg is equal to mass times acceleration due to gravity. So if we plug in all of our values and rearrange our equation, we find that the net force is equal to mg times sine of 35 degrees. And this is equal to 337 newtons, and we have our answer. Now, if we wanted to find the acceleration of the experimentalist, we know that the acceleration is equal to force divided by mass. So we plug in 337 newtons divided by 60 kilograms, and that gives us 5.6 meters per second squared. As a final thought, what do you think would happen if we doubled the steepness of the ramp? So is it 70 degrees instead of 35 degrees? You might think that since one of the numbers has doubled, the force would double, but that's actually not the case. We need to substitute the new value into the equation we derived to see the effect. Putting this in, we find that the net force is equal to mg times sine of 70 degrees, which is 553 newtons. As before, we can use our equation for acceleration to find that the experimentalist will accelerate at 9.2 meters per second squared. While this is bigger, it's not twice as big as our previous answer. The reason it's not twice as big is because when theta doubles, sine of theta doesn't double. In other words, sine isn't the linear function.