 In this problem we'll investigate a box sliding down a ramp. In previous videos we've shown that we can approximate the gravitational field of the Earth as a constant downwards acceleration. Now we're going to analyze a problem involving gravity on the Earth. A box has been placed on a ramp and is sliding down the ramp. Given that the box has a mass of 150 grams, the ramp is placed at an angle of 30 degrees to the horizontal, and the dynamic coefficient of friction between the box and the ramp is 0.4, then what is the net force acting on the box, and what is the acceleration of the box down the ramp? Firstly, let's start by drawing a diagram of the situation. We know that acceleration is the result of the net force applied to a body, so let's draw all the forces acting on the box, so that we can get the net force and the acceleration acting on the box. Remember that a diagram showing all the forces acting on an object and ignoring all the forces acting everywhere else is known as a free body diagram. These are very useful for solving mechanics problems involving forces. So what forces are acting on the box? There is definitely gravity pulling the box downwards towards the Earth, and there's also the normal force between the box and the ramp. Remember that the normal force describes the resultant force to electromagnetic interactions between atoms, resisting getting pushed together or apart. This resultant force is normal to the surface and is very short-ranged, and is only applied when two bodies are in contact. Finally, there's a friction force between the box and the surface. This force opposes the motion of the box, so it points up the ramp. These three forces are the only forces acting on the box. Let's quantify them so we can solve for the net force. Usually, when faced with these types of problems involving free body diagrams, it's helpful to choose some axes and add the components of the different vectors. In this case, it's easiest to choose our x-axis along the ramp, and our y-axis perpendicular to the ramp. This is a good choice of axes because two of our three forces lie directly along the axes, so we don't have to work out their components. The only vector we need to work out the components for is gravity. Let's break the gravitational force into our perpendicular components, the x-component along the ramp, and the y-component perpendicular to the ramp. You can show with geometry that the x- and y-component of the gravitational force meet at the same angle as the ramp and the ground, which is 30 degrees. Using trigonometry, this means that the x-component is fg times sin theta, and the y-component is fg times cos theta. Now we can look at the net force in the x- and y-directions. This net y-force is equal to the mass times the y-acceleration, but there's no acceleration in the y-direction at all. The box is sliding down the ramp. It's not lifting off away from the ramp or breaking down through the ramp. So the net force in the y-direction must be zero. And so the normal force is equal to the fg cos theta. Let's look at the x-direction now. Since there's no force in the y-direction, the total net force overall is going to be the sum of all the forces in the x-direction. This means that f-net is the x-component of gravity minus the frictional force. Now, the equation for frictional force is the coefficient of friction times the normal force. So we can use the relationship we worked out from the y-direction to write the frictional force as mu times mg cos theta. This means we can write the net force as mg sin theta minus mu mg cos theta. If we plug in the values from our question, we find that this gives us a net force downwards along the ramp of 0.226 Newtons. The friction that we looked at in this video was dynamic friction, which is also known as kinetic or sliding friction. This friction opposes the direction of motion with a force determined by the coefficient of friction and the normal force. This is different from static friction, which applies to a body at rest. Static friction only applies enough force to keep the body from starting to move. When solving problems involving friction, it's useful to identify which type of friction is acting so you know what kind of behavior to expect and you know which equations to use.