 The force problems that we've looked at so far have had the forces either parallel or perpendicular to each other. But now that we've learned a little bit about vectors, we can apply these new skills to look at problems where we have forces in any direction. We're going to look at a book sitting on a table. Now if we raise one side of the table, at some point the book will start to slide down the table. Let's consider the forces and figure out what that angle is that we need to raise the table to in order for the book to start sliding. The first step in a problem with forces is always to draw a free body diagram. In this case we have gravity acting downwards, the normal force acting perpendicular to the table, and the frictional force acting upwards along the slope of the table. The second step is to decompose the forces into an orthogonal coordinate system. To do this we need to choose a coordinate system. Here there are two obvious options. One is to have the x-axis parallel to the floor and the y-axis pointing vertically upwards. The other option would be to have the x-axis parallel to the surface of the table and the y-axis perpendicular to the surface of the table. It's up to you to choose whichever coordinate system will make your life easier. In this case we have two forces that are already in the slope-table coordinate system. We have friction parallel to the surface of the table, and we have the normal force perpendicular to the surface of the table. So for this problem we're going to choose the coordinate system that's in the reference frame of the table, because that means we'll only have to decompose one of the forces, not two. So now we need to decompose the gravitational force into components that are perpendicular and parallel to the table. Let's call these components A for the perpendicular component, and B for the parallel component. It's important to get the angles in here right. These two are right angles, and so this angle in here must be theta. We can now do some trigonometry and work out what A and B are. So sine theta is equal to the opposite over the hypotenuse, so this is B on MG, and cos theta is equal to the adjacent on hypotenuse, so this is A on MG. So now that we've decomposed the gravitational force into its parallel and perpendicular components, we can redraw our free body diagram with all of the forces now in our new coordinate system. So now we're ready to apply Newton's second law. We need to do this twice, one for the forces in the x-direction and one for the forces in the y-direction. In the y-direction, we have MG cos theta downwards and the normal force upwards. Now because the book isn't falling through the table or flying off the table, the sum of the forces in the y-direction is zero, and we can solve for the normal force. In the x-direction, let's first write down a bit more information about the frictional force. Now because our book isn't moving, it's static friction, not kinetic friction, that we need to think about here. The static frictional force is always less than or equal to the coefficient of static friction, mu s, times the normal force. And so here we're interested in the point where the book will start to slide. So this will happen when we tip the table enough that the component of the gravitational force that's in the direction of the table is large enough to overcome the maximum static frictional force. And so the point we're interested in is where the frictional force is maximum. So we're going to set the frictional force Ff equal to mu sn, its maximum value. So now we can solve Newton's second law for the x-direction. The sum of the forces is equal to ma. So here for the sum of the forces we have the frictional force acting upwards, and mg sin theta acting downwards in the x-direction. So just at the tipping point before the book starts to slide, the acceleration will be zero. And then we want to solve for the angle theta. Now depending what materials the book and the table are made from, we might have a static friction coefficient of around 0.40, which would give us a tipping angle of about 22 degrees.