Breaking Down Forces Into Components


Hello, my name is Walter Unglaub, and this is breaking down forces into components. A force is a vector, which means that it can be broken up into a set of orthogonal components. By orthogonal, we need to first define a set of dimensions. For this example I'm going to be looking at just two dimensions, so I'm going to have an x component and a y component. Here I have a block of mass m that's sliding along a frictionless ramp. This incline has an angle theta, so the first step is to determine all the forces that are acting on this mass to determine in which direction it will accelerate and by how much. So the first thing I'm going to draw is an arrow representing the force vector for gravity. So, I denote it by f sub g, and this is equal in magnitude to mg. Now, the other force is going to be the normal force, which is due to the surface of the incline acting on the mass keeping it on the surface. If I rotate this picture by this angle, then I can redraw my coordinate axis such that the normal force lies completely in the y-axis and the gravitational force, the weight of this mass, lies in this direction where we identify this angle as the angle of the incline, theta. So here I have two forces, and I want to break them up into their respective components in the x and y directions so that I can add up their respective components and determine what the resultant force will be. So on the y direction I have the normal force, and this is going to be completely balanced or counteracted by the y component of the gravitational force. So this length here is going to be equal to fg cosin theta, whereas the horizontal component is going to be equal to fg sin of theta. So, the sum of my forces in the y direction is going to be zero, 'cause the normal force is equal in magnitude to the vertical component of the gravitational force. But in the x direction, I see that I have a force pointing along this x direction. And that would be the x component of the gravitational force. So, I have f sub x, which is equal to the mass times the resultant acceleration, f sub x I know is mg sin theta, where fg is mg as I pointed out earlier, so I have mg sin theta is equal to mass times acceleration. The masses cancel out, so my acceleration is simply g times sin of theta. So now I can rewrite these various forces in terms of their components. So the normal force is simply going to be equal to mg times cosin of theta in the positive y direction, and my gravitational force is going to be equal to mg sin theta in the positive x direction minus mg cosin theta in the y direction. We have a minus because it's pointing down and I've chosen my coordinate system to follow this convention, whereas we move up, we go positive in y, and where we move from left to right we go positive in x. My name is Walter Unglaub, and this is breaking down forces into components.

Walter Unglaub graduated from the Colorado School of Mines with a B.S. in Engineering Physics and a M.S. Applied Physics.