Vectors and Parametric Equations
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
We're talking about finding components of a vector, and one of really common, somewhat easy problem is to find the horizontal and vertical components of a force.
So here I have a 2000 Newton force directed in an angle 25 degrees above the horizontal. So I want to find the horizontal and vertical components. Now we know that the horizontal and vertical components are going to be horizontal and vertical, so we don't have to worry about their directions. We just need to find their magnitudes.
Let me first draw them. So here is the horizontal component, and here is the vertical, and I'm going to call the horizontal one 'h' and the vertical one 'v', they're perpendicular to one another. Based on this picture, we have right triangle here, so I can find the magnitude of h by using cosine. Cosine of 25 will equal this length over this length, side adjacent over hypotenuse. So cosine of 25 degrees equals the magnitude of h divided by the hypotenuse which is 2000. That means the magnitude of h equals 2000 cosine 25. So I'll approximate that on my calculator.
You have 2000 times cosine 25 and I get 1812.6 and that will be in Newtons. Let's find the vertical component. We can use right angle trick for this one too, sine of 25 degrees is side opposite over hypotenuse, so it's going to be the magnitude of v over 2000. Sine 25 over magnitude of v over 2000, so 2000 times the sine of 25 equals the magnitude of v. So use my calculator 2000 sine 25, 845.2 Newtons.
So, my horizontal component is 1812 Newtons, and you can say to the left, and my vertical component is 845.2 Newtons upward. These are the horizontal and vertical components of the force.