Mechanical Engineering: Particle Equilibrium (2 of
19) Addition of Forces - Component
Mechanical Engineering:
Adjacent side relative to the hypotenuse just represents the hypotenuse this represents adjacent side and that's the angle between them I can say that this is f1 times the cosine of 30 degrees which is equal to 100 pounds times the cosine of 30 degrees which is 0.866 so this would be equal to 86 point 6 pounds all right f2 in the y-direction open up not yet f2 we're still on f1 that's equal to f1 times Z and now this would be the opposite side to the angle so that would be the sine of 30 degrees remember with vectors you could move vectors anyplace you like so I can take this component.
here and move with it over there to show that it's opposite to the angle right here so therefore it becomes a sine of 30 degrees that would be a hundred pounds times the sine of 30 which is one half which is 50 pounds so those are the two components the two the x and y components of force one we do the same for the second force so f2 in the X Direction is equal to f2 times the cosine of I believe it's the cosine because it's adjacent side to the angle so cosine of 45 degrees which is a hundred and fifty pounds times the cosine of 45 degrees which is 0.707 so let's get a read on that so 150 times 45 take the cosine equals and it would be 106 point one rounded off to one decimal place so a hundred and 6.1 pounds and we do the same for the y-direction F 2 in the Y Direction is equal to F two times the sine of 45 degrees notice that the y component would be opposite to the angle and that would be equal to 150 pounds times the sine of 45 degrees which would also be a hundred and 6.1 pounds now you may say well wait a minute isn't that force pointing in the negative direction the answer is yes it is but I'm only finding.
the magnitude so don't have to worry about it yet okay now we want to find the resultant in the x-direction so f resultant or F total in the x-direction is equal to f1 in the x-direction plus f2 in the x-direction now notice I'm going to make that a vector and now we do have to worry about the direction all right so this is equal to f1 in the X direction that would be eighty six point six pounds and that would be in the X direction so we can put X hat or I use EXO I doesn't matter plus f2 in the x-direction which would be a hundred and 6.1 pounds in the x-direction and we'll add those two together we get a hundred and ninety two point seven pounds one hundred and ninety two point seven pounds in the x-direction so this is how we make sure we keep the right sign in the correct direction okay now for the F total in the y direction that is equal to F one in the Y Direction plus F 2 in the Y direction again I say plus because at this point I don't know yet if it's positive negative direction I will worry about that when I plug in the components so this is F one in the Y direction which is positive so it'll be a positive 50 pounds in the y
direction but the second component F 2y is in the negative direction that will be a minus 106 point one pounds in the Y direction so 50 minus 106 that would be a minus 50 6.1 pounds in the Y direction so these are the x and y components of the resultant vector so finally can say that F total is equal to 192 point seven pounds in the X Direction minus 50 6.1 pounds in the Y direction so that would be in vector format what if I want to know the magnitude in the direction of the resultant force to find the magnitude I can simply say that F total is equal to the square root of the sum of the squares of the components that would be the F total in the X Direction squared plus F total in the Y Direction squared notice as I'm squaring this it doesn't matter if it's positive or negative so this is equal to the square root of the X direction would be 192 point 7
pounds we square that plus the in the Y direction would be a minus 50 6.1 pounds quantity squared so we get 192 point 7 we squared at plus 56 point 1 and we square it at equals then we take the square root and we get two hundred points seven pounds and that would be the magnitude of the resultant so rounded off to the nearest fam would be 201 pounds and finally I want to know the direction so
let's draw the resultant force the resultant force would look something like this that would be F total which means it would make an angle with the horizontal it's called that angle fee and to find the angle fee we can take the arctangent of the opposite over the adjacent so let's go ahead and do that so to find the angle feed that is equal to the arctangent of the opposite side divided by the adjacent side and of course the opposite side would be the Y component of the resultant force that Jason's I would be the X component of the resultant force so fee would be equal to he arctangent of the opposite side would be the Y component which would be 56.1 pounds divided by the X component