definitions over here of how to find the results. to keep you on your toes, I've

changed my coordinate system slightly. In my theory here I have x to the right, y

out of the page, and z down. in this problem that we're actually going to do, I

have x out of the board, y to the right, and z up. And so, the first thing we're

going to do is we're going to find the total force resultant. And so, the force

resultant is going to be equal to the sum of all of these individual forces, and so

I've got it. And all of these forces are parallel to each other. Some of them are

up, some of them are down, okay? And so, we have minus 20 in the

z-direction. So, minus 20k. And we have a minus 60 in the k-direction. We've got a

minus 80 in the k-direction. And we've got a plus 50 in theK k-direction. So, if I

add those altogether, I get the resultant force is going to have a magnitude of 110

pounds, and it's going to be in the negative k-direction. So, this is going to

be a 110 pounds down. But I'm not going to put it over here because my coordinate

system is actually separate. So, we'll leave that off. Our, my coordinate system

is different between these two so it would be down 110 pounds k. Alright. So, here's

my result in force magnitude 110 down, now I've got to find the distance in x and the

y-direction to that resultant force. And so, we'll use these definitions over here.

So first of all, I've got x equals the sum of the individual forces times there

distances in the x-direction divided by the resultant force. And so, in this case,

we have the minus 20 pound force and the distance in the x-direction to that force

is 1, 2 feet. This is times 2. And then, I've got my 60-pound force. This is going

to be, and it's down so it's going to be negative, so plus minus 60, for down,

times its distance, which is 2. And then, I've got a plus 50. 50 pounds is up, so,

it stays positive times the distance in the x-direction to its line of action,

which is one foot. And finally, I've got minus 80 times the distance in the

x-direction, which is 2. I divide all of that by the total magnitude, which is

minus 110 and I'll find out that the distance in the x-direction for my

resultant force is 2.45 feet. And then, I do the same thing in the y-direction. So,

y equals the sum of the individual forces times the distance to their line of

actions in the y-direction divided by the resultant force and magnitude. An d so, in

this case, the 20-pound force lies on the x-axis so it doesn't go out any distance

in the y-direction. We've got the minus 60-pound force and it goes out 2 units or

2 feet in the y-direction, so, and then we have the 50-pound force and it's up, so

it's positive. And we go 3 units or 3 feet to it, and then finally, the minus

80-pound force times its distance, which is 4, all divided by 110 minus 110 and

that equals 2.64 feet. So, our resultant force will be 2.45 feet in the x-direction

and 2.64 feet in the y-direction. So, it's going to be about here. It's going to be

down, it's going to be 110 pounds and so again, this distance and this distance,

distance in the x-direction is 2.45 feet and the distance in the y-direction is

2.64 feet, and that's the solution. So, you should be able to do any of these

single force resultants for a parallel force system. And I'll see you next

module.