Solving an Unknown 2D Force

by Patrick Ford
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Hey guys, We've already seen how to solve some two dimensional forces problems, but occasionally in some problems, you're gonna have to solve for a force without knowing against magnitude or direction from the problem. So what? I want to do this video to show you how we solve for an unknown two dimensional force. Let's get to the example so I can show you what I mean. We've got these three forces that are pulling this block. We know the magnitude and direction of two of them. This first force and the second force we want to do this problem is find the magnitude of the third Force. So basically want to find the magnitude of alcohol this F three so that the acceleration of the block is 2 m per second only along the X axis. What that means is that the acceleration in the X axis is two and a Y is equal to zero if it's only in the X axis. So we want to do is find this third force here, the magnitude. And if we want to figure out the magnitude of this third force, we have to use the Pythagorean theorem. We're gonna have to figure out its components f three x and F three y and then plug them basically back into this equation here. So I'm gonna go ahead and stick to the steps. I'm trying to sell for a missing force now, but I want to draw the free body diagram that's basically already drawn for us. So the question is, Do we have to draw this third force? And the answer is, it's kind of unclear which way it's gonna point. We know that the acceleration is going to be along the X axis here, but it's not really clear from the diagram, whether it's going to be up like this or down like that, below or above the positive X axis. So we're just gonna skip Skip that for now. We'll draw it later when we actually figure it out. So remember, if we want to figure out these components F three X and F three y, then we have to decompose any other two dimensional forces because basically we want to fill out this table here of all these components and then build out an equation. So I'm gonna do that. My only other two dimensional force here is this F one at 60 degrees above the X axis. So what I can do is break it up into its components. This is going to be my F one X, and this is gonna be f one y to calculate. I'm just gonna use my normal sine and CoSine so f one X is going to be 100 times the co sign of 60. And this is going to be 50 f one y is going to be 100 times the sine of 60 which is 86.6. Remember, the signs of components are really important. So here we have positive points along the positive X axis and positive y axis. So both of these are just gonna be positive. So this is 50 and 86 6 now. F two is actually a little bit simpler because we know that F two points 70 purely along the negative y axis. What that means is that the X component of F two is actually zero, and the y component is gonna be negative. 70. Not positive. 70. All right, so now we've got to figure out what this third force is. So we want to figure out the components of F three, and to do that, we're gonna actually have to write an equation. F equals M A in the X and the Y axis. Basically, we're going to do is build out an equation vertically using these numbers right here. We know that the sum of all forces, the net force is going to be F one plus F two plus F three. So when we write this in the X axis, your some of all forces in the X axis equal m a X. So what are all your forces in the X axis? You have F one X and then f two X over here. And then what about this third force? Do we write it as positive or negative? Well, that's the whole thing. If you ever are expanding, F equals M and your X and y axis, you're gonna assume the components of unknown forces are going to be positive. Meaning you wouldn't write them with, You know, we wouldn't write them with a negative sign or anything like that. So we basically just gonna assume that F three X is gonna be positive we won't stick a negative sign in there, and this is going to be m times a X. Now we just replace everything that we know. We know that F two X actually is going to be zero, and we know F one X is going to be 50. So 50 plus F three x is equal to and then the masses 40. And then the acceleration of the X axis is to remember that we said that X is equal to two. So basically, what we know is that all of the forces in the X axis have to equal times two, which is 80. Which means that we go Saul for this F three x What you're going to get f three X. It's just a T minus 50 and that's going to equal 30. So basically, we know that this X component has to be 30 because these things have to add up to which is mass times acceleration. Now you do the same thing in the Y axis, so the Y axis are f y equals m a Y, except the y axis actually easier, because we know that the acceleration of the Y axis is zero. So basically all the four step to cancel. So this is F one y plus f two y. And then again, we're just going to assume that the components of F three are going to be positive. So this is F three y here and this is equal m times a y. Actually, that's just zero, right? So that's zero here and now. We just replaced all the values that we know. You're basically just gonna drop them and plug them in from the table. So we know this f one y is 86.6 and this is negative. 17. That's the components. Plus F three y is equal to zero. And so we got 16.6 plus F three y secret zero. So that means that F three y has to be negative 16.6. So even though we just assumed that the letter F three y was positive, what we end up getting is a negative number. So it just means that we have to fill this out on the table. This is negative 16.6. That's what you need to cancel everything out on the Y axis 20 So now here we actually have our components. We know it's 30 and negative 16.6. So if you go back to the diagram, we can see that this force is going to look something like this, right? It's going to be 30 to the right, 16.6 down. So this is F three over here. So now the last thing we have to do is we have these components and we can figure out the magnitude by just using our Pythagorean theorem. So F three y is just gonna be square roots of 30 squared plus negative 16.6 squared and you'll get 34.3 Newtons. And that's the final answer. So you finally answers 34.3. And if you wanted to figure out the direction, you could just use the inverse tangent of your Y over X components, and that would be pretty easy to figure out. That's it for this one. Guys, let me know if you have any questions