Forces in 2D

Concept: Forces in Horizontal Plane

19m
Video Transcript

Hey guys. So in this video I want to talk about two dimensional forces in the horizontal plane and what that means is that we're going to have a flat surface kind of like the floor and all the forces are going to be along the flat surface so it could be this way or this way or this way, now because we are two dimensional forces they could also be at an angle instead of this way and this way it could be sort of somewhere in between as long as they are all along the floor. Right. Let me show the example real quick and it's something like this. If this is your top view so looking down this is what you see. But all these forces are flat against the ground. Before we jump into this let me give you a few reminders about forces. So forces are vectors and since they are vectors whenever we have a force at an angle whether it be like this for an angle sort of like this in a horizontal force we have to first decompose that force into x and y components that's the first the second thing I remind you of is that again since forces are vectors the net force that means the one force that could replace all the other forces you have is calculated by a process called vector addition which we've covered. So the net force right here is just the sum of all forces and this process is vector addition. OK and then one additional point here is that if you pulling parallel to a horizontal surface which is what we're doing here. Right. So let me put a little block m here on the surface pulling parallel to the surface means you're pulling something like this, it could also be a force this way as well. So if you're pulling parallel then the sum of all forces on the y axis has to be zero. Right because we can assume that this block isn't going to by itself break through the floor or just start flying up. Right. If you're not pulling in the y axis at all if all forces are flat in the x then no force go like this or no force go like this then it's guaranteed that the sum of all forces in the y axis is zero. And in this case because the only two forces you would have is mg and normal going up must also be that these forces have to be exactly the same magnitude that they cancel each other. So if the only two forces are these two on the y axis and they have to be the same. And that's what's going to happen here. So basically we're going to ignore those two two forces for these kinds of questions because the only thing that's going on in the y axis here is that normal equals mg and that doesn't really play into this question. All right. So let's see here I have a mass of fifteen hundred kilograms on a frictionless horizontal surface three forces pull on it as shown top view. And let me tell you what the forces are and what all the angles are, I'm gonna put these numbers in here. I just want to point out that a small mistake here this is supposed to be actually counter clockwise. CCW. So counterclockwise goes like this. OK. So once we get to F3 I'll show you what that looks like. So F1 is three hundred. Thirty seven above the positive x positive x is here thirty 37 above. Looks like it's going to put a 30 something over here kind of gets as small as possible so we can fill all the errors because there will be a bunch of them. F2 is 400 Newtons and it is at 53 below the negative x, negative x over here. So it's 53 below that it looks like this and F-3 is 500 counterclockwise counterclockwise from the negative y, negative is over here and counterclockwise clockwise would be this way counterclockwise is that way. So counterclockwise would look like this. So this is our 37 right here. OK so these are magnitudes and directions for all of these vectors. Right so the very start remember I want all my angles to be against the x axis is ideally right because I can decompose just remembering that Fx is Fcosine theta. And Fy is Fsine of theta. But that only works if the angle is against the x axis. So this angle is against the x axis that's good the angle at this one is against the x axis as well. And this one is not. This one is against the vertical against the y axis. So what I'm going to do is I'm gonna instead grab this angle from here and that's the one that I'm going to use. So if this is a 37, this has to be 53 ok and I'm going to gently scratch this out so I don't use it. OK. Now what we're looking for we're looking for the net force and the net force as I told you is the sum of all forces. So in this case is going to be just F1 plus F2 plus F3 but because these forces are at angels at different angles I can't simply add up all these numbers. I have to first I have to actually combine the components. So what I want you to remember. What I want you to remember is that every vector at an angle is made up of its two components. So the magnitude of Fnet that's what it looks like. Magnitude of the vector Fnet is given by the square root of fnet(x) plus Fnet(y) their square so basically the Pythagorean of these two. The angle is also going to depend on those two variables. It's the arctangent of Fnet(y) over Fnet(x). So long story short what you need actually is to find these two numbers and then you'll be able to get this as well as the angle right. So that's going to be our goal to find these two numbers. So if Fnet is F1 plus of F2 plus F3. Then it must be that Fnet(x) is F1(x), F2(x) plus F3(x)2 right all we do is repeat this equation and just put x's everywhere and the same thing happens for the y axis as well. So I'm going to wright here Fnet(y) equals F1(y) plus F2(y) plus F3(y) and all I have to do is find the six values and plug it in and then we'll be done. So let's do that. We have to decompose F1, F2 and F3 into x and y. So here is F1 I'm gonna draw these with a thick green line. Here's F1(x), F1(y) then here is F2(x), F2(y) and here I have F3(x) and F3(y). All right so let me decompose those. All the angles are already with the x axis so F1(x) will simply be F1 cosine of theta right so it's 300 cosine of it's theta which is 37 and then F2(x) and F3(x) I'm gonna find them here as well. F2(x) is F2 is over here, 400 cosine of 53 and F3(x) is 500 cosine of 53.

Now most of that I'm using 53 thirty seven not negative 53 or negative 37 and that's because I'm going to plug in the absolute angle here, well not the absolute angle, the relative angle I'm just gonna get positive because I'm going to check for signs the very end. Right. So once I get a number I leave a little space here to put whether it's positive or negative which I can do by simply looking to diagram. So 300 cosine of 37 gives me 240 Newtons. F2(x) gives me 240 Newtons as well. And then this one gives me 300 Newtons and then lastly let's check the signs don't forget to do that. F1(x) is going to the right so it's positive. F2(x) is going to the left so negative. And F3(x) is going to the right so it is positive ok. Now if I were to add these over here. I would get positive 240, negative 240 and positive 300. So the final answer is positive 300 Newtons, not the final answer but Fnet(x) right. So let's do Fnet(y) I'll try to squeeze in space over here. F1(y) is just F1 which is 300 sine of 37. And I can just kind of look at the numbers here and just copy that except that it's going to be sines instead, so 400 sine of 53 and F3 is 500, son of 53. Ok and I have these numbers here this is 180. This is 320 and this is 400. Now let's look at the signs. EF 1 y is positive and the other two guys are negative so positive negative negative. K. When I combine all the stuff in here I have. 180 positive, 320 negative 400 negative. And if I combine everything I have. Negative 540 Newtons oK I can now combine these two. Now that I have these two numbers here. These are sort of the preliminary answers here. I can find Fnets because Fnet is simply the magnet the square root I'm sorry the Pythagorean of two legs. So 300 square plus negative 540 Square and if you do this and you come to three significant figures this is a 540 over you get 680 Newtons. I can do I can get the angle as well. Theta for my net force the angle of my net force is the arctangent of y over x so negative 540 over 300 ok negative 540 over 300 if you do this the angle is the answer is negative 61 degrees. Now remember with the arctangent you have to make sure it's in the right place and the way we can do that is by drawing. So if the net is 300 on the X and 540 in the y looks like this Fnet(x) is 300 then I'm a little longer Fnet(y) is 540 I didn't put a negative because it's already point now. And that means that my net force actually looks like this and the angle is going to be against the x axis over here 61 degrees and we have to put a negative because I'm showing that the exact position of that angle and this is the fourth quadrant and the arctangent function works just as well in the fourth quarter. And it doesn't require any touchups. So we will leave it alone. Right all this was part A only but part B is pretty easy part B is just asking for the acceleration. I wanna know the acceleration. And I want to know what is the direction of the acceleration so I'm gonna call this theta of acceleration theta a, how do I find acceleration of force problem F equals ma so sum of all forces equals and ma. Sum of all forces is the same thing as net force which in this case is 618 divided by the total the mass which is fifteen hundred. And if you do this you get 0.412 Meters per second square. That's the answer finally answer for acceleration although I have to find the final answers here. Now what about the direction. Well F equals ma have these two vectors here and you should remember that it was in the direction of a is the same as the direction of the net force. Because if my net force is this way that I accelerate this way. So the angle for a this is the magnitude of a so I can do this the angle for a is simply the same one, negative 61 degrees ok fourth quadrant. All right that's it for this one. So here have three friends pull horizontally on ropes that are connected to the same box on the box here and the box sits at rest on a frictionless horizontal surface, it's a very similar setup to before is just we gonna ask for something different here.

So the box is initially at rest and guy A is going to pull with 30 towards the negative x axis and looks like this I'm gonna call this a equals 30 and guy B pulls with 40 towards the negative y axis. So B looks like this. Make that a little longer. Notice to put negatives because the arrows indicate the direction. And the question is what mostly the magnitude and direction of the force exerted by Guy C. So the box doesn't move. So the box starts at equilibrium starts with the rest of the initial velocity is zero. And I wanted to not move. In other words I want the velocity not to change. So I want the acceleration to be zero which is equilibrium. Which means that all the forces have to cancel to be able to piece all that together basically I want all forces to cancel. So I want to know what is the magnitude of C. And what is the angle of C so that the forces cancel and if I want the forces to cancel and these I want the sum of all forces to be zero. So the sum of all forces which is A plus B plus C I want that to be zero. So I want these 3 guys to add up to zero. Right now this problem is simple enough that you can kind of you might just have seen that you can eyeball this and realize that well I just need C to first of all be somewhat in this direction so it cancels both. And all you gonna need is for your Cx to cancel out to the 30 and for your Cy to cancel out the 40. And then now that you know Cx and Cy you can calculate C and the magnitude of C ok, So this question is simple enough that you could have just done that. But in the more general terms if you had a problem there was a little bit of this but a little more complicated which you would do is if the sum of all forces equals zero then because you have two dimensional forces you will do this in the x axis and in the y axis. So you'd set it up that Ax plus Bx plus Cx equals zero. And this is your equation the x axis and you would set it up that the same will happen in the y axis so Ay, By plus Cy equals zero. And by doing this you're going to be able to find Cx and Cy which remember every vector is composed of it's is made up of its legs. So to find the magnitude of C, I need Cx and Cy and to find the angle of C I need Cx and Cy and the way you will do it the long way if you will that's going to work for any question is any problem is this way. So let's check this out Ax is A in the x axis which is just 30, 30 to the left. Here we don't have to decompose because they're saying flatten the x and flatten the y. Bx, B is flat on the y axis so Bx is zero. Right. If it's going straight down there's no x component either way. And then Cx is what I'm looking for. So look what I get I get that Cx equals 30 which is what we could have quickly done by just looking at the problem on the y axis Ay, A is flat the x axis so it doesn't have a y component it's just flat square with zero. By is 40 but it's going down so it's negative 40 plus Cy and Cy is what I'm looking for and look what I get here I get that Cy equals. 40. So it's like something to be more elaborate more complicated problem maybe you wouldn't be able to right away see how forces combine. Right. And now I can just plug in these numbers here. So a 30 with a 40 gives me a 50. So the magnitude of 50 of C is 50 Newtons and the angle of C is the arctangent of y over x. So it's the arctangent of 40 over 30 and that is fifty three degrees. Okay so you do want C to B 50 and 53 so it looks something like this. C equals 50 at an angle of 53 degrees and that's the final answer alright, so hope this makes any sense, let me know if you guys have any questions.

Concept: 2D Push / Pull

8m
Video Transcript

Hey guys in this video I want to talk about two dimensional push and pull problems these are going to be problems where you have an object and you're going to pull it not flatten the X. axis but you can pull the down the angle or you can pull down like this so let's get started.Right so remember forces are vectors so whatever you have for set an angle we must decomposing into X. and Y..And we're going to do is we're going to draw all forces in the X. axis and they were going to write the sum of all forces in the X. axis equals MA X. This is basically just f equals MA.

But the sum of all forces in the X. are responsible for causing the acceleration in the X. axis and the sum of all forces in the Y. axis are responsible for causing the acceleration in the Y. axis and those two are independent from each other OK So after you write these you would plug in your forces and find your accelerations but what you should also do you would ask is this acceleration zebra is this object at equilibrium either in the X. axis or on the Y. axis right and one thing I will point out here is that if you are if you're pulling in a direction that is not parallel to a horizontal surface so let me show you. This is the result of surface parallel to the result a surface would be pulling or pushing in this direction but if you pull at say an angle like this.

Then you are not parallel to result the surface in a normal force will not be equal to your M G. right it's not going to be equal MG now in this problem let's say the object moves this way right so the accelerates this way as an AX in this problem the acceleration in the Y. axis will be zero because this object is expected to not break through the ground or fly up right so that mean. The forces in the Y. axis will cancel however there are more than just normal Y. axis so I cannot say that they equal each other in fact I have to decompose this guy into F Y. and F X and you can see here how there are more than just normal in M.G. therefore I cannot say it normally equals M.G.normal be different than MG right that's what's sort of the big deal about these problems you have to take care of normal careful all right let's do an example here and then I have two practice problems for you guys so an 8 kilogram block is initially at rest on horizontal frictionless surface so using a kilogram block and its originally a rest OK and you going to pull the block with a constant one hundred Newtons directed at thirty seven above the horizontal so its going to look like this we would sound a bit.

its gonna look like this F. equals one hundred directed at thirty seven above the horizontal Firstly what you do is you decompose this so I'm going to call this this in going to be FY and this is going to be F X. and if you remember F. X. is F cosine of theta as long as there is a right place with the X. axis which it is so this is one hundred cosine of thirty seven which is eighty it's going to the right so it is positive eighty on the Y. axis is one hundred sine of thirty seven and that is sixty it's going up so it's positive sixty newtons ok so I draw all my force on the diagram what else do i have here I have M G. pulling down OK and just to make life a little simpler here only is badly as ten not nine point eight And so eight times ten is just eighty and I have been pushing against the floor notice that this eighty is stronger than the sixty so this object isn't going to. fly up the eighty stronger than the sixty therefore I'm still pushing a little bit against the floor so I still have a little bit of a normal force how much of a normal force twenty right and how I was able to figure that out I got that because since the pulling force is not strong enough to remove to lift you from the ground. I know that this object is going to be in equilibrium in the Y. axis the forces would cancel on the Y. axis they won't accelerate so that's because since F Y is less than mg I know this block is not gonna fly up and certainly not going to break to the ground so I can say that the acceleration in the Y. axis will be zero and because of that that means that the forces will cancel, some of the force in theY. axis because they are zero so its gonna be I can say that force is up cancel the forces down so these two guys have to add up to eighty so I added sixty so all I need is an ab Sothat's the quick way of doing it the only way you remember is to mobile force on the Y. axis equals zero and then you do plus twenty.Plus F. YPlus negative MG equals zero and then you can see how N end up being this guy goes over here as a positive and this guy goes over there as a negative to the right side and you can see how its just eighty minus sixty which equals twenty OK So anyway that's what's going on the Y. axis. There's no acceleration to that slowly just a kind of as a reminder.

The first part of this question is actually to draw a free body diagram this is not a free body diagram because for by that you have to be dots with arrows only right so to pick a definition. I have to draw big dark here I will force cannot of it so N equals twenty. With i am going to draw it a little bigger FY equal sixty.M.G. equals eighty and F. X. equals eighty as well.Notice how I didn't draw the original F. your professor may or may not want you to draw the original forces at an angle but you certainly have to draw a decomposed ones right so that's the free body diagram right there and if I want to calculate the acceleration I already know that acceleration the Y. axis zero so I'm really looking for the acceleration in the X. axis and this is pretty simple if I don't have a force problem to give you a force and I'm asking for accelaration is just if it was just F equals MAand I'm lookingfor a force on acceleration the X. axis sum of all forces near X. equals m a X. and I'm looking for a X. the only force in the X. axis is the eighty and it's going to the right from the so that's a positive eighty.The mass is an eight so the acceleration will just be over eight or ten.Meters per second squared.and That's it for this one all right so what I want to do now is I want you to try the next two practice problems and remember what I've been doing you have to decompose your forces into X. and Y. first and then look at what happened to the X. axis look what happened to the Y. axis you do that independent from each other and you have to think Is this going to is the acceleration in the x or acceleration in the y is going to be zero and if you think it is in your forces are so give us a shot and hope you get it .

Problem: A 6-kg block is initially at rest on a horizontal frictionless surface. You push on the block with a constant 50 N directed at 53° below the horizontal. Calculate the block’s aX and aY.

4m

Problem: You push a 10-kg block against a wall with a constant 200 N at an angle (shown below). Calculate aX, aY.

5m