Forces in 2D - Online Tutor, Practice Problems & Exam Prep

In two-dimensional force problems, understanding vector components is crucial. For example, when analyzing a block pulled by forces at angles, the net force is determined by decomposing forces into their x and y components using trigonometric functions. The normal force can be calculated by balancing vertical forces, while acceleration is derived from Newton's second law, $F=ma$. This approach helps in solving for unknown forces and understanding equilibrium conditions effectively.

2D Forces in Horizontal Plane

Hey, guys. So up until now, all of our problems have only involved forces acting in the horizontal plane or the vertical plane. But now we're going to start to see some problems that are going to combine the two. So we're going to talk about some problems that are in two dimensions but where the forces are in a horizontal plane. Let's go ahead and check it out. We're really just going to combine a lot of things we know about forces with a lot of things we know about vectors in 2D. Alright? So we've got this 5-kilogram block here. We're just going to jump straight into the example. It's pulled by 2 horizontal forces. And we want to figure out the net force and then the acceleration in later parts. We're just going to stick to the steps. We're going to draw the free body diagram and the first thing we need is the weight force. Now, if you were to look at this block that's on a tabletop and see the weight force, you kind of have to look at the tabletop from the side like that. So your weight force comes down like this. This is \( w = mg \). Now the next thing we look for is applied forces and tensions. So we have one that's along the x-axis and one that's at some angle. So what does that mean? Does that mean that you have 2 forces, one like this and the other like this? Well, no. Because remember that these are horizontal forces. So what happens is when you're looking at it from the side view, if they're horizontal, you actually can't see them. So what we have to do is we have to look at this tabletop from the top. So imagine now from this table, you're looking at it from the top down and your forces are going to be along this plane. So here, what happens force. This is my \( f_1 \), and I know this is equal to 2 Newtons. And I have another force that's at some angle. This \( f_2 \) is equal to 5, and I know this is at 37 degrees. The last thing we check for is if 2 surfaces are in contact to see if there's any normal and friction. So we do have 2 surfaces in contact in the side view, so our normal force is going to point up like this. And there's no friction or anything like that, so we ignore that.

So here's what's going on in these problems. Right? We have these 2 different views here. So if all of your applied forces, right, so your applied forces here like \( f_1 \) and \( f_2 \) act only in the horizontal plane, meaning from the top view they're acting like this, then that means that your normal force is going to be equal to mg because your object is going to be at equilibrium in the vertical plane. Right? So basically the box is sliding only along the tabletop, So that means your vertical forces like weight and normal are going to cancel. So what happens in these problems is that really these vertical forces like weight and normal aren't really important, and the ones that we're going to focus on are going to be the \( f_1 \) and \( f_2 \), your applied forces. Alright? So let's go ahead and get started.

So in part a, we have to, figure out the net force. Right? We have net force like this. And so in previous videos, what we've done is that's just going to be sigma f, which means you're going to add your forces like \( f_1 \) and \( f_2 \). And in previous videos, if you have, like, 5 in this direction and 5 in this direction, then you just add them straight up like numbers and that makes 10. But the problem is we do that here because we have vectors. Right? We have these two arrows that point in different directions. So in this problem, what happens is that because forces are vectors, then when a force acts at some angle in two dimensions, we just treat it like any other vectors. And we're just going to decompose it into its x and y components. And then if you have multiple forces that are acting like we do in this problem here, then your net force is calculated by using vector addition. And that's exactly what we have to do here. So really just using a lot of the same vector addition that we've seen before with forces. So if we want to figure out the net force, our net force is actually going to be like this. So this is going to be our \( f_{\text{net}} \). And so what we're going to have to do is we're going to have to add these things by their components. So I'm going to have to figure out \( f_x \) and \( f_y \) and then use the Pythagorean theorem. So my magnitude of my net force is just going to be the Pythagorean theorem of my \( f_x^2 \) plus my \( f_y^2 \). So I just need to figure out those two numbers here and I can figure out the answer. So this is going to be square roots. I've got some number squared plus some number squared and that'll equal my net force. And to do this, we just built out a little table to keep track of all of our components. So that brings us to the second step after we've drawn the free body diagram. We have to decompose all of the 2-dimensional forces by using sine and cosines and things like that. Alright? And we have this table here to keep track of everything that's going on. So our first force, our x and y components, let's see. This first force is just 2 Newtons. It's purely along the x-axis. That's very straightforward. That just means that the x component of 2 and the y component is 0. For the second force, this is one where we have 5 at 37 degrees. So what happens is our \( f_{2x} \) is going to be 5 times the cosine of 37 and that makes 4. So this is going to be 4 here. And then our \( f_{2y} \) is going to be 5 times the sine of 37 and that's going to make 3. So this is going to be 3 years. Now you just add them straight up and your \( f_x \) component is going to be 6 and your \( f_y \) component is going to be 3 because all you're doing is you're just adding \( f_1 \) and \( f_2 \), the components straight down. So that means I have these numbers here. I have 6 squared plus 3 squared, and if you work this out, you're going to get 6.7 Newtons. So that's the answer to the first part. We got 6.7 Newtons, that is the magnitude of our net force in 2 dimensions. So let's move on to part b now.

Now what we're going to do is we're going to calculate the acceleration in the x-axis, that's \( a_x \). So the way we've done this before is by using \( f = ma \). If we have the net force, we can calculate the acceleration. But this net force here is in 2 dimensions, which means we're going to get a 2-dimensional acceleration. If we want to get the \( a_x \), the acceleration just in the x-axis, then we're going to have to use \( f = ma \), but we're going to be using the net force in the x-axis, and that's going to be an acceleration in the x-axis. So we're just really going to be looking at one of the axes, and that brings us to the 3rd step. We're going to write \( f = ma \) in the x and the y axis. So in part c, what we're going to be doing is calculating \( f_{\text{net}} \) sorry, the acceleration in the y-axis. So we're going to use \( f_{\text{net}} = ma_y \). Alright. We'll get to that in just a second. So the net force in the x-axis is actually very straightforward because we already know what that is. In the x-axis, it's just going to be the combination of the x component of the first force and the second force. So really our \( f_{\text{net x}} \) in the x-axis is 6, and this is going to be 5. And so what happens is whoops. So we have 6 equals 5 \( a_x \). And so therefore, your acceleration is going to be 1.2. That's going to be \( a_x \). Now we're just going to do the same exact thing in the y-axis. So here our \( f_{\text{net y}} \). So that number is from our table. It's going to be this one right here. So this is our \( f_{\text{net x}} \). That was this guy. And so this is going to be our \( f_{\text{net y}} \). So we got 3 is equal to 5 times \( a_y \). And so now what we've got is 3 over 5 which equals 0.6. And that is equal to our acceleration in the y-axis. Right? So now we're looking for the acceleration in the y-axis. We just use the net force in the y-axis. And now finally, what we want is we want the acceleration \( a \). So in parts b and c, we calculated \( a_x \) and \( a_y \), but now we want the acceleration which is really just going to be the 2-dimensional acceleration. Right? So this net force, \( f_{\text{net}} \) here, is going to mean that the object is going to accelerate in this direction, and this is our two-dimensional acceleration. So here, we used \( f_{\text{net x}} \). Here, we used \( f_{\text{net y}} \). To calculate the 2-dimensional acceleration, we're just going to use \( f = ma \). So this is \( f = ma \). In other words, what we have here is our \( f_{\text{net}} \) is equal to \( ma \). And so what we've got here is we already have that number. It's 6.7, which is equal to 5 times \( a \). And so that means that we've got 6.7 divided by 5 which is equal to 1.34 and that is equal to our acceleration. So we've got 1.34. There's actually another way you could have gotten this number. Remember that if you have both the components of the acceleration, then you can figure out the acceleration because it's all it's all vector addition. Right? So what I mean by that is that your acceleration is really just the hypotenuse of your \( a_x^2 \) plus \( a_y^2 \). So you could have also done this by doing 1.2 squared plus 0.6 squared. And if you go ahead and work this out, you're also going to get the same number which is 1.34 meters per second squared. So there are two different ways to get to that same number. Alright. So that's it for this one, guys. Let me know if you have any questions.

Three horizontal forces act on a box (mass = 8 kg) sitting on a smooth surface. F₁ is 30 N acting at 53° counterclockwise from the +x axis, F₂ is 13 N acting at 67.4° clockwise from the +x axis, and F₃ is 20 N directly along the -x axis. What are the magnitude and direction of the box's acceleration?

• A. 1.5 m/s², 14° ccw from +x-axis
• B. 1.5 m/s², 76° ccw from +x-axis
• C. 6.9 m/s², 1.1° cw from -x-axis
• D. 6.9 m/s², 0.46° cw from -x-axis

Solving an Unknown 2D Force

Hey, guys. So we've already seen how to solve some 2-dimensional forces problems. But occasionally, in some problems, you're going to have to solve for a force without knowing its magnitude or direction from the problem. So what I want to do in this video is 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 magnitudes and directions of 2 of them: the first force and the second force. What we want to do in this problem is find the magnitude of the third force. So, basically, we want to find the magnitude of, I'll call this f₃, so that the acceleration of the block is 2 meters per second only along the x-axis. What that means is that the acceleration in the x-axis is 2 and a_y is equal to 0 if it's only in the x-axis. So what 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 going to have to figure out its components, f_3x and f_3y, and then plug them basically back into this equation here. So I'm going to go ahead and stick to the steps. I'm trying to solve 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? The answer is it's kind of unclear which way it's going to point. We know that the acceleration is going to be along this 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 going to skip, you know, 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_3x and f_3y, 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 going to do that. My only other two-dimensional force here is this f₁ 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_1x and this is going to be f_1y. To calculate I'm just going to use my normal sine and cosine. So F_1x is going to be a 100 times the cosine of 60 and this is going to be 50. F_1y is going to be a 100 times the sine of 60 which is 86.6. Now remember that 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 going to be positive. So this is 50, 86.6. Now f₂ is actually a little bit simpler because we know that f₂ is purely along the negative y-axis. What that means is that the x component of f₂ is actually 0 and the y component is going to be negative 70, not positive 70. Alright. So now we've got to figure out what this third force is. So we want to figure out the components of f₃. And so to do that, we're going to actually have to write an equation f equals ma in the x and the y axes. 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 f₁ plus f₂ plus f₃. So when we write this in the x-axis your sum of all forces in the x-axis equals m a_x. So what are all your forces in the x-axis? You have f_1x and then f_2x 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 ma in your x and y axes, you're going to 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 basically, just going to assume that f_3x is going to 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_2x actually is going to be 0 and we know f_1x is going to be 50. So 50+f_3x is equal to, and then the mass is 40 and then the acceleration in the x-axis is 2. Remember that we said that a_x is equal to 2. So basically, what we know is that all of the forces in the x-axis have to equal 40 times 2, which is 80, which means that when we go solve for this f_3x, which you're going to get, f_3x is just 80 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 80 which is mass times acceleration. Now I just do the same thing in the y axis. So in the y-axis are f_y equals m a_y. Except the y-axis is actually easier because we know that the acceleration in the y-axis is 0. So basically all the forces have to cancel. So this is f_1y+f_2y and then again, we're just going to assume that the components of f₃ are going to be positive. So this is f_3y here and this is equal m times a_y. Actually, that's just 0. Right? So that's 0 here, and now we just replace all the values that we know. You're basically just going to drop them in, plug them in from the table. So we know this f one y is 86.6, and this is negative 70, that's the components, plus f_3y is equal to 0. And so we get, 16.6+f_3y is equal to 0. So that means that f_3y has to be negative 16.6. So even though we had just assumed that the letter f_3y was positive, what we end up getting is a negative number. So it just means that we have to fill this out in the table. This is negative 16.6. That's what you need to cancel everything out in the y-axis to 0. 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 f₃ 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 f₃ is just going to be the square root of 30 squared plus (-16.6) squared, and you'll get 34.3 newtons. And that's the final answer. So your final answer is 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.

2D Forces in Horizontal and Vertical Planes

Hey, guys. So in the last couple of videos, we were looking at these 2-dimensional forces problems where we had 2D forces on a horizontal plane only. So if you were looking at a tabletop and you were looking down like this, those forces would be pointing just along the horizontal. We're going to keep on going with that, but now we're going to take a look at some problems where you have 2-dimensional forces in the horizontal and vertical planes. It's very similar. We're just going to jump straight into the problem here. So we have a block that's on the floor. It's being pulled by some force. It's 37 degrees above the horizontal and we're going to assume there's no friction, but we want to figure out the normal force in the block. So we just stick to the steps, right? We're going to draw the free body diagram. So the first thing we do is check the weight force. So if we have this block that is sitting on the table like this, you're going to look at it from the side and the weight force is going to be acting vertically. So it's going to point down like this. So we've got our weight force which is equal to mg. We can actually calculate that real quick. It's \(5.1 \times 9.8\). If you work this out, you're going to get exactly 50. However, because this point's downwards, it actually has to pick up a negative sign. It's negative 50 newtons.

We know that there's an applied force. This is 10 newtons at 37 degrees above the horizontal. So what does that mean? Well, if you were looking at this object from the side like this, the horizontal is going to be this way. So 37 degrees above the horizontal means an applied force that looks like this. So our applied force actually points in this direction here. We know this is \(F_A\) and this is going to be our 10 newtons and we know this is 37 degrees.

Alright, and then there are no tensions, but we do have 2 surfaces in contact. This block is on the floor, so that means there is going to be some normal force like this. So there's a normal force. Alright. And then there's no friction. Right? So we're done with the free body diagram. Now before we get into the second step, we know we have to decompose our 2-dimensional forces. I just want to point out one thing. In previous problems, we had forces in the horizontal plane. We had to look at these 2 different views, the side view and then the top view. Well, in this case, where you have forces in the horizontal and vertical, you actually don't need this top view. We've accounted for all of our forces in just one diagram. So that's really good about these problems. They're a little bit simpler.

Alright. So let's decompose our 2-dimensional forces. We've got this applied force at 37 degrees. So I'm just going to draw its components. So this is going to be my \(F_{AX}\), and this is my \(F_{AY}\). Alright. And we can calculate these because we have the magnitude and the angle. Right? So \(F_{AX}\) - remember X goes with cosine, so this is going to be \(10 \times \cos(37)\) and it's eights. We know this is 8. And then our \(F_{AY}\), we know Y goes with a sin. So this is going to be \(10 \sin(37)\), and that's going to be 6. So we know that this is equal to 6. And both of these are positive because they point up and to the right. Right? So these are positive.

Okay. So now we just have to write out our \(F = MA\). Remember, we're trying to figure out the normal force that's acting on the block. So we've got to write it out in the x and the y-axis. So in the x-axis, we've got the sum of all forces in the x = \(M A_X\). We've got the sum of all forces in the y-axis = \(MA_Y\). Now if we're trying to look for the normal force, right, remember that this normal force points purely along the y-axis. Right? This normal force acts only along the y-axis.

So we're just going to start with F = MA in the y-axis because that's where that force belongs. Right? So we've got our normal force which points up, we've got the components of our applied force in the y-axis, that's \(F_A_Y\) that points up. Remember, we're not going to use \(F_A\) because that's a 2-dimensional force. We have to use this component that's just in the y-axis. And then we have our mg which points downwards and this is \(MA\). So we're trying to figure out our normal force, then we're going to have to figure out everything else in the equation. We know what the applied force component is. We know this is 6.

We know our mg is 50 down, so we just need to figure out what's the acceleration. Well, just like we did for the equilibrium problems, we're going to take a look at our upward forces and downward forces and see which one wins. So here we've got our components of our applied force, the 6 that's basically pulling upwards, so that means that in the y-axis only, this object is at equilibrium. So this acceleration is equal to 0. It doesn't go flying upwards and it definitely wouldn't go crashing through the floor. That wouldn't make any sense.

So what that means is that this object is at equilibrium and all the forces are going to cancel. So we have our normal force is equal to, once you move everything over to the other side, \(mg - F_{AY}\). So this is going to be our \(50 - 6\), and this is going to be \(44\) Newtons. So we actually just seen that, this looks very similar to what we've done in our equilibrium problems. So what in these kinds of problems, if your applied forces are going to act partially or completely vertically, right? We've got this force that is acting partially vertically. It's ataxial dimensional angle. Then your normal force as we've seen is not going to be equal to your \(mg\). I'm just going to quickly go through these different scenarios that you might see. We've actually seen a lot of this before. There's really just a couple of things you might see. You might see a situation where \(F_A\) is partially pushing the block into the ground. And so what we've seen here is that when you break up all of these forces into their components, then your normal force has to cancel out both of these forces. So if you're pushing down, your normal force is going to be greater than \(mg\). We've seen that before.

And if you're pulling up but not enough to lift just like we did in this example over here, right, a situation like this, then what happens is that your normal force is going to be less than \(mg\). That's exactly what we saw here. Your normal is \(44\) where your \(mg\) was \(50\). And then finally, if you're pulling up and it's enough to lift, meaning your upward forces are greater than your downward forces, then that means that there is no surface push anymore. The block is going to go flying like this, and so your normal force is equal to 0. Alright? So in most problems though, what's going to happen is in most problems, your upward forces are going to be less than your downward forces. Basically, these two scenarios like this. And so that means that the object is going to be in equilibrium just on the y-axis. So that just means that your forces on the y-axis are going to cancel, and your acceleration's going to be 0 on the y-axis. Alright? So just a little comment there. Let's go back and figure out the second part of this problem here. So in the second part, what we're trying to do is figure out the acceleration. So what does that mean? If we're trying to figure out \(a\), well, we just finished saying that the acceleration in the y-axis is going to be 0 because it's in equilibrium. So what that means here is we still have some leftover force. Right? We still have some components of this applied force that's going to pull this block to the right. So that means what we're really trying to find here is the acceleration in the x-axis. So here in part a, we used \(F = MA\) in the y-axis. Now we're just going to use \(F = MA\) in the x-axis, right? So we've got our \(F = MA\). And there's really only one force to account for, right? There's one force and that's the \(F_{AX}\) and this is equal to \(MAX\). So we've got this as 8, we know this is going to be \(5.1 \times a_x\). And so what this means is that your acceleration in the x-axis is \(1.57\) meters per second squared. And that's the answer. Alright, guys. So that's it for this one. Let me know if you have any questions.

You push a 5.1kg cart along the floor with an unknown force F at 30° below the horizontal. Using a scale, you know the Normal force is 70N. What is the horizontal acceleration of the cart?

• A. 7.84 m/s²
• B. 3.92 m/s²
• C. 6.79 m/s²
• D. 2.26 m/s²

Wind Force on a Dropped Box

Hey, guys. Let's check this one out. So you've got this 2 kilogram box. Right? So this 2 kilogram box like this, we're on a rooftop or something like this. We're basically going to drop it straight down. So there's basically just going to be some weight force once you drop it. However, there's also going to be a horizontal wind force that pushes this thing horizontally like this. We know this wind force is going to be 3 newtons, and we want to figure out the direction of the box's acceleration. So, I've got one force that acts horizontally, another one acts vertically, and I want to figure out what is the acceleration of this box.

Well, basically, if you think about it, these 2 forces, one to the right and one down, are going to produce an acceleration that acts this way, like this, and I want to figure out what is the direction of this theta here or what is the direction of this a. Alright. So, what I want to do is first, I want to draw the free body diagram. I've got this like this. I've got my weight force, this is my mg. If I go ahead and figure that out, this is 2 times 9.8, and this is actually going to be 19.6. We also have our wind force, our F_w, which is to the right, and that's 3 newtons. And that's basically it. Right? There's no tensions. And because this thing is traveling in the air, there's no normal force or friction.

The next thing I would have to do is I would want to decompose all my 2-dimensional forces, but I actually don't have any, so I can just skip that step. What I want to do is I want to write F = ma in the x and y axis. So, if you want to figure out the acceleration here, the direction of this acceleration, this is theta a, then the way I would do this is by getting the tangent inverse of the y components over the x components. So in order to do this, and to get a_y and a_x, I'm going to have write F = ma in both x and y axis. Right? So, on my x axis, I've got F = ma_x, and in the y axis, I'll have F = ma_y.

So now on the x axis, the only force that I have is F_w. So, this is F_w = m a_x. Really, this is just 3 newtons equals 2 times a_x. So a_x is equal to 1.5 meters per second squared. And that's one part of the puzzle. Now, I just need to figure out what a_y is equal to. So, remember that this thing is going to be pushed in the horizontal and the vertical so you can assume that the acceleration is 0 on either axis. What are my only forces that are acting in the y axis? I basically just have my negative mg. And that's just because I'm going to use the convention that up and to the right is positive, as I usually do.

So, I've got negative mg equals m a_y. You'll see that the m's cancel and basically your a_y is equal to negative g, which is just negative 9.8 meters per second squared. This makes sense because if the only downward force that's acting on this object is the weight force, then you're just going to accelerate at 9.8. That's what everything does. Right? So, basically, I've got those two numbers. So, whoops. Then I've got my negative 9.8 and now I can just go ahead and solve for theta a. So theta a is just going to be tangent inverse. Now I've got my a_y component, which is negative 9.8, divided by my x component, which is 1.5. And if you go ahead and work this out, you're going to get negative 81 degrees. This negative just means that it's below the horizontal, and that's what we should expect because it points downwards like this. So that's our answer, 81 degrees below the horizontal. Let me know if you guys have any questions.