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

Understanding gravitational forces in two dimensions involves vector addition and decomposition into components. By applying Newton's law of gravitation, the net force can be calculated using the equation $F=G(m$₁⁢m₂)/r^2. Utilizing symmetry can simplify calculations, allowing for the cancellation of equal and opposite components, ultimately leading to a clearer understanding of the net gravitational force's magnitude and direction.

Gravitational Forces in 2D

Hey, guys. So now that we've gotten a good feel for the universal law of gravitation in one dimension, it's time to look at how it works in 2D. But we're going to see very similar stuff that we've done before. So, we're working with net forces in one dimension; then, if we have multiple forces on an object, for instance, like this force right here that was, I don't know, 3 newtons, and we had another force that was acting on it like negative 4 newtons, then we can figure out what the net gravitational force is just by subtracting or adding these forces together. So I'd get a net force of negative one newton that points in that direction. It's simple addition, right? Well, in this case, let's check it out. So we have a sort of triangle of masses, so all these things are little m's. Newton's laws say they're all pulling on each other. So this bottom left one feels a force that points in that direction and it also feels a force on it that points in that direction. So let me just make up some numbers here. I'm just making this up. So let's say this was like 3 newtons and now this was like 4 newtons. How would I figure out the net gravitational force? So these questions will ask you what the net gravitational force is, but you can't just add these things together because they point in different directions. And so to solve for net forces in non-linear arrangements, you have to use vector addition. And remember that the force of gravity is a force and forces are vectors, So we can always take a vector like this and break it up into components. Alright. So the point is, if you have this vector here that points in some off angle, you can always decompose it into its x and y components just using your sine and cosine stuff from trig before. So we'd have to know some angle theta and stuff like that, right? So now what we can do is now we can take these components, add them up, and then figure out what the net gravitational force is like that. Okay. To sort of refresh ourselves with all of this stuff, let's go ahead and just do a quick example.

We're supposed to calculate what the magnitude and the direction of this net force is on the bottom mass, the m1 in the figure. So the first thing is we just have to label the forces that are acting on this. So I've got a force that points in this direction between these two masses and then we've got a force that force that points to the right from those two masses. Now this is between m1 and m2, so I'm going to call this f12. And then this is f13, so I'm going to call this f13. Okay. And so how is what is the net gravitational force gonna be? Well, I can't just add these things together so I have to use vector addition. If I add these 2 vectors tip to tail, what I'm gonna get is I'm going to get a net gravitational force that points in that direction. Right? So if these things pointed this way, the net gravitational force has to be between them. So this is the net gravitational force, and I'm to figure out the magnitude of that. So how do I figure out the magnitude? Well, remember that if we're figuring out the magnitude of a force and we have these sort of components like the x and y components, we do that using Pythagoras' theorem. So we've got f13 is the x component, so we've got to square that. And then we've got f12, that's the y component, and then we've got to square that. So this is like the hypotenuse of this triangle, right? So if we're trying to figure out what this distance is right here, then we have to figure out what sort of like the hypotenuse of this triangle that it makes, right? Okay. So now we actually have to figure out what these two forces are; what is f13 and what is f12? Well, I can figure out f13 just by using Newton's law of gravitation for point masses. So I've got big G times mass 1, mass 3 divided by the distance between r1 and 3 or m13 squared. So that distance between m1 and m3 is equal to 0.1. And then I have both of those masses, so I can go ahead and just figure that out. We've got 6.67 × 10 - 11 and the product of the 2 masses, 25 and 20 divided by 0.1 2 . So what you get for that is you get 3.34 × 10 - 6 . So I've got 10 - 6 there. Alright. And if I do the same exact thing for f1 and 2, I've got g m1 m2 divided by the distance between 12 squared, which is at 0.15. So just got the same setup here, 6.67 × 10 - 11 , then I've got 25, and now 30, and now divided by 0.15 2 . So what I get is 2.23 × 10 - 6 . And both of these things are in newtons. Alright. So now that I actually have the components of these and the actual forces, now I can go ahead and just figure out the net gravitational forces. So the magnitude of f net is going to be the square roots and then I've got just basically plugging these values. I've got 3.34 × 10 - 6 + 2.23 × 10 - 6 . If you go ahead and plug that in, you should get a net force of 4.02 × 10 - 6 and that's in newtons. How about the direction? How do I figure out what the direction is? Remember that that direction is going to be this angle theta, right? And theta, how do we relate theta? From the tangent. So we relate the tangent theta is equal to the y component divided by the x component. So in other words, it is the f12 divided by f13. So to go ahead and figure out this angle here, I've just got to take the inverse tangent. So that's going to be the inverse tangent of f12, which is going to be 2.23 × 10 - 6 ÷ 3.34 × 10 - 6 . If you do that in your calculator, what you should get is 33.7 degrees. Alright, guys. So that is the magnitude and the direction of this net gravitational force and, let's keep going on. Let me know if you have any questions.

Using Symmetry in 2D Gravitation

Hey, guys. So often problems that you'll see in 2D gravitation will have a lot of big numbers and a lot of steps to follow. In this video, I want to give you a shortcut in order to minimize the amount of work that you have to do. So, let's check it out. Let's say I was trying to figure out what the gravitational force is on this top mass right here. I want to find f_net. Now, how do I do that? Well, first, I have to go ahead and label my forces. There are 2 masses on this triangle, and they're both exerting gravitational forces on this top mass right here. So we've got a gravitational force from this guy that points in that direction and then f_g over here. Okay. So now that I've labeled the forces, I have to calculate them. But notice how if these masses are the same, the little m's are the same, and I haven't given numbers for those. I've just said that they're both little m's, and the distances between them are the same; so this is the distance between them. Then these things have to have the same gravitational force. These have to have the same f_g. Remember that f_g is just big G times the mass of both objects divided by their distance squared. So if you have the same m's and the same r's, the gravitational force is the same.

Now that we've calculated what the gravitational forces are if I actually had numbers for this, I'd have to decompose the vectors into its components. Just to refresh that, the way I would do that is if you have this vector that points off in this direction, then you can always split it off into its components. So I have an x-component which is f_x and this is a y-component f_y, if this is a force or a vector like f. Right? And we do that using trig. Right? As long as I have the angle theta relative to the x-axis, I can get its components. So the next step for this is, now that I have the forces, I can split it off into its components. So if this is the angle relative to the x-axis, then that means it's going to have an x-component here, and that x-component is going to be related to the cosine of theta, whereas the y-component down here that points in this direction is going to be related to the sine of theta.

But I have the same exact forces, and if these things have the same angles and the same forces, then you're going to end up with the same exact components in the x- and the y-directions. So the tip is, if you end up with 2 components in the y in either direction that are equal but opposite, you can always cancel them out if they're opposite. So this is another way where you can use symmetry to reduce the amount of work that you have to do. Let me show you. So I have these x- and y-components, but this force over here on the left is going to break up into the same exact components, f_x and f_y, except the y-components are going to add together here, whereas the x-components are equal, but they point in opposite directions. So we don't even have to do this in our calculators. We won't have to worry about it because the net force is going to end up canceling both of those out. Instead, what's going to happen is that the f_y components are going to add together and your net force is going to be pointing down here. That net force, which I wanted, is just going to be equal to 2 times f_y.

This is one way that symmetry can reduce the amount of work that you have to do. So I want to attach some numbers to this stuff. Let's go ahead and work out this quick example. Go ahead and pause the video after looking at this figure and see if you can figure out what the net gravitational force is on this little m from these two forces to these two masses over here on the right. Okay, cool. Let's get into it.

The first thing we have to do is we have to label our forces and calculate them. So I've got a force that points off in this direction, and I'm told that that f_g is equal to 5 newtons. And I have another force that points off in this direction, so that f_g is equal to 5 newtons. And I'm told it feels a 5-newton force from each m on the right, and that makes sense because we have the same exact r and the same m for both of them. So we know that those forces are equal. So now we just have to split it up into its components. So I've got this theta angle here that's relative to the x-axis. So that means that this component here is going to be f_x, and then I've got f_y that points off over here.

But this angle right here relative to the x-axis is going to be the same as this angle right here because I have the same distances involved r, and we can assume that these masses right here are sort of like connected by that vertical line. So that means that the angles for both of them are the same, which means that the y-components are going to be pointing in opposite directions, but they're going to be canceling each other out. So that I can cancel out my f_y with this f_y right here, and the f_x components are actually going to add together. That means that the net gravitational force is going to be pointing off in this direction. What are the x-components? So I actually can put, I could put numbers to that because I know what the force is. I know that it’s 5 newtons. As for the angle, I'm given that the angle is 53.1 degrees. So that means that each f_x component is 3 newtons. And as for the net force, these things are gonna add together, and they're gonna be perfectly equal to each other in terms of the components. So that means that the f_net is going to be 2 times f_x. So that means that the net gravitational force is 2 times 3, which is just 6 newtons.

Alright, guys. Let me know if you have any questions with this stuff.

Finding Net Forces in 2D Gravitation

Hey, guys. So for this video, I want to pull together everything that we've learned about gravitation in 2D. So Newton's law of gravity, vector components, and using symmetry to solve a real problem that you guys might see. Let's check it out.

So I've got three 50 kilogram masses. I'm told what the masses and the lengths of all of these things are in an equilateral triangle. And I'm supposed to be figuring out the magnitude and the direction of the net gravitational force on the bottom mass. So we're going to be taking a look at this guy right here. But we know the steps to solve any 2D gravitation problem. We've got to label the forces and then calculate them. Then we've got to decompose them into their components and then see if we can use some symmetry. And then finally, we're going to add those components and figure out what the net force is. Let's go ahead and take a look.

So first things first, I want to label these forces. So on the bottom mass, I have a gravitational force that points in that direction. And because it's between m₁ and m₃, I'm going to call this F₁₃. And then I've got another gravitational force that acts between m₁₃ß12, so I'm going to call F₁₂. Alright. So I've labeled them. Now it's time to actually calculate them. How do I calculate what the gravitational force is? I go back to my Newton's law of gravity. So the, gravitational force between 13 is going to be I'm going to treat them as point masses. So I've got GM₁M₃ divided by r squared. Do I have everything I need? Well, I have the masses of both of the objects and I have the center of mass distance between them. So I've got everything I need to go ahead and solve for this. So doing that, I get 6.67 × 10^-11, then times 50 times 50 and then divided by 0.6². And if you do that, you should get F₁₃ is equal to 4.63 × 10^-7 Newtons. So we've labeled the forces, now we have to actually calculate the other force, right? We have to figure out what F₁₂ is. So what is F₁₂? Well, notice here I have the same exact masses involved, so I have the same masses, so same m's, and the distance between them, so I've got let's see. I've got the same exact masses right here, and I also have the same distances between their centers. So because I have the same m's and the same r's here, then I know that F₁₂ is also just going to be 4.63 × 10^-7. I don't have to do that over again because I've already done it. Okay. So now I've calculated the forces. Now it's time to decompose them and then use symmetry. So if I'm ever using vector decomposition, I need a coordinate system first. So let's go ahead and draw that. I've got a y coordinate system right here, y-axis, and I've got an x-axis right over here. In order to get components, I need an angle, but not just any angle. I need an angle with specifically with respect to the x-axis. So I need that angle theta. How do I go about getting that angle? Well, we've looked at everything about this problem. We've got the masses and we've got the sides between them. But the one thing we don't know is, the angle. And the one thing we haven't used yet is that equilateral triangles have 60-degree angles between their sides. So that means that these angles right here are 60 degrees. How can I go with how can I go from using this information here, these 60-degree angles, to figure out what this thing is right here? There's a couple of different ways I could do it using geometry. Remember that if these are parallel lines, so this bottom and top line are parallel lines, then these angles right here are called opposite or alternating interior angles, and they're the same. Another way you can think about this is you can sort of draw like a perpendicular line that goes down between both of these lines here. This is a 90-degree angle and this is also a 90-degree angle. So that means that this has to be 30 degrees because 60:30 makes 90, which means that 30:90 means that this is a 60-degree angle. I know that's a whole lot of geometry stuff, but there are a couple ways you could sort of convince yourself that the angle that we're working with is actually 60 degrees as well. So now that we have that angle, we can figure out what the components are. We can split this thing up into Fₓ and Fᵧ. And we know that the relationship between Fₓ and Fᵧ is F cosine theta and F sine theta. So I've got F cosine theta and then I've got F sine theta right here. Okay. And now for the other angle right here, for the other vector, F₁₂, I've got this angle. Now what's that angle? If this is 60 degrees and these were 60 degrees, then it means that these two angles also have to be the same. So let's take a look here. We have the same exact force and we have the same exact theta between them. So that means that when we split this up into its components here, this Fₓ is going to be the same and this Fᵧ is going to be the same. So the question is now can we use symmetry to eliminate those things? And yes, we can, because we have the same exact force and we have the same exact angle. So these x components are just going to go away. They're going to cancel each other out. So I'm going to write that here. So these things are going to cancel because we have the same f g's. So we have same fg's, same theta, but opposite direction. So we end up canceling those things out. Alright, So now that we're done with the decomposition and using symmetry, it's time to basically just add up our final, components and then figure out what the net force is. We know the net force is going to be pointing straight up. So that actually takes care of the direction of the gravitational force. Now as for the magnitude, let me go ahead and write that down here. The magnitude of that net force is going to be 2 times Fᵧ. Now what is Fᵧ? Well, remember that Fᵧ using our decomposition was F times the sine of theta. I have what the force is, that's the gravitational force that we figured out in step 2, and then I have the angle now that we've done the decomposition. So I've got that, F sine theta is equal to 4.63 × 10^-7 times the sine of 60 degrees. And if you go ahead and do that, you should actually get 4 × 10^-7. That's the y-component. But remember that this is just one of the Fᵧ's. Right? This is just one of them. And the, net force is going to be 2 times Fᵧ. So we get 2 times this number right here, 4 × 10^-7. So that means that the net force is 8 × 10^-7, and that's in Newtons. So that's for the magnitude of the gravitational force. And as for the direction, we know it points in the plus y direction.

Alright, guys, let me know if you have any questions with this stuff.

Find the magnitude & direction of the net gravitational force on the center 5-kg mass in the rectangle below.

• A. 8.6×10^-8 N
• B. 4.28×10^-7 N
• C. -1.7×10^-7 N
• D. 6.5×10^-7 N

Here’s what students ask on this topic:

How do you calculate the net gravitational force in a 2D system with multiple masses?

To calculate the net gravitational force in a 2D system with multiple masses, you need to use vector addition. First, label all the forces acting on each mass. Then, calculate the gravitational force between each pair of masses using Newton's law of gravitation: $F=G(m{1}_{}m{2}_{})/r^2$. Decompose each force into its x and y components using trigonometric functions. Finally, sum the components in each direction to find the net force's magnitude and direction using the Pythagorean theorem and inverse tangent function.

What is the role of symmetry in simplifying gravitational force calculations in 2D?

Symmetry can significantly simplify gravitational force calculations in 2D by allowing you to cancel out equal and opposite components. If the system has symmetrical properties, such as equal masses and distances, the forces in opposite directions can cancel each other out. This reduces the number of components you need to consider, making the calculations easier. For example, in an equilateral triangle of masses, the horizontal components of the forces on a mass at the bottom can cancel out, leaving only the vertical components to be summed.

How do you decompose gravitational forces into components in 2D?

To decompose gravitational forces into components in 2D, you need to use trigonometric functions. First, identify the angle θ between the force vector and the reference axis (usually the x-axis). Then, use the cosine function to find the x-component and the sine function to find the y-component. The equations are: $F_x=F\cos \left(\theta \right)$ and $F_y=F\sin \left(\theta \right)$. This allows you to break down the force vector into its horizontal and vertical components, which can then be summed to find the net force.

What is the significance of using vector addition in calculating gravitational forces in 2D?

Vector addition is crucial in calculating gravitational forces in 2D because forces are vector quantities, meaning they have both magnitude and direction. When multiple forces act on an object, you cannot simply add their magnitudes; you must consider their directions. By using vector addition, you can accurately determine the net force acting on an object. This involves breaking each force into its x and y components, summing these components separately, and then using the Pythagorean theorem and trigonometric functions to find the resultant vector's magnitude and direction.

How do you find the direction of the net gravitational force in a 2D system?

To find the direction of the net gravitational force in a 2D system, you first need to determine the x and y components of each individual force. After summing these components, you can find the net force's direction using the inverse tangent function. The angle θ of the net force relative to the x-axis is given by: $\theta =\tan {-}^1\left(\frac{F_y}{F_x}\right)$. This angle tells you the direction of the net gravitational force relative to the horizontal axis.