What is the net gravitational force on the 20.0 kg mass in FIGURE P13.36? Give your answer using unit vectors.

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, consider a system of clay balls of different masses placed at various positions in space. As shown in the figure in the given scenario, calculate the net gravitational force on the 25 kg mass express your answer in unit vectors. OK. So our end goal is to calculate the net gravitational force on the 25 kg mass. OK. And then we need to express our answer in unit vector. So we're given some multiple choice answers, some multiple choice answers to look at and they're all in the same units of newtons. And they're also shows it as the same vector unit. So the unit vector as vector J. So let's read them off to see what our final answer might be. A is 2.52 multiplied by 10 to the power of negative five B is 3.73 multiplied by 10 to the power of negative six C is negative 2.21 multiplied by 10 to the power of negative five and D is negative 3.84 multiplied by 10 to the power of negative six. OK. So we're given a diagram here on the right and it shows the 25 kg mass, it shows the 15 kg mass and then another 15 kg mass. So I've gone ahead in drawing up a more detailed version of the diagram that was provided to us by the problem itself. And I defined the 25 kg mass as M one as mass one. And I called M two for the 15 kg mass. That's on the left side of the, on the negative side, I should say on the X axis or AX C. And then we have M three, which I called the other 15 kg mass on the right side on the positive side of the X axis. And then M one's on the Y axis. And as you can see, since I put it into a coordinate plane, it's very easy to see that we're dealing with like two separate triangles and R one and R two are both the hypotenuse. And we're gonna be using these two separate triangles to help us determine how to find the final answer for the net gravitational force on the 25 kg mass. And it's gonna be a little bit of like vector math and some trig here. But don't worry, I'll make some, it'll make a lot more sense as we go along. But just remember just, you know, to get in your mind, we're dealing with triangles and that's how we're gonna solve this problem. So first off to start solving this problem, we need to determine the forces acting on the 25 kg mass. So let's start by recalling and using the gravitational force equation. And let's start with the first one, let's call it F one is equal to the gravitational force constant multiplied by M one multiplied by M two, all divided by R one squared multiplied by negative sign the I hat minus cosine theta J hat where I hat and J hat are both vectors. OK. And then for F two, F two is equal to the gravitational force constant multiplied by M one multiplied by M three all divided by R two squared all multiplied by positive sine theta I hat minus O sine theta J hat. OK. And just to be clear, so we don't get confused. So F one is for, is focusing on R one, the R one triangle where R one is the hypotenuse. And then F two is focusing on R two, which is the hypotenuse for our second triangle. So just keep it separate, keep it in mind that we're dealing with two separate triangles here. OK. So using our figure from above, we can state that R one is equal to R two, which is equal to R and we can also say that M two is equal to M three because they're both, 15 kg is equal to M. So we can then go ahead and write that F one is equal to the gravitational force constant multiplied by M one multiplied by M divided by R squared multiplied by negative sign, the I hat minus cosine theta J hat and F two is equal to G multiplied by M one multiplied by M all divided by R squared, multiplied by sine I hat minus cosine theta J hat. OK. So the net force on the 25 kg mass is going to be F net, F subscript net. So the net force is equal to F one plus F two. So we need to take this a step further and we need to state that F net, the, the net force is equal to the gravitational force constant multiplied by M one multiplied by M divided by R squared, multiplied by negative sign data I hat minus O sign beta J hat plus the gravitational force constant multiplied by M one multiplied by M divided by R squared. And I'm gonna bring it down below here, but it's all of that multiplied by sign the I had minus sine theta J hack. So this is all for F net when we plug in what we know for F one and F two, which we just wrote above. OK. So let's start simplifying a little bit here. So F net, when we start to simplify the net force will equal two multiplied by the gravitational force constant multiplied by M one multiplied by M divided by R squared multiplied by negative cosine data J hat. So this is where all your trig identities come in handy is when we go to simplify. So if you need, are struggling with that, probably watch some of the math videos to help you get more confident with your trig identities. OK. So in order to find the value of F net, we need to determine the values for R and theta. So using the figure above, we determine that tangent of theta is equal to four divided by 10, which as you can recall, tangent is opposite over adjacent. So when we rearrange to isolate and sol for theta, we get that theta is equal to 10 inverse of 4/10 or divided by 10th, which is equal to 21.8 degrees. So we can use I ari theorem to solve for R. So remember it's A squared plus B squared equals C squared. So R would equal the square root of 0.04 which we have to convert centimeters to meters. So that's where that came from. So 0.04 m squared plus 0.10 means you have to convert 10 centimeters to meters squared. So when we plug that into a calculator, we should get R equals 0.11. Um OK. So at this point. We can now substitute all of our known variables to help us solve for F net. So let's do that. So the net force F net is equal to two multiplied by the gravitational force constant which is 6. multiplied by 10 to the power of negative 11. And its units are newtons multiplied by meters squared per kilogram squared multiplied by kilograms, multiplied by kg, all divided by 0.11 meters square. And all of this is multiplied by negative co sign of 21.8 degrees J hat. OK. So when we plug all of that into a calculator, we should get that the net force or the net gravitational force is equal to negative 3. four multiplied by 10 to the power of negative six vector JJ hat newtons. Awesome. And that is our final answer. We did it. OK. So let's go look at our multiple choice answers to see what the correct answer will be. So looking at our multiple choice answers, the correct answer has to be the letter D negative 3. multiplied by 10 to the power of negative six vector J newtons. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

What is the net gravitational force on the 20.0 kg mass in FIGURE P13.36? Give your answer using unit vectors.

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Key Concepts

Gravitational Force

Unit Vectors

Net Force