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Find the magnitude and direction of the net gravitational force on mass $A$ due to masses $B$ and $C$ in Figure $6.27 .$ Each mass is 2.00 $\mathrm{kg} .$

0.0261 $\mathrm{mm}$

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Simon Fraser University

University of Sheffield

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said Ah, the positive exact sis from left to right. So from the point where s O This is Figure E and this is figure piece of the point where a and see objects are all the spheres Are we take that as our initial or the origin? And then if we go on right, we'll say that we're going going to the positive direction off X Now if we want to draw the free world diagrams for both of us here one of the figures. So for figure, eh? We'll have the force that's acting on a select said This is a on the force acting on and you to be will be directed on the right and for sea It's the same exclusively on the right Let's call it F C and B. And of course, as we mentioned that this is our positive extraction on DH. We can said this as our right now for and ah, right and for b isin between CNB. So if we said he around, let's see here. So that means there's if Spain there's FC, which is acting on the left and we call it F c. And then this F B, which will be acting on the right now notice that I drew FC bigger than F B. The reason for that is, which is when we look at the in gravitation force. It says, if which is equal to Z times and one. So that's the first mass times the second mass divided by the distance between them. So we see that the force is inversely proportional to the square off the distance, A weapons. If the distance is bigger, that means the force is smaller. So since the distance between A and B is bigger than and see So that's why Ah, the gravity the gravitational force for in between a and B will be smaller than and see and similar thing applies figure. Be a swell. Okay, so let's now solve the problem part by part. So for part a Ah, if we tried to figure out if the that should be the gravitational constant times Mass eight times must be because we're dealing with We're not considering the third mouse, our mouse. See yer and then the distance between and B. So here we know the gravitation constantly. 6567 times into bar minus 11 Newton meters squared part, you know, grand squired. And then we know that these two masses are equal. So that's two points. Here's your programs. And then we divide that by 0.5 meter squared. That's the distance. Those in its 0.5 is because we have 10 centimeter, which is the distance between eight ends E N c. And then we have four centimeters, which is that this is in C and B. So we add bother them together and we get 50 centimeters there. So that's why well, 0.5 meter here. Now, from here we get the force as 1.61 point 06 nineteen 19 times 10 to the power minus nine. And yours on DA. Similarly, for eternal assault for FC will have G. And then we'll have a C divided by our A C squared. And if we have been plugged in the numbers here are a sea will be 10 centimeters. So we get the total force as to point 66 nine times into the part Bayous aid noon. So, since they're acting on the same directions so they will add Ah, get the net. So to get the net force. We need to add one of the forces together so F necked will be f B plus FC. So notice, since it's ah, linear a diagram or there's no contribution since they're aligned in the same line. So that means there's no why. Directional force. So we're only considering the extra in motion. So that a straight edition here and if we put the numbers, it's going to be 0.1 point 69 times 10 to the bar minus nine union plus 2.669 times and bar minus, ain't you? And that gives us a total of 2.8 times into far minus eight. Newman. Okay, on. Of course, the direction should be right. Okay, so for part B, we will do the same thing. But here, since F. C's on the left and F B is on the right and we considered from left to right, it's positive. So we will consider FC as negative here on DH if people positive because it's going thrice on the right. So if be here, you will be Jeanne times and a times and be over our A B squared. And here are a B is 40 centimeter. So again, if we plug in, all the numbers will get the gravitational forces 1.668 times and four minus fine. And we know that FC will be 2.669 10 isn't far minus eight. Newman. The reason we don't have to do the calculation is because this and this ah, these two diagrams are same. The only difference is the positions are switched, so it means the gravitational force should be equal. But the direction will change. So for F Net will have again f b plus F C. But in our case, FC will be negative. So that means we'll have 1.668 times. Tend bar minus nine Newton minus 2.669 times and far minus eight Newton. And from there we get the net force as negative 2.5 Time's tended bar Mine s minus eight noon on DH. We can see there's a negative sign and recon. We understand that since we took the left to right, direction is positive. So negative sign will be from right to left direction. So that means our net force is 2.5 time stands by minus eight. Newton, it's so are definite is 2.5 times 2.5 10 to 4 minus eight Newton and the direction is on the left. Thank you.

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