FIGURE P3.46 shows four electric charges located at the corners o...

Question

FIGURE P3.46 shows four electric charges located at the corners of a rectangle. Like charges, you will recall, repel each other while opposite charges attract. Charge B exerts a repulsive force (directly away from B) on charge A of 3.0 N. Charge C exerts an attractive force (directly toward C) on charge A of 6.0 N. Finally, charge D exerts an attractive force of 2.0 N on charge A. Assuming that forces are vectors, what are the magnitude and direction of the net force Fₙₑₜ exerted on charge A?

Accepted Answer

Hey, everyone in this problem, we're told that electric forces have their sources on charge. The basic law of electrostatics tells us that the light charges are gonna work out and unlike charges are going to attract. OK. So a group of students charged four spheres and assembled them in this rectangular shape. So we have sphere one in the bottom left, sphere, two in the bottom right sphere, three in the top, right and sphere four in the top left. OK. The width of this rectangle is 625 millimeters and the height is 415 millimeters. We're told that the force on sp sphere one is a 12 Newton attractive force from sphere two directed at word from sphere one to sphere two. OK. So before we finish with the rest of this question, let's draw that on our diagram is we have this attractive force from sphere two that is directed from sphere one two sphere two and that is 12 notes. And then we have a 22 new repulsive force from sphere three. OK. So it's directed from sphere three, two sphere one newtons. And finally, we have this 17 Newton attractive force from sphere four be directed from sphere one two sphere four. And that is a 17 new, we've drawn those on our diagram. Now, what is this question asking? The question is saying to treat the forces like vectors and to determine the magnitude and direction of the sum of forces on sphere. What? All right. So we're given five answer choices here. All of them have a different magnitude and direction for the sum of forces. OK? With the force in newtons, the direction in degrees. And we're gonna come back to those answer choices as we work through this problem. So if we're looking for the magnitude and direction of the sum of forces, let's go ahead and break these forces down into their X and Y components, then we can add by components. So we're gonna have our X component, we're gonna have our Y component and that's gonna be for each of our three forces. OK? So we're gonna have the force F 12. So that's gonna be the force between sphere one and two. We're gonna have the force F 13. So the force between sphere one and three, then of course, F 14. OK. So between one and four, and finally, we're gonna sum these all up to get our sum of courses at the end. OK? So we've made this little table and we're gonna fill it in. Let's start with, let me do a little dotted line. So we can separate these, let's start with the force from sphere, one to sphere two. We're gonna say that up into the right is our positive directions. So this vector for this force is pointing in our positive direction. It's pointing to the right, it has a magnitude of 12 noons and it's pointing perfectly horizontal. So it's gonna be completely in the X direction. So the X component is going to be 12 nodes that entire magnitude of the force and the Y component is just zero. S. Now we're gonna skip one and three for right now, we're gonna go to one in four and similarly to one and two. OK. The force between one and four that's acting upwards, that's our positive direction and it's perfectly vertical. So it's entirely in the Y direction. So the Y component is gonna be 17 units in the entire magnitude of our force and the X component is zero in this case. All right. So we've done those two that are a little bit easier. Now, we're gonna move to the force between sphere one and three. We know the hypo, OK. 22 noons, we need to break it into the opposite and adjacent sides. Now, in order to do that, we need to know the angle. OK. And so this angle in the top right between the horizontal and our hypo, we're gonna call up and we can find alpha because we know the dimensions of this rectangle, we know the opposite and adjacent side of this rectangle. So let's use the tangent to relate and we're called the tangent of alpha is gonna be the opposite side divided by the adjacent side. OK. So in this case, the tangent of alpha is going to be equal to 415 millimeters divided by 625 millimeters. If we take the inverse tangent of this, that is going to be our alpha value. So alpha is gonna be the inverse tangent of 415 divided by 625. Those units of millimeters will divide it. All right. So we're gonna leave that in that form for right now while we break this down our X component then of this force between one and three. Well, it's gonna be pointing to the left, we can break that force down. Let me do it on the left hand side here. If we break this force down, the X component is gonna be pointing to the left, the one component is gonna be pointing down. OK. So both are going to be negative. OK. They're in our negative directions. So the X component is negative and it's gonna be negative 22 noons multiplied by cosine of because we're talking about the adjacent su similarly, the Y component is gonna be negative 22 nodes multiplied by a sign of OK. So we're just breaking that vector down into its X and Y components and we've just left it as alpha right now. We know the value of alpha. We're just gonna leave it so that we don't have any round up there or we can avoid some round up there. All right. And now we're gonna, OK. So we're gonna take the X component and we're gonna add up this entire column. Yeah. So we have 12 noons plus negative 22 newtons times cosine of alpha plus zero newtons. And we get a sum of negative 6. 8516 noons approximately. And similarly, for the Y component, we sum up that entire column. Then we got 4. 83 newtons. OK? That was zero newtons plus negative 22 newtons multiplied by sine of alpha plus 17 newtons. So we have a negative X component, a positive Y component. Now remember the question was asking for the magnitude and direction. So we have the X and Y components. Now we need to combine those to find the magnitude of the total force. So let's move down here and let's draw out a little triangle. So our X component is negative. So we're gonna have some force pointing to the left and our one component is positive. So we're gonna have some force pointing up. Now, the X component we know is 6.3285169. OK? It's negative, but we're talking about the magnitude here. And then the right component 4.831783 news, the magnitude of the force we're looking for is going to be the hypotenuse. And we're gonna call this the magnitude of that. And this is actually, you know what, let's stick with constant notation. This is gonna be the magnitude of the sum of the forces. Now we have a right angle triangle. So just like before we can use Python theorem, we get that the sum of the forces, the magnitude squared, it's gonna be equal to 6.3285016 Newton squared plus 4.831783 Newton squared. If we simplify the radiant side and take the square root, OK, we're taking the positive route because we're talking about the magnitude, we get that the magnitude of the forces is about 7. nodes. OK. So that is the magnitude of the force. Now, looking at our answer choices, we can narrow this down to two options. Option A and option D both have the magnitude of the force being 7.96 newtons. OK. So we round to two decimal places. That is exactly what we found. Option BC and E have that the magnitude of the force is 1.5 newtons, which is not correct. So we can eliminate BC and E and now we're left with A or D and we have to figure out the direction in order to choose the correct one. So recall that if we're talking about the direction of a vector, we wanna measure from the positive x axis. OK. So the angle we wanna measure is actually this angle outside of our triangle. And that's gonna be data that's going to be the direction of this. Now the vector or sorry, the angle we can't calculate is going to be the angle inside the triangle. We're gonna call that theta prime. OK. Now, recall again, we're relating the opposite and adjacent side. So we're using the tangent again, tangent of theta prime is going to be the opposite side. 4.831783 news divided by the adjacent side 6.328516 newtons. If we take the inverse tangent, we get that theta prime, it's gonna be about 37.36 degrees. All right. So we found the angle inside of the triangle again to look at the direction of a vector, we want the angle from the positive X axis. And we can see that theta plus data prime is going to be equal to degrees because they form a straight line that tells us that the is going to be equal to 180 degrees minus the prime. We just found data prime was 37.36. And so our angle, theta is going to be equal to 142.64 degrees. And that is the direction of our vector. OK. So we have the magnitude, we have the direction of the sum of forces. Let's go back to our answer choices. We're going round to the nearest degree and we can see that the correct answer is option D, we had the force of 7.96 newtons in a direction of 143 degrees. Thanks everyone for watching. I hope this video helped see you in the next one.

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