To extricate an SUV stuck in the mud, workmen use three horizonta...

Question

To extricate an SUV stuck in the mud, workmen use three horizontal ropes, producing the force vectors shown in Fig. E4.2. (b) Use the components to find the magnitude and direction of the resultant of the three pulls. (image)

Accepted Answer

everyone in this problem. We have an old relic buried in the ground and we have three workmen trying to unearth it. They're applying three horizontal vector forces on the relic using ropes as shown below. Okay. And we're asked to find the magnitude and direction of the resultant force from those three pulls. Okay. And we're told to use components. Alright. So we want to use components. What we wanna do is draw a little table. So we're gonna have the X. Component and the Y. Component. And along the side here we have our three pulse A. B. And see. And to get the result, we're just going to add up each component of the three pulse A plus B plus C. Okay. Alright. So now what do we put in each of these? Well in each of these we're gonna put the X. And Y. Component of the vector. Okay. So let's start with a. Okay, A. is 57° from the Y. Axis. Let's just fill in what it is from the X axis. Um Let's keep everything related to the X axis. It'll be a little bit simpler just to see kind of the pattern and keep things consistent. Okay, so if we have 57 degrees from the y axis then this angle is going to be 33 degrees. Okay. They have to add up to 90 degrees. Alright. Now for the X. Component of a, the X. Component is going to be on the adjacent side. So it's gonna be related through coast. Okay so we're gonna have coasts of 33° times the Hippopotamus or the magnitude of that Pull A. Which is 746 Newtons. Okay. And what we're gonna get here is .648 Newtons. Okay. So that's four a. In the extraction. Now let's do the Y direction. Well in the Y direction. The Y component is going to be on the opposite side of our angle. So that's going to be related through sign. So we're gonna have sign of our angle 33° eight times 746 Newtons. The magnitude of that force or the hypotenuse of that triangle? You're thinking about triangles and we get .301. Newtons. Okay. So we have our X and Y components for a Let's do the same for B. No, when we're doing B okay we have the angle to the X axis. So that's great. But we have to think about the sign. Ok. A was in quadrant one. So both X and Y components were positive for B. We're in quadrant two. So the Y component is still positive but the X component is going to be negative. Okay, so for the X component we're going to have to have negative and again the X component is on this adjacent side to our angle. So that's related through coast. So we have coast of our angle 49 degrees and the magnitude 862 newtons. Okay. And this gives us an X. Component of b minus .5-3. Okay. It's really important you get that sign right? So that when you add up the vectors you get the resultant force. Okay? Otherwise you're gonna end up with some force that's very, very positive in one direction. Okay. Same thing for the Y. Again, the Y component is positive and why it's going to be related through sign? Because it's on the opposite side of our angle 49 degrees in terms of magnitude 862 newtons, which gives us a Y component of B 650. newtons. Okay, So we've done a we've done b Now all we have left is C. Alright, so first see again we have this angle that's related to the y axis in order to keep it consistent. We're going to put it to the X axis and you can do it either way, you can leave it as the angle 28 degrees. Just have to pay attention to whether you should be using sine or cosine. Ok, so we have 28 degrees here. That means that this remaining angle is 62 degrees. Okay, the to have to add up to 90 because they make a 90 degree angle between the X and Y axis. Now in this case and see we are in quadrant three. So both the X and Y component is negative. Okay, So for the X component we're gonna have negative and again on the adjacent sides, we're using coast of that angle 62 degrees. And the magnitude of the force 532 newtons, Which gives us a component negative. 0.759 newtons. And similarly for why? Again, it's negative because we're in quadrant three. We get sine of the angle 62 degrees times and magnitude 532 Newtons. Okay? And we get a Y component of negative. 469.7 to 8. Newtons. Right? So we've figured out the X and Y components of all three of the polls we were given To determine the result. We're going to add them up. Okay, so we're gonna take 625.648 newtons minus 565.5 to 3. Newtons minus 249.759 newtons. Which gives us a result of minus 189. newtons. Okay. And that's going to be the X component of our resulting force. Same for y. 406.301. Newtons plus 650.56 Newtons minus 469.7 to 8. Newtons. We get a resulting Y component of 587.133 Newtons. Okay. And so these are resulting components. What the question is asking is for the magnitude and direction of the resulting force. Okay, so in order to find the magnitude and direction, let's go ahead and draw a little diagram of what these components tell us. Okay, so I'm gonna move up so we can still see our components but that we have a bit of room to work. Okay, so our x component is negative. Are y component is positive? So we're gonna be in quadrant two. Okay, so we can draw Like this, we're gonna be in quadrant two. I sort of something like this is going to be a resulting force. Okay, we're this X component here And let me maybe draw in blue. So you can see this x component here is 100 has a magnitude of 189.634. Okay, this y component here has a magnitude of 587.133. What we want to find is the force. The magnitude of the force. Okay. Which is going to be the hypotenuse of this triangle. And we want to find the angle theta. The angle of data is from the positive X axis. So that means that the angles data is going to be this angle here. Okay, that's the angle we want to find but we know based on the two dimensions we have that the angle we're going to be able to find first, is this angle here let's call it theta prime. So we're gonna need to find data prime first and then relate the two. Okay, How are they going to be related? Well, we can write that data Is going to be equal to 180° minus data prime. Okay. Data plus data prime. Make this straight line. They go all the way around from one side to the other. So it's gonna be 100 and 80 degrees. Okay, So in order to find data, we take 100 and 80 degrees. We subtract that data prime that we find and we'll get the answer we're looking for. Alright, so what is data prime? Well, theta prime. We have the opposite and adjacent side. So we're going to relate through tan. So we're going to get tan inverse Of 587 133, divided by .634. And this is going to give us a theta prime of 72.1°. Okay, so this data prime here 72.1 degrees, which means that data, the value we're looking for 100 and 80 minus 72.1 degrees gives us 107.9 degrees. Okay, So this is the direction of that force that we were looking for 107.9°. Now I need to find the magnitude of that force F. Now we see this as a right triangle. We can use Pythagorean theorem. So, we have the f squared will be equal to 189.634. Newton squared Plus .133 Newtons all squared. This gives us an F squared, 380, 0.213645. Okay. Four and F. Taking the square root. And again, when you take the square root, you're going to get the positive and negative route. Ok. Here, we're just going to take the positive route because we know we're looking for the magnitude of the force. Okay? We're just gonna take the positive route. We get 617 newtons. Okay. And so, using the components of the three poles, A, B and C, we were given, we found that the force has a magnitude of 617 newtons and a direction of 100 and eight degrees. So, we're going to have answered D Thanks everyone for watching. I hope this video helped see you in the next one

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