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The {2-0 boom ABhas ned end steel cable Slrelcned from the Iree end Bof the boom polnt - 'located an the verical itlL ihe tension Ihe cable 470 Ib, determine t...

Question

The {2-0 boom ABhas ned end steel cable Slrelcned from the Iree end Bof the boom polnt - 'located an the verical itlL ihe tension Ihe cable 470 Ib, determine the peipendicular distance_ Irom polnt Ato cable &CNumerle Response7.37 * 2%

The {2-0 boom ABhas ned end steel cable Slrelcned from the Iree end Bof the boom polnt - 'located an the verical itlL ihe tension Ihe cable 470 Ib, determine the peipendicular distance_ Irom polnt Ato cable &C Numerle Response 7.37 * 2%



Answers

The 12 -ft boom $A B$ has a fixed end $A$. A steel cable is stretched from the free end $B$ of the boom to a point $C$ located on the vertical wall. If the tension in the cable is 380 Ib, determine the moment about $A$ of the force exerted by the cable at $B .$ (Figure cant copy)

Question Number 22. This is our boom, and our weight is hanging like this 2000 Newton and the weight off the boom is 1200 Newton. This angle is 65 degrees, and attention force is acting like this. This tension pools is making an angle with the horizontal 25 degrees. So from jeopardy, this Angelis, also 65 degree, does. The tension force is perpendicular to the length off the boom. The reaction forces at the base are are y and are now Let us take the talk off all the forces about the base and equip them 20 Because for the static equilibrium, the network on the boom should be zero, so we can write he into three by four times the length off. The boom is equal to that talk. By the 2000 Newton Force 2000 l cause 65 degrees plus 1200 into al by two in the course off 65 degrees. Solving this 40 we will get the attention for stay equal to or in 65 Newton. This is the required tension force on the boom, and the reaction forces are our X is equal to be cost 25 degrees or 1465 into course, off 25 degrees. This is our is equal to our X is equal to 13 Tow it Newton. Now for the vertical little room off the boom we can write, the reaction falls. Our wives should be equal to 1200 Newton plus 2000 Milton minus E sign. 25 degrees. Such shooting the value off key we will get our wife will too. 25 aired. Zero Newton. This is that required reaction post.

Hello. So this problem is about movements, If you recall moment, is they put out of force under perpendicular distance. So for the first part of this question, we want to find attention. We're going to use moment the laws off Acqua Librium. So for a moment, I want to recall that the some off the clockwise movement should we go to the some of the anti clockwise moment. So I'm gonna find moment about eight and for a moment, about a I have three forces to keeps five kids and attention. I remember the moment the distance should be the perpendicular distance. So, um, you realize that about a the two clips and five kids were going clockwise on the T went anti clockwise. So for the clock, at this moment, I have forced on resistance of my first forces to non distance. So for the two keeps the distance between a onda. Let's see, G, that's the center of gravity is 20 ft. If we resolve that horizontally because we need a perpendicular distance. So it has to be this distance and not the inclined. No, we want the horizontal this and that's perpendicular to the direction of the way. So if we resolve that using an angle 20 course 30 that the particular distance. Okay, then we have another wait five keeps. Um, Now, also, be times the distance. The distance between a. B is 40. So if you resolved that like we did 42 this time from a to that will be 40 custard for it. Austerity. Now we go toe see times per nuclear disarm between tea and ate this t That's a So I'm going to say that this is my problem. You distance right here. 90 degrees. So because the length off a B s 40 then this distance right here, this right on the triangle. So they're just going to be, um that would just be ties. 40 Sign terror. But the distance so t just gonna be 29 for 24 kiss. So that's what the first part. So let's go to the second part, which is to find the reaction on a so at a the reaction has two components. There's a X going to the right. Then there is a y great upwards. So you're going to resolve forces for the second part. So there are three forces. I'm going to start by dream my heart, you know, resolutions Check. Horizontal. All right, so the two forces those five kids and two kids, their vehicles. So General Motors. And also we forget about that. Then we have a explains the right. So we have a X. They have a why t is going to the left, but it is not, um it is not horizontal, so we need to resolve it. So we need to do some small geometry. Um, so the angle in here is 35 Standard stand down. This is 90 degrees. And so this angle right here at B on B is gonna be so nine. You have 90. We have 30 does 1 20. So there's going to be 60. Okay, If I draw the horizontal component off T this way. Small angle here. That's going to be 20. So this this t X going to the left. This is T why we're very good upwards. So this it's just going to be so t X is just, um that will be t cost 20 before, So I just see course 20 which is just 29. 9 to 4. That's cause 20 and that gives me 28.11 My kids? What? I would go to the vehicle. So for the forces moving upwards we have Why there? We have to see why. So that's that, then, for forces going downwards, we have two plus five. Um, so let's see. Uh, yeah. No, I think there's a mystic right here. Let me correct that. Uh, the t y because attention is just with t y should be done. What? Instead, see what is dull. So I was going to change everything. I will do this again. So for the forces going up, I'm just going to be just a y forces going down will be, um, plus two plus five plus t y. But see why it is just, uh, t son 20 which is just 29 points. 924 Sign 20. So why record two plus five plus 29 point like to four. Okay, sun 20 e dot Don't give me up. 17.234 Uh, I'm just making 35. So that's 235 kiss. So, to find the reaction, this is just gonna be a fight over a story. Um X squared because you one squared on, uh, just gonna be square root. 28.19 squared, plus 17 points to you're five. Spot the reaction of a It's gonna be, uh, too for my age. Okay. To fire back eight tips. So that's there for this question. Thank you.

This problem. We have a boom, um, that's carrying a load p supported by cables. And we know that the tension in one of the cables is 138 pounds and that the results of the load P and the force exerted at a by the two cables must be directed along the axis. Um oh. A So we need to determine the tension in cable a the other cable A c. So again, I'm not gonna I don't, um you know, they have the skills to draw things. Figure, uh, but you could look at it in the book. And if we look at references way, have our point stuff interests are a is 48 inches in the X direction R B is 29 inches in the UAE and 24 inches in the Z and R C is 25 inches in the UAE and minus 36 inches in the Z direction. We can then get used these points to get unit vectors from one from A to B and one from A to C on these air. Given by these factors here to this points from A to B and this points from A to C. Now we know that the forces in the two cables this one points from a is a B. Let me actually maybe write that down here, so f one equals f Uh, Let's see, I used a t t a c e um, outside a t a b e a b n f two because t a c e a city Now we know we know what was you know, a B You know this one way? No, this value here way No, the vector the direction In this case, we don't know we know the direction, but we don't know the value, but we do know that the result of these two needs to be purely in the Z direction cannot have a component in the Z direction because we know the result has to be a long Oh, A and P is in the y direction. So this will obviously have a component in the Y direction toe offset p. But we need to figure out we need to figure out what this value is such that the Z component is zero. Well, the Z component if we, you know, plug in the values. For these things. Z component is out 183. The value of T A e times 0.393 Here minus t a C time 0.554 with this value here, and that has to be zero. Which tells us that T A. C has to be very close to 130 on that shouldn't be Newton's. That should be pounds pounds.

This problem. We have a a basically a crane. Um oh, whom it's a pinned at this point here. And then it's extended out over so a wall here and it's carrying a lower way demolition ball, wrecking ball here or something. And then there's another cable that was attached at this point. That then comes back. Attach is to a point to the left. So we have our reaction forces at the pivot, their reaction for strong, capable of weight off wrecking ball and weakens, not writing foursome position vector. So here's the force and position of pivot, which I've taken to is this force here, and we know that it makes a on an angle of 22 degrees, uh, with your my 22 degrees with the scores on or the horizontal here. And so we know then that this has a magnitude up to on the direction is given by this and fears the 22 degrees and the position is the distance l from the origin and in this direction, where in this problem data is five degrees and, um, three is the weight, and it's also located position to so we could go through and start calculating forces and doing are forced balance and movement balance. We get a in an expression, we get expressions for the the force in the pivot reaction force and the force and the cable. And I left them symbolically here. Just Teoh, I wanna think about what? How this changes. You change the angle. You see that they're all proportional to the weight, which makes sense. Um, and they are have this term co seeking off data minus e, and Corsican is one over the sine. So if this goes to zero that these things go to infinity because, yeah, that makes them. Because if this if this table were attached right here than problems, um basically wouldn't hold itself. So you can kind of play around, you know, if you want. But if you plug in the values for all this stuff that were given, get the enough one. X is £1688 1 why Reaction force in the vertical direction is £1182 and the the attention in the cable IHS £1821


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