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Question 16 (6.66662 points)Thc spacc truss shown supports a 150-Ib cratc at point € Assumc that Fravity acts In the ncgative dircctionTnc unit vectors in the...

Question

Question 16 (6.66662 points)Thc spacc truss shown supports a 150-Ib cratc at point € Assumc that Fravity acts In the ncgative dircctionTnc unit vectors in the directions 0f mcmbers CAand CB #rc a5 follok5"ucA [+0.3541 0.7071-0.612k] ucB [0 3541 0.707i-0.612k/If mcmbers CA and CB cach carry 122.5-Ib Ot compressivc torcc; #hich ot thc following Is closcst to thc force in member CD?a) 95 Ib (b) 245 Ib ((c) 22016 M0) 173 Ib (T)

Question 16 (6.66662 points) Thc spacc truss shown supports a 150-Ib cratc at point € Assumc that Fravity acts In the ncgative dircction Tnc unit vectors in the directions 0f mcmbers CAand CB #rc a5 follok5" ucA [+0.3541 0.7071-0.612k] ucB [0 3541 0.707i-0.612k/ If mcmbers CA and CB cach carry 122.5-Ib Ot compressivc torcc; #hich ot thc following Is closcst to thc force in member CD? a) 95 Ib ( b) 245 Ib ( (c) 22016 M 0) 173 Ib (T)



Answers

The vertices of a quadrilateral are $A(1,2,-1)$, $B(-4,2,-2), C(4,1,-5)$ and $D(2,-1,3)$. Forces of magnitudes $2 \mathrm{~N}, 3 \mathrm{~N}, 2 \mathrm{~N}$ are acting at point $A$ along the lines $A B, A C, A D$ respectively. Their resultant is: (a) $\frac{10 \hat{i}-9 \hat{j}+6 \hat{k}}{\sqrt{26}}$ (b) $\left(\frac{\hat{i}-9 \hat{j}-6 \hat{k}}{\sqrt{26}}\right)$ (c) $\frac{\hat{i}+9 \hat{j}+16 \hat{k}}{\sqrt{26}}$ (d) $\frac{\hat{i}-19 \hat{j}+6 \hat{k}}{\sqrt{26}}$

In program to 44. We're solving a question Where a somewhere to 2.3. Except we have different magnitudes for the two forces. P and cute. So first, let's look at this question from using the parallelogram hole. First we draw four signs off the parallelogram. Sorry, looks a little bit bad. And in a diagonal, the parallelogram is gonna be the result in force, which will name fr and the angle we're looking for off the force. We can name it Fada. So to solve this triangle, the one I'm hiding here. First, we need to find the angle that I'm pointing out here. And we know that the size for and sins on the two of Jason angles in the parallelogram at up to 1 80 degrees and we know one of them. It's 75 degrees, which is from 25 plus 50. We know that this angle would be 1 80 minus 75 degrees, which give us +105 degrees. And then since we know the two sides and the angle between them, we can use the coastline long too soft for the side. And to do that, we ride it out as such. It's a simple question, too soft, and then we get possibly a P 10 Kips for the result in force. So to find the angle fatal, we can use the sign all since we already know all the angles. I mean one angle and many two of the two sides. So let's read it out like this five over a 0.3 which we just calculated a buff. It's equal to angle sine of the angle that we're looking for and over four. And then after that, we can get they That was 28.8 degrees. And if you want to write it out in terms off its relation with the horizontal line, we can subtract 28.8 by five degrees, which give us before date. So there you go. You have your answers, the magnitude of the force and also the direction of the force

We're going to use the distance formula to prove or disprove concurrency. So we have J k que el in drill if peak que Q r pr So looking at J. K, we have six minutes. Once the square root of six minus one squared, that's five and four minus negative C squared Justin squared. That's going to be square root of 1 25. I'm gonna leave it at that for Kael. We have one minus negative nine squared, which is 10 squared in a negative six minus five, which is 11 squared. You know what I I can on that to 21 we have G l. That's six minus negative nine square, which is 15 squared and four minus five square, which is one squared. I'm gonna leave it at that for now. I'm going to see if we have similar numbers for this next section before continue and expand on that. So we have why in peak, which is going to be zero minus five square That's five five squared and seven minus negative three School, which is 10 squared. It's the same as G K. We'll leave the math, Eric, you are. We have five minus 15 square. That's 10 squared and we have negative three in minus eight square. That's 11 square, same SQL now PR. We have zero minus 15 square. That's 15 squared. We I did. I say that right? P aria and seven minus eight square. That's one squared, which just check out so I can see that triangle J k l is congruent to triangle p que or.

In order to find the vector that is represented by Peak You here first, let's find Peak you So, p thank you are here. So the vector P Q. Since it ends in Q, we're going to do three minus negative for So this first dr here we have three minus negative for and in comma. Then we'll have negative for minus one. So our vector here is one of equal to this is going to be seven comma, Negative five. So now remember, I have is equal to this vector here 10 and then J hat is equal to 01 So this I here means that we are or the one here means that we this is going to be our coefficient of the I. So this is going to be able to Now we have here seven high, like so and then minus five J like that. This, uh, minus five corresponds to the, uh, coefficient in front of the J. So minus five here 01 So this is gonna be our final answer here

Hi. It is exercise. We have the following points on the plane and we need to resolve the following vector. So the first vector is a vector you given by 2.5. And the vector the distance vector between A. And B minus 0.8. The distance vector between CMT. So the The vector A. B. It's the vector from eight that starts from A. And B. These vector city is the vector that point two d. From C. Okay, so we need to define these vectors first. So the vector A. B. You can observe that these jails given by this obstruction of the .2 to -100. And that is just a vector to I plus two. J. Similarly for the vector city we take first the point D. That is 34 and then we subtract the 340.70 From this. We have changed the vector -4. I plus four G. Okay so we have our two vectors that we need to distance vectors. And from that we can resolve the vector. So the vector U. We're result Multiplying 2 5 to the vector to I plus j minus here -0.8 times a vector -4 I plus four J. We resolve this some component by component. We obtain the following picture, 8.2 I. They were in the I direction plus 1.8 in the J direction. So this is a vector U. Now we need to calculate another factory and the picture is given by to the vector W. In this case is equal to 2.5. But now the distance vector, the A. Technically that means that we're going to invert the the vector, so we are going to change the components. And if this vector is a vector to to this is going to be the vector -2, I -J. Because we are inverted -0.8 times a vector CD This is the same vector that looked at here. We have minus side. So this is going to be important later. And this minus sign with this minus sign will cancel. And we have just plus your .8 times this electricity. So we have these vectors already calculated the distance vectors, the only one that we need to recalculate it is the distance vector B. A. But as I just mentioned is just inverting the the vector just putting the minus sign in front of it, so minus two I minus two J. And that's so from that we obtained that this vector will be 2.5 times the victory minus two. I minus T. J Plus 08 here. And the electricity is the same. So -4. I plus four jake. When we resolve this we obtain the vector W. Is equal to -8, 2, I -18 J. And you kind observed that there is some relation between them. So that corresponds to the second part of this exercise. So the second part of this exercise is explain what is the relationship between the view and vector vector W. In the victory. So the vector W. Let me regret it here is 8.2. I minus 18 day. And the vector V is a vector 8.2 I Plus 1.8 J. And you can observe that the vector V is equals two minus a vector W. But this relations have a name and they are called anti parallel and why this is expected? Well because you can observe that here. So I'm going to write again here. The vector W. And the vector V. The whole expression So the victory be was equals 2-5. The distance back third between a and B -08. The vector the vector city. Okay now how is to find the veteran W. Veteran W. Here was defined as 25 times the victory be A. So here we change the direction of them Plus 0.8. The Vector City. And you can see that this picture the inversion of the direction of the vector will just put a minus sign in front of it. So this will result in -2.5. The vector A. B Plus 0.8. The Vector City. And this is just equal to minus The vector to five times a victory. A. B minus your point A. The distance back. Third between C and D. But this is what? Well this is just the vector V. That we have here. So we have that director W was equals two minus the vector B. So that's why this was expected because by definition the vector W is minus two, and we obtain an anti part factors.


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