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Alung the fixed right-angle rocls and connected (2/181 from book) Collars and slide Bo slac ki Delermine Ihe horizontal acceleration cord constant length (and cloes...

Question

Alung the fixed right-angle rocls and connected (2/181 from book) Collars and slide Bo slac ki Delermine Ihe horizontal acceleration cord constant length (and cloes collar , given constaini UDTd velocity vA collar function 0l Y

alung the fixed right-angle rocls and connected (2/181 from book) Collars and slide Bo slac ki Delermine Ihe horizontal acceleration cord constant length (and cloes collar , given constaini UDTd velocity vA collar function 0l Y



Answers

Collars $A$ and $B$ slide along the fixed right-angle rods and are connected by a cord of length $L$ Determine the acceleration $a_{x}$ of collar $B$ as a function of $y$ if collar $A$ is given a constant upward velocity $v_{A}$

In this question. Questions off motions, everything. The moment of equation off motion about 0.8. So some off moments around a equals m kay around eight. So m g sign Seton Times. Improvers, too, equals M p because science sita times over to so Sita in this case equal then in verse, e over ji.

This question. We have this set up. There's a 16 ft long court attached at point C going through to police here. Here. Okay, then, uh, at point B Uh ah. What s B is going to 6 ft. Ah, the point B is being pulled down with 4 ft per second. Accelerate a 3 ft per second squared downward. We want to find a velocity and acceleration of point A at this instance. Mhm. So to do this problem first, we need to express the line off the court in terms off S a N S b and some other constants given the question and then also going to put the origin at the okay. And then I'm also going to define my positive directions. Going to the right is positive. Going up. It's positive. Okay, so in that case s a will be negatives. SB is going to be negative. Er why co ordinates below pointy? So, uh, they'll be they'll take negative numbers. Okay, so the length of the court is equal to, um you look at this. So it starts from C passed through here and then be right. So, uh, this land would be a z a square slice, three square square root. And then you have to of this high partners. And then where would they go? Seven SB. Okay, so you have to as a square plus tree square minus SB mhm. So this is our and then we take behind derivative. And this is what we get to s a B A do right by s a square last night. Uh, minus G B goes to zero. Okay, so we be is equal to, um to s a e a You Bye bye s a square. Last night may take another time. Derivative. You get acceleration. Okay, so we have a b is equal to two square You. Bye. Bye. Yes, a square last night plus two s a acceleration of a You are by s a square square square last night minus, um, to s a square You Bye bye. As a square last night did the tree house me. So we have our last DRB and acceleration of B. So the next we want you actually, you need to find the a m b a s a p a. So, uh, when s a Is he go to, uh, minus 4 ft. Okay, because SB six So the Okay, so this is 6 ft, So this will be five and five. That means as it will be, four. But the Y coordinate off essay will be minus four. Okay, so we be and will be is given to B minus 4 ft per second because it's pointing downward. So we have, uh my next fall is to go to to times minus four v a C by by five. Okay, so, uh, okay. So if you look at this, you can cancel the are minus four. So we A is just by over 22.5 ft per second. Positive. So this is upward. Okay, so we filed b a the next. We want to find the acceleration off a eso. Yes. When you have given that a Z a minus 12 ft, bebe is minus 4 ft per second. Acceleration of B is minus. Tree feeds both against way. Okay, so we just look at the expression for acceleration, and then we just substitute the relevant numbers. Okay, So ministry is you go to, um, to times 2.5 square by five plus to s a a would be, uh, minus four s. A is a term that we want to find about five and minus to times four square. Ah, and 2.5 square. You are by five Cube. Okay, so if you re arrange, you have negative eight. Over five times a subway is going toe. Uh, negative tree minus minus 2.5. So you can two times 2.55 So you are 11. 2.5 in the first time. And then for the second term, you can sell one off the four. In order to 40.5 square. You take care it off to five. So, um and you're like this, Uh, there were five. Okay. And so this is equal to Yeah. Negative, Chief for nine. Okay, so a some A is five times 3.9. You are by eight. Then you get 2.44 Thanks. Her second square. Okay, this is positive. So a sub is also upward. Yeah. So this is the acceleration appointing. Right? So, again, um, I fixed the origin. I chose my positive directions I put down my accordion is properly, uh, do the calculation and, uh, velocity and the acceleration off 0.8. We will come out correctly with the proper sign, Okay? And so that za reason why I chose to coordinate system and the origin. Okay. Yeah, and so that's all for this question.

Free body diagram of the question. We can write that N equals two minus MG plus f 70 to. Okay, we're tita is given as Katie. So from the contribution of linear momentum from zero equals to zero degree two Theta equals to 90 degree. We can write that M. V. One plus one by K interrogation from zero to bye bye. To that is 90 degree F course Katie plus um UK multiplied by n. Multiplied by Dita. And this will be equal to M. V. Okay, where we do is the speed at T equals to 90 degree. Okay, so from here we can write that M. V. One plus one by K interrogation from zero to pi by two. F course tita plus new K. And N is equal to this value. So we can write that F scientist to minus Milky MG. And Duty to this is equal to M week. Okay, so from here after solving further we get and this we get we two equals two. We do equals to even plus F by mt bracket scientist to plus Milky course tita From 0 to Pi by two limits minus mu K pi G divided by two. Okay so now from here after solving further we get we two equals two. B one plus F by MK Bracket one plus new K minus new K by G. Divided by two. Okay so this is a speed When data is equal to 90°.. Okay This is the speed at close to 90°.. Okay so now moving to the next in which uh we have a condition when we want to become a veto so we can write that F by MK one plus Milky minus mu Cape I G by two. Okay, this will become zero because women will be equals to re to. Okay so from here forth F equals to M U. K by M. G divide by Two kinds of one plus music. Okay so this is done for the force. Okay so this is done for the 4th 1.

So let's first right down the masses off the two colors and be the mass of a on the massive Be identical. So Emma is equal to M B, and this is their weight. £2.6 over G, which is 32.2 ft per square second. And so the masses workout to zero 0.8 0752 Key Pa's many decimal points as possible. And that's pound second squid her foot. Now the first thing we'll do is apply the conservation of angular momentum for color A. This means that the angular momentum in state two, when the court breaks, should be equal to the angular momentum off the system in ST one, which is the initial state of the system, so H 02 must be able to age or one. So you write this in equation form. This is in equation form. This is M A. Are one v dita. One is equal to m A are to and the transfers component of the Velocity V Peter in state two. And so we could rearrange this and we confined vita in state to so witty danced a to is equal toe are one VP toe one over our two, and this is equal to are one squared will replace V Tita by our data dot So that's our one squared detail. Not one over our two. And so this is we could substitute our values in here. 0.6 squared multiplied. Bye. The radio speed 12 radiance per second over to 1.2. And hence we get that. The radio component off velocity is three point six feet the second. So that's the final radio component off the velocity. Now, from here, since we have V tita, we can calculate eat a dot So dita dot to is equal to Vita in the second state over our A and this is equal to 3.6. She just calculated over 1.2. So that's three radiance per second, and we'll use that later on Yet. So now, for part B. Yeah, let why be the position. Coordinate off B and we'll take it as positive upward with the region at Oh, a constraint off the rod is that are minus y is equal to a constant. And this implies that the second derivative of why with respect to time is simply equal to the second derivative of our with respect to time. So using some kind of Matics mhm, you can see that the why component off the acceleration of the he's obviously, why double dot and this is our double dot, and we also know that the our component of the radio component off the acceleration of A is equal to our double dot minus are peter dot squared. So now that we have the accelerations, let's apply them to each color. Let's apply Newton's second law. So for color be you must have that all the forces in the Y direction when summed. Let's give the mess of B time to its final acceleration a B. And so this is the tension in the card. T minus. The weight of B must equal to the massive B times. Why double dot, which we said is the mass of B times are double dot and we'll call this equation one so T minus. W B is equal to m b r. Double dot now for color A with a blind Newton's second law again, and the some off all the radio forces must equal to the mass of a times the acceleration of a in Israel component. So the radio forces, due to the tension minus t so minus T must equal to me and the acceleration is our double dot minus are detail dot squared, and this will call the creation to. So if we add equation 12 equation to we can eliminate t So creation one plus equation two gives us minus the weight off BWB is equal to M A plus. M b times are double dot de tu rd t squared plus m a are d d d d t squared or deter dot squared. And so what we want to calculate is first attention and also the acceleration of a letter to the Rod. So the first thing to calculate acceleration off a relative to our rod instantly are double dot So this we can find from rearranging the above equation to be m A are dita dot squared minus the weight of B over come a plus M B. And we can put some values into this. So we get 0.0 80 75 times are 1.2 times theta dot, which is three not squared minus the weight of B, which is 2.6. And this is all over the some of the masses of A and B, which is 0.8075 plus another 0.8075 since we get the acceleration off the off a right into the rod to be 10 point seven feet per square. Second on this acceleration is a Lady Lee inward. What So essentially, that's the victor. A off point A or collar? A relative to the rod Detention T we can calculate as M B into our double dot plus g. So again we can substitute these values into this. So that's zero point 08075 into minus 10.7 for our double dot last 32 0.2 ft square. Second, Fergie since we get the attention T to be one point 736 pounds


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