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FIGURE P1A39(C14p39) 29.00 object hangs equiiunum Irom Strino distance (FlO . PI4 39a) . (a) Determine the TensionIcnaci sttlngJncar mass density0.0040 kalm; The st...

Question

FIGURE P1A39(C14p39) 29.00 object hangs equiiunum Irom Strino distance (FlO . PI4 39a) . (a) Determine the TensionIcnaci sttlngJncar mass density0.0040 kalm; The string Wrapped arounolight; frictionless pullevs uat are separated by theSubmn AnswerTries 0/10what frequency must the strlng between the pulleys vibrate Submil AnSMNUM Tries 0/10orderthe standing-wave dattem shownAigure P14.39b?

FIGURE P1A39 (C14p39) 29.00 object hangs equiiunum Irom Strino distance (FlO . PI4 39a) . (a) Determine the Tension Icnaci sttlng Jncar mass density 0.0040 kalm; The string Wrapped arouno light; frictionless pullevs uat are separated by the Submn Answer Tries 0/10 what frequency must the strlng between the pulleys vibrate Submil AnSMNUM Tries 0/10 order the standing-wave dattem shown Aigure P14.39b?



Answers

A $12.0$ -kg mass hangs in cquilibrium from a string with a total length of $L=5.00 \mathrm{~m}$ and a linear mass density of $\mu=0.00100 \mathrm{~kg} / \mathrm{m}$. The string is wrapped around two light, frictionless pullcys that arc separated by a distance of $d=2.00 \mathrm{~m}$ (Fig. $\mathrm{P} 18.70 \mathrm{a})$. (a) Determine the tension in the string. (b) At what frequency must the string between the pulleys vibrate to form the standing-wave pattern shown in Figure P18.70b?

Mhm This problem covers the concept of the wave on a string. And to solve this problem first, we need to find the length of the rope at as to the block. So the daughter length of the rope is five veto. And the length between the jubilee is two m. So you can see the length of the robot has to, The block on each side is 1.5 m. Now we have to calculate the angle theater. So if we draw a perpendicular on the string attached to the fully, then this land will be one m so we can see the angle theater equals signing was off one meter upon 1.5 m. Or the Angle Theatre comes out to be 41 eight degrees. Now let's show the free body diagram of torture, the forces acting on the objectives, the weight acting vertical downward. That tension force in the string addressed at an angle theater from the vertical. Since the blog is in a Librium, so we can write the summation of all external photos along the right direction is zero from Northern Second Love motion are two times Chico's Theatre equals MG. Are that ancient equals MG upon two goals theater. So for partly the tension in the string equals the moss. Uh there is 12 Kg And to 9.8 May just four seconds square upon two times the cause of forgiven .8 decrease of the tension in the strings comes out to the 79 note. Now party for a string fix at the both ends. The harmonic frequency of an equals and upon to well into square out of the tension in the string upon the linear mass density. Uh huh. And in the previous part we have give a symbol of T. To the ancient. So the tension is T. Okay, so from this we can say the third harmonic frequency have trees 3.2 and into the square root of the tension upon the linear mass density. Now substitute the values of the Total harmonic frequencies three into 2 into balance. two m into square out of 79 Milton upon 0.0010 kg go meet Are the frequency of trees, 2.1 in two and 3 Hotch. So 2.1 into it and there's two hutch. Our 210 Hotch. I think it was.

So for party, we can see this is be a quick sketch of figure p 14.49 a and we see that when d is equaling 2.0 meters, we can see that the length Ellis equaling 5.0 meters minus d divided by two. This is equaling 1.50 meters. And so Fada would be equaling arc sine uh d over too divided by L. And in this case this would be 41 0.5 degrees. So we can say that when evaluating the net vertical force on the lowest bid of string, we can apply Newton's second law in the UAE direction. This would be equaling two f co sign of theta minus m g. This is equaling, of course, zero because the system is an equilibrium and the tension in the string after is then found to be mg over to co sign of Fada. So this would be 12.0 kilograms multiplied by 9.80 meters per second squared. This would be divided by two co sign of 41.5 degrees and we find that then the tension is going to be 78.9 Newton's. This would be our final answer for part A for part B, then the speed of the transverse waves in the string would be the square root of the force divided by mu, the lean year density. This would be equaling the square root of 78 0.9. Newton's divided by a point 001008 kilograms per meter, and we found the speed to be approximately 281 meters per second. So for the pattern shown, we can say that three times slammed over two equals. D. So we can say that the wavelength lambda is to de over three since me too times d, which is 2.0 meters and this would be divided by three. And so we can then say the frequency f would be equaling the speed the wave speed divided by Lambda. And so this would be equaling. Ah, three multiplied by 281 meters per second, divided by two times two. So 4.0 meters and this is gonna be equaling. Approximately 211 hurts. This would be our frequency for part B. That is the end of the solution. Thank you for watching

Everybody. So I drew a little image for everybody here and for a um it's asking us determined attention in this strange And so what is gonna be doing some simple trade here? And so let's find the sign data, and that's gonna be wide over 1.5. OK, And that's because here we have worn and one okay, we need to find out what data is equal to and to do that we need. So what we're doing is we need to find the sine function here, okay? And because of that, we do this side and this side is simple tray. Okay. And so we have to over three. Technically, if we will NYSE, you don't have to do that. You can use Internet into calculator. That is totally fine. OK? And so data equals sign um, to me of one, 23 or a member put or 1/1 0.5. Whatever one you want putting calculator, it doesn't matter, and it's gonna evil. Teoh, 41.8 degrees. Okay, so he found the data so now will weaken. Dio is we need to find the tension in the string and the tension that drink is mass times gravity, which is this mass here. And it's gonna be two times. And the co signer dig it up. Um, that's just because we have 12 And here we have 1 20 for the mass, 9.81 meters per second. Squared times two times 41.8 groups I am was forgot. The coastline there they really right co sign at times 41.8. Okay. And you put it onto calculator and you get attention of 78.9 Newton's now, um B is asking us here. Um what at what pregnancy must string between the police, vibrate to form a steaming the wave. Okay, so we have abrasions or the velocity as t equals new in the new Israeli mass densities, we have 78.9 linear Mastin. See, the string is 0.1 square root and we get to 81 meters. Suck it. And now we have our d as me 3/2 terms. Lambda Lambda equals two D over three and R D is two which is given given to us, and that's been equal to 1.33 mirrors Time three and R. G is this guy right? And once cooled down. Now we do the frequency, which is the velocity times La Bamba. When we get to 81 um, area and we get to 80 one meters per second, divided by 1.33 meters and we get to 11 hurts. Right? That's it. Thank you, guys.

Okay, so in this problem, we have kind of a contraption, um, set up, which you can see by looking at the picture in your textbook, I will draw some diagrams to kind of help with the problem, so don't worry about that. Um, but we have a box that is, uh, hanging off of a rope that's tied around to police. It makes a triangle. Um, and the mass of this box is 12 0.0 kg. Um, the total length of the rope l is, uh, 5.0 m. Ah. The linear mass density of the rope mu is 0.100 kilograms per meter. And we're told that the distance between the two police eso the side of the top of the length of the top side of the triangle is, uh, d, which is equal to 2.0 m. Um, and in part A, we want to find the tension in the rope and in part B. We want to figure out, um, the frequency of the standing wave that shown in the book. Um, so I will go ahead and start by drawing a quick force diagram off the box. The box has, um, three forces acting on it. There is the force of gravity just pulling straight down. Obviously, it has to tensions which are equal to each other. Um, going up into either side like this. So t and t, they're equal to each other because they're supplied by the same rope. So the road has experienced the same tension, and both undeliverable pull with the same tension on the box. Um, it's also worth noting that there at a specific angle, um, and I'm going to call this angle from the vertical seda. Um, so we want to figure out what the tension in the rope is. Um and so the first way that we can really think about this is by comparing it with gravity. So there's gravity pulling down on the rope, and then both tensions are pulling upward on it. They're also pulling to the side. But the horizontal components of the tension cancels each other out. The vertical components duct, um, and so it's a little bit. And because we know what the force of gravity is, since we know the mass of the box, um, that will get us you know, one step closer to figuring out the tension, so ah, gravity pulls downward. So the force of gravity MGI is going to be opposite the for the vertical force of tension on the box. And we're going to double the force of tension going upward because, um, each side of the rope provides an equal not going upward. So, um, and each one hat is from, you know, one amount of the force of the tension, so we have to tensions pulling upward, but it's only the vertical component. So we want just the y component off the force attention. So if we look at our diagram that just drew the vehicle, force is going to be adjacent to this angle state up If we draw like, if we were to decompose this tension angle, um, and make a right triangle with the vertical plane, um, the y component would be along adjacent to the angles data, and then the horizontal coat horizontal component would be opposite the angle fail with t being the high pop news. So we can say that T y is going to be equal to tee times coastline of data, so mg then becomes equal to two. T co sign Seita. We can go ahead and solve this equation for tea which will help us out later. So just divide both sides by two coastline data. So then t becomes equal to mg, divided by two co sign data. So we have MGI is just a constant, Um but we don't know Seita. So that's the only thing that we actually need to find in order to figure out what the tension is. So we don't even need to really consider the horizontal components of the tension. We just need to figure out whatever state it is. And here it's helpful to consider the actual triangle that is made by the rope. Um, so we have Ah, this rope is put in a tranq information at the top two corners. There are police that's wrapped around, but for simplicity sake, I'm going to just try it as a triangle. We're told that the topside has a length of D, which is equal to two meters. Um and we know that the total length of the rope is 5 m, um, and where it's implied because the box has to be centered in order for the tensions to balance. Um, we can assume that these two sides are equal. Um, which you know, by doing some quick geometry. In reasoning, you can conclude that both of these sides are 1.5 m. So we have We know the three lengths of the struggle, but we don't know any of the angles. So in order to find angles, the easiest way to do this is to use the law of CO signs. Ah, which is of course, C squared is equal to a squared plus B squared Ah, minus to A B co sign Capital C. Um, where capital C is just the angle that is opposite, uh, whatever side, you just designate as lower case seat. So the angle that were interested in Is this angled down here? Because if we figure out this angle, we know that fado will be one half of whatever that angle is. So if we find the angles, see which I'm gonna label here, Um, then we confined theta and then we confined our attention so we'll go ahead and solved. We're going to do some algebra to rearrange this equation to solve for Capital C and then we will put in the values and will sell for seat. So we'll go ahead and rearrange this equation a little bit. Um, so we can say that Ah, if we can subtract ah, this term from both sides and this term from both sides to get the equation to a B Times Co. Sign of capital sees equal to a squared plus B squared minus C squared and then divide both sides by two a. B and take the inverse coastline of the results and you'll get an equation for Capital C, which is, um, co sign Converse closer of a squared plus B squared my A C squared all over to a B. And so that is our equation for the angle, see, and so we can go ahead and plug in our side lengths. Ah, so see is people to co sign in verse of and then A and B are both 1.5 m. So 1.5 zero square plus 1.5 years squared and then see is the is deep, which is the 2 m at the top. So 2.0 squared all that's divided by two times 1.50 times, 1.50 And if you plug that into your calculator for see, you should get an angle of 83.6 degrees. And then we know that state A is going to be equal to see divided by two, which, if you calculate that you should get 41 41.8 degrees. So that is our theta. So we can return to our equation that we had before for tea. So T is equal to MGI, divided by two co sign Seita, and so we can plug in all the known values here. So M is 12 kg, 12.0 g, of course is 9.80 m per second squared, and in the denominator, we have to times co sign of 48 41.8 degrees. And if you plug that into a calculator, you should get attention force of 78.9 kittens. So that's how you do part A in part B. We want to figure out what the frequency of the ah shown standing wave pattern is, um, up at the top part of De um of the triangle. So I'm if I quickly, like, redraw the triangle. The way that they're showing it is that there's three anti notes for their standing wave. Um, so it looks a little bit like that. And what that tells us is that the wavelength of this wave, uh, lambda is going to be two thirds of D because if you look at it, one wavelength happens in this area. That I just, uh, delineated is two thirds of D. Um, and then we need to figure out the frequency so we can go ahead and set up the equation that f is gonna be equal to be over Lambda. We just gave an equation for Lambda, um, and then V because we're given, um that it's a string under tension and we know it's mu because we're given that right at the beginning, we can say that V is going to be equal to the tension force divided by mu. So then our equation for F very quickly becomes, um, three over to d times the square root of tea over you. And we have tea. We have d, and we have muse. So we're all set to sell this cell for this value so F becomes three over two times 2 m. Uh, times 78.9 Newtons for attention and then mu is 0.1 year zero And if you plug that into your calculator, you will get a frequency of 211 hurts and that's how you do this problem.


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