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Enenae Morcer thee and string applving = end massless and equilibrium Arom ditance passc: over held mjs Hlorco and The string Vength _ {upportink horuontal bar of t...

Question

Enenae Morcer thee and string applving = end massless and equilibrium Arom ditance passc: over held mjs Hlorco and The string Vength _ {upportink horuontal bar of the left; with uniform directed- angle force makine mass horvontal end hanring = right = 15.0 kB: thet attached and = 102 N Jached pulley - Jnd 35.0 . frictionles: 22,0 KB; 42 4N, M= 118 N 32 0 Let T1.1N, 131 N, 6.67 m Find 120 N 73.5 N; 5.72 m, Find d) 49.0 N 104 8.18 m, 78.4 N,; Find d) 132 N 7.00 m; 6) 118 = 261 N, D) Find /3) 147

enenae Morcer thee and string applving = end massless and equilibrium Arom ditance passc: over held mjs Hlorco and The string Vength _ {upportink horuontal bar of the left; with uniform directed- angle force makine mass horvontal end hanring = right = 15.0 kB: thet attached and = 102 N Jached pulley - Jnd 35.0 . frictionles: 22,0 KB; 42 4N, M= 118 N 32 0 Let T1.1N, 131 N, 6.67 m Find 120 N 73.5 N; 5.72 m, Find d) 49.0 N 104 8.18 m, 78.4 N,; Find d) 132 N 7.00 m; 6) 118 = 261 N, D) Find /3) 147 N, 84,9 Find a) 204 N;



Answers

Threc equal weights $A, B$, and $C$ of mass $2 \mathrm{~kg}$ each are hanging on a string passing over a fixed frictionless pulley as shown in the Fig. $7.308$. The tension in the string connecting weights $B$ and $C$ is a. zero b. $13 \mathrm{~N}$ c. $3.3 \mathrm{~N}$ d. $19.6 \mathrm{~N}$

In this connection, we have given pulley block system and it is told that we have to find the tension in the string A. B. So we can see that if we brought the free body diagram for this solution. So we can see that as the weight dirties 0.3 into 10 which is three Newton trying to flyer the two blocks downward so to block system. So we can say that the value of maximum frictional force that is F max is equal to which is 0.5 into that is normal force, which is one into 10 this season. First block, nothing. Second blog. This is also 0.5 into one into 10. And this will come out to be 10 newton as we can say that the value of maximum frictional forces greater than this. So we can say that the system is in equilibrium and friction of block B is sufficient to balance the weight. Yes, we can say the attention between A&B. s0. So this is the answer of arguing cushion for that option. This is the correct joy. That is zero. Thank you.

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.

For this problem on the topic off rotational dynamics and static equilibrium, we're told that a string passes over a pulley and has two masses attached to either end were given both those masses and were given the radius off the disk at the center off the pulley. We know that it has fiction in its excel, and we asked to calculate the magnitude of the fictional talk that has to be exerted by the axle to keep the system in static equilibrium. So here we have a diagram for Situation On will use Newton's second law flotation to find the fictional talk that would make the angular acceleration off the system equal to zero. Now, in each case, the talk exerted on the pulley by the hanging masses is the weight of the mass multiplied by the radius of the fully. We'll take the top u T M one to be clockwise and in the negative direction. Now, if we write Newton's second law for rotation, this is the sum off the talks in the system, which is minus are times M. One times G, the weight off block one, Lastly talk you too mass to which is our into two times G. This is in the opposite direction, so this is positive plus the friction talk. But you wish to calculate for fr must equal to zero since the system is in static equilibrium. So by rearrange, we confined these this fictional talkto fr so two f r is our kinds g into M one minus m two are Here is the radios off the disc, which were given to be 0.94 m G acceleration due to gravity even know is 9.81 meters per square. Second, any difference in masses in one minus M two is zero point 635 minus zero point 321 that's in K g. So calculating we get the frictional talk Torre fr which keeps the system in equilibrium to be zero point 29 Newton meters that

I am prevent a surrender figure next to work on the police minus arm and many people are and to G due to yeah, friction. That must be true. So targeting to friction. You will get rtg Mm minus and substitute. The value of the values are given in the problem. So just only we have to put values. And finally you will get started to construction to me. Quite Nigeria New Tarantula meter. Mm. Mhm. That so? Thanks for watching it.


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