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[16.) A hydraulic lift system utilizes 2-Inch diameter cylinder. Worker must determine how much the system capable of lifting before deciding how much weight can be...

Question

[16.) A hydraulic lift system utilizes 2-Inch diameter cylinder. Worker must determine how much the system capable of lifting before deciding how much weight can be lifted safely The svstem runs at 800 Psi: How much the system capable of lifting?17.) At 659 and 120 psi pressure, gas occupies Volume of 4.5 in3 . How many inJ will occupy at 250 Fiand 150 pSi?18.) A hydraullic system uses an input and output plston cylinder with 3000 psl The input piston has an area 0f 3.14 In? and needs force of 9

[16.) A hydraulic lift system utilizes 2-Inch diameter cylinder. Worker must determine how much the system capable of lifting before deciding how much weight can be lifted safely The svstem runs at 800 Psi: How much the system capable of lifting? 17.) At 659 and 120 psi pressure, gas occupies Volume of 4.5 in3 . How many inJ will occupy at 250 Fiand 150 pSi? 18.) A hydraullic system uses an input and output plston cylinder with 3000 psl The input piston has an area 0f 3.14 In? and needs force of 9420 Ibs_ The output piston has an area 0f 19,6 In" How much force does the second (output) piston cylinder apply? What the mechanical advantage of this system? 19.) A flow meter attached hydraulic Iine that measures 18 gal/lmin_ The Iine has an inside diameter of inches. What the flow velocilty where the fluid passes the meter?



Answers

For the system shown in Fig. $13-4$, the cylinder on the left, at $L$, has a mass of $600 \mathrm{~kg}$ and a crosssectional area of $800 \mathrm{~cm}^{2}$. The piston on the right, at $S$, has a cross-sectional area of $25 \mathrm{~cm}^{2}$ and a negligible weight. If the apparatus is filled with oil $(\rho=0.78$ $\mathrm{g} / \mathrm{cm}^{3}$ ), find the force $F$ required to hold the system in equilibrium as shown.
The pressures at points $H_{1}$ and $H_{2}$ are equal because they are at the same level in a single connected fluid. Therefore, (Pressure due to $8.0 \mathrm{~m}$ of oil) $$ \frac{(600)(9.81) \mathrm{N}}{0.0800 \mathrm{~m}^{2}}=\frac{F}{25 \times 10^{-4} \mathrm{~m}^{2}}+(8.0 \mathrm{~m})\left(780 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) $$ from which $F=31 \mathrm{~N}$.

Good day. The topic is about pascal's principle pascal's principle states that the pressure Applied to L fluid is transmitted undiminished all points of the fluid. Suppose we have we apply pressure at .1 and then the last section of the column. And we know that by pascal's principle this pressure is transmitted to the right section Of the column such that at the same level as .1 which is point to the pressure is the same since pressure is the ratio of force over area. Then it follows that the ratio of force and area at .1 must be equal to the ratio of course. And area at .2. Suppose we have a given system that contains a A cylinder which mass is 600 kg and uh cross sectional area of internet centimeter squared. We wish to find the force that must be exerted at the other section of the setup. We see here S. F. In order to keep the system at equilibrium. We know that this piece stein at where we accept this pressure is it's a okay, relatively massiveness compared to someone And that it is situated eight m above where above M. one. And that the column where they set up is filled with Oil which has a density of .78 g per centimeter cube. So now let us note that At point each one and H two at the two sections of the fluid are at the same height. So it follows that The pressure at each one is equal to the pressure at H. two. So we write p one is equal to P. H. Two. The pressure at point. Each one is equal to the force that it supports over the area and the pressure at each to that will be equal to die. Force that the pressure that is exerted at the surface of the fluid due to the force that it supplied there. Plus the the pressure that the certain uh depth or the certain height of column exerts. So that will be a combination of the force. So we say this is forced each one area H1 or let's speak it more Similar that a one and then the force Over area two plus as we said, The pressure that is exerted by the column of liquid, which is eight m deep. So that will be the pressure due to hydrostatic pressure. And resolve that as the product of density acceleration due to gravity and the height. So let us just define first. What are those needed to be defined if FH one is equal to the force that each at that point each one supports. So this supports the weight of mass one. So we saw this as the weight of mass one. So that will be must one times gravity acceleration due to gravity. That will be 600 Times 9.8. So area one is given there but we need to express this into meter or meter squared, since we want to express our pressure in pasqua. That's the combination of force in newton and area in meters squared. So this will be nothing that 100 centimeters 100 square centimeters squared is equal to one m square. So that will be 800 divided by 100 square. Or that 0.08 meter square. So we might also want to express a two also two m squared, since we need that also in Our solution. And so this will be 25 over 100 squared. And that will be 0.0025 meter squared. Right? So Area one is given their above that. Put it down here. So the density also the density must be in kilogram metre squared or rather kilograms per meter cubed. So that will be expressed in kilograms per meter cubed. By multiplying this conversion factors. So we have 100 Cube meter Cube or rather centimeter cube or one m cube. And this will give us 780 kilogram per meter cube. So now it has have let's plug in here. Are you've been And 0.08 it was f. So it's f that we wish to find here. And then we have the density which is 780 9.8 times it. Right? So let us just rearrange this one to solve for F. We have F equals 0.025 600 times 9.8 over 0.08 -780. Mhm Times 9.8 times eight solving Should give us force that is equal to 31 newton. That's of course of 31 Utah and must be applied at the peace point at the pistone in order to keep the system at equilibrium. So what was the tendency if this force was not there? So, since uh this this mass here is supported by the fluid then it experts fresher in the fluid. It has a tendency to kick the fluid ah above or it tends to dip it up so that force their will cause it. We'll keep the fluid on the right side of the right section of the column from moving up. So I hope everything is clear.

Told. The hydraulic system of backhoe is used to lift a load as shown in figure 11.48 So I've reproduced the backhoe here. There's a force from the hydraulic system acting here. That's a distance L one from our pivot point at a There's the weight of the arm, which, actually with center of mass. That's a distance l two from a. And then there's the way of the payload in the bucket. That accident distance L three from the pivot point. No, the point of a is fixed, so there's some of the moments about that. Point is zero. And if we take counterclockwise is positive, we wind up with our moment. Balance is the force from the hydraulic system times L and the force from the hydraulic system X perpendicular to the moment arm for the weight and the load. Do not at public a killer, so we need to take the side beginning to get the component that is perpendicular by taking a sign of this angle. So we get the weight of the arm times. A sign of that angle is the component in this direction. Kind 02 It's in the clap eyes direction. So we have a minus sign, likewise, for out for the weight of the payload. But now we have all three. And again we have a minus. Sign here because it's acting to rotate the bucket, the arm clockwise so we can solve for elf. And we get just this simple equation for our the force acting on hydraulic cylinder. We can start plugging in some numbers, so we have the have the length ratio l one over to over one. The weight of the arm is its mass times gravity and then sign the angle that the arm is acting, um, relative to the article is 30 degrees. And so we have our three over l one times the weight of the payload, which is 400 kilograms times t and then again, sign of 30 degrees. Now, if you plug everything in, we wind up with a force of 13.8 killings. If we figure out what the pressure in the slaves under is, well, that's just the force acting on the bucket arm divided by the area of the slave so under. So we have given that the the radius of the diameter of the slave cylinder is 2.5 centimeters, so it's one over its area. Is this value here multiplying it times the force and we get 28.1 killer pass scales. Now we're told that we have a a lever arm pushing on the master cylinder and that this lever arm has a mechanical advantage of five, meaning that this distance here is gonna be 1/5 of this distance here. You know the pressure in our hydraulic fluid In here? There's just 28.1 killer Pascal's so we can figure out what forces acting on our are master cylinder. And that is simply the pressure in the hydraulic fluid times the area of the master cylinder were given that the diameter of the master cylinder it is 2.8 centimeters. And so you plug that in and we get that their force acting on the master cylinder is 14. Point is one is 1413 Newtons, And so then, if we have a mechanical advantage of five, the force acting at our at the other end of the lever arm is 1/5 of that, and that is 283 Newtons and in fact this guy should be drawn the other way

So this is the leg with pressure loan on. This is the heart we know. What is that does not be in here. So the be is given by get a hotdog Pique was now does the age. The high difference over here is the report Five major times Desk Mark Uli ease density off the blood pressure at the hardest 1 20/80 The density of that is 1.5 times 10 to the three kilogram part of me that you and the gravity's 9.8 made up for a second squared. No, this will talk to be Newton per meter square and one may be better off mark Your is 1 33 Newton per meter squared. So this comes out to be 38.7 millimeter off Mikey. So now, pressure of the leg. He was pressured at the heart. Thus don't be pressure that the heart is 1 20 millimeter off Mercury. Thus fresh does be east 38.7 millimeter off mockery. So that comes up 31 58.7 me dio off Mercury. So that's the pressure at the lead

So we want to find out one diameter would need in making a piston cylinders. That kid left a certain mass. We know that when we have this piston at rest, when we balance the forces, we have the force provided by the hydraulic lift. So it's pressure times the area, and that's gonna be equal to the force of gravity that's pushing down on, uh, this weight that's being lifted. So what we can do is we can rewrite the area in terms of the diameter by taking pie and multiplying it by he divided by two squared. Then if we simplify this a little bit, just a pi times d squared over four. Since ah, we're plugging in D over two for the radius. And now we plug this ah, expression for the area back in. What we get is pressure times pi times d squared before is equal to the four student gravity. Andi, since where we want to know the diameter that this piston and cylinder has to be, we're gonna be arranged to sell for D. So then, when we get is this expression here we have two times the square root of sometimes gene departed by pressure times pi. And in this case, I'm gonna use, uh, five were given the pressure and kill a pascal's. But I'm converting that Teoh Pascal so that our units are consistent. So then when we get when we plug in with the values that we know to this equation for D here we end up getting it's a value of 0.146 meters.


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