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Hydraulic lift. You are designing a hydraulic lift for an automobile garage. It will consist of two oil-filled cylindrical pipes of different diameters. A worker pushes down on apiston at one end, raising the car on a platform at the other end. (See Figure $13.41 .$ ) To handle a full range of jobs, you must be able to lift cars up to $3000 \mathrm{kg},$ plus the 500 $\mathrm{kg}$ platform on which they are parked. To avoid injury to your workers, the maximum amount of force a worker should need to exert is 100 $\mathrm{N}$ (a) What should be the diameter of the pipe under the platform? (b) If the worker pushes downwith a stroke 50 $\mathrm{cm}$ long, by how much will he raise the carat the other end?

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Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Rutgers, The State University of New Jersey

Simon Fraser University

University of Winnipeg

Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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Okay, So in this problem, we have ah, hydraulic lift. So let's draw here the simplify hydraulic lift. So here we have one end of the trolley clears, and here we have the order and of the trial that lift. So in this point in here, we can put the car to lift it up. And here we are doing a force downwards to lift his car the weight of his car. So according to the Bosco Low, we can say that the pressure Let's call this 0.1 in this point to the pressure 1.1 should be equal The pressure in points to which means that the force apply in 0.1, divided by the cross sectional area off 0.1 should be equalled the force on points to divided by the cross sectional area off point to Okay, So what we know is that the Force one is equal 100 Newtons in this problem, we know that we can calculate the force to because the weight off the car. So let's calculate this is going to be 3000 and 500 kilograms. That multiplies gravity's a celebration which is 9.8, which gives us a force. True off. Let's see three point for three times. 10 to the four Newton's. Okay, so we have the force to We have to force one. Now let's look to the area. The area is a cylindrical. Gasoline is a just is a political pipe. Therefore we can see that the area. Whoops. Let's write this. We can say that the area is a circle with area pie are square, so just simplify the relation that bus Carlo. We can actually see that Force one divided by the radio's one square is going to be equal. They're forced to divided by their radios. True, it's cracked. And what this problem once is just the radius off the second point, which is the area that is lifting up the car. So let's calculated swell. If we rearrange the secretion going to have that are to going to be go, they are one square roots off F two divided by F one. Okay, so we can calculate this and let's see, we're going to find that the radius is going to be 0.1 25 is square root off three point for three time. Stand to the fore divided by 100. Okay, so the radius of the second cylinder is going to be two point dirty, too. Meters. Okay, so that's the first answer. Now we have to calculate what is the diameter? Ah, let's see. Okay, so we calculate what is the diameter of the pipe under the platform. Now we have to calculate how much we raised a car. If if the worker bushes down with a stroke 50 centimeters long. So let's see, to calculate this, I can be off this problem. We can actually use a type off continuity equation. So we can say that the distance, one times the area one is going to be called the distance two times the area to. So if we want to lift it up 50 centimeters, 50. Saying team interest is the D one. So, actually, we want to lift it up. We actually want to know how much we raised a car. We do not know how much were raised, but we are using 50 centimeters long stroke. So the D one is the 50 centimeters, and the D two is going to be how much the car is lifted up. So we can say that the justice too going to be a distance one plus the area one divided by the area to So this is just distance one by are one square divided by far are two square can cross I in here and we have at a distance to it's going to be 50 70 metres and multiplies zero point 1 25 divided by two point 32 a square. So if recalculate this, we have that the bottom farm will raise up a distance off one point 45 millimeters okay. And asked the final answer to this problem. Thanks for watching.

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