(II) A hydraulic press for compacting powdered samples has a l...

Question

(II) A hydraulic press for compacting powdered samples has a large cylinder which is 10.0 cm in diameter, and a small cylinder with a diameter of 2.0 cm (Fig. 13–55). A lever is attached to the small cylinder as shown. The sample, which is placed on the large cylinder, has an area of 4.0 cm² . What is the pressure on the sample if 320 N is applied to the lever?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A heavy toolbox that has a bottom surface area of 20.0 centimeters squared is accidentally pressed inside of an automotive garage using a hydraulic lift. A hydraulic lift consists of a large piston that has a diameter of 30.0 centimeters and a small piston that has a diameter of 5.0 centimeters. The lift is operated by applying force on the small piston using a lever mechanism that is similar to the one shown in the diagram below. The lever has a length of capital L. What is the pressure applied on the toolbox if 400 newtons is applied to the lever? So that's our end goal is our end goal is we're ultimately trying to figure out what the pressure applied on the toolbox is. So with that in mind, let's quickly look at the diagram that's provided to us by the prom itself. As you could see using an orange like rectangular shaped box, we have our toolbox. So the orange rectangular prism shaped box, it almost looks kind of cube like is our toolbox. And it's located on top of this piston here because it's inside of an automotive garage using a hydraulic lift. And you could see in gray, it denotes the hydraulic fluid. And the diameter of this larger piston is 30.0 centimeters. And the toolbox is located on top of the larger piston. And then we have our smaller piston which has a diameter of 5.0 centimeters. And of course, we have the length L capital L which is the length of the lever, which is denoted in blue arrows pointing towards the larger piston going beyond the smaller piston following this like empty white bo like rectangular box side of a black bolded line. And then we have the lever which there is 400 newtons of force being pushed downwards denoted by our red arrow pointing down on this lever. OK. So now that we have a visualization of what's going on for this particular problem, let's read off our multiple choice answers to see what our final answer might be. Now let us note that they're all in the same units of newtons per meter squared. So A is 1.1 multiplied by 10. The power is seven B is 1.4 multiplied by 10 to the power of seven C is 2.8 multiplied by 10 to the power of seven and D is 3.2 multiplied by 10 to the power of seven. OK. So first off, let us write down all of our known and unknown variables. So we know that the diameter of the large piston, which we're gonna call D two, D two is equal to 30.0 centimeters, which is also equal to 0.300 m, which I've gone ahead and performed the conversion for you. But you could use dimensional analysis or you could look up this value yourself. But to save time, I just gave it to you. And we know that the diameter of the small piston which we're gonna call D one is equal to 5.0 centimeters, which is also equal to 0.050 m. OK? And we know that the force applied on the lever which we're gonna call F or specifically capital F is equal to 400 newtons. We also know that the surface area of the toolbox, which we're gonna call a subscript T which is the area of the toolbox is equal to 20.0 centimeters squared, which is equal to 2.0 multiplied by 10 to the power of negative 3 m squared. And we also know that the length of the lever is L but we do not know what L is and we do not know what the pressure of the toolbox is because we're trying to figure out what the pressure applied on the toolbox is. So we do not know what PT is since we're gonna call the pressure of the toolbox PT and we do not know what the value of PT is. So that's our value that we're ultimately trying to solve for is PT. So now at this point, in order to help us solve for this problem, let us draw a diagram to help us better visualize our given variables. So with that in mind, so using the diagram that's provided to us by the prom itself, we'll see that we have our F two force that's pointing upwards towards the toolbox. And then we have our diameter value D two for the larger piston. And then we have our D one value for our smaller piston. And then we have vector F one similar to vector F two pointing upwards in the larger piston. For the smaller piston, we have vector F one pointing upwards too. And then we still have the length of our lever. But in this case, it's noted by a red arrow. And then we have our force vector F pointing downwards on the lever itself. We also need to enlarge our diagram to make an analysis of the lever to help us find the force F one on the smaller piston. So using our lever analysis to find the force F one on the small piston, we could see that vector F one is pointing upwards through what appears to be the center of our lever. And then we have our downwards force which is the force being applied on the lever itself, which is vector F. So the distance between vector F, one to the edge of our lever is L divided by two, denoted by a smaller set of red arrows below our initial lever, which has two longer red arrows denoting from where the lever begins and ends. OK. So now at this point, we must note that the leather, the lever system is in equilibrium. Thus, the net to is zero. Thus, we can recall and write that the sum of torque which we're gonna use Tau to denote torque is equal to negative F multiplied by capital L plus F one multiplied by L divided by two which is all equal to zero. So L capital L divided by two. So capital L divided by two is all equal to zero. So let us note and recall that the torque causing. And so therefore, we need to recall and note that the torque in this case is causing clockwise rotation. Therefore, it will be negative. Thus, we can recall and write that F multiplied by L is equal to F one multiplied by L divided by two. So now we need to rearrange this equation to isolate and sulfur F one. So when we do that, we will find that F one is equal to two multiplied by F which is equal to when we plug in our known values, two multiplied by 400 newtons, which is equal to 800 newtons. So now we must recall and use Pascal's principle. So let us note that the pressure exerted by the small piston is equal to the pressure in the large piston. So let us also note that the pressure is given by the force over the area. And we need to recall that the area of a cylinder is. And let's write this as a side note in blue, that the area of a cylinder, which we're gonna say A is equal to pi multiplied by diameter divided by two squared. And once again, this is the area of a cylinder. So with that in mind, we can go ahead and write that P one, the pressure, the initial pressure is equal to P two, the final pressure. So the initial pressure is equal to the final pressure. So we could expand this expression to write that F one divided by a, one is equal to F two divided by A two, which we can expand this even further to write that F one divided by I multiplied by D one, divided by two square is equal to F two, divided by pi multiplied by D two divided by two squared. So now at this point, we need to rearrange this equation to isolate and solve for F two. And when we do that, using a little bit of algebra you will find that F two is equal to F one multiplied by D two divided by D one, which is equal to 800 newtons multiplied by awesome, multiplied by 0.300 m divided by 0.050 m squared. Which when we plug that into a calculator, we will get 28, eight 00 newtons. So now we could solve for the pressure on the toolbox, which is our final answer that we're ultimately trying to solve for by recalling and using the following equation that the pressure exerted on the toolbox is equal to the force on the toolbox divided by the area of the toolbox, which is equal to two eight 800 newtons divided by 2.0 multiplied by 10 to the power of negative 3 m squared. So when we plug that into our calculator and we round to one decimal place, we will get 1.4 multiplied by 10 to the power of seven and its units will be newtons per meters squared. And that's it we've saw for our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter B which states once again 1.4 multiplied by 10 to the power of seven newtons per meter squared. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video I

Key Concepts

Hydraulic Pressure

Area of Cylinders

Mechanical Advantage

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