A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction.
This is College Physics Answers with Shaun Dychko. In part A of this question we'll find the ratio of the areas of the slave cylinder to the master cylinder. So we know that the pressure in the two cylinders has to be the same, that's Pascal's principle. We'll replace pressure with force of the master cylinder divided by area of the master cylinder equals force of the slave cylinder divided by area of the slave cylinder. We'll get a ratio of slave cylinder to master cylinder areas by multiplying both sides by A s over F m. F m cancels on the left and we're left with A s over A m which is the question, what is the ratio of the areas On the right hand side we have the areas canceling and we're left with the ratio of the forces. Now we're told that the force on the slave cylinder has to be 100 times that in the master cylinder. So we'll replace F s with 100 F m and the F m's cancel. So we're left with 100 is the factor by which the area of the slave cylinder has to be greater than the area of the master cylinder. Then we're asked to figure out what is the ratio of their diameters? Well we know the ratio of the areas and the area of one of them is pi times diameter squared over four. So we have pi d slave squared over four divided by pid master squared over four and this is the square of the ratio of the diameters. Now to solve for the ratio of the diameters we take the square root of both sides. So that is d s over d m is the square root of A s over A m. So that's the square root of 100 because 100 is the ratio that we found in part A and that is 10. So the ratio of the diameters is 10 so the slave cylinder has a diameter ten times that of the master cylinder. Now part C, we're asked to compare the vertical distance through which the two cylinders will move. Now we know that the volume by which the slave -- oh sorry, the volume by which the master cylinder decreases, is equal to the amount by which the slave cylinder increases its volume because these-- where is the better picture? I should draw you a better picture. We have a slave cylinder here, sorry a master cylinder there connected through a tube to a slave cylinder which is a lot bigger, diameter is 10 times bigger as we've discovered already, and when the piston in the master cylinder goes down some height delta h, that's going to cause a change in volume, call that delta v master and the amount by which this volume decreases, well all that fluid that was in that volume is now gone, has to be pushed through this tube into the slave cylinder. So the slave cylinder's volume is going to increase by the same amount. All right, so that's what this says and the volume of a cylinder is its area times its height. So we're trying to find the change of volume of the slave is the area of the slave cylinder times the change in height of the slave cylinder. That equals the area of the master cylinder multiplied by the change in height of the master cylinder. So the ratio of the changes in height is the ratio of the areas -- well, it's not quite that simple though because the subscripts are on opposite sides here. So we'll -- what do we do here? We're going to multiply this by one over A s h s. Sorry, I didn't do that right did I? We want to divide by delta h m that's for sure in order to get delta h s over delta h m. Then we also want to divide by A s. So delta h s over delta h m, the ratio of the distances through which the cylinders move, is equal to the opposite ratio of their areas. So h s over h m is A m over A s which is the reciprocal of A s over A m. A s over A m is something we've already found, which is 100. So the reciprocal of 100 is 0.01 and so the slave cylinder will move only one percent the distance that the master cylinder moves.