Hydraulic Lift II.The piston of a hydraulic automobile lift is 0.30 m in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 kg? Also express this pressure in atmospheres.

Video transcript

Welcome back everybody. We have a nurse that is using a hydraulic lift to transfer a patient from a bed to a chair, which I'm gonna represent by this rectangle right here. We're told a couple different things. We're told that the area. Sorry, let me change color here real quick. We're told that the area of the cross section of the piston is 0.02 m squared. We're told that the mass of the patient is 80 kg and we need to find what the pressure on the gauge is of the hydraulic lift. Well, when we are dealing with a hydraulic lift, we are dealing with a couple of different pressures here. So let me go ahead and draw those out first and foremost. Going upward. We are going to have the force applied by the hydraulic lift itself, which is going to be the pressure of the hydraulic fluid times the area of the piston. Now acting against that. We are going to have the force of the atmosphere acting against the atmospheric pressure acting against that, which is going to be the pressure of the atmosphere times on the area of the the hydraulic lift there and then we are just going to have the weight of the patient. Also weighing down on the hydraulic lift using Newton's second law, we know that the sum of all forces is going to equal the mass of the piston times the acceleration. Well, let's fill in some things here, all the forces present here, we're going to have um the pressure of the hydraulic fluid times the area minus the pressure of the atmosphere. Times times the area minus the weight of the patient equal to this right side here. Well we are just assuming that the mass of the piston is negligible which in a hydraulic lift typically is very, very small. So we're just going to set this right side to zero, rearranging some terms here. We we're going to have that the pressure of the hydraulic fluid minus the atmospheric pressure times the area of the hydraulic lift is equal to the weight of the patient. Well, what is the difference in this pressure right here? Well that is going to be the exact pressure read on the pressure gauge. We're gonna have that. The pressure gauge is equal to the weight of the patient. And if I divide the area on both sides this will be divided by the area. Now we have a formula. So let's just go ahead and plug in our numbers here we have that. The pressure on the gauge is equal to the mass of the patient which is 80 times the acceleration due to gravity, which is 9.81 divided by the area which is 0.2 m squared equal to 3.92 times 10 to the fourth pascal's corresponding to our answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one