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What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000 -kg car (a large car) resting on the slave cylinder? The master cylinder has a 2.00 -cm diameter and the slave has a 24.0 -cm diameter.
$F_{1}=136 \mathrm{N}$
Physics 101 Mechanics
Chapter 11
Fluid Statics
Fluid Mechanics
Rutgers, The State University of New Jersey
Simon Fraser University
University of Winnipeg
Lectures
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Pressure and Force An auto…
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Hydraulic Lift I. For the …
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Two hydraulic piston/cylin…
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A hydraulic lift is to be …
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this problem will ask to determine what force must be exerted on the master cylinder cylinder of a hydraulic lift to support the weight of the 2000 kilograms card. Investing on the slave so under the master cylinder has a 2.0 centimeter diameter and the slave cylinder has a 24 centimeter diameter. So I've drawn a picture of a schematic of the set up here. So we have our car appear, which has a massive 2000 kilograms so we can determine its weight. The slaves. So it's resting on the slave still under here so we can figure out what the pressure in this fluid is because you know the cross sectional area that this is a circular cross section, you know, the cross sectional area and we know the force. So the pressure is the force divided by area. We also know that the pressure here, I mean, the fluid is the same in all parts of the fluid, so it's also the same here. We know the cross sectional area of this, the master's longer, so we can try to figure out what the force on that is to create to cause that pressure. So we go through that and again we can figure out the pressure from the master cylinder is the weight of the car times the pressure from this. Sorry, The slave cylinder is the weight of the car times the area of the slave cylinder. I am using the formula for area with the amateur. We have that. Then we also know that the pressure on the master cylinder down here has to be equal to the pressure on the sleeve. Still under up here. Pressure on the master cylinder is the force on the master cylinder devised by the area of the master cylinder. And we can write that it's this Now we can we know that then this. These two things have to be equal. So we set those equal to each other pi over four. Cancel out and we can then solved for F, we find that f is the ratio of the diameter of the master so under to the diameter of the slaves under squared times the weight of the vehicle. So the weight of the vehicle is 2000 times 9.8 and this ratio is 1/12. So we have to get divided by 1 12 square bugging, You know, over numbers. You get that. The force on the Master Phil under here is 136 Newton's.
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