Engineering

You are given a hydraulic jack which you'd like to keep in your car in case you have a flat. Your vehicle weighs 4500lbs, and you know the jack will have to lift 1500 lbs when you put it under the axle. The hand end cylinder of the jack is 1/2inch in diameter and the "work" end of the cylinder is 3.5 inches in diameter. It has an 8 inch long handle; the distance from the attachment point of the handle to the small piston of the jack is 1.5 inches. How much force will you have to apply to the end of the handle to lift the car and change a tire?


Force= 1500*/(MAjack*MaHandle)
Force= 1500/[(3.5/.5)^2 *(6.5/1.5)]

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check that


i don't understand that formula can you elaborate a little bit i came out with 7.07lbs of force and i don't think that is enough. Thanks


Car A has a speed of 20 km/h, which is being increased at the rate of 300 km/h2 as the car enters an expressway. At the same instant, car B is decelerating at 250 km/h2 while traveling forward at 100 km/h. Determine the velocity and acceleration of A with respect to B.

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