A 12000N car is raised using a hydraulic lift, which consists of a U-tube with a

Question

A 12000N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m^3 and capped at both ends with a tight-fitting pistons. The wider arm of the U-tube has a radius of .18m and the narrower arm has a radius of .05m. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20m? For purpose of this problem neglect the weight of the pistons.

Explanation / Answer

Force = Pressure * area area = p * r^2 The pressure in both arms of the U-tube is the same! Wide arm Pressure = Force ÷ area Wide arm Pressure = 12,000 ÷ (p * 0.18^2) = 117,892.55 N/m^2 The pressure in both arms of the U-tube is the same! Narrow arm pressure = 117,892.55 N/m^2 Force = 117,892.55 * (p * 0.05^2) Force = 925.9 N This is the force to keep the pistons at the same level! The ratio of the forces = Mechanical advantage Mechanical advantage = 12,000 ÷ 925.9 = 12.96 : 1 Ratio of areas = p * 0.18^2 ÷ p * 0.05^2 = 12.96 : 1 What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? oil with a density of 800 kg/m^3 volume of oil in wider arm = p * 0.18^2 * 1.2 = 0.122 m^3 The weight of the oil in the wider arm = volume * weight density weight density = 9.8 * 800 Weight = p * 0.18^2 * 1.2 * 9.8 * 800 = 957.6 N Mechanical advantage = 12.96 : 1. The force on the narrow arm = 1/12.96 * 957.6 = 73.9 N Total force on the narrow arm = 925.9 + 73.9 = 999.8 N (ans) hope it helps

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