Question:
Hydraulic Lift Problem?
A 12,000-N car is raised using a hydraulic lift which consists of a U-tub with arms of unequal area, filled with oil with a density of 800 kg/m^3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially 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 1.20 m? For purposes of this problem, neglect the weight of the pistons
Answer:
start by making sure wipers are completely off.at the base of wiper am is a plastic cover that needs to come off. there is a nut.13 or 15mm.place wipers all the way down to stop,then tighten nuts as tight as you can.(p.s.may have to loosen a bit first to get them to move down.999% sure on this. ex GM tech.be careul ad good luck.
Force Area X Pressure I always work in PSI and Lbs, and Inches (I'm old school and nver use metric, so I always convert them over) You can convert the weight to lbs and area to inches and modify the formula to Pressure Force divided by Area From what you have here, I kind of assume that the 36cm is 36cm squared and is the area of the cylinders piston? (length doesn't matter) You should clarify that.
Force Pressure * area area π * r^2 The pressure in both arms of the U-tube is the same! Wide arm Pressure Force ÷ area Wide arm Pressure 12,000 ÷ (π * 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 * (π * 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 π * 0.18^2 ÷ π * 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 π * 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 π * 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 Pressure weight density * depth Weight density 9.8 * mass density Weight density 9.8 * 800 kg/m^3 7840 N/m^2 Force ÷ area 7840 * depth
with wiper switch off push wiper down as for as they go and tighten nut and you should be good to go
with wiper switch off push wiper down as for as they go and tighten nut and you should be good to go
start by making sure wipers are completely off.at the base of wiper am is a plastic cover that needs to come off. there is a nut.13 or 15mm.place wipers all the way down to stop,then tighten nuts as tight as you can.(p.s.may have to loosen a bit first to get them to move down.999% sure on this. ex GM tech.be careul ad good luck.
Force Area X Pressure I always work in PSI and Lbs, and Inches (I'm old school and nver use metric, so I always convert them over) You can convert the weight to lbs and area to inches and modify the formula to Pressure Force divided by Area From what you have here, I kind of assume that the 36cm is 36cm squared and is the area of the cylinders piston? (length doesn't matter) You should clarify that.
Force Area X Pressure I always work in PSI and Lbs, and Inches (I'm old school and nver use metric, so I always convert them over) You can convert the weight to lbs and area to inches and modify the formula to Pressure Force divided by Area From what you have here, I kind of assume that the 36cm is 36cm squared and is the area of the cylinders piston? (length doesn't matter) You should clarify that.
Force Pressure * area area π * r^2 The pressure in both arms of the U-tube is the same! Wide arm Pressure Force ÷ area Wide arm Pressure 12,000 ÷ (π * 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 * (π * 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 π * 0.18^2 ÷ π * 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 π * 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 π * 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 Pressure weight density * depth Weight density 9.8 * mass density Weight density 9.8 * 800 kg/m^3 7840 N/m^2 Force ÷ area 7840 * depth
Force Pressure * area area π * r^2 The pressure in both arms of the U-tube is the same! Wide arm Pressure Force ÷ area Wide arm Pressure 12,000 ÷ (π * 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 * (π * 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 π * 0.18^2 ÷ π * 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 π * 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 π * 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 Pressure weight density * depth Weight density 9.8 * mass density Weight density 9.8 * 800 kg/m^3 7840 N/m^2 Force ÷ area 7840 * depth
with wiper switch off push wiper down as for as they go and tighten nut and you should be good to go
