A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled?
A 12,000-N 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/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm 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 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 1.20 m? For purposes of this problem, neglect the weight of the pistons. I used F1/A1=F2/A2 and found F1 to be 925.93N But I found another formula A1*D1=A2*D2 but I am not sure how to put this all together.
A mechanic used a hydraulic lift to raise a 12,054 N car 1.89 m above the floor of a garage. CONTinued...?
12054*1.89/4.75 Watts c) 4796 Watts __________________-
Hydraulic Lift Problem?
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
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal?
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm 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 at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car? For purposes of this problem, neglect the weight of the pistons. A) 727 N B) 926 N C) 2900 N D) 3330 N E) 1.20 kN
What is the initial force that must be applied to the smaller piston in order to start lifting the car?
Mine would be off too, if you don't give a radius for the smaller arm! All you need do is find the cross- sectional area of the 2 sides of the tube (or the pistons, same thing). Then, divide the larger by the smaller area, to find their ratios. Let's say it's 3:1. All you then do, is divide the 12,000N by 3. You will have the small piston force in N to just balance the car. In reality, you would need to exceed that force very slightly, in order to raise the car.
Physics problem need help ASAP!!!!!!!?
You have left two unknowns. By not giving the size of the smaller piston I can make the answer of the force anything I like simply by altering the area of that piston. For example if the piston is 1cm radius the area is (1/18)^2 of that of the larger piston so the force is (1/18)^2 * 12000N = 37 N But if I choose a piston of 4 cm then the force must be (4/18)^2 * 12000 =593 N Just note that the force is proportional to the AREA. and area is proportional to the square of the radii hence always (R1/R2)^2 = F1/F2 in any such system.