Class 11 Project For Hydraulic lift

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Physics Project HRITAM SHARMA ROLLNO. 9 XI A

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Acknowledgement

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Index page 4 - Objective, Introduction, Materials and Equipment.

page 5 - Theory. page 6 - Experimental Procedure, Calculations.

page 7 - Result, Precautions, Sources of error.

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Objective To apply the principles of hydraulics in building a working model of a hydraulic lift.

Introduction What do a wheelchair lift, an excavator at a construction site, and a dentist's chair have in common? The ability to lift a heavy load. And in all of these examples, the ability is due to the power of hydraulics. Hydraulics is the study of liquids and their mechanical properties: for instance, how they move, resist movement, act when subject to pressure, and so forth. In engineering, one application of hydraulics is using liquids, like water and oil, to move things. Why use a liquid to move things, instead of, say, air? One important reason is that a liquid is incompressible, which means that if you press on it, you cannot change its volume

Materials and Equipment 1. small container of radius 3 cm (A) 2. another pipe like container of radius 0.5 cm (B) 3. connecting pipe of plastic of radius 1 cm (C) 4. epoxy(M-Seal) 5.piston or radius 0.4 cm (D) 6. round 'thermocol' piece 'or radius 2.98 cm (E)

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7. Glycerin

Theory Using Pascal's law which states that if gravity effects could be neglected, then pressure at all points in a liquid container is the same.

Let P1,P2,F1,F2,A1,A2,X1,X2,W1,W2 be the pressure, force applied, area, distance moved, work done by piston(a) and piston (b) respectively. P1 = P2 (Pascal's law) F1/A1 = F2/A2 F2/F1 = A2/A1 ............................(i) A2 >> A1 F2 >> F1 A1X1 = A2X2 A2/A1 = X1/X2............................(ii) A2 >> A1 X1 >> X2 W1 = F1X1 &W2 = F2X2 W1/W2 = F1X1/F2X2 from (i) and (ii) W1/W2 = A1A2/A2A1

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W1 = W2 Therefore work done by piston(A) is equal to work done by piston(B).

Experimental Procedure 1. Assemble the material as shown in figure (1) and seal the joints with epoxy to make the model of hydraulic lift. 2. Pour Glycerin through (a) until it reaches level (b). 3.Put payload on (E). 4.Insert piston (D) in (B) shown in figure (2) and push gently. 5.(E) will rise up due to the force exerted by piston (D) through the liquid.

Calculations P1 = P2 (Pascal's law) F1/A1 = F2/A2 F1 = F2.A1/A2 ............................(iii) F2 is the force exerted by piston (D), F1 is the force exerted on (E).

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A2 is the area of pipe like container (B) i.e. 0.78 cm^2, and A1 is the area of small container (A) i.e. 28.27 cm^2. after putting values of (A) and (B) in (iii) F1 = F2 28.27/0.78 F1 = 36.F2 (approx)

Result The force exerted by liquid on (E) is 36 times greater than that of the force exerted by piston (D) on liquid.

Precautions 1. No air bubbles should be present. 2. Piston should be pressed gently. 3. Give enough time to epoxy to harden completely.

Sources of error

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1. Joints may not be air tight. 2. Leaks in containers. 3. Presence or air bubbles.