Example 14.2 Help with end of the problem please 2. 0.66/1 points | Previous Ans

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Example 14.2 Help with end of the problem please

2. 0.66/1 points | Previous Answers SerPSL9 14.Ab.002 Example 14.2 The Car Lift In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of radius 5.63 cm. This pressure is transmitted by a liquid to a piston that has a radius of 13.7 cm (A) What force must the compressed air exert to lift a car weighing 13,500 N? (B) What air pressure produces this force? SOLVE IT (A) What force must the compressed air exert to lift a car weighing 13,500 N? Conceptualize Review the material about Pascal's law to understand the operation of a car lift Categorize This example is a substitution problem. Solve FJA, = FJA, for F1: Fi-(A2)F2- r(5.67 × 10-2 m)2 (1.35 × 102 N) T(13.7 x 10-2 m)2 2279.87 (B) What air pressure produces this force? F. Use the definition of pressure to find the air pressure that produces this force: (5.63 × 10-2 m)2 A = 228951 This pressure is greater than the atmospheric pressure MASTER IT HINTS GETTING STARTED I IM STUCKI Piston 1 has a diameter of 0.113 m and piston 2 has a diameter of 0.315 m. In the absence of friction, determine the magnitude of the force F necessary to support the large weight, W = 2,248 N

Explanation / Answer

Here you have two things to consider: the pressures on the pistons, and the torque due to the force applied on the lever.

Sum of the torques = 0
Sum of the pressures = 0

The pivot point is at poiston 2.

Let R1 be the distance from the pivot to F (12")
Let R2 be the distance from the pivot to piston 1 (2")

R1F = R2F1

where F1 is the net force on piston 1

Sum of the pressures = 0 implies that

P2 = P1

F2/A2 = F1/A1

2248 N/ (0.315)^2 = F1 / (0.113)^2

F1 = 2248 * 0.113^2 /0.315^2

= 289 N

F = (R2/R1)F1

= 2*289/12 N

= 48.2 N

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