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Solnce55 [7]
3 years ago
7

Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to another (called

the secondary). Since the pressures will be equal, if the surface areas are different then the forces applied to the cylinders' pistons will be different. Suppose in a hydraulic lift, the piston of the primary cylinder has a 1.85-cm diameter and the piston of the secondary cylinder has a 24.5-cm diameter. What force, in newtons. must be exerted on the primary cylinder of this lift to support the weight of a 2200-kg car (a large car) resting on the secondary cylinder?
Physics
1 answer:
Paul [167]3 years ago
3 0

Answer:

F1= 122.93 N

Explanation:

Pascal´s Principle

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

P=F/A

P1=P2

\frac{F_{1} }{A_{1}} = \frac{F_{2} }{A_{2}}   Formula (1)

Data

D1 =  1.85 cm : primary cylinder piston diameter

D2 = 24.5 cm : secondary cylinder piston diameter

m = 2200-kg  : car mass

Piston area calculation

A_{1} = \frac{\pi *(D_{1})^{2}  }{4}

A_{1} = \frac{\pi *(1.85)^{2}  }{4}

A1= 2.688 cm²

A_{2} = \frac{\pi *(D_{2})^{2}  }{4}

A_{2} = \frac{\pi *(24.5)^{2}  }{4}

A2 = 471.435 cm²

Calculating of the weight of the car (W)

W = m*g = 2200-kg * 9.8 m/s² = 21560 N

Calculation of the force in Newtons to be exerted on the primary cylinder piston

Data:

A1= 2.688cm²

A2= 471.435 cm²

F2 = W=  21560 N

We replace data in the formula (1)

\frac{F_{1} }{A_{1}} = \frac{F_{2} }{A_{2}}

F_{1} =\frac{F_{2}*A_{1}  }{A_{2} }

F_{1} =\frac{21560N*2.688 cm^{2} }{471.435 cm^{2} }

F1= 122.93 N

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T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

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     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

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We solve this system of equations

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In this initial case

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Let's find the tension

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    T = ½ M g

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We seek tension for this case

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Let's look for the relationship between the tensions of the two cases

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