Physics, 19.11.2019 03:31 Truezoe09

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?

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Answer from: 0465834wu1

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

Answer from: Quest

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hope this

Answer from: Quest

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west