Physics, 16.11.2019 00:31 priscillavaladez1112

In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radius of r1 = 5.05 cm. this pressure is transmitted by an incompressible liquid to a second piston of radius 14.7 cm. what force must the compressed air exert on the small piston in order to lift a car weighing 13,300 n?
(a) neglect the weights of the pistons. n
(b) what air pressure will produce a force of that magnitude? pa
(c) show that the work done by the input and output pistons is the same. (do this on paper. your instructor may ask you to turn in this work.)
(d) a hydraulic lift has pistons with diameters 7.90 cm and 35.3 cm, respectively. if a force of 825 n is exerted at the input (smaller) piston,
(e) what maximum mass can be lifted at the output piston? m = kg

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Answer from: trvptierra

(a) 1,569.63 N

(b) 195,933.99 Pa

(d) 1,647.47 kg

Explanation:

Let the cross-sectional area (CSA) of small piston = A₁

Let the cross-sectional area (CSA) of the bigger piston = A₂

Let the Force applied at the smaller piston = F₁

Let the Force applied at the bigger piston = F₂

The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂ (1)

(a) F₁= F₂ xA₁ /A₂

F₂ = 13,300 N

A₁ = π r₁²

=π x (0.0505)²

= 0.008011 m²

A₂= π r₂²

=π x (0.147)²

= 0.06788 m²

Substituting into (1)

F₁ = 13,300 x 0.008011/0.06788

= 1,569.6272

≈ 1,569.63 N

(b) Air pressure = Force/Area

= F₁/A₁

= 1,569.6272/ 0.008011

= 195,933.99 Pa

Workdone = force x distance (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d) F₂ = F₁ x A₂/ A₁ (3)

A₁ = π r₁²

=π x (0.079)²

= 0.01960 m²

A₂= π r₂²

=π x (0.353)²

= 0.3914 m²

Substituting into (2)

F₂= 825 x 0.3914/ 0.01960

= 16,474.7448

≈ 16,474.74 N

= 1,647.47 kg

Answer from: zahriamarie10

(a) The magnitude of force on smaller piston is 1569.63 N.

(b) The magnitude of air pressure on smaller piston is 19.59 Pa.

(d) The magnitude of force at output is 16472.10 N.

(e) The maximum mass lifted at the output piston is of 1680.82 kg.

Given data:

The radius of smaller piston is, $r_{1}=5.05 \;\rm cm$ .

The radius of larger piston is, $r_{2}=14.7 \;\rm cm$ .

Weight of car is, $W= 13300 \;\rm N$ .

(a) Neglecting the weight of piston.

Let F and F' be the force on smaller and larger piston respectively. Such that force on larger piston is equal to weight lifted.

F' = W

Use pascal's law to obtain the force on the smaller piston as,

$\dfrac{F}{A}=\dfrac{F'}{A'} \\F=\dfrac{F'}{A'} \times A$

Here, A and A' are the areas of smaller and larger piston respectively. Solving as,

$F=\dfrac{W}{\pi \times r^2_{2}} \times \pi \times r^2_{1}\\F=\dfrac{13300}{ 14.7^2} \times 5.05^2\\F=1569.63 \;\rm N$

Thus, the force on smaller piston is 1569.63 N.

(b) Then, the magnitude of air pressure is,

$P=\dfrac{F}{A} \\P=\dfrac{1569.63}{\pi \times r^2_{1} }\\P=\dfrac{1569.63}{\pi \times 5.05^2}\\P= 19.59 \;\rm Pa$

Thus, the magnitude of air pressure on smaller piston is 19.59 Pa.

(c)

The work done by the input and output piston is,

$W = P \times V$

Here, P is the pressure and V is the volume on each piston.

The pressure and volume for both the piston is same.

Thus, we can conclude that work done by input and output of both pistons are same.

(d) Again apply the pascal's law as,

$\dfrac{F}{A}=\dfrac{F'}{A'} \\F'=\dfrac{F}{A} \times A'\\F'=\dfrac{F}{\dfrac{\pi}{4}d^{2} } \times {\dfrac{\pi}{4}d'^{2}} \\F'=\dfrac{825}{7.90^{2} } \times 35.3^{2}}\\F'=16472.10 \;\rm N$

Thus, force at output is 16472.10 N.

(e) And the mass lifted as output is,

$F' =m' \times g$

Here, g is the gravitational acceleration.

$16472.10 =m' \times 9.8\\m'=1680.82 \;\rm kg$

Thus, the maximum mass lifted at the output piston is of 1680.82 kg.

Learn more about pascal's law here:

link

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