Respuesta :
Answer:
a) [tex]\frac{A_{s}}{A_{p}}= 100[/tex]
b) [tex]\frac{d_{s}}{d_{p}} = 10[/tex]
c) [tex]\frac{L_{s}}{L_{p}}=\frac{1}{100}[/tex]
Explanation:
Applying the Pascal principle:
[tex]F=P*A[/tex] Equation (1)
F: Force
A: Area
P: Pressure
Nomenclature:
Fp: Force on the primary piston
Fs: Force on the secondary piston
Ap: Primary piston area
As: Secondary piston area
a) Calculation of the relationship of areas:
The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.
In the equation (1):
[tex]P = \frac{F}{A}[/tex]
Pp = Ps
[tex]\frac{F_{p}}{A_{p}}=\frac{F_{s}}{A_{s}}[/tex] Equation(2)
[tex]\frac{A_{s}}{A_{p}}=\frac{F_{s}}{F_{p}}[/tex]
We know that Fs = 100Fp, then we replace Fs in equation (2)
[tex]\frac{A_{s}}{A_{p}}=\frac{100*F_{p}}{F_{p}}[/tex] We eliminate Fp, then:
[tex]\frac{A_{s}}{A_{p}}= 100[/tex] Equation (3)
b)Calculation of the relationship of diameters (d)
In the Equation (3) we replace: [tex]A_{s}= \frac{\pi*d_{s}^{2}}{4}[/tex], [tex]A_{p}= \frac{\pi*d_{p}^{2}}{4}[/tex]
[tex]\frac{\frac{\pi*d_{s}^{2}}{4}}{\frac{\pi*d_{p}^{2}}{4}}=100[/tex] ; we eliminate π
[tex]\frac{d_{s}^2}{d_{p}^2} = 100[/tex]
[tex]\frac{d_{s}}{d_{p}} = \sqrt[2]{100} = 10[/tex]
c) Reduction factor of the distance that the force moves at the exit (Ls) with respect to the distance that the force moves at the entrance (Lp):
The volume that moves at the entrance (Vp) is equal to the volume that moves at the exit (Vs).
[tex]V=A*L[/tex]
Vp = Vs
Ap * Lp = As * Ls
[tex]\frac{L_{s}}{L_{p}}=\frac{A_{p}}{A_{s}}[/tex] , of the equation (3) [tex]\frac{A_{s}}{A_{p}}= 100[/tex] then, [tex]\frac{A_{p}}{A_{s}}= \frac{1}{100}[/tex]
[tex]\frac{L_{s}}{L_{p}}=\frac{1}{100}[/tex]
The ratios of the area, diameter of the hydraulic system cylinder is 100 and 10 respectively. The factor by which the distance move from output to input force is 1/100.
What is pascal law?
Pressure applied to a closed system of fluid will be transfer at each point of the fluid and the boundaries of the system.
Let suppose at the two point of such system the input force apllied is F₁ and output force we get is F₂. The area of this points is A₁ and A₂ respectively. As the pressure at both point is same. Then by the pascal law,
[tex]\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}[/tex]
- (a) The ratio of the area of the cylinder that is being controlled to the area of the master cylinder -
A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. Therefore,
[tex]\dfrac{F_1}{A_1}=\dfrac{100F_1}{A_2}\\\dfrac{A_2}{A_1}=100[/tex]
- (b) The ratio of their diameters.
For the cylindrical hydraulic system, the ratio of area in terms of diameter can be given as,
[tex]\dfrac{A_2}{A_1}=100\\\dfrac{\pi \dfrac{d_2^2}{4}}{\pi \dfrac{d_1^2}{4}}=100\\\dfrac{d_2}{d_1}=\sqrt{100}=10[/tex]
- (c) Factor by which the distance through which the output force moves reduced relative to the distance through which the input force moves-
For this case, the ratio of distance moved from exit to entrance will be equal to the ratio of areas at the entrance to the exit. Therefore,
[tex]\dfrac{L_2}{L_1}=\dfrac{A_1}{A_2}\\\dfrac{L_2}{L_1}=\dfrac{1}{100}[/tex]
Thus, the ratios of the area, diameter of the hydraulic system cylinder is 100 and 10 respectively. The factor by which the distance move from output to input force is 1/100.
Learn more about the pascal law here;
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