Question

what should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of it's brake shoe by exerting a force of 0.5 Newton on pedal?​

Answer 1

Explanation:

Consider thew ratio of cross-section of the master cylinder and wheel cylinder be X

1

:X

2

Let the force applied on pedal be F

1

=0.5N

And the force applied on brake shoe

We know from the hydraulic machine,

Pressure on narrow piston = pressure on broader piston

x

1

f

1

=

x

2

f

2

⇒

f

2

f

1

=

x

2

x

1

⇒

X

2

X

1

=

15

0.5

⇒

X

2

X

1

=1/30

⇒ Hence, the ratio is 1:30

Answer 2

We know from the Pascal's law - fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container.

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

Here the force applied F₁ = 0.5 N ;

with correspondig area of master cylinder -A₁ ;

force to be obtained F₂ = 15 N ;

with correspondig area of break cylinder - A₂ .

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

⇒ $\frac{A_{1} }{A_{2} } = \frac{0.5 N}{15 N}$

⇒$\frac{A_{1} }{A_{2} } = \frac{1}{30}$

∴ The ratio of area of cross section of the master section and the wheel cylinder of require hydraulic brake is 1 : 30 .

~hope it helps

@Manogna