Question

in a hydraulic press a force of 200N is applied to a master piston of area 25cm2.If the press was designed to produce a force of 5000N determine a)The area of the slave piston b) The radius of the slave piston

Answer 1

Pressure= force/area

F1/A1=F2/A2

200/25=5000/x

Area of slave cylinder is 625 cm^2

Answer 2

Answer:

The area of the slave piston is 625sq.cm. and its radius is 14.1cm.

Explanation:

Hydraulic press:

• Hydraulic press consists of a small piston of cross sectional area $A_1$ and a large piston of cross sectional area $A_2$. When a force $F_1$ is applied on small piston. the pressure exerted by the force is transmitted throughout the liquid to the large piston.
• The forces and areas of the two pistons are related through the equation
• $\dfrac{F_1}{F_2} =\dfrac{A_1}{A_2}$

Given the force by master piston, $F_1=200N$

Cross sectional area of master piston, $A_1=25cm^2=25\times 10^{-4} m^{2}$

The force produced on slave piston, $F_2=5000N$

a) Substituting the given data in the formula,

$\dfrac{200}{5000} =\dfrac{25\times 10^{-4} }{A_2}$

$A_2 =\dfrac{25\times 10^{-4} \times5000}{200}$

    $=25\times 10^{-4} \times25=625\times 10^{-4} m^{2}=625cm^{2}$

Therefore, the area of the slave piston is 625sq.cm.

b) The piston has circular cross-sectional area.

The area of the circle is $\pi r^{2}$.

$A_2=\pi r^{2}\implies 625=3.14\times r^{2}$

$\implies r^{2}=\dfrac{625}{3.14} =199.04$

$\implies r=\sqrt{199.04} =14.1 cm$

Therefore, the radius of the slave piston is 14.1cm.

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