a effort of 100n can raise a load of 2000n in a hydraulic press. calculate the cross-sectional area of a small piston in it. The cross-sectional area of a large piston is 4m^s​

Respuesta :

Answer:

[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]

Explanation:

The pressure on the pistons is given as;

Pressure = [tex]\frac{Force}{Area}[/tex]

So that,

Pressure on the small piston = [tex]\frac{F_{1} }{A_{1} }[/tex] and Pressure on the large piston = [tex]\frac{F_{2} }{A_{2} }[/tex]

Thus,

[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]

Given that: [tex]F_{1}[/tex] = 100 N, [tex]F_{2}[/tex] = 2000 N, [tex]A_{2}[/tex] = 4 [tex]m^{2}[/tex].

[tex]\frac{100}{A_{1} }[/tex] = [tex]\frac{2000}{4}[/tex]

[tex]A_{1}[/tex] = [tex]\frac{100*4}{2000}[/tex]

    = [tex]\frac{400}{2000}[/tex]

    = 0.2

[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]

The area of the small piston is 0.2 [tex]m^{2}[/tex].

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