A piston with a cross-sectional area [tex]a[/tex] is used in a hydraulic press to exert a small force of magnitude [tex]f[/tex] on the enclosed liquid. A connecting pipe leads to a larger piston with a cross-sectional area [tex]A[/tex].

If the piston diameters are 2.99 cm and 65.9 cm, what force magnitude on the small piston will balance a 38.2 kN force on the large piston?

Pascal's law, a force of approximately 55.313 N on the small piston balances the 38.2 kN force on the large piston by adjusting the ratio of their respective areas.

The formula for pressure is:
Pressure = Force / Area

Therefore, the pressure exerted on the small piston can be expressed as:
[tex]P_small = F_small / a[/tex]

And the pressure exerted on the large piston can be expressed as:
[tex]P_large = F_large / A[/tex]

Since the pressure is the same for both pistons, we can equate the two expressions:
[tex]F_small / a = F_large / A[/tex]

We are given that F_large is 38.2 kN (kilonewtons), A is the cross-sectional area of the large piston (65.9 cm), and a is the cross-sectional area of the small piston (2.99 cm).

Substituting the given values:
F_small = (38.2 kN / 65.9 cm) * 2.99 cm

Make sure to convert the units to match.

Now, let's calculate the force on the small piston using the given values:
F_small = [tex](38.2 * 1000 N) / (65.9 * 10^-4 m^2) * (2.99 * 10^-4 m^2)[/tex]

After performing the calculations, we find:
F_small ≈ 55.313 N

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