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Forces, Density and Pressure: Question 4

Syllabus 4.3

Structured AS 6 marks

A solid engineering component is made from a uniform metal alloy. The component has a mass of 3.6 kg3.6\text{ kg} and a volume of 4.0×10−4 m34.0\times10^{-4}\text{ m}^3.

(a) Calculate the density of the alloy. [2]

(b) The component is used as a piston with a cross-sectional area of 5.0×10−3 m25.0\times10^{-3}\text{ m}^2. Calculate the force the piston must exert to produce a pressure of 2.4×105 Pa2.4\times10^{5}\text{ Pa} on the fluid beneath it. [2]

(c) The component is later fully submerged in a tank of oil of density 850 kg m−3850\text{ kg m}^{-3}, at a depth of 1.2 m1.2\text{ m} below the oil's surface. Using g=9.81 m s−2g = 9.81\text{ m s}^{-2}, calculate the hydrostatic pressure due to the oil at this depth. [2]

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Worked solution

Part (a): Density of the alloy

Density is mass per unit volume: ρ=mV=3.64.0×10−4\rho = \frac{m}{V} = \frac{3.6}{4.0\times10^{-4}} ρ=9.0×103 kg m−3\rho = 9.0\times10^{3}\text{ kg m}^{-3}

Part (b): Force needed for the required pressure

Pressure is force per unit area, p=F/Ap = F/A, so rearranging for FF: F=pA=(2.4×105)(5.0×10−3)F = pA = (2.4\times10^{5})(5.0\times10^{-3}) F=1.2×103 NF = 1.2\times10^{3}\text{ N}

Part (c): Hydrostatic pressure in the oil

The hydrostatic pressure difference at depth Δh\Delta h in a fluid of density ρ\rho is: Δp=ρgΔh\Delta p = \rho g \Delta h

Substituting the oil’s density, g=9.81 m s−2g = 9.81\text{ m s}^{-2}, and Δh=1.2 m\Delta h = 1.2\text{ m}: Δp=(850)(9.81)(1.2)\Delta p = (850)(9.81)(1.2) Δp=8338.5×1.2\Delta p = 8338.5 \times 1.2 Δp≈1.00×104 Pa\Delta p \approx 1.00\times10^{4}\text{ Pa}

Final answers

  • (a) ρ=9.0×103 kg m−3\rho = \boxed{9.0\times10^{3}}\text{ kg m}^{-3}
  • (b) F=1.2×103 NF = \boxed{1.2\times10^{3}}\text{ N}
  • (c) Δp≈1.00×104 Pa\Delta p \approx \boxed{1.00\times10^{4}}\text{ Pa}