Pressure is a fundamental concept in physics with many practical applications. A student studies three situations involving pressure. First, she calculates ...
Question 1 Report
Pressure is a fundamental concept in physics with many practical applications. A student studies three situations involving pressure. First, she calculates the pressure exerted by a concrete column on the ground. The column has a square cross-section of side 0.30 m and a height of 2.5 m. The density of concrete is 2400 kg/m³ and g = 9.8 N/kg. Second, she considers the water pressure at the bottom of a swimming pool 3.0 m deep, with water density 1000 kg/m³ and atmospheric pressure 1.0 × 10⁵ Pa. Third, she analyses a simple hydraulic system where a force of 50 N is applied to a piston of area 2.0 cm² connected by oil to a larger piston of area 80 cm². She writes up her findings in a laboratory report.
(a) Calculate the mass and weight of the concrete column. [2]
(b) Calculate the pressure the column exerts on the ground. [2]
(c) Calculate the total pressure at the bottom of the swimming pool. [2]
(d) Calculate the force exerted by the large piston in the hydraulic system. [3]
(e) The student notices that the concrete column exerts a much higher pressure on the ground than the water does on the pool floor. Explain why, even though the water covers a much larger area. [2]
Answer Details
Part (a): Mass and weight of the concrete column
Volume of the column:
\(V = 0.30 \times 0.30 \times 2.5 = 0.225\text{ m}^3\)
Mass:
\(m = \rho V = 2400 \times 0.225 = 540\text{ kg}\)
Weight:
\(W = mg = 540 \times 9.8 = 5292\text{ N}\)
Part (b): Pressure on the ground
The base area of the square column:
\(A = 0.30 \times 0.30 = 0.090\text{ m}^2\)
\(p = \frac{F}{A} = \frac{5292}{0.090} = 58\,800\text{ Pa} \approx 5.9 \times 10^4\text{ Pa}\)
Part (c): Total pressure at the bottom of the pool
Pressure due to the water column:
\(p_{\text{water}} = \rho g h = 1000 \times 9.8 \times 3.0 = 29\,400\text{ Pa}\)
Total pressure includes atmospheric pressure pressing on the water surface:
\(p_{\text{total}} = 1.0 \times 10^5 + 29\,400 = 129\,400\text{ Pa} \approx 1.29 \times 10^5\text{ Pa}\)
Part (d): Force from the large piston
First find the pressure in the oil. Convert the small piston area: 2.0 cm\(^2\) = 2.0 \(\times\) 10\(^{-4}\) m\(^2\).
\(p = \frac{F}{A} = \frac{50}{2.0 \times 10^{-4}} = 2.5 \times 10^5\text{ Pa}\)
This pressure is transmitted through the oil to the large piston (area = 80 cm\(^2\) = 80 \(\times\) 10\(^{-4}\) m\(^2\)):
\(F = p \times A = 2.5 \times 10^5 \times 80 \times 10^{-4} = 2000\text{ N}\)
The force multiplication ratio is \(\frac{80}{2.0} = 40\), turning 50 N into 2000 N.
Part (e): Why the column exerts higher pressure despite the pool covering more area
Pressure is force per unit area, not total force. The concrete column concentrates a large weight (5292 N) on a small base (0.09 m\(^2\)), producing about 59 000 Pa. The swimming pool water at only 3.0 m deep produces just 29 400 Pa from the water alone, because the liquid pressure formula \(p = \rho g h\) depends only on depth, density and \(g\), not on the width or total area of the pool. A wider pool has more total weight but also more area, so the pressure per unit area stays the same.
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