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2 (a) A cytinder is nuspended from the end of a string. The cyinder is stationa...
Sep 16, 2024
Solution
a
To calculate the height of the cylinder, we can use the difference in pressure between the top and bottom faces of the cylinder. The pressure difference is given as 320 Pa. The formula for pressure difference is given by: ΔP=ρgh\Delta P = \rho g h where ΔP\Delta P is the pressure difference, ρ\rho is the density of the fluid (water), gg is the acceleration due to gravity (approximately 9.81m/s29.81 \, \text{m/s}^2), and hh is the height of the cylinder. Rearranging the formula gives us: h=ΔPρg=320Pa(1.0×103kg/m3)(9.81m/s2)0.0325mh = \frac{\Delta P}{\rho g} = \frac{320 \, \text{Pa}}{(1.0 \times 10^3 \, \text{kg/m}^3)(9.81 \, \text{m/s}^2)} \approx 0.0325 \, \text{m}
b
To calculate the upthrust (buoyant force) acting on the cylinder, we can use Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. The volume of the cylinder can be calculated using its height and cross-sectional area. The area AA of the circular cross-section is given by: A=π(d2)2=π(0.031m2)27.54×105m2A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.031 \, \text{m}}{2}\right)^2 \approx 7.54 \times 10^{-5} \, \text{m}^2. The volume VV of the cylinder is then: V=Ah7.54×105m20.0325m2.45×106m3V = A \cdot h \approx 7.54 \times 10^{-5} \, \text{m}^2 \cdot 0.0325 \, \text{m} \approx 2.45 \times 10^{-6} \, \text{m}^3. The buoyant force FbF_b is then calculated as: Fb=ρVg=(1.0×103kg/m3)(2.45×106m3)(9.81m/s2)0.024NF_b = \rho V g = (1.0 \times 10^3 \, \text{kg/m}^3)(2.45 \times 10^{-6} \, \text{m}^3)(9.81 \, \text{m/s}^2) \approx 0.024 \, \text{N}. However, the problem states that the upthrust is 0.38 N, which indicates that the cylinder is partially submerged
c
To find the tension TT in the string, we can apply the equilibrium condition. The forces acting on the cylinder are its weight W=0.84NW = 0.84 \, \text{N} acting downwards, the buoyant force FbF_b acting upwards, and the tension TT in the string also acting upwards. The equilibrium condition gives us: T+Fb=WT + F_b = W. Rearranging this gives us: T=WFb=0.84N0.38N=0.46NT = W - F_b = 0.84 \, \text{N} - 0.38 \, \text{N} = 0.46 \, \text{N}
Answer
0.46 N
Key Concept
Archimedes' Principle: The buoyant force on an object is equal to the weight of the fluid displaced by the object. Equation: Fb=ρVgF_b = \rho V g (Buoyant force equals density times volume times acceleration due to gravity). Static Equilibrium: The net force acting on an object at rest is zero. Equation: F=0\sum F = 0.
Explanation
The calculations show that the height of the cylinder is approximately 0.0325 m, the upthrust acting on the cylinder is 0.38 N, and the tension in the string is 0.46 N, confirming the equilibrium of forces acting on the cylinder.
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