Question 2.8: A circular cylinder of 1.8 m diameter and 2.0 m long is acte...
A circular cylinder of 1.8 m diameter and 2.0 m long is acted upon by water in a tank as shown in Fig. 2.31a. Determine the horizontal and vertical components of hydrostatic force on the cylinder.
Let us consider, at a depth z from the free surface, an elemental surface on the cylinder that subtends an angle d \theta at the centre. The horizontal and vertical components of hydrostatic force on the elemental area can be written as
d F_{ H }=9.81 \times 10^{3}\{0.9(1+\cos \theta)\}(0.9 d \theta \times 2) \sin \thetaand d F_{V}=9.81 \times 10^{3}\{0.9(1+\cos \theta)\}(0.9 d \theta \times 2) \cos \theta
Therefore, the horizontal and vertical components of the net force on the entire cylindrical surface in contact with water are given by
F_{H}=\int_{0}^{\pi} 9.81 \times 10^{3}\{0.9(1+\cos \theta)\} 1.8 \sin \theta d \theta N = 31.78 kN
F_{V}=\int_{0}^{\pi} 9.81 \times 10^{3}\{0.9(1+\cos \theta)\} 1.8 \cos \theta d \theta N
=9.81 \times 10^{3} \times 0.9 \times 1.8\left[\int_{0}^{\pi} \cos \theta d \theta+\int_{0}^{\pi} \cos ^{2} \theta d \theta\right]
=9.81 \times 10^{3} \times 0.9 \times 1.8\left[0+\frac{\pi}{2}\right] N
= 24.96 kN
Alternative method:
The horizontal component of the hydrostatic force on surface ACB (Fig. 2.31b) is equal to the hydrostatic force on a projected plane area of 1.8 m high and 2 m long.
Therefore, F_{H}=9.81 \times 10^{3} \times 0.9 \times(1.8 \times 2) N = 31.78 kN
The downward vertical force acting on surface AC is equal to the weight of water contained in the volume CDAC. The upward vertical force acting on surface CB is equal to the weight of water corresponding to a volume BCDAB.
Therefore the net upward vertical force on surface ACB
= Weight of water corresponding to volume of BCDAB Weight of water in volume CDAC
= Weight of water corresponding to a volume of BCAB (half of the cylinder volume)
Hence, F_{V}=9.81 \times 10^{3} \times \frac{1}{2}\left\{3.14 \times(0.9)^{2} \times 2\right\} N = 24.96 kN
