Question 3.8: Determine the force exerted by sea water (sp. gravity = 1.02......
Determine the force exerted by sea water (sp. gravity = 1.025) on the curved portion AB of an oil tanker as shown in Fig. Ex. 3.8. Also determine the direction of action of the force.
Consider 1 m width perpendicular to paper,
The horizontal component of the force acting on the curved portion AB
= γ A \overline{h} = (1025 × 9.81) (4 × 1)(15 + 4/2) \\ = 683757 N
Line of action of this horizontal force = \overline{h} + (I_{G} /\overline{h} A) \\ = 17 + [1 × 4³/12] [1/(17 × 4 × 1)] = 17.0784 m
from top and towards left
The vertical force is due to the volume of sea water displaced.
Vertical force = [volume BCDE + volume ABE] γ \\ = [(15 × 4 × 1) + (4² × π × 1/4)] [1025 × 9.81] \\ = 729673 N (acts upwards).
To find the location of this force : Centre of gravity of the column BCDE is in the vertical plane 2 m from the edge.
Centre of gravity of the area ABE = (4 – 4R/3π) from the edge
= (4 – 4 × 4/3π) = 2.302 m from the edge.
Taking moments of the area about the edge, the line of action of vertical force is
= [(2.302 × 4² π /4) + 2 × (15 × 4)] / [(4² π/4) + (15 × 4)] = 2.0523 m
from the edge.
The resultant force = (683757^{2} + 729673^{2})^{0.5} = 999973.16 N
The direction of action to the vertical is,
\tan θ = 683757 / 729673 = 0.937 ∴ θ = 43.14°
The answer can be checked by checking whether the resultant passes through the centre of the circle (as it should) by taking moments about the centre and equating them.
729673(4 – 2.0523 ) – 683757 × 2.078 = 337
Compared to the large values, the difference is small and so the moments are equal and the resultant can be taken as zero. Hence the resultant can be taken to pass through the centre of the cricle.