Determine the total pressure and position of centre of pressure on an isosceles triangular plate of base 5m and
altitude 5m when the plate is immersed vertically in an oil of specific gravity 0.8. The base of the plate is 1m
below the free surface of oil.
Answers
Answer:
261927N, 3.19m
Explanation:
I don't knwo the explanation can anybody tell me the answer
For the given question, refer to the diagram given below:
Concept:
We need to apply the concept of a line equation.
Given:
Base of Isosceles triangle = 5m
Altitude = 5m
Specific gravity = 0.8
Base of the plate = 1m
Find:
We need to determine the total pressure and position of the centre of pressure on an isosceles triangular plate.
Solution:
Specific gravity i.e density of oil = 0.8 × 1000 = 800 kg/m³
We have, Surface oil pressure equals zero
Oil pressure at 1 m deep is 800 kg/m2.
Using the base's intersection with altitude as the origin
Line AC equation, passing, y = -2x + c (2.5,0)
Therefore, 0=—5+c, c = 5
Line AC's equation is y = -2x + 5.
Therefore pressure at Y = 800 (1-y)
Therefore, Total force on plate can be formulated as,
Total force on plate = ∫800(1-y)xdy from y = 0 to y = 1.
= 800 ∫ (1-y)(5-y)/2 dy
= 400 ∫ (5-6y+y²) dy
= 400(5y-3y² + y³/3)
= 400(5-3+1/3)
= 933
Therefore total force applied to the plate from x=—1 to 1
= 2 × 933 = 1866 kg
pressure moment about the base from y=0 to 1
= 800 ∫ y(1-y)(5-y)/2 dy
= 400 ∫ (5y-6y²+y³)dy
= 400(5y²/2 — 2y³ + y⁴/4)
= 400(5/2-2+1/4)
= 400(3/4)
= 300
Total pressure moment from x = —1 to 1
= 2 × 300
= 600 kg m
Center of pressure above the base, therefore, equals 600/1866 = 0.32 m
Thus, the total pressure and position of the centre of pressure on an isosceles triangular plate is 1866 Kg and 0.32 m
#SPJ3
