Answer: 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
Step-by-step explanation:
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