HW2S - Missing water: The horizontal force is calculated...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
P2.70 Solution: For water, we take ± = 9790 N / m 3 . The hydrostatic force on the valve is : The center of pressure is slightly below the center line by an amount: Taking the moment about the hinge: F = p CG A = h ( ² 4 ) d 2 = (9790 N m 3 ) h ( 4 )(0.2286 m ) 2 = 401.8 h y CP = ± sin ³ I xx F = (9790)sin(30 ± )( ´ / 4)(0.1143) 4 401.8 h = 0.001633 h M hinge = Fl = (401.8 h )(0.15 + 0.00163 h ) = 50 ± h = 0.8187 m
Background image of page 2
P2.90 For water, we take ± = 9790 N / m 3 . The vertical force on AB is the weight of the missing water above AB-see dashed lines. Calculate this as a rectangle as follow:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Missing water: The horizontal force is calculated from the vertical projection of panel AB: v = (1.5 m )(0.75)(1.2 m ) + (1 4 )(0.75 m ) 2 (1.2 m ) = 1.35 + 0.145 = 1.495 m 3 F V = v = (9790 N / m 3 )(1.495 m 3 ) = 14636.1 N Vertical Force F H = p CG hA projection = (9790 N / m 3 )(1.5 m + 0.375 m )(0.75 m )(1.2 m ) = 16500 N Horizontal force...
View Full Document

Page1 / 5

HW2S - Missing water: The horizontal force is calculated...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online