Question

    A homogeneous, 4 ft wide, 8 ft long rectangle gate weighing 800 lbis held in...

   
A homogeneous, 4 ft wide, 8 ft long rectangle gate weighing 800 lbis held in place by a horizontal flexible cable. Water acts againstthe gate, which is hinged at point A. friction in the hinge isnegligible. Determine the tension in the cable.(answer=1350 lb)

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Answer #1
Concepts and reason

Pascal’s law states that at any point in a fluid, pressure is independent of motion and direction if there is no shear stress.

Consider the hydrostatic forces on an inclined plate submerged in a fluid.

N

Resultant force on the plate is given as follows:

Fr=yhA

Here, specific weight of the fluid is , location of centroid of the plate from the free surface of water is h, and area of the plate is A.

The resultant force acts perpendicular to the gate.

Write the point of action of the resultant force on the plate.

+a=*

Here, distance of the centroid along the inclined plate from the free surface is and moment of inertia of the plate along centorial axis is .

Write the point of action of the resultant force on the plate along the depth.

ho=ht Lxx sin’o
hA

Fundamentals

Moment of inertia of rectangular plate about its centroid is given by

1, без

Here, base length of the rectangle is b and height of the rectangle is h.

Specific weight ():

It is the weight of the fluid per unit volume and it is equal to product of density and acceleration due to gravity.

= pg

Specific weight of water is 62.41b/ft?
.

Draw free body diagram of the plate.

de
5
3
HR
30/
/ 8 ft
00W

Calculate the center of the plate from free surface of water, h.

h=3sin 60°
= 2.598 ft

Determine the center of pressure of the point from the free surface of water.

hp =h+ xx sino
hA
( bd
sin? 0
hp =hth(bxd)

Substitute for b, for d, 2.598 ft
for h, and for .

(4x 6
12
sin 60
sin 60
h, = 2.598+
2.598x4x6
= 3.46405 ft

Calculate the hydrostatic force on the plate.

F = yhA
F=yhbd

Substitute 2.598 ft
for h, 62.41b/ft?
for , for b, and for d.

F = 62.4x 2.598x4x6
= 3890.7648 lb

Consider moments about point A in the free body diagram.

MA=0
Tsin 60° (8) -W sin 30°(4)- Fr(6- b) = 0
sin 60°)

Substitute 800 lb
for W, 3890.7648 lb
for , and 3.46405 ft
for .

3.46405
Tsin 60°(8) -(800)sin 30°(4)-(3890.7648) 6-sin 609 ) = 0
6.9282T - 9381.76=0
T=1354.14 lb
T=1350 lb

Ans:

The tension in the cable is 1350 lb
.

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