Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 3, Problem 85P

A triangular-shaped gate is hinged at point A, as shown. Knowing that the weight of the gate is 100 N, determine the force needed to keep the gate at its position for unit width. The line of action of the weight of the gate is shown by the dashed line.

Expert Solution & Answer
Check Mark
To determine

The force needed to keep the gate at its position for unit width.

Answer to Problem 85P

The force needed to keep the gate at its position for unit width is 23070.21N.

Explanation of Solution

Given information:

The weight of the gate is 100N

The below figure shows the force acting on the gate to keep it at equilibrium position.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 85P

Figure-(1)

Write the expression for Pythagoras theorem in triangle ABC.

  BC=AB2AC2...... (I)

Write the expression for hydrostatic force acting on the member BC of the gate.

  FBC=ρgAx¯...... (II)

Here, the density of the water is ρ, the gravitational force is g, the area of the triangular gate is A and the location of the centroid from the free surface is x¯.

Write the expression for sinθ from the triangle ABC.

  sinθ=ACBC...... (III)

Write the expression for hydrostatic force on member AB.

  FAB=ρgAx¯AB...... (IV)

Here, the location of Centroid of the member AB from the free surface is. x¯AB.

Write the expression for point of application of force.

  hAB=x¯+IAx¯sin2θ...... (V)

Here, the moment of inertia of the triangular gate is I.

Write the expression for area of the member BC.

  ABC=BC×W

Here, the width of gate to be kept is 1m.

Write the expression for area of the member AB.

  AAB=BA×W

Here, the width of gate to be kept is 1m.

Write the expression for the location of centroid of the member AB from the free surface.

  x¯BC=(1+AB×sinθAC)...... (VI)

Write the expression for moment of inertia of triangular gate.

  I=AB×W312...... (VII)

Write the expression for moment about point A.

  MA=0FBC×BC2+w×1+F×ACFAB×(( AB+ 1 sinθ ) h AB sinθ)=0..... (VIII)

Here, the weight of the gate is w.

Calculation:

Substitute, 2.4m for AB and 2m for AC in Equation (I).

  BC= ( 2.4m )2 ( 2m )2=5.76m4m=1.76m=1.3266m

Substitute, 1000kg/m3 for ρ, 9.81m/s2 for g

  1.3266m2 for A and 1m for x¯ in Equation (II).

  FBC=1000kg/m3×9.81m/s2×1.3266m2×1m=9810kg/m2s2×1.3266m3=13013.946N

Substitute, 2m for AC and 2.4m for BC in Equation (III).

  sinθ=2m2.4m

Substitute, 2.4m for AB, 2m and sinθ in Equation (VI).

  x¯BC=(1+ 2.4m×( 2m 2.4m ) 2m)=(1m+1m)=2m

Substitute 1000kg/m3 for ρ

  9.81m/s2 for g

  2.4m2 for A and 2m for x¯AB in Equation (IV).

  FAB=1000kg/m3×9.81m/s2×2.4m2×2m=9810kg/m2s2×4.8m3=47088N

Substitute 2.4m for 1m for W in Equation (VII).

  I=2.4m× ( 1m )312=0.2m4

Substitute 2m for x¯

  0.2m4 for I

  2.4m2 for A and sinθ from above in Equation (V).

  hAB=2m+0.2m42.4m2×2m×( 2 2.4)2=2m+0.0289m=2.0289m

Substitute, 1.3266m for BC

  100N for w, 47088N for FAB, 2.4m for AB

  2.0289m for hAB in Equation (VIII).

  {13013.946N× 1.3266m 2+100N×1+F×2m47088N×( ( 2.4m+ 1 ( 2m/ 2.4m ) ) 2.0289m ( 2m/ 2.4m ) )}=0672.55Nm+100N+2Fm47088N(1.16532m)=046140.437N+2F=0F=23070.21N

Conclusion:

The force needed to keep the gate at its position for unit width is 23070.21N.

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Chapter 3 Solutions

Fluid Mechanics: Fundamentals and Applications

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