Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 15, Problem 8P

(a)

To determine

The force exerted by the water on the bottom.

(a)

Expert Solution
Check Mark

Answer to Problem 8P

The force exerted by the water on the bottom is 5.88×106 N, down.

Explanation of Solution

Write the equation for pressure on the bottom due to water.

  Pb=FbA

Here, Pb is the pressure on the bottom due to water, Fb is the force exerted by the water on the bottom and A is the area of the bottom.

Rewrite the above equation for Fb.

  Fb=PbA        (I)

Write the equation for the pressure on the bottom due to water.

  Pb=ρgh        (II)

Here, ρ is the density of water, g is the acceleration due to gravity and h is the depth of water.

Conclusion:

Substitute 1000 kg/m3 for ρ , 9.80 m/s2 for g and 2.00 m for h in equation (II) to find Pb .

  Pb=(1000 kg/m3)(9.80 m/s2)(2.00 m)=1.96×104 Pa

Substitute 1.96×104 Pa for Pb and (30.0 m)(10.0 m) for A in equation (I) to determine Fb .

  Fb=(1.96×104 Pa)(30.0 m)(10.0 m)=5.88×106 N, down

Therefore, the force exerted by the water on the bottom is 5.88×106 N, down.

(b)

To determine

The force exerted by the water on each end.

(b)

Expert Solution
Check Mark

Answer to Problem 8P

The force exerted by the water on each end is 196 kN, outward.

Explanation of Solution

Pressure varies with depth.

Write the equation for the force on a strip of height dz and length L.

  dF=PdA        (III)

Here, dF is the force on a strip of height dz and length L , P is the pressure on the strip and dA is the area of the strip.

Write the equation for P.

  P=ρgz

Write the equation for dA.

  dA=Ldz

Here, L is the length of the strip and dz is the height.

Put the above two equations in equation (III).

  dF=ρgzLdz=ρgLzdz

Integrate the above equation over the entire height of the strip.

  0hdF=0hρgLzdzF=ρgL[z22]0h=12ρgLh2=12ρghLh        (IV)

Write the equation for the total area.

  A=Lh        (V)

The pressure at the top of the pool is zero.

Write the equation for the pressure at surface of the pool.

  Pt=0        (VI)

Here, Pt is the pressure at surface of the pool.

Write the equation for the average pressure.

  Pavg=Pt+Pb2

Here, Pavg is the average pressure.

Put equations (II) and (VI) in the above equation.

  Pavg=0+ρgh2=12ρgh        (VII)

Put equations (V) and (VII) in equation (IV).

  F=PavgA        (VIII)

Put equation (II) in equation (VII).

  Pavg=12Pb        (IX)

Conclusion:

Substitute 1.96×104 Pa for Pb in equation (IX) to find Pavg .

  Pavg=12(1.96×104 Pa)=9.80×103 Pa

The length of each end will be 10.0 m.

Substitute 10.0 m for L and 2.00 m for h in equation (V) to find A .

  A=(10.0 m)(2.00 m)=20.0 m2

Substitute 9.80×103 Pa for Pavg and 20.0 m2 for A in equation (VIII) to find F.

  F=(9.80×103 Pa)(20.0 m2)=196×103 N1 kN1000 N=196 kN

The force will be directed outward.

Therefore, the force exerted by the water on each end is 196 kN, outward.

(c)

To determine

The force exerted by the water on each side.

(c)

Expert Solution
Check Mark

Answer to Problem 8P

The force exerted by the water on each side is 588 kN, outward.

Explanation of Solution

Equation (VIII) gives the force exerted on each side of the swimming pool by water.

Conclusion:

The length of each side will be 30.0 m .

Substitute 30.0 m for L and 2.00 m for h in equation (V) to find A.

  A=(30.0 m)(2.00 m)=60.0 m2

Substitute 9.80×103 Pa for Pavg and 60.0 m2 for A in equation (VIII) to find F.

  F=(9.80×103 Pa)(60.0 m2)=588×103 N1 kN1000 N=588 kN

The force will be directed outward.

Therefore, the force exerted by the water on each side is 588 kN, outward.

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Chapter 15 Solutions

Principles of Physics: A Calculus-Based Text

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