Chapter 13, Problem 68A

(a)

To determine

Pressure at the base of the dam.

Answer to Problem 68A

  $1.7\times {10}^5\text{}Pa$

Explanation of Solution

Given: Height of the reservoir, $h=17m$

Formula used:

Pressure at the depth h can be obtained by:

  $P=\rho hg$

where, $\rho =1.00\times {10}^{3}kg/m^3$ is the density of water and $g=9.8N/kg$ is the gravitational acceleration.

Calculation: The pressure P at the base of the dam is:

  $\begin{array}{l}P=(1.00\times {10}^{3}kg/m^3)(17m)(9.8N/kg)\\ P=1.7\times {10}^5\text{}Pa\end{array}$

Conclusion: Hence, pressure of reservoir at the base of the dam is $1.7\times {10}^5\text{}Pa$

(b)

To determine

Pressure 4.0 m from the top of the dam.

Answer to Problem 68A

  $3.9\times {10}^4\text{}Pa$

Explanation of Solution

Given: Height, $h=4.0m$

Formula used:

Pressure at the depth h can be obtained by:

  $P=\rho hg$

where, $\rho =1.00\times {10}^{3}kg/m^3$ is the density of water and $g=9.8N/kg$ is the gravitational acceleration.

Calculation: The pressure $4.0m$ from the top of the dam is:

  $\begin{array}{l}P=(1.00\times {10}^{3}kg/m^3)(4.0m)(9.8N/kg)\\ P=3.9\times {10}^4\text{}Pa\end{array}$

Conclusion: The pressure $4.0m$ from the top of the dam is $3.9\times {10}^4\text{}Pa$ .