Figure shows a vertical wall of horizontal length L = 5.0 m and depth H = 3.0 m that is fully submerged in water , with its top surface of the water . What is the force on the wall due to the water pressure ?
Figure shows a vertical wall of horizontal length L = 5.0 m and depth H = 3.0 m that is fully submerged in water , with its top surface of the water . What is the force on the wall due to the water pressure ?
Text Solution
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Let y represent the depth below the top water surface . Then the gauge pressure is
`p_(g) = rho gy " " (14-25)`
Note that because the expression contains a variable (namely , depth y) , we cannot use p = F/A to find the force on the wall because there is no one value of pressure . However , if we pick a small area dA on the wall Fig , we can relate the pressure p and force dF on the wall by
`p_(g) = (dF)/(dA) " " (14-26)`
(We use the "d" notation to indicate that we are dealing with tiny portions and differential amounts.) Equation 14-26 tells us that the force on dA is
`dF = p_(g) dA " " (14-27)`
For area dA, we could choose a small square at some depth y . However , note that the variable in Eq 14-25 is y and not x . Thus , the pressure at a given depth y is the same at any horizontal position along the wall . So , we can choose dA to be the area of a strip of length L and tiny height dy . That is ,
dA = Ldy `" " (14-28)`
Then the force on that strip is
`dF = p_(g) L dy, `
or , with Eq. 14-25 ,
`dF = rho gy L dy " " (14-29)`
To find the total force on the wall , we need to add the forces on all the horizontal strips from top to bottom of the wall . Of course , because each strip has the tiny height dy , there are a huge number of strips along the wall and adding up their forces would take "for ever" (much longer than our patience) . However , we can add them via integration rather quickly . Integrating both sides of Eq. 14-29 , we find
`int dF = int rho gy L dy " " (14-30)`
We want to include all the strips between y = 0 (at the top of the wall) and y = 3. 0 m (at the bottom of the wall) . To do so , we use those values as the limits to the integral on the right side of Eq. 14-30 . The integration on the left side yields the total force F we seek . We find
`F = rho g L int_(0)^(3.0 m) y dy `
`= rho g L (1)/(2) y^(2) |_(0)^(3.0 m)`
` = (1000 kg // m^(3) ) (9.8 m//s^(2)) (5.0 m) (1)/(2) (3.0 m)^(2)`
`2.21 xx 10^(5) N`.
`p_(g) = rho gy " " (14-25)`
Note that because the expression contains a variable (namely , depth y) , we cannot use p = F/A to find the force on the wall because there is no one value of pressure . However , if we pick a small area dA on the wall Fig , we can relate the pressure p and force dF on the wall by
`p_(g) = (dF)/(dA) " " (14-26)`
(We use the "d" notation to indicate that we are dealing with tiny portions and differential amounts.) Equation 14-26 tells us that the force on dA is
`dF = p_(g) dA " " (14-27)`
For area dA, we could choose a small square at some depth y . However , note that the variable in Eq 14-25 is y and not x . Thus , the pressure at a given depth y is the same at any horizontal position along the wall . So , we can choose dA to be the area of a strip of length L and tiny height dy . That is ,
dA = Ldy `" " (14-28)`
Then the force on that strip is
`dF = p_(g) L dy, `
or , with Eq. 14-25 ,
`dF = rho gy L dy " " (14-29)`
To find the total force on the wall , we need to add the forces on all the horizontal strips from top to bottom of the wall . Of course , because each strip has the tiny height dy , there are a huge number of strips along the wall and adding up their forces would take "for ever" (much longer than our patience) . However , we can add them via integration rather quickly . Integrating both sides of Eq. 14-29 , we find
`int dF = int rho gy L dy " " (14-30)`
We want to include all the strips between y = 0 (at the top of the wall) and y = 3. 0 m (at the bottom of the wall) . To do so , we use those values as the limits to the integral on the right side of Eq. 14-30 . The integration on the left side yields the total force F we seek . We find
`F = rho g L int_(0)^(3.0 m) y dy `
`= rho g L (1)/(2) y^(2) |_(0)^(3.0 m)`
` = (1000 kg // m^(3) ) (9.8 m//s^(2)) (5.0 m) (1)/(2) (3.0 m)^(2)`
`2.21 xx 10^(5) N`.
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