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Water and liquid is filled up behind a s...


Water and liquid is filled up behind a square wall of side `l`
Find out (a). Pressure at A, B and C
(b). Forces in part AB and BC
(c). Total force and point of application of force. (Neglect atmosphere pressure in every calculation.)

Text Solution

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(a). As there is no liquid above 'A'
So pressure at `A,p_(A)=0`
pressure at `B,p_(B)=rhogh_(1)`
Pressure at `C,p_(C)=rhogh_(1)+2rhogh_(2)`
(b). Force at A=0
Take strip of width dx at a depth x in part AB
Pressure is equal to `rhogx`
force on strip=pressure`xx`area
`dF=rhogxldx`
Total force upto `B:Funderset(0)overset(h_(1))intrhogxldx=(rhogxlh_(1)^(2))/(2)=(1000xx10xx10xx5xx5)/(2)=1.25xx10^(6)N`
In part BC for force take a elementary strip of width dx in portion BC.
pressure is equal to `=rhogh_(1)+2rhog(x-h_(1))`
force on elementary strip=pressure`xx`area `dF=[rhogh_(1)+2rhog(x-h_(1))]ldx`
Total force on part BC `F=int_(h_(1))^(l)[rhogh_(1)+2rhog(x-h_(1))]ldx`
`=[rhogh_(1)x+2rhog[(x^(2))/(2)-h_(1)x]]_(h_(1))^(l)l`
`=rhogh_(1)h_(2)l+2rhol[(l^(2)-h_(1)^(2))/(2)-h_(1)l+h_(1)^(2)]`
`=rhogh_(1)h_(2)l+(2rhogl)/(2)[l^(2)+h_(1)^(2)-2h_(1)l]=rhogh_(1)h_(2)l+rhogl(l-h_(1))^(2)`
`=rhogh_(2)l[h_(1)+h_(2)]=rhogh_(2)l^(2)=1000xx10xx5xx10xx10=5xx10^(6)N`
(c). Total force
Taking torque about A
Total torque of force in `AB=intdF*x=int_(0)^(h_(1))rhogxldx.x`
`=[(rhoghlx^(3))/(3)]_(0)^(h_(1))=(rhoglh_(1)^(3))/(3)=(1000xx10xx10xx125)/(3)=(1.25xx10^(7))/(3)N-m`
Total torque of force in BC `=intdF*x`
On solving we get `=rhogh_(1)h_(2)l[h_(1)+(h_(2))/(2)]+rhogh_(2)^(2)l[h_(1)+(2h_(2))/(3)]`
`=1000xx10xx5xx5xx10[5+2.5]+1000xx10xx25xx10[5+(10)/(3)]`
`=2.5xx7.5xx10^(6)+(62.5)/(3)xx10^(6)=(118.75)/(3)xx10^(6)`
total torque `=(11.875xx10^(7))/(3)+(1.25xx10^(7))/(3)=(13.125xx10^(7))/(3)`
total torque =total `xx` distance of point of application of force from top `=F.x_(p)`
`6.25xx10^(6)x_(p)=(13.125xx10^(7))/(3)`
`x_(p)=7m`
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