Water and liquid are filled up behind a square wall of side. Find out
a. pressures at `A, B` and `C`
b. forces in part `AB` and `BC`
c. total force and point of application of force (neglect atmospheric pressure in every calculation)
Water and liquid are filled up behind a square wall of side. Find out
a. pressures at `A, B` and `C`
b. forces in part `AB` and `BC`
c. total force and point of application of force (neglect atmospheric pressure in every calculation)
a. pressures at `A, B` and `C`
b. forces in part `AB` and `BC`
c. total force and point of application of force (neglect atmospheric pressure in every calculation)
Text Solution
Verified by Experts
As there is no liquid above `A`
pressure at `A` is `p_(A)=0`
Pressure at `B` is `p_(B)=rhogh_(1)`
pressure at `C` is `p_(C)=rhogh_(1)+2rhogh_(2)`
b. Force at `A` is zero. Take a strip a of width `dx` at a depth `'x'` in part `AB`. Pressure is equal to `rhogx`.
Force on strip `=` pressure `xx` area
`dF=rhogldx`
Total force up to `B`,
`F_(B)=int_(0)^(h)rhog ldx=(rhoglh_(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 givenby
`rhogh_(1)+2rhog(x-h_(1))`
Force on the elementary strip `=` pressure `xx` area
`:.dF=[rhogh_(1)+2rhog(x-h_(1))]ldx`
The limit of `x` will be `x=h_(1)` to `x=h_(1)+h_(2)=l`
The force on part `BC` is `F=int_(h_(1))^(h_(1)+h_(2))[rhogh_(1)=2rhog(x-h-10]l`
But `h_(1)+h-2=l=[rhogh_(1)x+2rhog[(x^(2))/2-h_(1)x]]_(h_(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. Method 1
Total `=5xx10^(6)+1.25xx10^(6)=6.25xx10^(6)N`
Taking torque about `A` the total torque of force in `AB` is
`intdx=int_(0)^(F)rhogxldxx`
`[(rhoglx^(3))/3]_(0)^(h_(1))=(rhoglh_(1)^(3))/3=(1000xx10xx10xx125)/3`
`=(1.25xx10^(7))/3Nm`
The total torque of force in `BC` about `A` is given by
`intdFx=int_(R)^(l)[rhogh_(1)+2rhog(x-h_(1))]ldx.x`
`rhogh_(1)h_(2)l[h_(1)+(h_(2))/2]+rhogh_(2)^(2)l[h_(1)+(2h_(2))/3]`
`=2.5xx7.5xx10^(6)+62.5/3xx10^(6)`
`118.75/3xx10^(6)Nm`
Total torque `=(11.875xx10^(7))/3+(1.25xx10^(7))/3`
`=(13.125xx10^(7))/3Nm`
Total torque `=` Total force `xx` Distance of point of application of force from top `= Fx_(p)`
`implies 6.25xx10^(6)x_(p)=(13.125xx10^(7))/3Nm`
`implies x_(p)=7m`
Method 2. Pressure diagram on the wall
Total force on the wall per unit width of the wall `=` Area of presure diagram.
`F/b=1/2(rhogh_(1))h_(1)+rhogh_(1)h_(2)+1/2(2rhogh_(2))h_(2)`
`=rhog[(h_(1)^(2))/2+h_(2)^(2)]+rhogh_(1)h_(2)`
`F_(del)=F=1/2rhog_(1)^(2)+rhoglh_(2)[h_(1)+h_(2)]`
`6.25xx10^(6)N`
Total force on `AB` part `F_(AB)=1/2rhoglh_(1)=1.25xx10^(6)N`
Total force on `BC` part `F_(BC)=rhoglh_(2)(h_(1)+h_(2))=5xx10^(6)N` The total force will act at the centroid of the pressure diagram. Let the distance of centroid of pressure diagram be `'r'` from `A`
`r=(rhogh_(1)^(2))/2xx2/3h_(1)+(2rhogh_(2)^(2))/2(h_(1)+2/3h_(2)) +(rhogh_(1)h_(2))(h_(1)+h_(2)/2))/((rhogh_(1)^(20)/2+rhoh_(2)^(2)+rhogh_(1)h_(2))`
`=((h_(1)^(3))/3+h_(2)^(2)(h_(1)+2/3h_(2))+h_(1)h_(2)(h_(1)+(h_(2))/2))/((h_(1)^(2))/2+h_(2)^(2)+h_(1)^(2))`
`r=([1/3+(1+2/3)+(1+1/2)]5)/(1/2+1+1)=(7/2)/(5/2)xx5=7m`
pressure at `A` is `p_(A)=0`
Pressure at `B` is `p_(B)=rhogh_(1)`
pressure at `C` is `p_(C)=rhogh_(1)+2rhogh_(2)`
b. Force at `A` is zero. Take a strip a of width `dx` at a depth `'x'` in part `AB`. Pressure is equal to `rhogx`.
Force on strip `=` pressure `xx` area
`dF=rhogldx`
Total force up to `B`,
`F_(B)=int_(0)^(h)rhog ldx=(rhoglh_(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 givenby
`rhogh_(1)+2rhog(x-h_(1))`
Force on the elementary strip `=` pressure `xx` area
`:.dF=[rhogh_(1)+2rhog(x-h_(1))]ldx`
The limit of `x` will be `x=h_(1)` to `x=h_(1)+h_(2)=l`
The force on part `BC` is `F=int_(h_(1))^(h_(1)+h_(2))[rhogh_(1)=2rhog(x-h-10]l`
But `h_(1)+h-2=l=[rhogh_(1)x+2rhog[(x^(2))/2-h_(1)x]]_(h_(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. Method 1
Total `=5xx10^(6)+1.25xx10^(6)=6.25xx10^(6)N`
Taking torque about `A` the total torque of force in `AB` is
`intdx=int_(0)^(F)rhogxldxx`
`[(rhoglx^(3))/3]_(0)^(h_(1))=(rhoglh_(1)^(3))/3=(1000xx10xx10xx125)/3`
`=(1.25xx10^(7))/3Nm`
The total torque of force in `BC` about `A` is given by
`intdFx=int_(R)^(l)[rhogh_(1)+2rhog(x-h_(1))]ldx.x`
`rhogh_(1)h_(2)l[h_(1)+(h_(2))/2]+rhogh_(2)^(2)l[h_(1)+(2h_(2))/3]`
`=2.5xx7.5xx10^(6)+62.5/3xx10^(6)`
`118.75/3xx10^(6)Nm`
Total torque `=(11.875xx10^(7))/3+(1.25xx10^(7))/3`
`=(13.125xx10^(7))/3Nm`
Total torque `=` Total force `xx` Distance of point of application of force from top `= Fx_(p)`
`implies 6.25xx10^(6)x_(p)=(13.125xx10^(7))/3Nm`
`implies x_(p)=7m`
Method 2. Pressure diagram on the wall
Total force on the wall per unit width of the wall `=` Area of presure diagram.
`F/b=1/2(rhogh_(1))h_(1)+rhogh_(1)h_(2)+1/2(2rhogh_(2))h_(2)`
`=rhog[(h_(1)^(2))/2+h_(2)^(2)]+rhogh_(1)h_(2)`
`F_(del)=F=1/2rhog_(1)^(2)+rhoglh_(2)[h_(1)+h_(2)]`
`6.25xx10^(6)N`
Total force on `AB` part `F_(AB)=1/2rhoglh_(1)=1.25xx10^(6)N`
Total force on `BC` part `F_(BC)=rhoglh_(2)(h_(1)+h_(2))=5xx10^(6)N` The total force will act at the centroid of the pressure diagram. Let the distance of centroid of pressure diagram be `'r'` from `A`
`r=(rhogh_(1)^(2))/2xx2/3h_(1)+(2rhogh_(2)^(2))/2(h_(1)+2/3h_(2)) +(rhogh_(1)h_(2))(h_(1)+h_(2)/2))/((rhogh_(1)^(20)/2+rhoh_(2)^(2)+rhogh_(1)h_(2))`
`=((h_(1)^(3))/3+h_(2)^(2)(h_(1)+2/3h_(2))+h_(1)h_(2)(h_(1)+(h_(2))/2))/((h_(1)^(2))/2+h_(2)^(2)+h_(1)^(2))`
`r=([1/3+(1+2/3)+(1+1/2)]5)/(1/2+1+1)=(7/2)/(5/2)xx5=7m`
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