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Question Number 131012 by mohammad17 last updated on 31/Jan/21

Commented by mohammad17 last updated on 31/Jan/21

help me sir

$${help}\:{me}\:{sir} \\ $$

Answered by mr W last updated on 31/Jan/21

Commented by mohammad17 last updated on 31/Jan/21

can you complete the solution sir

$${can}\:{you}\:{complete}\:{the}\:{solution}\:{sir} \\ $$

Commented by mr W last updated on 31/Jan/21

b=width of gate=1.5m  F_1 =((ρgbH^2 )/2)  F_2 =ρgbHL  the gate begins to open when  F_1 ×(H/3)≥F_2 ×(L/2)  ((ρgbH^2 )/2)×(H/3)≥ρgbHL×(L/2)  ⇒H≥(√3)L=20(√3)≈34.6 cm  ⇒h≥100+34.6=134.6 cm

$${b}={width}\:{of}\:{gate}=\mathrm{1}.\mathrm{5}{m} \\ $$$${F}_{\mathrm{1}} =\frac{\rho{gbH}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${F}_{\mathrm{2}} =\rho{gbHL} \\ $$$${the}\:{gate}\:{begins}\:{to}\:{open}\:{when} \\ $$$${F}_{\mathrm{1}} ×\frac{{H}}{\mathrm{3}}\geqslant{F}_{\mathrm{2}} ×\frac{{L}}{\mathrm{2}} \\ $$$$\frac{\rho{gbH}^{\mathrm{2}} }{\mathrm{2}}×\frac{{H}}{\mathrm{3}}\geqslant\rho{gbHL}×\frac{{L}}{\mathrm{2}} \\ $$$$\Rightarrow{H}\geqslant\sqrt{\mathrm{3}}{L}=\mathrm{20}\sqrt{\mathrm{3}}\approx\mathrm{34}.\mathrm{6}\:{cm} \\ $$$$\Rightarrow{h}\geqslant\mathrm{100}+\mathrm{34}.\mathrm{6}=\mathrm{134}.\mathrm{6}\:{cm} \\ $$

Commented by mohammad17 last updated on 31/Jan/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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