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Question Number 125858 by bramlexs22 last updated on 14/Dec/20

Commented by mr W last updated on 14/Dec/20

13×26×2×62.4×1=42 182 ft−lb

$$\mathrm{13}×\mathrm{26}×\mathrm{2}×\mathrm{62}.\mathrm{4}×\mathrm{1}=\mathrm{42}\:\mathrm{182}\:{ft}−{lb} \\ $$

Commented by bramlexs22 last updated on 14/Dec/20

sir how with integral method?

$${sir}\:{how}\:{with}\:{integral}\:{method}? \\ $$

Commented by mr W last updated on 14/Dec/20

dW=ρAhdh  W=ρA∫_0 ^H hdh=((ρAH^2 )/2)=ρV(H/2)  =62.4×13×26×2×(2/2)=42 182 ft−lb

$${dW}=\rho{Ahdh} \\ $$$${W}=\rho{A}\int_{\mathrm{0}} ^{{H}} {hdh}=\frac{\rho{AH}^{\mathrm{2}} }{\mathrm{2}}=\rho{V}\frac{{H}}{\mathrm{2}} \\ $$$$=\mathrm{62}.\mathrm{4}×\mathrm{13}×\mathrm{26}×\mathrm{2}×\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{42}\:\mathrm{182}\:{ft}−{lb} \\ $$

Commented by bramlexs22 last updated on 14/Dec/20

thank you sir

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

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