Procedimiento De Mecanica De Fluido.docx

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2) 𝑀 = π‘šπ‘” 𝑀2=(0.769)(9.8)

𝑀2 = 0.75362𝑁

𝑀3 = (0.2689)(9.8) 𝑀4 = (0.377)(9.8) 𝐹𝑝 = (𝜌 βˆ— 𝑔 βˆ— π‘Ž) (

𝑀3 = 2.63522𝑁 𝑀4 = 3.6946𝑁

𝑦2 ) 2 (100

𝐹𝑝 =

𝐾𝑔 9.8π‘š ) ( 2 ) (0.075π‘š)(0.052 ) π‘š3 𝑠 2

Puntualizando la carga β„Ž β†’ 1.67π‘π‘š 3 3) 𝑀2 𝑀𝑏 = 0 𝑀𝐿 βˆ’ 𝐹𝑝 𝐿𝑝 = 0 (0.75362𝑁)(0.30π‘š) βˆ’ 𝐹𝑝 (0.2283π‘š) = 0 𝐹𝑝 =

0.226086 0.2283

𝐹𝑝 = 0.99030𝑁

4) 𝑀3 𝐹𝑝 = (𝜌 βˆ— 𝑔 βˆ— π‘Ž)

𝑦2 2

(100π‘˜π‘”/π‘š3 )(9.8 π‘š/𝑠2 )(0,75π‘š)(0.1π‘š)2 2

𝐹𝑝 = (

𝑀𝐡 = 0 𝑀𝐿 βˆ’ 𝐹𝑝 𝐿𝑝 = 0 (2.63522)(0.30) βˆ’ 𝐹𝑝 (0.2117) = 0 𝐹𝑝 =

0.790566 0.2117

𝐹𝑝 = 3.734𝑁 5) 𝐹𝑝 = (𝜌 βˆ— 𝑔 βˆ— π‘Ž)

𝑦2 2

)

𝐹𝑝 = 3.675𝑁

𝐹𝑝 = 0.91875𝑁

𝐹𝑝 = (

(1000π‘˜π‘”/π‘š3 )(9.8 π‘š/𝑠 2 )(0,75π‘š)(0.124π‘š)2 ) 2

Puntualizando la carga β„Ž 0.124π‘š = = 0.0413π‘š 3 3

Σ𝑀𝑏 = 0 𝑀𝐿 βˆ’ 𝐹𝑝 𝐿𝑝 = 0 (3.6949𝑁)(0.30) βˆ’ 𝐹𝑝 (0.2037π‘š) = 0 1.10838 = 𝐹𝑝 0.20376

𝐹𝑝 = 5.439𝑁

𝐹𝑝 = 5.650𝑁

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