Fr.docx
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+
∑(𝐹𝑅 )𝑥 = ∑ 𝐹𝑅(𝑋)
→
(𝐹𝑅) = 26 ( 𝑥
5 ) − 30 sin 30 13
(𝐹𝑅) = 5 𝑘𝑁 𝑥
+
∑(𝐹𝑅 )𝑌 = ∑ 𝐹𝑅(𝑌)
↑
(𝐹𝑅) = −26 ( 𝑌
12 ) − 30 cos 30 13
(𝐹𝑅) = 49,96 𝑘𝑁 𝑦
la magnitud de la fuerza resultante : 𝐹𝑅 = √(𝐹𝑅 (𝑌) )2 + (𝐹𝑅 (𝑥) )2 𝐹𝑅 = √(5)2 + (49,98)2 𝐹𝑅 = 50,52 𝑘𝑁 El angulo de la fuerza resultante: 𝐹𝑅 (𝑌) 𝜃 = 𝑡𝑎𝑛−1 ( ) 𝐹𝑅 (𝑥) 49,98 𝜃 = 𝑡𝑎𝑛−1 ( ) 5 𝜃 = 84, 30 Momento de la fuerza resultante : +(𝑀𝑅 )𝐴=∑ 𝑀𝐴 (𝑀𝑅 )𝐴 = 30 sin 30 (0,3) − 30 cos 30 (2) − 6 (
5 12 ) (0,3) − 26 ( ) 6 − 45 13 13
(𝑀𝑅 )𝐴 = −239,46 𝑘𝑁. 𝑚 → 239 𝑘𝑁. 𝑚~𝑔𝑖𝑟𝑜 ℎ𝑜𝑟𝑎𝑟𝑖𝑜
∑(𝐹𝑅 )𝑥 = ∑ 𝐹𝑅(𝑋)
→
(𝐹𝑅) = 26 ( 𝑥
5 ) − 30 sin 30 13
(𝐹𝑅) = 5 𝑘𝑁 𝑥
+
∑(𝐹𝑅 )𝑌 = ∑ 𝐹𝑅(𝑌)
↑
(𝐹𝑅) = −26 ( 𝑌
12 ) − 30 cos 30 13
(𝐹𝑅) = 49,96 𝑘𝑁 𝑦
la magnitud de la fuerza resultante : 𝐹𝑅 = √(𝐹𝑅 (𝑌) )2 + (𝐹𝑅 (𝑥) )2 𝐹𝑅 = √(5)2 + (49,98)2 𝐹𝑅 = 50,52 𝑘𝑁 El angulo de la fuerza resultante: 𝐹𝑅 (𝑌) 𝜃 = 𝑡𝑎𝑛−1 ( ) 𝐹𝑅 (𝑥) 49,98 𝜃 = 𝑡𝑎𝑛−1 ( ) 5 𝜃 = 84, 30 Momento de la fuerza resultante : +(𝑀𝑅 )𝐴=∑ 𝑀𝐴 (𝑀𝑅 )𝐴 = 30 sin 30 (0,3) − 30 cos 30 (2) − 6 (
5 12 ) (0,3) − 26 ( ) 6 − 45 13 13
(𝑀𝑅 )𝐴 = −239,46 𝑘𝑁. 𝑚 → 239 𝑘𝑁. 𝑚~𝑔𝑖𝑟𝑜 ℎ𝑜𝑟𝑎𝑟𝑖𝑜
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