estatica

1. 𝐴𝑦
𝐿𝑦
𝐴𝑥
𝑀𝐴 = 0
− 1 𝑘𝑁 6 𝑚 − 2 𝑘𝑁 4 𝑚 − 2 𝑘𝑁 2 𝑚 + 𝐿𝑦 12 𝑚 = 0
−6 𝑘𝑁 ∗ 𝑚 − 8 𝑘𝑁 ∗ 𝑚 − 4 𝑘𝑁 ∗ 𝑚 + 𝐿𝑦 12 𝑚 = 0
−18 𝑘𝑁 ∗ 𝑚 + 𝐿𝑦 12 𝑚 = 0
𝐿𝑦 12 𝑚 = 18 𝑘𝑁 ∗ 𝑚
𝐿𝑦 =
18 𝑘𝑁 ∗ 𝑚
12 𝑚
= 1.5 𝑘𝑁
𝑓𝑦 = 0
𝐴𝑦 − 1 − 2 − 2 − 1 + 1.5 = 0
𝐴𝑦 − 4.5 = 0
𝐴𝑦 = 4.5 𝑘𝑁
A_x=0

2. 4.5 𝑘𝑁
1.5 𝑘𝑁
4.5 𝑘𝑛
1 𝑘𝑛
𝐹𝐴𝐵
𝐹𝐴𝐶
𝐹𝐴𝐵
𝐹𝐴𝐶
2 m
2 m
𝑐𝑜𝑠 =
𝑎
ℎ
=
2
2.8
= 𝐹𝑥(𝐹𝐴𝐵)
𝑠𝑒𝑛 =
𝑜
ℎ
=
2
2.8
= 𝐹𝑦(𝐹𝐴𝐵)
2 m
1 m
𝑐𝑜𝑠 =
𝑎
ℎ
=
2
2.24
= 𝐹𝑥(𝐹𝐴𝐶)
𝑠𝑒𝑛 =
𝑜
ℎ
=
1
2.24
= 𝐹𝑦(𝐹𝐴𝐶)
𝐹𝑌 = 0
𝐹𝐴𝐵 ∗
2
2.8
+ 𝐹𝐴𝐶 ∗
1
2.24
+ 4.5 𝑘𝑁 − 1 𝑘𝑁 = 0
2
2.8
𝐹𝐴𝐵 +
1
2.24
𝐹𝐴𝐶 = 1 𝑘𝑁 − 4.5 𝑘𝑁
2
2.8
𝐹𝐴𝐵 +
1
2.24
𝐹𝐴𝐶 = −3.5 𝑘𝑁 1)

3. 𝐹𝑥 = 0
2
2.8
𝐹𝐴𝐵 +
2
2.24
𝐹𝐴𝐶 = 0
2
2.8
𝐹𝐴𝐵 +
1
2.24
𝐹𝐴𝐶 = −3.5 𝑘𝑁 1)
2) Se va a resolver por el
método de eliminación
−1 ∗
2
2.8
𝐹𝐴𝐵 +
2
2.24
𝐹𝐴𝐶 = 0 = −
2
2.8
𝐹𝐴𝐵 −
2
2.24
𝐹𝐴𝐶 = 0
2
2.8
𝐹𝐴𝐵 +
1
2.24
𝐹𝐴𝐶 = −3.5 𝑘𝑁
−0.45𝐹𝐴𝐶 = −3.5 𝐾𝑛
𝐹𝐴𝐶 = −
3.5 𝐾𝑛
−0.45
= 7.8 𝐾𝑛

4. 2
2.8
𝐹𝐴𝐵 +
2
2.24
𝐹𝐴𝐶 = 0
𝐹𝐴𝐶 = −
3.5 𝐾𝑛
−0.45
= 7.8 𝐾𝑛
2
2.8
𝐹𝐴𝐵 +
2
2.24
(7.8 𝑘𝑁) = 0
2
2.8
𝐹𝐴𝐵 + 7 𝑘𝑁 = 0
𝐹𝐴𝐵 =
(−7 𝑘𝑁)(2.8)
2
= −9.9 𝑘𝑁