Vectores y Fuerzas Fisica 2014
Vectores y Fuerzas Fisica 2014
Vectores y Fuerzas Fisica 2014
TEMA:
CENTRO DE GRAVEDAD, CINEMÁTICA,
DINÁMICA, TRABAJO
CURSO : FÍSICA I
DOCENTE : Msc. MANUEL ESTEVES PAIRAZAMAN
ALUMNO
CICLO : II
SECCIÓN :B
UNIVERSIDAD NACIONAL DE HUANCAVELICA
EJEMPLO Nº 01:
Determinar el ángulo para conectar el elemento a la placa tal que la resultante
de las fuerzas FA Y FB esté dirigida horizontalmente a la derecha. Determine
además la magnitud de la fuerza resultante.
⃗⃗⃗⃗
𝐹𝐵 = 6𝐾𝑁 𝑠𝑒𝑛 40° 𝑖 − 6𝐾𝑁 cos 40° 𝑗
⃗⃗⃗⃗
𝐹𝑅 = (8𝐾𝑁 𝑠𝑒𝑛 𝜃 + 6𝐾𝑁 𝑠𝑒𝑛 40°)𝑖 + (8𝐾𝑁 cos 𝜃 − 6𝑁𝐾 cos 40°) 𝑗
Ángulo
8𝐾𝑁 𝑐𝑜𝑠 𝜃 − 6𝐾𝑁 𝑐𝑜𝑠 40° = 0
8𝐾𝑁 𝑐𝑜𝑠 𝜃 = 6𝐾𝑁 𝑐𝑜𝑠 40°
𝜃 = 54.93°
⃗⃗⃗⃗
𝐹𝐵 = 6𝐾𝑁 𝑠𝑒𝑛 40° 𝑖 − 6𝐾𝑁 cos 40° 𝑗
EJEMPLO:
UNIVERSIDAD NACIONAL DE HUANCAVELICA
Calcule las componentes horizontales y verticales de las fuerzas mostradas en
la figura:
8
tan 𝜃 = 15
8
𝜃 = tan−1 ( )
15
𝜃 = 28,07°
𝐹=𝐹 ⃗⃗⃗⃗𝑋 + 𝐹
⃗⃗⃗⃗𝑌
𝐹 = 𝐹𝑋 𝑖 + 𝐹𝑋 𝑗
𝐹 = 170𝑁 𝐶𝑜𝑠 28,07°𝑖 + 170𝑁 𝑆𝑒𝑛 28,07°𝑗
𝐹 = 150 𝑁𝑖 + 79,99 𝑁𝑗
tan 𝛼 = (
3√3
) 𝛼 + 90 + 𝛽 = 180°
6
40,89° + 90 + 𝛽 = 180°
3√3
𝛼 = tan−1 ( ) 130,89° + 𝛽 = 180°
6
𝛽 = 49,11°
𝛼 = 40,89°
⃗ = −𝑇
𝑇 ⃗⃗⃗⃗𝑋 − ⃗⃗⃗⃗
𝑇𝑌
⃗ = −𝑇𝑋𝑖 − 𝑇𝑋𝑗
𝑇
⃗ = −600𝑁 𝑆𝑒𝑛 49,11°𝑖 − 600𝑁 𝐶𝑜𝑠 49,11°𝑗
𝑇
⃗ = −453,58𝑁𝑖 − 392,77𝑁𝑗
𝑇
⃗⃗⃗⃗
𝐹𝑋 = 800𝑖𝑁
⃗⃗⃗⃗
𝐹𝑅 = (800 − 453.58)𝑁𝑖 − 392.77𝑁𝑗
⃗⃗⃗⃗
𝐹𝑅 = 346.42𝑁𝑖 − 392.77𝑁𝑗
⃗⃗⃗
𝐹2 = −800𝑁 𝑆𝑒𝑛 60°𝑖 + 800𝑁 𝐶𝑜𝑠 60°𝑗
√3 1
⃗⃗⃗
𝐹2 = −800𝑁 𝑆𝑒𝑛 𝑖 + 800𝑁 𝐶𝑜𝑠 𝑗
2 2
⃗⃗⃗
𝐹2 = −400√3𝑁 𝑖 + 400𝑁 𝑗
⃗⃗⃗
𝐹3 = −450𝑁 𝑆𝑒𝑛 75°𝑖 + 450𝑁 𝐶𝑜𝑠 75°𝑗
⃗⃗⃗
𝐹3 = −434,67𝑁𝑖 + 116,47𝑁𝑗
UNIVERSIDAD NACIONAL DE HUANCAVELICA
Magnitud:
⃗⃗⃗⃗𝑅 | = √(−703,23)2 + (707,79)2 𝑁
|𝐹
⃗⃗⃗⃗𝑅 | = 997,75𝑁
|𝐹
Dirección:
𝑒 (−703,23𝑖 + 707,79𝑗)𝑁
𝑒= =
|𝑒| 997,75𝑁
−703,23
𝜃𝑋 = cos−1 ( 997,75 ) = 134,81°
707,79
𝜃𝑌 = cos −1 (997,75) = 44,82°
3. Expresar la fuerza F de 36 kN en función de los vectores unitarios i, j y k.
Hallar la proyección sobre el eje x.
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 𝑒
𝐹𝑅 = |𝐹|
𝑆 ⃗ )𝑚
(6𝑖 + 12𝑗 + 12𝑘
𝑒= =
|𝑆| √62 + 122 + 122 𝑚
⃗
6𝑖 + 12𝑗 + 12𝑘
𝑒=
18
𝐹 = ⃗⃗⃗⃗⃗
|𝐹| 𝑒
⃗
6𝑖 + 12𝑗 + 12𝑘
𝐹 = 36𝑁 ( )
18
⃗
𝐹 = 12𝑁𝑖 + 24𝑁𝑗 + 24𝑁𝑘
⃗
7𝑖 − 6𝑗 + 6𝑘
𝑒=
11
⃗⃗⃗⃗⃗
𝐹 = |𝐹| 𝑒
⃗
7𝑖 − 6𝑗 + 6𝑘
𝐹 = 110𝑁 ( )
11
⃗ )𝑁
𝐹 = (70𝑖 − 60𝑗 + 60𝑘
a) b)
∑⃗⃗𝐹𝑋 = 0 ∑⃗⃗𝐹𝑋 = 0
∑𝐹 = 0 { } ∑𝐹 = 0 { }
∑ ⃗⃗𝐹𝑌 =0 ∑ ⃗⃗𝐹𝑌 =0
∑ 𝐹𝑋 = 0 ∑ 𝐹𝑋 = 0
𝑇𝐴 − 𝑇𝐶 𝐶𝑜𝑠 10° = 0 𝑇𝐵 𝑆𝑒𝑛 20° − 𝑇𝐴 = 0
𝑇𝐴
𝑇𝐵 =
𝑆𝑒𝑛 20°
𝑇𝐴 − 𝑇𝐶 𝐶𝑜𝑠 10° … … … … 1
2835,65𝑁
∴ 𝑇𝐵 = = 8290,89𝑁
𝑆𝑒𝑛 20°
∑ 𝐹𝑌 = 0
𝑇𝐶 𝑆𝑒𝑛 10° − 𝑤1 = 0 ∑ 𝐹𝑌 = 0 𝑇𝑂 = 𝑤2
500 𝑁 𝑇𝐵 𝐶𝑜𝑠 20° − 𝑤2 = 0
𝑇𝐶 =
𝑆𝑒𝑛 10° 8290,89𝑁 𝐶𝑜𝑠20° = 𝑤2
∴ 𝑇𝐶 = 2879,39 𝑁 … … … 2 ∴ 𝑇𝑂 = 𝑤2 = 7790,89 𝑁
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨 𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑎 ⃗ )𝑚
(1,2𝑖 − 1,5𝑗 + 2,4𝑘 𝑏⃗ ⃗ )𝑚
(−1,8𝑖 − 2,1𝑗 + 2,1𝑘
𝑒𝐴 =
⃗⃗⃗ = 𝑒𝐵 =
⃗⃗⃗⃗ =
|𝑎| √(1,2)2 + (−1,5)2 + (2,4)2 𝑚 |𝑏⃗| √(−1,8)2 + (−2,1)2 + (2,1)2 𝑚
⃗
−1,8𝑖 − 2,1𝑗 + 2,1𝑘
⃗
1,2𝑖 − 1,5𝑗 + 2,4𝑘 𝑒𝐵 =
⃗⃗⃗⃗
𝑒𝐴 =
⃗⃗⃗ √12,06
√9,45
𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨
−1,8𝑖 − 2,1𝑗 + 2,1𝑘⃗
⃗
1,2𝑖 − 1,5𝑗 + 2,4𝑘 ⃗⃗⃗⃗
𝑇𝐵 = 𝑇𝐵 ( )
⃗⃗⃗⃗
𝑇𝐴 = 𝑇𝐴 ( ) √12,06
√9,45 ⃗
⃗⃗⃗⃗
𝑇𝐵 = −0,52𝑇𝐵 𝑖 − 0,60𝑇𝐵 𝑗 + 0,60𝑇𝐵 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,39𝑇𝐴 𝑖 − 0,49𝑇𝐴 𝑗 + 0,78𝑇𝐴 𝑘
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
𝑐 ⃗ )𝑚
(−1,2𝑖 + 1,8𝑗 + 0,9𝑘
𝑒𝐶 =
⃗⃗⃗⃗ =
|𝑐 | √(−1,2)2 + (1,8)2 + (0,9)2 𝑚
⃗
−1,2𝑖 + 1,8𝑗 + 0,9𝑘
𝑒𝐶 =
⃗⃗⃗⃗
√5,49
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
⃗
−1,2𝑖 + 1,8𝑗 + 0,9𝑘
⃗⃗⃗⃗
𝑇𝐶 = 𝑇𝐶 ( )
√5,49
⃗⃗⃗⃗ ⃗
𝑇𝐶 = −0,51𝑇𝐶 𝑖 + 0,77𝑇𝐶 𝑗 + 0,38𝑇𝐶 𝑘
⃗
⃗⃗⃗ = −𝒘 𝒆
𝒘 ⃗⃗⃗ = −𝟓𝟎𝟎𝑵 ⃗𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗
𝑻𝑨 + ⃗⃗⃗⃗⃗
𝑻𝑩 + ⃗⃗⃗⃗
𝑻𝑪 + 𝒘 = 𝟎
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,39𝑇𝐴 𝑖 − 0,49𝑇𝐴 𝑗 + 0,78𝑇𝐴 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,52𝑇𝐵 𝑖 − 0,60𝑇𝐵 𝑗 + 0,60𝑇𝐵 𝑘
UNIVERSIDAD NACIONAL DE HUANCAVELICA
⃗⃗⃗⃗ ⃗
𝑇𝐶 = −0,51𝑇𝐶 𝑖 + 0,77𝑇𝐶 𝑗 + 0,38𝑇𝐶 𝑘
𝑤 ⃗
⃗⃗ = 0𝑖 + 0𝑗 − 500𝑁𝑘
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨 𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑎 ⃗ )𝑚
(1,2𝑖 − 0,9𝑗 + 1,8𝑘 𝑏⃗ ⃗ )𝑚
(−1,2𝑖 − 1,8𝑗 + 1,2𝑘
𝑒𝐴 =
⃗⃗⃗ = 𝑒𝐵 =
⃗⃗⃗⃗ =
|𝑎| √(1,2)2 + (−0,9)2 + (1,8)2 𝑚 |𝑏⃗| √(−1,2)2 + (−1,8)2 + (1,2)2 𝑚
⃗
−1,2𝑖 − 1,8𝑗 + 1,2𝑘
⃗
1,2𝑖 − 0,9𝑗 + 1,8𝑘 𝑒𝐵 =
⃗⃗⃗⃗
𝑒𝐴 =
⃗⃗⃗ √6,12
√4,38
𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨
−1,2𝑖 − 1,8𝑗 + 1,2𝑘⃗
⃗
1,2𝑖 − 0,9𝑗 + 1,8𝑘 ⃗⃗⃗⃗
𝑇𝐵 = 𝑇𝐵 ( )
⃗⃗⃗⃗
𝑇𝐴 = 𝑇𝐴 ( ) √6,12
√5,49
⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,49𝑇𝐵 𝑖 − 0,73𝑇𝐵 𝑗 + 0,49𝑇𝐵 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,51𝑇𝐴 𝑖 − 0,38𝑇𝐴 𝑗 + 0,77𝑇𝐴 𝑘
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
𝑐 ⃗ )𝑚
(0𝑖 + 1,5𝑗 + 0𝑘
𝑒𝐶 =
⃗⃗⃗⃗ =
|𝑐 | √(0)2 + (1,5)2 + (0)2 𝑚
0𝑖 + 1,5𝑗 + 0𝑘⃗
𝑒𝐶 =
⃗⃗⃗⃗
1,5
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑻𝑪 = |𝑻 𝑪 | ⃗⃗⃗⃗
𝒆𝑪
⃗
0𝑖 + 1,5𝑗 + 0𝑘
⃗⃗⃗⃗
𝑇𝐶 = 𝑇𝐶 ( )
1,5
⃗⃗⃗⃗ ⃗
𝑇𝐶 = 0𝑇𝐶 𝑖 + 1𝑇𝐶 𝑗 + 0𝑇𝐶 𝑘
𝒘 ⃗
⃗⃗⃗ = −𝒘 𝒆 ⃗
⃗⃗⃗ = −𝟏𝟓𝒌𝑵 𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗
𝑻𝑨 + ⃗⃗⃗⃗⃗
𝑻𝑩 + ⃗⃗⃗⃗
𝑻𝑪 + 𝒘 = 𝟎
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,51𝑇𝐴 𝑖 − 0,38𝑇𝐴 𝑗 + 0,77𝑇𝐴 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,49𝑇𝐵 𝑖 − 0,73𝑇𝐵 𝑗 + 0,49𝑇𝐵 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐶 = 0𝑇𝐶 𝑖 + 1𝑇𝐶 𝑗 + 0𝑇𝐶 𝑘
𝑤 ⃗
⃗⃗ = 0𝑖 + 0𝑗 − 15𝑘𝑁𝑘
0,51 0 0 0,51 0
−0,38 0 1 −0,38 0
| 0,77 15 0 | 0,77 15 −7,65
𝑇𝐵 = = = 10,76𝑘𝑁
| 0,51 −0,49 0| 0,51 −0,49 −0,6272
−0,38 −0,73 1 −0,38 −0,73
0,77 0,49 0 0,77 0,49
⃗
−4𝑖 − 1𝑗 − 10𝑘
𝑒𝐴 =
⃗⃗⃗
√117
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨
⃗
−4𝑖 − 1𝑗 − 10𝑘
⃗⃗⃗⃗
𝑇𝐴 = 𝑇𝐴 ( )
√117
⃗⃗⃗⃗ ⃗
𝑇𝐴 = −0,37𝑇𝐴 𝑖 − 0,09𝑇𝐴 𝑗 − 0,92𝑇𝐴 𝑘
𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩 𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
𝑏⃗ ⃗ )𝑚
(2𝑖 + 2𝑗 − 10𝑘 𝑐 ⃗ )𝑚
(3𝑖 − 2𝑗 − 10𝑘
𝑒𝐵 =
⃗⃗⃗⃗ = 𝑒𝐶 =
⃗⃗⃗⃗ =
|𝑏⃗| √(2)2 + (2)2 + (10)2 𝑚 |𝑐| √(3)2 + (2)2 + (−10)2 𝑚
⃗
2𝑖 + 2𝑗 − 10𝑘 ⃗
3𝑖 − 2𝑗 − 10𝑘
𝑒𝐵 =
⃗⃗⃗⃗ 𝑒𝐶 =
⃗⃗⃗⃗
√108 √113
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑻𝑩 = |𝑻 𝑩 | ⃗⃗⃗⃗
𝒆𝑩 𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
UNIVERSIDAD NACIONAL DE HUANCAVELICA
⃗
2𝑖 + 2𝑗 − 10𝑘 3𝑖 − 2𝑗 − 10𝑘⃗
⃗⃗⃗⃗
𝑇𝐵 = 𝑇𝐵 ( ) ⃗⃗⃗⃗
𝑇𝐶 = 𝑇𝐶 ( )
√108 √113
⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,19𝑇𝐵 𝑖 + 0,19𝑇𝐵 𝑗 − 0,96𝑇𝐵 𝑘 ⃗⃗⃗⃗ ⃗
𝑇𝐶 = 0,28𝑇𝐶 𝑖 − 0,19𝑇𝐶 𝑗 − 0,94𝑇𝐶 𝑘
𝒘 ⃗
⃗⃗⃗ = −𝒘 𝒆 ⃗
⃗⃗⃗ = 𝟐𝟓𝒌𝑵 𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗
𝑻𝑨 + ⃗⃗⃗⃗⃗
𝑻𝑩 + ⃗⃗⃗⃗
𝑻𝑪 + 𝒘 = 𝟎
⃗⃗⃗⃗ ⃗
𝑇𝐴 = −0,37𝑇𝐴 𝑖 − 0,09𝑇𝐴 𝑗 − 0,92𝑇𝐴 𝑘
⃗⃗⃗⃗
𝑇𝐵 = −0,19𝑇𝐵 𝑖 + 0,19𝑇𝐵 𝑗 − 0,96𝑇𝐵 𝑘 ⃗
⃗⃗⃗⃗ ⃗
𝑇𝐶 = 0,28𝑇𝐶 𝑖 − 0,19𝑇𝐶 𝑗 − 0,94𝑇𝐶 𝑘
𝑤 ⃗
⃗⃗ = 0𝑖 + 0𝑗 + 25𝑘𝑁𝑘
−0,37 0 0 −0,37 0
−0,09 0 1 −0,09 0
| −0,92 −25 0 | −0,92 −25 1,96
𝑇𝐵 = = = 5,31𝑘𝑁
| 0,51 0,19 0,28 | −0,37 0,19 0,223844
−0,38 0,19 −0,19 −0,09 0,19
0,77 −0,96 −0,94 −0,92 −0,96
9. El disco circular de la figura pesa 2,5 kN. Determinar las tensiones de los cables
A, B y C.
UNIVERSIDAD NACIONAL DE HUANCAVELICA
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨 𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑎 ⃗ )𝑚
(1,3𝑖 − 1𝑗 + 1,3𝑘 𝑏⃗ ⃗ )𝑚
(−1𝑖 − 1𝑗 + 2𝑘
𝑒𝐴 =
⃗⃗⃗ = 𝑒𝐵 =
⃗⃗⃗⃗ =
|𝑎| √(1,3)2 + (−1)2 + (1,3)2 𝑚 |𝑏⃗| √(−1)2 + (−1)2 + (2)2 𝑚
⃗
−1𝑖 − 1𝑗 + 2𝑘
⃗
1,3𝑖 − 1𝑗 + 1,3𝑘 𝑒𝐵 =
⃗⃗⃗⃗
𝑒𝐴 =
⃗⃗⃗ √6
√4,38
𝑻 ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗𝑩 = |𝑻𝑩 | 𝒆 ⃗⃗⃗⃗𝑩
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨
−1𝑖 − 1𝑗 + 2𝑘⃗
⃗
1,3𝑖 − 1𝑗 + 1,3𝑘 ⃗⃗⃗⃗
𝑇𝐵 = 𝑇𝐵 ( )
⃗⃗⃗⃗
𝑇𝐴 = 𝑇𝐴 ( ) √6
√4,38
⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,41𝑇𝐵 𝑖 − 0,41𝑇𝐵 𝑗 + 0,82𝑇𝐵 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,62𝑇𝐴 𝑖 − 0,48𝑇𝐴 𝑗 + 0,62𝑇𝐴 𝑘
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗
𝒆𝑪
𝑐 ⃗ )𝑚
(−1,3𝑖 + 1𝑗 + 1,3𝑘
𝑒𝐶 =
⃗⃗⃗⃗ =
|𝑐 | √(−1,3)2 + (1)2 + (1,3)2 𝑚
⃗
−1,3𝑖 + 1𝑗 + 1,3𝑘
𝑒𝐶 =
⃗⃗⃗⃗
√4,38
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
−1,3𝑖 + 1𝑗 + 1,3𝑘⃗
⃗⃗⃗⃗
𝑇𝐶 = 𝑇𝐶 ( )
√4,38
⃗⃗⃗⃗ ⃗
𝑇𝐶 = −0,62𝑇𝐶 𝑖 + 0,48𝑇𝐶 𝑗 + 0,62𝑇𝐶 𝑘
𝒘 ⃗
⃗⃗⃗ = −𝒘 𝒆 ⃗
⃗⃗⃗ = −𝟐, 𝟓𝒌𝑵 𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗
𝑻𝑨 + ⃗⃗⃗⃗⃗
𝑻𝑩 + ⃗⃗⃗⃗
𝑻𝑪 + 𝒘 = 𝟎
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,62𝑇𝐴 𝑖 − 0,48𝑇𝐴 𝑗 + 0,62𝑇𝐴 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,41𝑇𝐵 𝑖 − 0,41𝑇𝐵 𝑗 + 0,82𝑇𝐵 𝑘
UNIVERSIDAD NACIONAL DE HUANCAVELICA
⃗⃗⃗⃗ ⃗
𝑇𝐶 = −0,62𝑇𝐶 𝑖 + 0,48𝑇𝐶 𝑗 + 0,62𝑇𝐶 𝑘
𝑤 ⃗
⃗⃗ = 0𝑖 + 0𝑗 − 2,5𝑘𝑁𝑘
10. El fardo representado en la figura tiene una masa de 500 kg. Determinar las
tensiones de los cables A, B y C utilizados para soportarlos.
UNIVERSIDAD NACIONAL DE HUANCAVELICA
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨 𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑎 ⃗ )𝑚
(2𝑖 − 3𝑗 + 8𝑘 𝑏⃗ ⃗ )𝑚
(−4𝑖 − 3𝑗 + 8𝑘
𝑒𝐴 =
⃗⃗⃗ = 𝑒𝐵 =
⃗⃗⃗⃗ =
|𝑎| √(2)2 + (−3)2 + (8)2 𝑚 |𝑏⃗| √(−4)2 + (−3)2 + (8)2 𝑚
⃗)
(−4𝑖 − 3𝑗 + 8𝑘
⃗
2𝑖 − 3𝑗 + 8𝑘 𝑒𝐵 =
⃗⃗⃗⃗
𝑒𝐴 =
⃗⃗⃗ √89
√77
𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨
⃗)
(−4𝑖 − 3𝑗 + 8𝑘
⃗
2𝑖 − 3𝑗 + 8𝑘 ⃗⃗⃗⃗
𝑇𝐵 = 𝑇𝐵 ( )
⃗⃗⃗⃗
𝑇𝐴 = 𝑇𝐴 ( ) √89
√77
⃗⃗⃗⃗ ⃗ ⃗⃗⃗⃗ ⃗
𝑇𝐵 = −0,42𝑇𝐵 𝑖 − 0,32𝑇𝐵 𝑗 + 0,85𝑇𝐵 𝑘
𝑇𝐴 = 0,23𝑇𝐴 𝑖 − 0,34𝑇𝐴 𝑗 + 0,91𝑇𝐴 𝑘
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
𝑐 ⃗ )𝑚
(0𝑖 + 3𝑗 + 8𝑘
𝑒𝐶 =
⃗⃗⃗⃗ =
|𝑐 | √(0)2 + (3)2 + (8)2 𝑚
⃗
0𝑖 + 3𝑗 + 8𝑘
𝑒𝐶 =
⃗⃗⃗⃗
√73
𝑻 ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗𝑪 = |𝑻𝑪 | 𝒆 ⃗⃗⃗⃗𝑪
0𝑖 + 3𝑗 + 8𝑘⃗
⃗⃗⃗⃗
𝑇𝐶 = 𝑇𝐶 ( )
√73
⃗⃗⃗⃗
𝑇𝐶 = 0𝑇𝐶 𝑖 + 1𝑇𝐶 𝑗 + 0𝑇𝐶 𝑘⃗
𝒘 ⃗
⃗⃗⃗ = −𝒘 𝒆 ⃗
⃗⃗⃗ = −𝟓𝟎𝟎 𝑲𝒈 𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗
𝑻𝑨 + 𝑻 ⃗⃗⃗⃗⃗𝑩 + ⃗⃗⃗⃗
𝑻𝑪 + 𝒘 = 𝟎
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,23𝑇𝐴 𝑖 − 0,34𝑇𝐴 𝑗 + 0,91𝑇𝐴 𝑘
⃗⃗⃗⃗
𝑇𝐵 = −0,42𝑇𝐵 𝑖 − 0,32𝑇𝐵 𝑗 + 0,85𝑇𝐵 𝑘 ⃗
⃗⃗⃗⃗ ⃗
𝑇𝐶 = 0𝑇𝐶 𝑖 + 1𝑇𝐶 𝑗 + 0𝑇𝐶 𝑘
𝑤
⃗⃗ = 0𝑖 + 0𝑗 − 500𝐾𝑔 𝑘 ⃗
0 −0,42 0 0 −0,42
0 −0,32 1 0 −0,32
| 500 0,85 |
0 500 0,85 −210 −210
𝑇𝐴 = = = = 321,002𝐾𝑔
| 0,23 −0,42 0| 0,32 −0,42 −0,382 − 0,272 −0,6542
−0,34 −0,32 1 −0,34 −0,32
0,91 0,85 0 0,91 0,85
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0,23 0 0 0,23 0
−0,34 0 1 −0,34 0
| 0,91 500 0 | 0,91 500 −115𝐾𝑔
𝑇𝐵 = = = 175,78𝐾𝑔
| 0,23 −0,42 0| 0,32 −0,42 −0,6542
−0,34 −0,32 1 −0,34 −0,32
0,91 0,85 0 0,91 0,85
11. Un cilindro homogéneo de acero que pesa 500 N pende de un hilo flexible y se
apoya sobre un plano inclinado liso en la forma que se indica en la figura.
Determinar tensión T del hilo y la fuerza R que ercer el plano inclinado sobre el
cilindro.
∑ 𝐹𝑥 = 0 𝑌 ∑ 𝐹𝑦 = 0
+ → ∑ 𝐹𝑥 = 𝑇𝐴 − 𝑇𝐶 𝐶𝑜𝑠 10° = 0
12. Una esfera homogénea que pesa 50N se apoya sobre dos planos lisos que
forman una V según se indica en la figura. Determinar las fuerzas que dichos
planos ejercen sobre la esfera en los puntos de contacto A y B.
∑ 𝐹𝑥 = 0 ∑ 𝐹𝑦 = 0
𝐹𝑁𝐴 𝐶𝑜𝑠 45° − 𝐹𝑁𝐵 𝐶𝑜𝑠 60° = 0 𝐹𝑁𝐵 𝑆𝑒𝑛60° + 𝐹𝑁𝐴 𝑆𝑒𝑛 45° − 𝑤 = 0
1 1
𝐹𝑁𝐴 = 𝐹𝑁𝐵
√2 2 √3 √2 1
1 𝐹𝑁𝐵 + 𝐹𝑁𝐵 ( ) = 50
𝐹𝑁𝐴 = 𝐹𝑁𝐵 √2 2 2 √2
2
√3 1
√2 𝐹𝑁𝐵 = + 𝐹𝑁𝐵 = 50
𝐹𝑁𝐴 = 𝐹𝑁𝐵 ∙∙ ⋯ ⋯ ⋯ ⋯ 𝟏 2 2
2
Reemplazando 2 en 1 √3 + 1
𝐹𝑁𝐵 = = 50
2
√2 2
𝐹𝑁𝐴 = 𝐹𝑁𝐵 𝐹𝑁𝐵 = 50 ( )
2
√3 + 1
√2
𝐹𝑁𝐴 = 36,60 ( )
2
∴ 𝐹𝑁𝐵 = 36,60 𝑁 … … … . . 𝟐
∴ 𝐹𝑁𝐴 = 25,88 𝑁
UNIVERSIDAD NACIONAL DE HUANCAVELICA
∑ 𝐹𝑥 = 0 ∑ 𝐹𝑦 = 0
𝐹𝑁 𝐶𝑜𝑠 60° − 𝑇 = 0 𝐹𝑁 𝑆𝑒𝑛 60° − 𝑤 = 0
√3
1 𝐹𝑁 = 200
𝐹𝑁 =𝑇 2
2 2
𝐹𝑁 = 200 ( )
𝐹𝑁 = 𝑇(2) ⋯ ⋯ ⋯ (𝟏) √3
400
𝐹𝑁 = 𝑁 ⋯ ⋯ ⋯ (𝟐)
√3
Reemplazando 2 en 1
𝐹𝑁 = 𝑇(2)
400
=𝑇
√3
1 400 200
2
( 3) = 𝑇 ∴ 3𝑁=𝑇
√ √
VECTORES SOLUCIONARIO
UNIVERSIDAD NACIONAL DE HUANCAVELICA
2. Es posible aplicar a un cuerpo simultáneamente una fuerza de 6 kN y otra
de 8 kN de modo que produzcan el mismo efecto de una sola fuerza.
Determinar la magnitud de dicha fuerza (kN).
Por propiedad:
𝑹=𝑭
⃗𝑨 = −𝟐𝒋 ⃗𝑩
⃗ = 𝟐𝒋 ⃗𝑪 = 𝟒𝒋
𝑅⃗ = 𝐵 + 𝐶 − 𝐴
𝑅⃗ = 4𝑖 + 4𝑗
𝑅⃗ = 10√2
∴ 𝑅⃗ = 10√2
(4𝑖−3𝑗) (4𝑖−3𝑗) 𝟒𝒊 𝟑𝒊
𝑒= 𝑒= ⃗ =
𝒆 −
√(4)2 +(−3)2 5 𝟓 𝟓
̅ = 𝟐𝒋 ; 𝑩
𝑺𝒊 𝑨 ̅ = ⃗⃗⃗⃗
̅ = 𝟒𝒋 − 𝟑𝒊 ; 𝑪 𝟐𝒊
𝐴+𝐵 ⃗ + 𝐶 = 2𝑗 + 4𝑗 − 3𝑖 + 2𝑖
𝐴+𝐵 ⃗ + 𝐶 = −𝑖 + 6𝑗
𝐴+𝐵 ⃗ + 𝐶 = √(−1)2 + (6)2
⃗⃗ + 𝑩
𝑨 ⃗ = √𝟑𝟕
⃗⃗ + 𝑪
Solución:
⃗𝑩⃗ + ⃗𝑬
⃗ =𝟎
⃗ + ⃗𝑪 = 𝟏𝟎𝒖
⃗𝑩
Solución:
𝐷𝐵 + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ 𝐹𝐺 + ⃗⃗⃗⃗⃗
𝐷𝐸 = 0
⃗⃗⃗⃗⃗⃗ = −2𝑏𝑖 + 𝑎𝑗
𝐷𝐵
⃗⃗⃗⃗⃗ = −𝑎𝑖
𝐹𝐺
⃗⃗⃗⃗⃗
𝐷𝐸 = −2𝑏𝑖 + 𝑎𝑖 − 𝑎𝑖
⃗⃗⃗⃗⃗
𝐷𝐸 = −2𝑏𝑖
⃗⃗⃗⃗⃗⃗
𝑭 𝒙𝒚 = 450𝑁 𝐶𝑜𝑠 62°𝑖 + 450𝑁 𝑆𝑒𝑛 62°𝑗
⃗⃗⃗⃗⃗⃗
𝑭 𝒙𝒚 = 211,26𝑁°𝑖 + 397,33𝑁 𝑗
En el sistema x’ y’
a) Los ángulos x , y , z. ⃗
⃗ = 12,86𝑘𝑁𝑖 + 17,15𝑘𝑁𝑗 + 12,86𝑘𝑁𝑘
𝑭
⃗ = |𝐹 | 𝑒
𝑭 ⃗ 3
𝜃𝑥 = cos −1 ( ) = 59,04°
⃗ )𝑚
(3𝑖 + 4𝑗 + 3𝑘 √34
𝐹
⃗ =
𝒆 = 3
|𝐹 |
√32 + 42 + 32 𝑚 𝜃𝑦 = cos −1 ( ) = 46,69°
√34
⃗
3𝑖 + 4𝑗 + 3𝑘 3
⃗ =
𝒆 𝜃𝑧 = cos −1 ( ) = 59,04°
√34 √34
⃗
3𝑖 + 4𝑗 + 3𝑘
⃗𝑭 = 𝟐𝟓𝒌𝑵 ( )
√34
Forma vectorial de R:
⃗⃗⃗⃗⃗
𝑭𝑹 = ⃗⃗⃗⃗
𝑭𝟏 + ⃗⃗⃗⃗
𝑭𝟐 + ⃗⃗⃗⃗
𝑭𝟑
⃗
⃗⃗⃗1 = −25𝑘𝑁 𝐶𝑜𝑠26° 𝑆𝑒𝑛30°𝑖 + 25𝑘𝑁 𝐶𝑜𝑠26° 𝐶𝑜𝑠30°𝑗 + 25𝑘𝑁 𝑆𝑒𝑛26°𝑘
𝐹
⃗⃗⃗
𝐹 = −11,23𝑘𝑁𝑖 + 19,46𝑘𝑁𝑗 + 10,96𝑘𝑁𝑘 ⃗
1
⃗⃗⃗ ⃗
𝐹2 = 10𝑘𝑁 𝐶𝑜𝑠60° 𝐶𝑜𝑠60°𝑖 − 10𝑘𝑁 𝑆𝑒𝑛60° 𝐶𝑜𝑠60°𝑗 + 10𝑘𝑁 𝑆𝑒𝑛60°𝑘
1 1 √3 1 √3
⃗⃗⃗
𝐹2 = 10𝑘𝑁 ∙ 𝑖 − 10𝑘𝑁 ∙ 𝑗 + 10𝑘𝑁 ⃗𝑘
2 2 2 2 2
1 √3 ⃗
⃗⃗⃗
𝐹2 = 10𝑘𝑁 𝑖 − 10𝑘𝑁 𝑗 + 5√3𝑘𝑁𝑘
4 4
⃗⃗⃗
𝐹2 = 2,5𝑘𝑁𝑖 − 2,5√3𝑘𝑁𝑗 + 5√3𝑘𝑁𝑘 ⃗
⃗⃗⃗ ⃗
𝐹3 = 15𝑘𝑁 𝐶𝑜𝑠16° 𝐶𝑜𝑠50°𝑖 + 15𝑘𝑁 𝐶𝑜𝑠16° 𝑆𝑒𝑛 50°𝑗 + 15𝑘𝑁 𝑆𝑒𝑛16°𝑘
⃗⃗⃗
𝐹2 = 9,27𝑘𝑁𝑖 + 11,05𝑘𝑁𝑗 + 4,13𝑘𝑁𝑘⃗
⃗⃗⃗⃗
𝐹𝑅 = 0,54𝑘𝑁𝑖 + 26,18𝑘𝑁𝑗 + 23,75𝑘𝑁𝑘 ⃗
Los ángulos x , y , z .
0,54 26,18 23,75 ⃗
𝑒=( 𝑖+ 𝑗+ 𝑘)
35,35 35,35 35,35
0,54
𝜃𝑥 = cos −1 ( ) = 89,12°
35,35
UNIVERSIDAD NACIONAL DE HUANCAVELICA
26,18
𝜃𝑦 = cos−1 ( ) = 42,21°
35,35
23,75
𝜃𝑧 = cos −1 ( ) = 47,79°
35,35
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨 𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑎 ⃗ )𝑚
(1,2𝑖 − 1,5𝑗 + 2,4𝑘 𝑏⃗ ⃗ )𝑚
(−1,8𝑖 − 2,1𝑗 + 2,1𝑘
𝑒𝐴 =
⃗⃗⃗ = 𝑒𝐵 =
⃗⃗⃗⃗ =
|𝑎| √(1,2)2 + (−1,5)2 + (2,4)2 𝑚 |𝑏⃗| √(−1,8)2 + (−2,1)2 + (2,1)2 𝑚
⃗
−1,8𝑖 − 2,1𝑗 + 2,1𝑘
⃗
1,2𝑖 − 1,5𝑗 + 2,4𝑘 𝑒𝐵 =
⃗⃗⃗⃗
𝑒𝐴 =
⃗⃗⃗ √12,06
√9,45
𝑻𝑩 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |𝑻𝑩 | ⃗⃗⃗⃗
𝒆𝑩
𝑻𝑨 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑨 | ⃗⃗⃗⃗𝒆𝑨
−1,8𝑖 − 2,1𝑗 + 2,1𝑘⃗
⃗
1,2𝑖 − 1,5𝑗 + 2,4𝑘 ⃗⃗⃗⃗
𝑇𝐵 = 𝑇𝐵 ( )
⃗⃗⃗⃗
𝑇𝐴 = 𝑇𝐴 ( ) √12,06
√9,45 ⃗
⃗⃗⃗⃗
𝑇𝐵 = −0,52𝑇𝐵 𝑖 − 0,60𝑇𝐵 𝑗 + 0,60𝑇𝐵 𝑘
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,39𝑇𝐴 𝑖 − 0,49𝑇𝐴 𝑗 + 0,78𝑇𝐴 𝑘
𝑻𝑪 = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
𝑐 ⃗ )𝑚
(−1,2𝑖 + 1,8𝑗 + 0,9𝑘
𝑒𝐶 =
⃗⃗⃗⃗ =
|𝑐 | √(−1,2)2 + (1,8)2 + (0,9)2 𝑚
−1,2𝑖 + 1,8𝑗 + 0,9𝑘 ⃗
𝑒𝐶 =
⃗⃗⃗⃗
√5,49
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑻𝑪 = |𝑻𝑪 | ⃗⃗⃗⃗𝒆𝑪
UNIVERSIDAD NACIONAL DE HUANCAVELICA
−1,2𝑖 + 1,8𝑗 + 0,9𝑘⃗
⃗⃗⃗⃗
𝑇𝐶 = 𝑇𝐶 ( )
√5,49
⃗⃗⃗⃗
𝑇𝐶 = −0,51𝑇𝐶 𝑖 + 0,77𝑇𝐶 𝑗 + 0,38𝑇𝐶 𝑘⃗
⃗⃗⃗ = −𝒘 𝒆
𝒘 ⃗ ⃗⃗⃗ = −𝟓𝟎𝟎𝑵 ⃗𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗
𝑻𝑨 + ⃗⃗⃗⃗⃗
𝑻𝑩 + ⃗⃗⃗⃗
𝑻𝑪 + 𝒘 = 𝟎
⃗⃗⃗⃗ ⃗
𝑇𝐴 = 0,39𝑇𝐴 𝑖 − 0,49𝑇𝐴 𝑗 + 0,78𝑇𝐴 𝑘
⃗⃗⃗⃗
𝑇𝐵 = −0,52𝑇𝐵 𝑖 − 0,60𝑇𝐵 𝑗 + 0,60𝑇𝐵 𝑘⃗
⃗⃗⃗⃗ ⃗
𝑇𝐶 = −0,51𝑇𝐶 𝑖 + 0,77𝑇𝐶 𝑗 + 0,38𝑇𝐶 𝑘
𝑤
⃗⃗ = 0𝑖 + 0𝑗 − 500𝑁𝑘⃗
⃗⃗⃗⃗⃗⃗⃗⃗
𝑴𝑨𝟏 = ⃗⃗⃗⃗⃗⃗
𝑹𝟏 × ⃗⃗⃗⃗
𝑭𝟏
𝑟1 = (6𝑖 + 3𝑗)𝑚
⃗⃗⃗
⃗⃗⃗
𝐹1 = 600𝑁 𝐶𝑜𝑠45°𝑖 + 600𝑁 𝑆𝑒𝑛 45°𝑗
1 √2
⃗⃗⃗
𝐹1 = 600𝑁 𝑖 + 600𝑁 𝑗
√2 2
⃗⃗⃗
𝐹1 = 300√2𝑁𝑖 + 300√2𝑁𝑗
𝑖 𝑗 ⃗
𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑀𝐴1 = | 6 3 0| 𝑚𝑁
300√2 300√2 0
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐴1 = 𝑂𝑚𝑁 𝑖 + 𝑂𝑚𝑁 𝑗 + (2545,58 − 1272,79)𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐴1 = 1272,79𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗⃗ 𝒓𝟐 × ⃗⃗⃗⃗
𝑴𝑨𝟐 = ⃗⃗⃗⃗ 𝑭𝟐
𝑟2 = 12𝑚𝑖
⃗⃗⃗
⃗⃗⃗
𝐹2 = 750𝑁 𝐶𝑜𝑠60°𝑖 + 750𝑁 𝑆𝑒𝑛 60°𝑗
1 √3
⃗⃗⃗
𝐹2 = 750𝑁 𝑖 + 750𝑁 𝑗
2 2
⃗⃗⃗
𝐹2 = 375𝑁𝑖 + 375√3𝑁𝑗
𝑖 𝑗 ⃗
𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑀𝐴2 = | 12 0 0| 𝑚𝑁
375 375√3 0
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐴2 = 𝑂𝑚𝑁 𝑖 + 𝑂𝑚𝑁 𝑗 + 7794,23𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐴2 = 17794,23,79𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗⃗ 𝒓𝟑 × ⃗⃗⃗⃗
𝑴𝑨𝟑 = ⃗⃗⃗⃗ 𝑭𝟑
UNIVERSIDAD NACIONAL DE HUANCAVELICA
𝑟3 = (9𝑖 − 3𝑗)𝑚
⃗⃗⃗
⃗⃗⃗
𝐹3 = 900𝑁 𝐶𝑜𝑠30°𝑖 − 900𝑁 𝑆𝑒𝑛 30°𝑗
√3 1
⃗⃗⃗
𝐹2 = 900𝑁 𝑖 − 900𝑁 𝑗
2 2
⃗⃗⃗
𝐹3 = 450√3𝑁𝑖 − 450𝑁𝑗
𝑖 𝑗 𝑘⃗
⃗⃗⃗⃗⃗⃗⃗
𝑀𝐴3 = | 9 −3 0| 𝑚𝑁
450√3 −450 0
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐴3 = 𝑂𝑚𝑁 𝑖 + 𝑂𝑚𝑁 𝑗 + (−4050 + 2338,27)𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐴2 = −1711,73𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗⃗ 𝒓𝟐 × ⃗⃗⃗⃗
𝑴𝑩𝟐 = ⃗⃗⃗⃗ 𝑭𝟐
𝑟2 = −6𝑚𝑖
⃗⃗⃗
⃗⃗⃗
𝐹2 = 750𝑁 𝐶𝑜𝑠60°𝑖 + 750𝑁 𝑆𝑒𝑛 60°𝑗
1 √3
⃗⃗⃗
𝐹2 = 750𝑁 𝑖 + 750𝑁 𝑗
2 2
⃗⃗⃗
𝐹2 = 375𝑁𝑖 + 375√3𝑁𝑗
𝑖 𝑗 ⃗
𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑀𝐵2 = | −6 0 0| 𝑚𝑁
375 375√3 0
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐵2 = 𝑂𝑚𝑁 𝑖 + 𝑂𝑚𝑁 𝑗 − 3897,11𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑀𝐵2 = −3897,11𝑚𝑁 𝑘 ⃗
⃗⃗⃗⃗⃗⃗⃗⃗ 𝒓𝟑 × ⃗⃗⃗⃗
𝑴𝑩𝟑 = ⃗⃗⃗⃗ 𝑭𝟑
UNIVERSIDAD NACIONAL DE HUANCAVELICA
𝑟⃗⃗⃗3 = (−9𝑖 − 3𝑗)𝑚
⃗⃗⃗3 = 900𝑁 𝐶𝑜𝑠30°𝑖 − 900𝑁 𝑆𝑒𝑛 30°𝑗
𝐹
√3 1
⃗⃗⃗
𝐹2 = 900𝑁 𝑖 − 900𝑁 𝑗
2 2
⃗⃗⃗
𝐹3 = 450√3𝑁𝑖 − 450𝑁𝑗
𝑖 𝑗 ⃗
𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑀𝐵3 = | −9 −3 0| 𝑚𝑁
450√3 −450 0
⃗⃗⃗⃗⃗⃗⃗ ⃗
𝑀𝐵3 = 𝑂𝑚𝑁 𝑖 + 𝑂𝑚𝑁 𝑗 + (4050 + 2338,27)𝑚𝑁 𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑀 𝐵3 = 6388,27𝑚𝑁 𝑘
⃗
⃗⃗⃗⃗⃗⃗⃗⃗
𝑀𝑅𝐵 = (−6363,96𝑘 ⃗ − 3897,11𝑘
⃗ + 6388,27𝑘
⃗ )𝑚𝑁
∴ ⃗⃗⃗⃗⃗⃗⃗⃗
𝑀𝑅𝐵 = −3872,8𝑚𝑁𝑘⃗
⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑩 = 𝑻𝑨𝑩 𝒆 𝑨𝑩
⃗⃗⃗⃗⃗⃗
𝑆𝐴𝐵 ⃗ )𝑚
(−16𝑖 + 11𝑗 + 8𝑘
𝑒𝐴𝐵 =
⃗⃗⃗⃗⃗⃗ =
⃗⃗⃗⃗⃗⃗
|𝑆 𝐴𝐵 | √(−16)2 + (11)2 + (8)2 𝑚
⃗
−16𝑖 + 11𝑗 + 8𝑘
𝑒𝐴𝐵 =
⃗⃗⃗⃗⃗⃗
21
⃗
−16𝑖 + 11𝑗 + 8𝑘
⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐵 = 840𝑘𝑔 − 𝑓 ( )
21
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐴𝐵 = −640𝑘𝑔 − 𝑓𝑖 + 440𝑘𝑔 − 𝑓𝑗 + 320𝑘𝑔 − 𝑓𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑪 = 𝑻𝑨𝑪 ⃗⃗⃗⃗⃗⃗
𝒆𝑨𝑪
⃗⃗⃗⃗⃗⃗
𝑆𝐴𝐶 ⃗ )𝑚
(−16𝑖 − 16𝑗 + 8𝑘
𝑒𝐴𝐶 =
⃗⃗⃗⃗⃗⃗ =
⃗⃗⃗⃗⃗⃗
|𝑆 𝐴𝐶 | √(−16)2 + (−16)2 + (8)2 𝑚
⃗
−16𝑖 − 16𝑗 + 8𝑘
𝑒𝐴𝐶 =
⃗⃗⃗⃗⃗⃗
24
⃗
−16𝑖 − 16𝑗 + 8𝑘
⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐶 = 1200𝑘𝑔 − 𝑓 ( )
24
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐴𝐶 = −800𝑘𝑔 − 𝑓𝑖 − 800𝑘𝑔 − 𝑓𝑗 + 400𝑘𝑔 − 𝑓𝑘
⃗⃗⃗⃗
𝑇𝑅 = ⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐵 + ⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐶
⃗⃗⃗⃗ ⃗ )𝑘𝑔 − 𝑓
𝑇𝑅 = (−1440𝑖 − 360𝑗 + 720𝑘
⃗⃗⃗⃗𝑅 | = √(−1440)2 + (−360)2 + (720)2 𝑘𝑔 − 𝑓
∴ 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒏𝒅𝒐 𝑴𝒂𝒈𝒏𝒊𝒕𝒖𝒅: |𝑇
⃗⃗⃗⃗⃗𝑹 | = 𝟏𝟔𝟒𝟗, 𝟕𝟑𝒌𝒈 − 𝒇
|𝑻
∴ 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒏𝒅𝒐 𝑫𝒊𝒓𝒆𝒄𝒄𝒊ó𝒏:
−1440
𝜃𝑥 = cos−1 ( ) = 150,79°
1649.73
−360
𝜃𝑦 = cos −1 ( ) = 102,60°
1649,73
UNIVERSIDAD NACIONAL DE HUANCAVELICA
720
𝜃𝑦 = cos −1 ( ) = 64,12°
1649,73
⃗⃗⃗⃗⃗⃗
𝑴𝑨 = 𝒓 ⃗⃗⃗⃗⃗⃗ ⃗
𝑨𝑪 × 𝑭
𝑨𝑪 = (𝟖𝟎𝒊 + 𝟑𝟎𝟎𝒋)𝒎𝒎
⃗⃗⃗⃗⃗⃗
𝒓
⃗𝑭 = |𝑭
⃗ | ⃗⃗⃗⃗⃗⃗⃗
𝒆𝑪𝑫
⃗⃗⃗⃗⃗⃗
𝑆𝐶𝐷 (−320𝒊 − 𝟑𝟎𝟎𝒋 + 𝟑𝟐𝟎𝒌⃗ )𝑚𝑚
𝑒𝐶𝐷 =
⃗⃗⃗⃗⃗⃗ =
⃗⃗⃗⃗⃗⃗
|𝑆 𝐶𝐷 | √(−320)2 + (300)2 + (320)2 𝑚𝑚
⃗
−320𝑖 − 300𝑗 + 320𝑘
𝑒𝐶𝐷 =
⃗⃗⃗⃗⃗⃗
√294800
⃗
−320𝑖 − 300𝑗 + 320𝑘
𝐹 = 200𝑁 ( )
√294800
⃗
𝐹 = −117,87𝑖 − 110𝑗 + 117,87𝑘
⃗⃗⃗⃗⃗
𝑀𝐴 = −117,87𝑖 − 110𝑗 + 117,87𝑘⃗
𝑖 𝑗 ⃗
𝑘
⃗⃗⃗⃗⃗
𝑀𝐴 = | 80 300 0 | 𝑚𝑚𝑁
−117,87 −110,51 117,87
⃗⃗⃗⃗⃗ ⃗ )𝑚𝑚𝑁
𝑀𝐴 = (35361𝑖 − 9429,6𝑗 + (−8840,8 + 35361)𝑘
⃗⃗⃗⃗⃗ ⃗ )𝑚𝑚𝑁
𝑀𝐴 = (35361𝑖 − 9429,6𝑗 + 26520,2𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑫𝑩 = 𝑻𝑫𝑩 ⃗⃗⃗⃗⃗⃗⃗
𝒆𝑫𝑩 ⃗⃗⃗⃗⃗⃗⃗
𝑻𝑫𝑪 = 𝑻𝑫𝑪 ⃗⃗⃗⃗⃗⃗⃗
𝒆𝑫𝑪
𝑆 ⃗ )𝑝𝑖𝑒𝑠
(0𝑖 − 1,5𝑗 + 0𝑘 𝑆 ⃗ )𝑝𝑖𝑒𝑠
(−1,5𝑖 + 1𝑗 + 3𝑘
𝑒𝐷𝐵 =
⃗⃗⃗⃗⃗⃗ = 𝑒𝐷𝐶 =
⃗⃗⃗⃗⃗⃗ =
|𝑆| √(0)2 + (−1,5)2 + (0)2 𝑝𝑖𝑒𝑠 |𝑆| √(−1,5)2 + (1)2 + (3)2 𝑝𝑖𝑒𝑠
⃗
0𝑖 − 1,5𝑗 + 0𝑘 ⃗
−1,5𝑖 + 1𝑗 + 3𝑘
𝑒𝐷𝐵 =
⃗⃗⃗⃗⃗⃗ 𝑒𝐷𝐶 =
⃗⃗⃗⃗⃗⃗
√2,25 √12,25
⃗
0𝑖 − 1,5𝑗 + 0𝑘 ⃗
−1,5𝑖 + 1𝑗 + 3𝑘
⃗⃗⃗⃗⃗⃗
𝑇𝐷𝐵 = ⃗⃗⃗⃗⃗⃗
𝑇𝐷𝐵 ( ) ⃗⃗⃗⃗⃗⃗
𝑇𝐷𝐶 = ⃗⃗⃗⃗⃗⃗
𝑇𝐷𝐶 ( )
√2,25 √12,25
⃗⃗⃗⃗⃗⃗
𝑇 ⃗
𝐷𝐴 = 0𝑇𝐷𝐵 𝑖 − 1𝑇𝐷𝐵 𝑗 + 0𝑇𝐷𝐵 𝑘
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐷𝐶 = −0,43𝑇𝐷𝐶 𝑖 + 0,29𝑇𝐷𝐶 𝑗 + 0,86𝑇𝐷𝐶 𝑘
⃗
⃗⃗⃗ = −𝒘 𝒆
𝒘 ⃗⃗⃗ = −𝟐𝟎 𝒍𝒃 − 𝒇 ⃗𝒌
𝒘
∑𝐹 = 0
⃗⃗⃗⃗⃗⃗
𝑇 ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗
𝐷𝐴 = 𝑇𝐷𝐵 + 𝑇𝐷𝐶 + 𝑤 ⃗⃗ = 0
⃗⃗⃗⃗⃗⃗
𝑇 ⃗
𝐷𝐴 = 0,67𝑇𝐷𝐴 𝑖 − 0,33𝑇𝐷𝐴 𝑗 + 0,67𝑇𝐷𝐴 𝑘
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐷𝐵 = 0𝑇𝐷𝐵 𝑖 − 1𝑇𝐷𝐵 𝑗 + 0𝑇𝐷𝐵 𝑘
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐷𝐵 = −0,43𝑇𝐷𝐶 𝑖 + 0,29𝑇𝐷𝐶 𝑗 + 0,86𝑇𝐷𝐶 𝑘
𝑤 ⃗
⃗⃗ = 0𝑖 + 0𝑗 − 20𝑙𝑏 − 𝑓𝑘
𝑖 = 0,67𝑇𝐷𝐴 + 0𝑇𝐷𝐵 − 0,43𝑇𝐷𝐶 = 0
𝑗 = −0,33𝑇𝐷𝐴 − 1𝑇𝐷𝐵 + 0,29𝑇𝐷𝐶 = 0
⃗ = 0,67𝑇𝐷𝐴 + 0𝑇𝐷𝐵 + 0,86𝑇𝐷𝐶 = 20 𝑙𝑏 − 𝑓
𝑘
0 0 −0,43 0 0
0 −1 0,29 0 −1
| 20 0 0,86 | 20 0 −8,6
𝑇𝐷𝐴 = = = 29,89𝑙𝑏 − 𝑓
| 0,67 0 −0,43| 0,67 0 −0,58 − 0,29
−0,33 −1 0,39 −0,33 −1
0,67 0 0,86 0,67 0
UNIVERSIDAD NACIONAL DE HUANCAVELICA
0,67 0 −0,43 0,67 0
−0,33 0 0,29 −0,33 0
| 0,67 200 0,86 | 0,67 20 −284 − 5,89 −6,73
𝑇𝐷𝐵 = = = = 7,74𝑙𝑏 − 𝑓
| 0,67 0 −0,43 | 0,67 0 −0,87 −0,87
−0,33 −1 0,29 −0,33 −1
0,67 0 0,86 0,67 0
0,67 0 0 0,67 0
−0,33 −1 0 −0,33 −1
| 0,67 0 20 | 0,67 0 −13,4
𝑇𝐷𝐶 = = = 15,4𝑙𝑏 − 𝑓
| 0,67 0 −0,43| 0,67 0 −0,87
−0,33 −1 0,29 −0,33 −1
0,67 0 0,86 0,67 0
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑫 = 𝑻𝑨𝑫 ⃗⃗⃗⃗⃗⃗⃗
𝒆𝑨𝑫
⃗⃗⃗⃗⃗⃗ ⃗ )𝑝𝑖𝑒𝑠
𝑇𝐴𝐷 = (−12𝑖 − 4𝑗 + 6𝑘
𝑆 ⃗ )𝑝𝑖𝑒𝑠
(−12𝑖 − 4𝑗 + 6𝑘
𝑒⃗⃗⃗⃗⃗⃗
𝐴𝐷 = =
|𝑆| √(−12)2 + (−4)2 + (6)2 𝑝𝑖𝑒𝑠
⃗
−12𝑖 − 4𝑗 + 6𝑘
𝑒⃗⃗⃗⃗⃗⃗
𝐴𝐷 =
√196
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑫 = 𝑻𝑨𝑫 ⃗⃗⃗⃗⃗⃗⃗
𝒆𝑨𝑫
⃗
−12𝑖 − 4𝑗 + 6𝑘
⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐷 = ⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐷 ( )
14
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐴𝐷 = −0,86𝑇𝐴𝐷 𝑖 − 0,29𝑇𝐴𝐷 𝑗 + 0, 43𝑇𝐴𝐷 𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑪 = 𝑻𝑨𝑪 ⃗⃗⃗⃗⃗⃗
𝒆𝑨𝑪
𝑆 ⃗ )𝑝𝑖𝑒𝑠
(−12𝑖 + 9𝑗 + 8𝑘
𝑒𝐴𝐶 =
⃗⃗⃗⃗⃗⃗ =
|𝑆| √(−12)2 + (9)2 + (8)2 𝑝𝑖𝑒𝑠
UNIVERSIDAD NACIONAL DE HUANCAVELICA
⃗
−12𝑖 + 9𝑗 + 8𝑘
𝑒𝐴𝐶 =
⃗⃗⃗⃗⃗⃗
17
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑪 = 𝑻𝑨𝑪 ⃗⃗⃗⃗⃗⃗
𝒆𝑨𝑪
⃗
−12𝑖 + 9𝑗 + 8𝑘
⃗⃗⃗⃗⃗⃗⃗
𝑻𝑨𝑪 = 𝑻𝑨𝑪 ( )
17
⃗⃗⃗⃗⃗⃗ ⃗
𝑇𝐴𝐶 = −0,71𝑇𝐴𝐶 𝑖 + 0,53𝑇𝐴𝐶 𝑗 + 0, 47𝑇𝐴𝐶 𝑘
⃗
⃗⃗⃗ = −𝒘 𝒆
𝒘 ⃗⃗ = −𝟔𝟎 𝒍𝒃 − 𝒇 ⃗𝒌
⃗𝒘
∑𝐹 = 0
⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐵 = ⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐶 + ⃗⃗⃗⃗⃗⃗
𝑇𝐴𝐷 = 0
𝑖 = 𝑇𝐴𝐵 − 0,71𝑇𝐴𝐶 − 0,86𝑇𝐴𝐷 = 0
𝑗 = 0𝑇𝐴𝐵 − 0,53𝑇𝐴𝐶 − 0,29𝑇𝐴𝐷 = 0
⃗ = 0𝑇𝐴𝐵 + 0,45𝑇𝐴𝐶 + 0,43𝑇𝐴𝐷 = 60 𝑙𝑏 − 𝑓
𝑘
1 0 −0,86 1 0
0 0 −0,29 0 0
| 0 603 0,43 | 0 60 17,4
𝑇𝐴𝐶 = == = 48,33𝑙𝑏 − 𝑓
|1 −0,71 −0,86| 1 −0,71 0,36
0 0,53 −0,29 0 0,53
0 0,45 0,43 0 0,45
1 −0,71 0 1 0
0 0,53 0 0 −1
| 0 0,45 60 |
𝑇𝐴𝐷 = 0 0 = 31,8 = 88,33𝑙𝑏 − 𝑓
|1 −0,71 −0,86| 1 −0,71 0,36
0 0,53 −0,29 0 0,53
0 0,53 0,43 0 0,45