1. The document is an exam with multiple questions about static forces and cables sustaining weights in equilibrium. 2. The first question involves calculating the total force on a point from three cables pulling with 600N of force each. The total force is found to be 1572.87N. 3. The second question involves finding the net force on a system of eight forces. 4. The third question involves calculating the tensions in three cables (A, B, C) necessary to sustain a 500N cylinder in equilibrium. The tensions are found to be TA=832.91N, TB=59.65N, and TC=577.28N.

1. EXAMEN T1 DE ESTATICA
DEPARTAMENTO DE INGENIERIA
Duración: 90min. Calificación
ALUMNO
DOCENTE FECHA: CLASE
1. Las tres fuerzas tienen magnitud 600N. Determine la magnitud de la fuerza
total ejercida en el punto A por los dos cables.
A (0,0,12) B(0,-10,0) C(6,0,0) D(-4,3,0)
𝑇𝐴𝐵= 600
(0,−10,0)−(0,0,12)
15.62
= (0i-384.12j+460.95k)
𝑇𝐴𝐶= 600
(6,0,0)−(0,0,12)
13.42
= (268.26i+0j-536.51k)
𝑇𝐴𝐵= 600
(−4,3,0)−(0,0,12)
13
= (-184.62i+138.46j-553.85k)
𝐹𝑅
⃗⃗⃗⃗⃗ = 83.64i – 245.66j – 1551.31k
|𝐹𝑅| = 1572.87 N

2. 2. Obtener la resultante del sistema de fuerzas que se muestra en la figura
F1=50N 𝐹1 = (-30i , 0j , 40k)
F2=50N 𝐹2 = (30i , 0j , -40k)
F3=20N 𝐹3 = (0i , 20j , 0k)
F4=60N 𝐹4 = (0i , -60j ,0k)
F5=40N 𝐹5 = (0i , 40j , 0k)
F6=10N 𝐹6 = (0i , 0j , 10k)
F7=30N 𝐹7 = (0i , 0j , -30k)
F8=20N 𝐹8 = (0i , 0j , 20k)
0i, 0j, 0k
A (3,0,0) 𝐹1= 50 (
−3+0+4
5
) = (-30.35i+0j+40k)
C (0,0,4)
E (0,4,4) 𝐹2= 50 (
−3+0−4
5
) = (30i+0j-40K)

3. 3. Determine la tensión desarrollada en los cables A, B y C que es necesaria para
lograr el equilibrio del cilindro de 500N.
𝐴 =1.2i -1.5j+2.4k
⋎𝐴= [
1.2𝑖+(−1.5)𝑗+2.4𝑘
√(1.2)2+(1.5)2+(2.4)2
] = 0.39i-0.49j + 0.82k
𝐹𝐴 = |𝑇𝐴| ⋎𝐴 = 0.39𝑇𝐴𝑖 - 0.49𝑇𝐴𝑗+0.82𝑇𝐴𝑘
𝐵
⃗ =-1.8i -2.1j+2.1k
⋎𝐵= [
−1.8𝑖−2.1𝑗+2.1𝑘
√(−1.8)2+(−2.1)2+(2.1)2
] = -0.51i - 0.61j +0.61k
𝑇𝐵 = -0.51𝑇𝐵i - 0.61𝑇𝐵j +0.61𝑇𝐵k
𝐶 =-1.2i +1.8j-0.9k
⋎𝐶= [
−1.2𝑖+1.8𝑗−0.9𝑘
√(−1.2)2+(1.8)2+(−0.9)2
] = -0.51i + 0.77j – 0.38 k
𝑤
⃗⃗ = -500k
∑𝐹𝑋=0; 0.39𝑇𝐴 - 0.51𝑇𝐵 – 0.51𝑇𝐶= 0 (1)
∑𝐹𝑌=0; -0.49𝑇𝐴 - 0.61𝑇𝐵 + 0.77𝑇𝐶= 0 (2)
∑𝐹𝑍=0; 0.82𝑇𝐴 + 0.61 𝑇𝐵 – 0.38𝑇𝐶 -500 = 0 (3)
𝑇𝐴= 832.91 𝑇𝐵= 59.65 𝑇𝐶= 577.28
IFRI=1015.16N

4. 4. Tres cables sostienen una caja como se muestra en la figura. Determine el
peso de la caja, si se sabe que la tensión en el cable AD es de 616 lb.
𝐴𝐶
⃗⃗⃗⃗⃗ =0i +60j+32k
⋎𝐴𝐶= [
−0𝑖+60𝑗+32𝑘
√(0)2+(60)2+(32)2
] = 0.882j + 0.470k
𝐹𝐴𝐶 = |𝑇𝐴𝐶| ⋎𝐴𝐶 = 0.882𝑇𝐴𝐶j + 0.470𝑇𝐴𝐶k
𝐴𝐵
⃗⃗⃗⃗⃗ =-36i +60j-27k
⋎𝐴𝐵= [
−36𝑖+60𝑗−27𝑘
√(−36)2+(60)2+(−27)2
] = -0.48i + 0.8j -0.36k
𝑇𝐴𝐵 = -0.48𝑇𝐴𝐵i + 0.8𝑇𝐴𝐵j -0.36𝑇𝐴𝐵k
𝐴𝐷
⃗⃗⃗⃗⃗ =40i +60j-27k
⋎𝐴𝐷= [
−40𝑖+60𝑗−27𝑘
√(40)2+(60)2+(−27)2
] = 0.519i + 0.799j – 0.350 k
𝑇𝐴𝐷 = 616 lb*⋎𝐴𝐵 = (319.704 lb I + 479.864 lb j -215.6 lb k
𝑤
⃗⃗ = -Wj
∑𝐹𝑋=0; -0.48𝑇𝐴𝐵 + 319.7 = 0 (1)
∑𝐹𝑌=0; 0.882𝑇𝐴𝐶 + 0.8𝑇𝐴𝐵 + 479.864 lb -w= 0 (2)
∑𝐹𝑍=0; 0.470𝑇𝐴𝐶 – 0.36 𝑇𝐴𝐵- 215.6 lb = 0 (3)
𝑇𝐴𝐵= 666.05 𝑇𝐴𝐶= 968.889 lb w= 1867.64 lb