1. The document is about solving physics problems involving forces and vectors in Spanish. It includes calculating the net force and angle of a system of forces, determining if vectors are parallel or perpendicular, and solving problems involving tension, normal force, friction, and inclines. 2. Diagrams and equations are provided to solve for variables like tension, normal force, angle of incline, and components of forces. 3. The problems involve systems of forces, vectors, inclines, and friction. Diagrams, variables, and equations are used to solve for unknown values.

1. UNIVERSIDAD FERMÍN TORO
SISTEMA DE APRENDIZAJES INTERACTIVOS A DISTANCIA
CABUDARE
Actividad 2
FÌSICA I
INTEGRANTE:
FERNANDO GONZÁLEZ
C.I.: 21.037.695
INGENIERÍA EN TELECOMUNICACIÓN

2. 1
1 Km
1,25 Km
825 m
Σ Fx = F1x + F2x + F3x Σ Fy = F1y + F2y + F3y
Σ Fx = 825 m x cos 270° + 1250m x cos (120°) + 1000m x cos (42°)
Σ Fx = 26,17 m
Σ Fy = 825 m x sen 270° + 1250 m x sen (125°) + 1000m x sen (42°)
Σ Fy = 868,07 m
Fr = √( 𝐹𝑥)2 + (𝐹𝑦)² Fr = √(26,17𝑚)2 + (868,07)² Fr = 868,46mts
Θ = tan−1
(
𝐹𝑦
𝐹𝑥
) θ = tan−1
(
868,07𝑚
26,17𝑚
) θ = 88,27°
48°
35°

3. 2
A = 6i + 2j – 4k B = gi + N + 6k
(6i + 2j – 4k) . (-gi + Nj + 6k) = 0 PARALELO
- 54 + 2N + 24 = 0 2N = 78 N= 39
(6i + 2j – 4k) . (-gi + Nj + 6k) = -1 PERPENDICULAR
-54 + 2N – 24 = -1 2N = 77 N =
77
2

4. 3
a)
= 750 kgr
Σ Fx = 0 Rx - Tcos∝ = 0 Σ Fx = 0 Tsen∝ - Wb = 0
1600Kgr . sen∝ - 750Kgr = o
sen∝ =
750𝐾𝑔𝑟
1500𝐾𝑔𝑟
h =
𝐶𝐴
𝑐𝑜𝑠∝
h =
80𝑐𝑚
𝑐𝑜𝑠30°
h = 92,37 cm
∝ = sen−1
(
1
2
) ∝ = 30°
80cm
h = ?
T = 1500 kgr
M
T
Wb
Rx
Wc

5. 3
b)
Tan∝ =
𝐶𝑂
𝐶𝐴
Tsen 37,15° - Wb = 0
T =
750 𝑥 9,8 𝑚/𝑠𝑒𝑔2
𝑠𝑒𝑛 37,15°
T = 12170,81 New
∝ = tan−1
(
70𝑐𝑚
92,37𝑐𝑚
) ∝ = 37,15°
92,37
7cm

6. 4
80Kgr . 9,8m/seg
784 New
Tx = 0 Tax + Tbx + Px =0
Ta . cos48° + Tb . cos145° + 784 cos270° = 0
0,6691 Ta – 0,8191 Tb = 0
Ty = 0 Tay + Tby + Py = 0
Ta . sem(48°) + Tb . sen(145°) + 784 sen270° = 0
0,7431 Ta + 0,5735 Tb = 784
0,6691 Ta - 0,8191 Tb = 0
0,7431 Ta + 0,5735 Tb = 784
Ta = 647,09 New
Tb = 528,59 New
48°35° 48°35°
TB TA