This document solves a physics problem involving forces and torques. It provides the given values of mass and gravity. It then sets up two equations of torque balance and solves them to find the horizontal distance (Dx) and normal force (P). Dx is calculated to be 31 meters and P is calculated to be 186.2 Newtons.

1. Carlos Colmenarez III-142-00125p
Datos:
Mb= 25 Mc=13 G = 9,8 m/s2
Solución:
∑ Mc = 0 ; Dy ( 4m) – Dx ( 3m) = 0 Dy= ¾ Dx
1. ∑ MA = 0 ; ¾ Dx(14m) – Wb (4m) - Wc (10m) – P(5m) = 0
2. ∑ MB = 0 ; ¾ Dx (10m) - Dx(5m) – Wc (6m) = 0
Solución 2 ecuacion Dx = 2, 4 Wc
Mb= Wb = 25 Kg
Mc= Wc = 13 Kg
Dx = 2, 4 Wc Dx = 2, 4 x 13 Dx = 31
Solucion ecuacion 1
¾ Dx(14m) – Wb (4m) - Wc (10m) – P(5m) = 0
¾ 31 (14m) – 25(4m) - 13 (10m) – P(5m) = 0
P = 19 x G P = 19 x 9,8
P = 186,2 N