This document shows calculations to find the tension (T) in a system with two blocks connected by a rope on a ramp. It first calculates the normal force (N2) on the 6kg block, then the frictional force (Fr2). Using Newton's Second Law and setting the sums of forces equal, it derives an expression for the acceleration (a) and substitutes back to find the tension (T) is 23N.

1. :

9. N2 = m2gCos30°
N2 = 6kg*9.81m/s^2Cos30°
N2 = 50.97N
Fr2 = 50.97N * 0.36
Fr2 = 18.35 N
sustituyendo e ∑Fx
T + 18.35N = a6kg
T = a6kg - 18.35N (B)
igualamos ecuaciones A y B
a2kg + 9.2214N = a6kg - 18.35N
despejando a
a2kg - a6kg = -18.35N - 9.2214N
a = 6.89 m/s^2
teniendo la aceleracion sustituimos en A o en B
T = a2kg + 9.2214N = 6.89m/s^2*2kg + 9.2214N = 23N
T = a6kg - 18.35N = 6.89m/s^2*6kg - 18.35N = 23N
T =23N