find the tension in each cord in the figure if the weight of the suspended object is w = 210n
a) find the tension in the cord a for system (b).
b) find the tension in the cord b for system (b).

We have to solve the system of equations:
- T a * cos 60° + T b * sin 45° - 210 N = 0
- T a * sin 60° + Tb cos 45° =  0

- 1/2 T a + √2 / 2 T b = 210 N
- √3 / 2 T a + √ 2 / 2 T b = 0     / * ( - 1 )

- 1/2 T a + √2 / 2 T b = 210 N
√ 3 / 2 T a - √2 / 2 T b = 0

T a ( √3 / 2 - 1 / 2 ) = 210 N
0.366 T a = 210 N
T a = 210 N : 0.366
1 ) T a = 573.77 N
Ta * √3 / 2 = T b * √ 2 / 2
T b = ( 573.77 * 1.732 ) : 1.4142
2 ) T b = 702.7 N 

The car would move to the right with 150 n of force. the 150 n of force from friction and air resistance slows the car from moving to the right with 300 n of force.