The pendulum bob B has a weight of 10 lb and is released from rest in the position shown, q = 0^o. Determine the tension in string BC just after the bob is released, q = 0^o, and also at the instant the bob reaches point A, q = 45^o. Take r = 4 ft.

1) 21.2 N
2) 25 .5 N
3) 15.2 N
4) 20 N

Equation of motion: Since the bob is just being released, v = 0. Summing up the forces applied in n direction at B we have
S F_n = ma_n
T = 10/32.2(0²/4) = 0 N
At A we have
S F_t = ma_t
10 cosq = 10/32.2 a_t  and a_t = 32.2 cosq
S F_n = ma_n
T - 10sinq = 10/32.2(v²/4)

Kinematics: The speed of the bob at the instant when q = 45^o can be determined using vdv = at ds.
However, ds = 4dq, then vdv = 4a_t dq

Substituting q = 45^o and v² = 182.14 ft²/s²
T - sin45^o = 10/32.2(182.14/4) so, T = 21.21 N

So the tension in the string T = 21.2 N