→
34. (a) Using notation suitable to a vector capable calculator, the Fnet = 0 condition becomes →
→
→
→
F1 + F2 + F3 = (6.00 ∠ 150º) + (7.00 ∠ −60.0º) + F3 = 0 . →
Thus, F3 = (1.70 N) ^i + (3.06 N)j^. (b) A constant velocity condition requires zero acceleration, so the answer is the same. → → G (c) Now, the acceleration is a = (13.0 m/s 2 ) ˆi − (14.0 m/s 2 ) ˆj . Using Fnet = m a (with m = 0.025 kg) we now obtain →
F3 = (2.02 N) ^i + (2.71 N) ^j.