Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P092

Transcript

92. (a) With SI units understood, the net force is G G G Fnet = F1 + F2 = ( 3.0 N + ( −2.0 N ) ) ˆi + ( 4.0 N + ( −6.0 N ) ) ˆj G which yields Fnet = (1.0 N) ˆi − ( 2.0 N) ˆj. G (b) The magnitude of Fnet is Fnet = (1.0 N) 2 + (− 2.0 N) 2 = 2.2 N. G (c) The angle of Fnet is § − 2.0 N · ¸ = − 63 ° . 1.0 N © ¹ θ = tan − 1 ¨ G (d) The magnitude of a is a = Fnet / m = (2.2 N) /(1.0 kg) = 2.2 m/s 2 . G G (e) Since Fnet is equal to a multiplied by mass m, which is a positive scalar that cannot G affect the direction of the vector it multiplies, a has the same angle as the net force, i.e, G θ = − 63 ° . In magnitude-angle notation, we may write a = ( 2.2 m/s 2 ∠ − 63° ) .