G G 5. We denote the two forces F1 and F2 . According to Newton’s second law, G G G G G G F1 + F2 = ma , so F2 = ma − F1 .
b
g
G (a) In unit vector notation F1 = 20.0 N i and
G a = − (12.0 sin 30.0° m/s 2 ) ˆi − (12.0 cos 30.0° m/s 2 ) ˆj = − ( 6.00 m/s 2 ) ˆi − (10.4m/s 2 ) ˆj. Therefore, G F2 = ( 2.00kg ) ( −6.00 m/s 2 ) ˆi + ( 2.00 kg ) ( −10.4 m/s 2 ) ˆj − ( 20.0 N ) ˆi = ( −32.0 N ) ˆi − ( 20.8 N ) ˆj. G (b) The magnitude of F2 is G | F2 |= F22x + F22y = (− 32.0 N) 2 + (− 20.8 N) 2 = 38.2 N. G (c) The angle that F2 makes with the positive x axis is found from tan θ = (F2y/F2x) = [(–20.8 N)/(–32.0 N)] = 0.656. Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the x and y components are negative, the correct result is 213°. An alternative answer is 213 ° − 360 ° = − 147 ° .
