Three forces are applied to a 10 kg box. The magnitude and direction of the forces are:
1) 5.0N due North
2) 3.0N due West
3) 6.0N @ 30 degrees S of E
Determine the net force acting on the box (include both magnitude and direction). What is the net acceleration of the box?
VECTOR Add East and South Components of 6.0 N force, respectively, to 3N and 5N forces.
ie
East Component of 6N = 6 cos 30° = 5.2 N => added to 3.0 N W = 2.2 N, E
South Component of 6N = 6 sin 30° = 3.0 N => added to 5.0 N N = 2.0 N, N
Size of Fnet = √2.2²+2.0² = √8.84 = 3.0 N ANS
Direction of Fnet = arctan 2/2.2 = 42° ANS
Net Accelertion on box = Fnet/m = 0.3 m/s² at 42° ANS