A large box is being pushed by three separate people. The first person is pushing the box at a rate of 6 m/s directly north. The second person is pushing the box at a rate of 6 m/s directly to the northeast. The third person is pushing the box at a rate of 6 m/s directly east. Determine the magnitude and direct of the resultant vector of he box.
We can find the east component of the velocity.
v_east = 6 m/s + (6 m/s) (cos (45))
v_east = 6 m/s + 4.24 m/s
v_east = 10.24 m/s
We can find the north component of the velocity.
v_north = 6 m/s + (6 m/s) (sin (45))
v_north = 6 m/s + 4.24 m/s
v_north = 10.24 m/s
We can find the magnitude of the velocity.
v = sqrt { (v_east)^2 + (v_north)^2}
v = sqrt { (10.24 m/s)^2 + (10.24 m/s)^2}
v = 14.5 m/s
We can find the angle theta north of east.
tan(theta) = 10.24 / 10.24
theta = arctan (1)
theta = 45
Note that this angle is directly northeast.
The magnitude of the velocity is 14.5 m/s and the box is moving directly to the northeast.
