Question

Two forces (25.0 N at 30.0° south of east, and 68.0 N at 70.0° north of west) are applied to a 280 kg object. If the net force is determined by these two forces, then find the acceleration of the object

Answer 1

Here , let east is the positive x direction

north is the positive y direction

F1 = 25 * (cos(30) i - j * sin(30))

F2 = 68 * (- cos(70) i + j * sin(70))

for the net force

net force = F1 + F2

net force = 25 * (cos(30) i - j * sin(30)) + 68 * (- cos(70) i + j * sin(70))

net force = -42.4 i + 51.4 j

net force = sqrt(42.4^2 + 51.4^2) at arctan(51.4/42.4) degree north of west

net force = 66.6 N at 50.5 degree north of west

acceleration = net force/mass

acceleration = (66.6)/280

acceleration = 0.24 m/s^2 at 50.5 degree north of west

the acceleration of the object is 0.24 m/s^2 at 50.5 degree north of west