Question

Three forces act on an object. the first force is 37.5 lbs at 130.0 degrees, the second force is 28.0 lbs at 30.0 degrees east of south, and the third force is 11.0 lbs due south. What is the net force (magnitude and direction) exerted on this body?

Answer 1

Given that

F1 = 37.5 lbs at 130.0 deg

F1_x = 37.5*cos 130 deg = -24.1 lbs

F1_y = 37.5*sin 130 deg = 28.7 lbs

F2 = 28.0 lbs at 30.0 deg east of south = 28.0 lbs at 300 deg CCW from +ve x-axis

F2_x = 28.0*cos 300 deg = 14 lbs

F2_y = 28.0*sin 300 deg = -24.25 lbs

F3 = 11.0 lbs due south = 11.0 lbs at 270 deg CCW from +ve x-axis

F3_x = 11.0*cos 270 deg = 0

F3_y = 11.0*sin 270 deg = -11.0 lbs

Now net force will be:

F_net = F1 + F2 + F3

F_net = (F1_x + F2_x + F3_x) i + (F1_y + F2_y + F3_y) j

F_net = (-24.1 + 14 + 0) i + (28.7 - 24.25 - 11.0) j

F_net = (-10.1) i + (-6.55) j

Now magnitude of net force will be:

|F_net| = sqrt ((-10.1)^2 + (-6.55)^2)

|F_net| = 12.0 lbs

Direction of net force will be:

Direction = arctan (Fy_net/Fx_net)

Direction = arctan (6.55/10.1)

Direction = 33.0 deg below +ve x-axis = 33.0 deg South of east = 57 deg East of south = 327 deg CCW from +ve x-axis

(All angles are same so use the one which is required by your instructor)

Let me know if you've any query.