Three forces act on an object at the same time. F1 = 10.0 N 50.0degrees north of east, F2 = 15.0 N 25.0degrees south of west, and F3 = 15.0 N 38.0degrees east of south. What is the magnitude and direction of the net force?
To find the magnitude and direction of the net force acting on the object, we need to add the three given forces vectorially.
Step 1: Convert all the forces into their x and y components.
For F1 = 10.0 N at 50.0 degrees north of east,
Fx1 = 10.0 N * cos(50.0) = 10.0 N * 0.6428 ≈ 6.428 N (positive x-direction)
Fy1 = 10.0 N * sin(50.0) = 10.0 N * 0.7660 ≈ 7.660 N (positive y-direction)
For F2 = 15.0 N 25.0 degrees south of west,
Fx2 = 15.0 N * cos(180 - 25.0) = 15.0 N * cos(155.0) = 15.0 N * (-0.5736) ≈ -8.604 N (negative x-direction)
Fy2 = 15.0 N * sin(180 - 25.0) = 15.0 N * sin(155.0) = 15.0 N * 0.8192 ≈ 12.288 N (positive y-direction)
For F3 = 15.0 N 38.0 degrees east of south,
Fx3 = 15.0 N * cos(180 - 38.0) = 15.0 N * cos(142.0) = 15.0 N * (-0.7660) ≈ -11.490 N (negative x-direction)
Fy3 = 15.0 N * sin(180 - 38.0) = 15.0 N * sin(142.0) = 15.0 N * (-0.6428) ≈ -9.643 N (negative y-direction)
Step 2: Add the x and y components of all forces to find the net force components.
Fx_net = Fx1 + Fx2 + Fx3
= 6.428 N + (-8.604 N) + (-11.490 N)
= -13.666 N (to the left)
Fy_net = Fy1 + Fy2 + Fy3
= 7.660 N + 12.288 N + (-9.643 N)
= 10.305 N (upward)
Step 3: Calculate the magnitude and direction of the net force using the net force components.
Magnitude of net force (F_net) = sqrt(Fx_net^2 + Fy_net^2)
= sqrt((-13.666 N)^2 + (10.305 N)^2)
= sqrt(186.901 N^2 + 106.194 N^2)
= sqrt(293.095 N^2)
≈ 17.1 N
Direction of net force = arctan(Fy_net / Fx_net)
= arctan(10.305 N / (-13.666 N))
= arctan(-0.754)
≈ -37.3 degrees (measured from the positive x-axis counterclockwise)
Therefore, the magnitude of the net force is approximately 17.1 N directed at an angle of -37.3 degrees counterclockwise from the positive x-axis.