a force f1 of 18.8 newtons pulls at an angle of 12 degrees above due east.

Question

a force f1 of 18.8 newtons pulls at an angle of 12 degrees above due east. A force f2 of 3.2 newtons pulls at an angle of 88 degrees below due east. find the direction of the resultant force as a standard angle

Answer 1

The angle of f1 = 12 degrees

The angle of f2 = -88 degrees because it is below.  Let's turn this into a positive angle.

The angle of f2 = 360 - 88 = 272 degrees

 

Next we break each force down to the x and y components.

 

f1_x = 18.8 cos(12) = 18.39 N

f1_y = 18.8 sin(12) = 3.91 N

 

f2_x = 3.2 cos(272) = 0.11 N

f2_y = 3.2 sin(272) = -3.20

 

Now, we add combine our x components, and our y-components.  After that, we use the Pythagorean theorem to find our resultant.

 

R_x = f1_x + f2_x = 18.39 N + 0.11 N = 18.5 N

R_y = f1_y + f2_y = 3.91 N - 3.20 N = 0.71 N

 

R = √(R_x² + R_y²)

R = √((18.5)² + (0.71)²)

R = 18.51 N

 

The resultant force is 18.51 N.

 

Next we determine the direction.  Calculate the angle using inverse tangent.

 

θ = tan^-1(R_y / R_x)

θ = tan^-1(0.71 / 18.5)

θ = 2.198 degrees

 

The direction of the resultant is 2.198 degrees due east.