The magnitude and direction of two forces acting on an object are 50 pounds, S55°E, and 110 pounds, N68°E, respectively.

Question

Find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force. And find the direction angle in degrees.

Answer 1

S55°E means the primary direction is south and then I angle 55° toward east and N68°E means the primary direction is north and I angle 68° toward east. So a sketch of the forces would look like this:

To add these forces, we break each into its component forces in the x and y directions:

Let F₁ be the force that is the 50 pounds acting in the S55°E direction:

Notice that 55° is equivalent to 25° below the "due east" direction so

F_1-x = F₁cos(θ) = 50cos(25°) = 45.315 lbs

F_1-y = -F₁sin(θ) = -50sin(25°) = -21.131 lbs (negative because it is going downwards)

Let F₂ be the force that is the 110 pounds acting in the N68°E direction:

Notice that 68° is equivalent to 12° above the "due east" direction so

F_2-x = F₂cos(θ) = 110cos(12°) = 107.596 lbs

F_2-y = F₂sin(θ) = 110sin(12°) = 22.870 lbs

Adding the two x-direction components:

F_x-Total = F_1-x + F_2-x = 45.315 + 107.596 = 152.9 lbs

Adding the two y-direction components:

F_y-Total = F_1-y + F_2-y = -21.131 + 22.870 = 1.739 lbs

To get the magnitude of the total resultant force, we use the Pythagorean Theorem:

F_Total = √(152.9² + 1.739²) = √23385 = 152.92 pounds

The direction of the total force is found using inverse tangent. The angle off the "due east" line is tan^-1(1.739/152.9) = tan^-1(0.011375) = 0.652°. Converting that back into a bearing it becomes N89.3°E

Answer 2

Draw and lable a diagram!

Place the 110 lb vector tail-to-tip of the 50 lb vector.

Using Law of Consines to determine magnitude of resultant vector.