The magnitude and Direction of two forces acting on an Object

Question

The magnitude and direction of two forces acting on an object are 35 pounds, N45E, and 55 pounds, S30E, respectively. 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.

Answer 1

In addition to Yefim's solution, you can also use the Law of Cosines. Sketching this out you get:

and that leads to the triangle:

Applying the Law of Cosines we get:

F_R² = 35² + 55² - 2(35)(55)cos(75°)

F_R² = 3253.55

F_R = 57.04 or rounding, F_R = 57 pounds

To find the angle, use the Law of Sines to find ∠α

sin(α)/55 = sin(75°)/57.04

sin(α) = 0.93138

α = 68.65°

θ = 68.65° - 45° = 23.65° but to put this in the format of bearing, we need to subtract this from 90° so we can get a SxxE designation:

90° - 23.65° = 66.35° so the angle bearing is S66.3°E

Answer 2

R_x = 35cos45° + 55cos60° = 52.25 lb; R_y = 35cos45°- 55cos30° = - 22.88 lb

R = (52.25² + (- 22.88)²)^1/2 = 57.04 lb

tanθ = F_y/F_x = - 22.88/52.25; θ = - tan^-1(22.88/52.25) = - 23.6°

Answer 3

N45E at 35

S30E at 55

construct a parallelogram and calculate the diagonal with cosines law

d^2= 35^2 +55^2 -2(35)(55)Cos75

d = sqr(1225 [phone]Cos75)

= about 59.04 lbs at about S66.35E

Answer 4