Question

Two forces, F⃗ 1 and F⃗ 2, act at a point. F⃗ 1 has a magnitude of 8.20 N and is directed at an angle of 64.0 ∘ above the negative x axis in the second quadrant. F⃗ 2has a magnitude of 5.20 N and is directed at an angle of 53.8 ∘ below the negative x axis in the third quadrant.

Part A

What is the x component of the resultant force?

Part B

What is the y component of the resultant force?

Express your answer in newtons.

Part C

What is the magnitude of the resultant force?

Answer 1

Given that :

magnitude of F'₁ = 8.2 N                                                    At an angle, ₁ = 64 degree

magnitude of F'₂ = 5.2 N                                                    At an angle, ₂ = 53.8 degree

Part-A : The x component of the resultant force which is given as -

F'_X = F'_1X + F'_2X                                                                { eq.1 }

F'_X = (8.2 N) Cos 64⁰ + (5.2 N) Cos 53.8⁰

F'_X = (8.2 N) (0.4383) + (5.2 N) (0.5906)

F'_X = (3.59 + 3.07) N

F'_X = 6.66 N

Part-B : The y component of the resultant force which is given as -

F'_Y = F'_1Y + F'_2Y                                                                { eq.2 }

F'_Y = (8.2 N) Sin 64⁰ + (5.2 N) Sin 53.8⁰

F'_Y = (8.2 N) (0.8987) + (5.2 N) (0.8069)

F'_Y = (7.36 + 4.19) N

F'_Y = 11.5 N

Part-C : The magnitude of the resultant force which is given as -

F_net = F'_X + F'_Y    { eq.3 }

inserting the values in eq.3,

F_net = (6.66 N)² + (11.5 N)²

F_net = (44.35 + 132.25) N²

F_net = 176.6 N²

F_net = 13.2 N