Question

Please answer the question below (ABOUT VECTORS AND MAGNITUDE)

Answer 1

(a) v appears to have a fixed direction along the positive x-axis. If ||u|| = 150 N, ||v|| = 220 N, then when θ = 30°, you have

u = (150 N) (cos(30°) i + sin(30°) j ) ≈ (129.904 i + 75 j ) N

v = (220 N) (cos(0°) i + sin(0°) j ) = (220 i ) N

(i and j are the unit vectors in the positive x and y directions)

and their sum is

u + v ≈ (349.904 i + 75 j ) N

with magnitude

||u + v|| ≈ √((349.904)² + (75)²) N ≈ 357.851 N ≈ 357.9 N

and at angle φ made with the positive x-axis such that

tan(φ) ≈ (75 N) / (349.904 N)   →   φ ≈ 12.098° ≈ 12.1°

(b) Letting θ vary from 0° to 180° would make v a function of θ :

u = (150 N) (cos(θ) i + sin(θ) j ) = (150 cos(θ) i + 150 sin(θ) j ) N

Then

u + v = ((220 + 150 cos(θ)) i + (150 sin(θ)) j ) N

→   M = ||u + v|| = √((220 + 150 cos(θ))² + (150 sin(θ))²) N

M = √(48,400 + 66,000 cos(θ) + 22,500 cos²(θ) + 22,500 sin²(θ)) N

M = 10 √(709 + 660 cos(θ)) N

(c) As a function of θ, u + v makes an angle α with the positive x-axis such that

tan(α) = (150 sin(θ) / (220 + 150 cos(θ))

→   α = tan⁻¹((15 sin(θ) / (22 + 15 cos(θ)))

(d) Filling in the table is just a matter of evaluating M and α for each of the given angles θ. For example, when θ = 0°,

M = 10 √(709 + 660 cos(0°)) N = 370 N

α = tan⁻¹((15 sin(0°) / (22 + 15 cos(0°))) = 0°

When θ = 30°, you get the same result as in part (a).

When θ = 60°,

M = 10 √(709 + 660 cos(60°)) N ≈ 323.3 N

α = tan⁻¹((15 sin(60°) / (22 + 15 cos(60°))) = 23.8°

and so on.