Answer:
D = 1.8677 miles
, θ = 24.28º at South of West
Explanation:
This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry
first displacement. d = 1.2 miles to 30º south of East
cos ( 360-30) = cos (-30) = x₁ / d
sin (-30) = y₁ / d
x₁ = d cos (-30)
y₁ = d sin (-30)
x₁ = 1.2 cos (-30) = 1,039 miles
y₁ = 1.2 sin (-30) = -0.6 miles
second shift. d = 2.0 miles to 20º West of South
cos (270-20) = x₂ / d
cos (250) = y₂ / d
x₂ = 2.0 cos 250 = -0.684 miles
y₂ = 2.0 sin250 = -1.879 miles
Third displacement. d = 1.6 miles to 30º South of West
cos (180 + 30) = x₃ / d
sin (210) = y₃ / d
x₃ = 1.6 cos 210 = -1.3856 miles
y₃ = 1.6 sin 210 = -0.8 miles
Fourth displacement. d = 2.6 miles to 15º West of North
cos (90 + 15) = x₄ / d
sin (105) = y₄ / d
x₄ = 2.6 cos 105 = -0.6729 miles
y₄ = 2.6 sin 105 = 2,511 miles
having all the components we add
x-axis (West-East direction)
X = x₁ + x₂ + x₃ + x₄
X = 1.039 -0.684 - 1.3846 - 0.6729
X = -1.7025 miles
Y = y₁ + y₂ + y₃ + y₄
Y = -0.6 -1.879 -0.8 +2.511
Y = -0.768
The modulus of this displacement is we use the Pythagorean theorem
D = √ (X² + Y²)
D = √ (1.7025² + 0.768²)
D = 1.8677 miles
let's use trigonometry to find the direction
tan θ = Y / X
θ = tan⁻¹ Y / x
θ = tan⁻¹ (0.768 / 1.7025)
θ = 24.28º
as the two components are negative this angle is in the third quadrant
therefore in cardinal direction form is
θ = 24.28º at South of West
is the normal force. So we say the friction force is directly proportional to the normal force with a constant of proportionality equal to the coefficient of friction.