Answer:
1. 50·√2·i - 50·√2·j
2. The magnitude is 220.49 pounds, the direction 16.7°
3. 197.9 N
4. The resultant speed is 778.31 km/hr
The true course of the plane is 27.72°
5. The direction of the boat is 354.315°
The speed of the boat is 34.19 knot
Step-by-step explanation:
1. Given that the airplane is descending at 100 mph at an angle 45° below the horizontal, we have;
Horizontal component = 100 × cos(45°) = 100 × (√2)/2 = 50·√2 mph
The vertical component = 100 × sin(-45°) = 100 × (-√2)/2 = -50·√2 mph
Which gives;
50·√2·i - 50·√2·j
2. The given forces are;
F₁ = 250 pounds at 25° with the horizontal
F₁ₓ = 250 × cos(25°) = 226.58 pounds
[tex]F_{1y}[/tex] = 250 × sin(25°) = 105.655 pounds
F₂ = 45 pounds at 250° with the horizontal
F₂ₓ = 45 × cos(250°) = -15.39 pounds
[tex]F_{2y}[/tex] = 45 × sin(250°) = -42.29 pounds
The magnitude of the resultant force F = √((F₁ₓ - F₂ₓ)² + ([tex]F_{1y}[/tex] - [tex]F_{2y}[/tex])²)
F = √((226.58 + (-15.39))² + (105.655 + (-42.29))²)
F = √(211.19)² + (63.365)²) = 220.49 pounds
The direction of the resultant force θ = [tex]tan^{-1} \left (\dfrac{F_y}{F_x} \right) = tan^{-1} \left (\dfrac{63.365}{211.19} \right) = 16.7^ {\circ}[/tex]
3. The horizontal force = 120 × cos(60°) + 180 × cos(40°) = 197.9 N
4. Direction of airplane = 30° North of East. we have;
Speed of airplane, [tex]S_{A}[/tex] = 724 km/hr
The x component of the speed of the airplane, [tex]S_{Ax}[/tex] = 724 × cos(30°) = 627 km/hr
The y component of the speed of the airplane, [tex]S_{Ay}[/tex] = 724 × sin(30°) = 362 km/hr
The wind speed, [tex]S_w[/tex] = [tex]S_{wx}[/tex] = 62 km/hr
The resultant speed in the x direction, Sₓ = [tex]S_{Ax}[/tex] + [tex]S_{wx}[/tex] = 627 + 62 = 689 km/hr
The resultant speed in the y direction, [tex]S_{y}[/tex] = 362 km/hr
The magnitude of the resultant speed, S = √(Sₓ² + [tex]S_{y}[/tex]²) = √(689² + 362²) = 778.31 km/hr
The direction of the airplane θ = [tex]tan^{-1} \left (\dfrac{S_y}{S_x} \right) = tan^{-1} \left (\dfrac{362}{689} \right) = 27.72^ {\circ}[/tex]
5. The x, [tex]S_{yx}[/tex] and y, [tex]S_{yy}[/tex] components of the speed of the yacht are;
[tex]S_{yy}[/tex] = 20×sin(185°) = -1.743 knots
[tex]S_{yx}[/tex] = 20×cos(185°) = -19.924 knots
The x, [tex]S_{wx}[/tex] and y, [tex]S_{wy}[/tex] components of the speed of the yacht are;
[tex]S_{wx}[/tex] = 15×sin(290°) = -14.1 knots
[tex]S_{wy}[/tex] = 15×cos(290°) = 5.13 knots
The resultant x, [tex]S_{x}[/tex], and y, [tex]S_{y}[/tex] are found as follows;
[tex]S_{x}[/tex] = [tex]S_{wx}[/tex] + [tex]S_{yx}[/tex] = -19.924 + -14.1 = -34.024 knots
[tex]S_{y}[/tex] = [tex]S_{yy}[/tex] + [tex]S_{wy}[/tex] = -1.743 + 5.13 = 3.387
The resultant speed is given by the relation;
S =√([tex]S_{x}[/tex]² + [tex]S_{y}[/tex]²) = √((-34.024)^2 + 3.387^2) = 34.19 knot
The direction of the boat, θ = [tex]tan^{-1} \left (\dfrac{S_y}{S_x} \right) = tan^{-1} \left (\dfrac{3.387}{-34.024} \right) = -5.685^ {\circ}[/tex]
The direction of the boat = -5.685 = 354.315°
The speed of the boat = 34.19 knot.
