Math Problem Statement

Given: CD = 0.45 AF = 2.38m² W = 13 500N Density = 1.2256kg/m³ frl = 0.0131 Nb = 0.9 \mu = 0.42 S = 62m V1 = 43.05556 m/s Yb = 1.04

Find final velocity considering aerodynamic resistance on level ground and at 3.4336°

Solution

To find the final velocity considering aerodynamic resistance on level ground and at a slope of 3.4336°, we'll need to apply principles of aerodynamics and mechanics. The aerodynamic drag force FDF_D and the force due to gravity FgF_g need to be calculated.

Here's a step-by-step breakdown:

1. Aerodynamic Drag Force FDF_D

The drag force can be calculated using the drag equation:

FD=12×Density×CD×AF×V2F_D = \frac{1}{2} \times \text{Density} \times \text{CD} \times \text{AF} \times V^2

Where:

  • Density = 1.2256 kg/m³
  • CD = 0.45
  • AF = 2.38 m²
  • VV = Final velocity (unknown)

2. Force Due to Gravity FgF_g

The gravitational force component acting down the slope is:

Fg=W×sin(θ)F_g = W \times \sin(\theta)

Where:

  • WW = 13,500 N (Weight)
  • θ\theta = 3.4336° (Slope angle)

3. Total Resistance Force FRF_R

Total resistance considering rolling resistance FrrF_{rr}, aerodynamic drag, and gravitational force is:

FR=Frr+FD+FgF_R = F_{rr} + F_D + F_g

Where rolling resistance FrrF_{rr} is:

Frr=μ×W×cos(θ)F_{rr} = \mu \times W \times \cos(\theta)

Where:

  • μ\mu = 0.42 (Coefficient of rolling resistance)

4. Final Velocity Calculation

We will assume that the final velocity VfV_f can be found by balancing the driving force with the total resistance. The equation for the balance of forces is:

W×Yb=FRW \times Yb = F_R

Where YbYb is the tractive force.

Let's calculate these values:

  1. Calculate the gravitational force component FgF_g: Fg=13,500×sin(3.4336)F_g = 13,500 \times \sin(3.4336^\circ)

  2. Calculate rolling resistance FrrF_{rr}: Frr=0.42×13,500×cos(3.4336)F_{rr} = 0.42 \times 13,500 \times \cos(3.4336^\circ)

  3. Calculate aerodynamic drag force FDF_D using the formula: FD=12×1.2256×0.45×2.38×Vf2F_D = \frac{1}{2} \times 1.2256 \times 0.45 \times 2.38 \times V_f^2

  4. Determine final velocity VfV_f by solving: 13,500×1.04=Frr+12×1.2256×0.45×2.38×Vf2+13,500×sin(3.4336)13,500 \times 1.04 = F_{rr} + \frac{1}{2} \times 1.2256 \times 0.45 \times 2.38 \times V_f^2 + 13,500 \times \sin(3.4336^\circ)

I'll perform the calculations now.The final velocity VfV_f, considering aerodynamic resistance on level ground and at a slope of 3.4336°, is approximately 107.41 m/s.

Would you like more details or have any further questions?

Here are five related questions you might consider:

  1. How does the slope angle affect the final velocity in this scenario?
  2. What would be the final velocity if the air density were higher or lower?
  3. How would changing the drag coefficient CDC_D affect the final velocity?
  4. What is the impact of the rolling resistance coefficient μ\mu on the final velocity?
  5. How does the frontal area AFA_F influence the aerodynamic drag?

Tip: The drag force increases with the square of the velocity, meaning that even small increases in speed can result in significantly higher resistance forces.

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Math Problem Analysis

Mathematical Concepts

Aerodynamics
Mechanics
Force Analysis

Formulas

Drag Equation
Force due to Gravity
Rolling Resistance

Theorems

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Suitable Grade Level

Advanced Engineering