Analysis of Double Inclined Plane System

Apr 5, 2025

Lecture Notes: Double Inclined Plane with a Car and Mass

Objective

  • Analyze the motion of a car on a double inclined plane connected to a mass.
  • Determine the average braking force needed to maintain constant speed descent.

Problem Setup

  • Mass of car (m1): 1700 kg
  • Mass of connected mass (m2): 1500 kg
  • Incline angle for car (theta): 30 degrees
  • Incline angle for mass (alpha): 20 degrees
  • Goal: Car descends at constant speed with applied braking force.

Initial Steps

  1. Draw Free Body Diagram (FBD)

    • Car on inclined plane:
      • Weight of the car (m1g) acts downward.
      • Resolving weight:
        • Parallel component: m1g * sin(theta)
        • Perpendicular component: m1g * cos(theta)
      • Normal force (perpendicular to the surface).
      • Braking force (opposite to motion).
  2. FBD for Connected Mass

    • Mass on incline:
      • Weight of the mass (m2g) acts downward.
      • Resolving weight:
        • Parallel component: m2g * sin(alpha)
        • Perpendicular component: m2g * cos(alpha)
      • Normal force (perpendicular to the surface).

Key Forces Analysis

  • Forces on car along incline:

    • m1g sin(theta) (down the slope)
    • Braking force (Fb) opposing m1 in descent
    • Force due to mass (m2g sin(alpha)) pulling the car
  • Total Force Equation:

    • m1g sin(theta) - Fb - m2g sin(alpha) = 0 (constant speed implies net force is zero)

Calculation

  • Using Newton’s Second Law (F = ma)

    • As acceleration (a) is zero, total force is zero.
    • Solving for braking force (Fb):
      • Fb = m1g sin(theta) - m2g sin(alpha)
  • Plugging in Values:

    • m1 = 1700 kg, m2 = 1500 kg, g = 9.8 m/s²
    • theta = 30°, alpha = 20°
    • Fb = 1700 * 9.8 * sin(30) - 1500 * 9.8 * sin(20)
    • Result: Fb = 3300 N

Conclusion

  • The braking force required to maintain a constant descent speed is 3300 N.

Additional Notes

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