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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