Torque and Lever Arm

Ease of Pushing a Door

• Applying force at different points on a door:
  • Position A: Easiest to push
  • Position B: Harder to push
  • Position C: Very difficult to push
• Difference is due to torque
• Torque ( au) = Force (F) × Moment Arm (r)
• Greater r means greater torque for the same force

Torque Calculation

• Force Perpendicular to Door:

  • Let L be the distance from force application point to axis of rotation
  • Moment arm r is the perpendicular distance from the axis of rotation to the line of action of the force
  • Torque ( au) = F × r
  • Equivalent equations:
    • Torque = F imes L (perpendicularly applied force)

• Force at an Angle ( heta):

  • Draw line of action and perpendicular line from axis of rotation
  • Create a right triangle:
    • r (moment arm) = L × sin( heta)
  • Torque = F × L × sin( heta)
  • Alternatively:
    • Decompose force into perpendicular component: F × sin( heta)
    • Torque = Perpendicular force component × L = F × L × sin( heta)

Direction of Torque

• Counterclockwise rotation: Positive torque
• Clockwise rotation: Negative torque

Example Problems

• Problem 1:

  • Apply a 200N force 3m from axis (+ direction):
    • Torque1 = 200N × 3m = +600 Nm
  • Apply a 400N force 1.5m from axis (- direction):
    • Torque2 = 400N × 1.5m = -600 Nm
  • Net Torque: 600 Nm - 600 Nm = 0 Nm (equilibrium)

• Problem 2:

  • Forces applied to a propeller at different angles and distances:
    • Torque1 = 300N at 4m (clockwise): -300 × 4 = -1200 Nm
    • Torque2 = 600N at 3m, 60° (counterclockwise): 600 × 3 × sin(60) ≈ +1558.8 Nm
    • Torque3 = 500N at 5m, 50° (counterclockwise): 500 × 5 × sin(50) ≈ +1915.1 Nm
    • Net Torque: -1200 + 1558.8 + 1915.1 ≈ +2274 Nm (counterclockwise rotation)

Leverage: Seesaw and Simple Machines

• Seesaw Example:

  • 300N applied 6m from the fulcrum (input torque):
    • Torque1 = 300N × 6m = 1800 Nm
  • Output torque should be equal:
    • Torque2 = F2 × 3m
    • 1800 Nm = F2 × 3m
    • F2 = 600N (output force)
  • Mechanical Advantage (MA) = Output Force / Input Force = 600N / 300N = 2

• Shovel Example:

  • 200N force applied at 1m from axis (input torque):
    • Torque1 = 200N × 1m = 200 Nm
  • Output arm distance = 0.1m
    • Torque2 = F2 × 0.1m
    • 200 Nm = F2 × 0.1m
    • F2 = 2000N (output force)
  • Mechanical Advantage: Output Force / Input Force = 2000N / 200N = 10
    • Smaller force applied further from the axis generates a larger force closer to the axis
    • Proper placement is key to maximizing force efficiency