What minimum force is needed to lift the piano mass using the pulley apparatus shown in Fig 4 66 720

Introduction

• Objective: To find the force F needed to lift the mass m using a system of pulleys.

System Description

• Diagram Components:

  • Fixed pulley connected to another pulley.

  • Force F is applied at one point to lift mass m (piano).

Forces Acting on the System

• Tension Forces:

  • Tension in the ropes is represented as T1, T2, T3, T4.

• Weight of the Block:

  • Weight (mg) of mass m acts downward.

Free Body Diagram (FBD) for Mass m

• Vertical Forces:

  • Downward: Weight (mg)

  • Upward: Tension (T4)

• Equilibrium Condition:

  • The net vertical force must balance:

    • T4 = mg

FBD for Second Pulley

• Forces Acting on Second Pulley:

  • Tension forces: T2 (from one side) and T1 (from the other side), T4 acting downward

• Equilibrium Equation:

  • T4 = T1 + T2

  • Since T1 = T2 (same rope), we can express this as:

    • T4 = 2T1

Solving for Tension T1

• Replacing T4:

  • From T4 = mg, equating gives:

    • mg = 2T1

• Rearranging:

  • T1 = mg/2

• Conclusion for Tensions:

  • T1 = T2 = F = mg/2

Finding Each Tension Force

• From earlier results:

  • T4 was mg

  • For T3 (tension for pulley 1):

    • Forces Acting:

      • T3 upwards, T1 and T2 downwards, and applied force F.

• Equilibrium for Pulley 1:

  • T3 = T1 + T2 + F

• Replacing T1 and T2:

  • Since T1 = T2 = F = mg/2:

  • T3 = (mg/2) + (mg/2) + (mg/2)

  • T3 = 3(mg/2)

Summary of Tensions

• T1 = T2 = F = mg/2

• T3 = 3(mg/2)

• T4 = mg

Conclusion

• Force F needed to lift mass m is mg/2.

• The tension forces are:

  • T1 = mg/2

  • T2 = mg/2

  • T3 = 3mg/2

  • T4 = mg

• This completes the problem.