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Assume the table is frictionless. Blocks 1 and 2 have coefficients μₛ and μₖ between them. Assume μₛ = 2μₖ....

Assume the table is frictionless. Blocks 1 and 2 have coefficients μₛ and μₖ between them. Assume μₛ = 2μₖ. The system is initially at rest.
a. Derive an equation for the acceleration of the system in terms of Block 5 and other variables and constants when Block 1 moves with Block 2.
b. For the conditions in (a), derive T₂₃, T₃₄, T₄₅.
c. Derive an equation for the minimum mass of Block 5 that will cause Block 1 to begin moving with respect to Block 2.
d. For the conditions in (c), derive T₂₃, T₃₄, T₄₅.
e. For the conditions in (c), derive the acceleration of Block 1 from the perspective of Block 2.

Answer

  1. To derive the acceleration of the system, consider the forces acting on Block 5. The force due to gravity on Block 5 is F = m₅g. This force causes the entire system to accelerate. The total mass of the system is 2M + 8M + 4M + 3M + m₅ = 17M + m₅. Therefore, the acceleration a of the system is given by:

    a = \( \frac{m₅g}{17M + m₅} \)

  2. For the conditions in (a), the tensions can be derived as follows:

    • T₂₃ = 2Ma
    • T₃₄ = (2M + 8M)a = 10Ma
    • T₄₅ = (2M + 8M + 4M)a = 14Ma
  3. To find the minimum mass of Block 5 that will cause Block 1 to begin moving with respect to Block 2, consider the static friction force between Block 1 and Block 2. The maximum static friction force is fₛ = μₛ(2Mg). For Block 1 to start moving, the force exerted by Block 5 must overcome this static friction:

    m₅g = μₛ(2Mg)

    Since μₛ = 2μₖ, we have:

    m₅ = 2μₖ(2M)

  4. For the conditions in (c), the tensions are:

    • T₂₃ = 2M(a + aₛ)
    • T₃₄ = (2M + 8M)(a + aₛ)
    • T₄₅ = (2M + 8M + 4M)(a + aₛ)

    Where aₛ is the acceleration due to static friction.

  5. For the conditions in (c), the acceleration of Block 1 from the perspective of Block 2 is:

    a₁₂ = a - aₛ

    Where aₛ is the acceleration due to static friction.

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