In the system shown in the figure , block A has mass m_A = 2.08 kg, block B has mass m_B = 0.370 kg, and the rope connecting them has a nonzero mass 0.203 kg. The rope has a total length 1.03 m and the pulley has a very small radius. Let d be the length of rope that hangs vertically between the pulley and block B.

If there is friction between block A and the table top, with mu_k = 0.195 and mu_s = 0.255, find the minimum value of the distance d such that the blocks will start to move if they are initially at rest. Ignore any sag in the horizontal part of the rope.

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To find the minimum value of the distance (d) such that the blocks start to move, we need to consider the forces acting on the system.

Let's analyze the forces acting on each block individually:

For block A:
1. Weight (mg_A): The weight of block A acts vertically downward and has a magnitude of (m_A) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
2. Normal force (N_A): The normal force acts perpendicular to the table surface and opposes the weight of block A.
3. Friction force (f_A): The friction force opposes the motion of block A, and its magnitude depends on the coefficient of kinetic friction (mu_k) and the normal force (N_A). The friction force is given by f_A = mu_k * N_A.

For block B:
1. Weight (mg_B): The weight of block B acts vertically downward and has a magnitude of (m_B) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
2. Tension in the rope (T): The tension in the rope acts in both directions, upward on block B and downward on the pulley.

Considering the system as a whole:
1. Tension in the rope (T): Since the rope is inextensible, the tension in the rope is the same throughout its length.
2. Force due to the hanging rope (T * cosθ): This force acts horizontally and opposes the force of friction acting on block A, where θ is the angle between the horizontal part of the rope and the vertical part.

To determine the minimum value of d at which the blocks start to move, we need to find the point where the static friction force (fs) reaches its maximum value. At this point, the force due to the hanging rope will be equal to the maximum static friction force, causing the blocks to start moving.

The maximum static friction force (fs_max) is given by fs_max = mu_s * N_A, where mu_s is the coefficient of static friction.

Now, let's solve for the equilibrium condition where the maximum static friction force is equal to the force due to the hanging rope:

fs_max = T * cosθ

Since fs_max = mu_s * N_A, we can write:

mu_s * N_A = T * cosθ

Now, let's find expressions for N_A and T using the given information:

N_A = mg_A (since block A is not accelerating vertically)
T = mg_B + m_R * g + fs_max (where m_R is the mass of the rope)

Substituting these expressions into the equation above, we get:

mu_s * mg_A = mg_B + m_R * g + fs_max * cosθ

Now, let's solve for fs_max:

fs_max = mu_s * mg_A - mg_B - m_R * g

Since fs_max = T * cosθ, we have:

T * cosθ = mu_s * mg_A - mg_B - m_R * g

We also know that the horizontal component of the tension in the rope is given by T * cosθ. Since there is no acceleration in the horizontal direction, this force should be equal to the force due to the hanging rope:

T * cosθ = T * d / L

Here, L is the total length of the rope.

Now, we can substitute T * cosθ in terms of T and solve for d:

T * d / L = mu_s * mg_A - mg_B - m_R * g

d = (mu_s * mg_A - mg_B - m_R * g) * L / T

Finally, we can substitute the expressions for T, m_R, and L to obtain d in terms of the given quantities:

d = (mu_s * mg_A - mg_B - m_R * g) * L / (mg_B + m_R * g + mu_s * mg_A)