Pulley Problem?

Question

block A has mass m_A = 2.09 kg, block B has mass m_B = 0.450 kg, and the rope connecting them has a nonzero mass 0.159 kg. The rope has a total length 1.07 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. Ignore any sag in the horizontal part of the rope.

If there is friction between block A and the table top, with mu_k = 0.201 and mu_s = 0.248, 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.

Force of the hanging mass is that of the vertical portion of the rope + mass of Block B * gravity
(M_r + M_B)(g) +
(.159+.450)(9.8) = 5.9682N = -5.9682N = Fnet of Block B

Tension T from Block A equals to Fnet of Block B = 5.9682N
Static Friction = (0.248)(n) = -5.079N

n = Mass of Block A * gravity = (2.029)(9.8)

Do I need Kinetic Friction in this problem?

Please Guide me. I'm Lost...

Thank You

Answer 1

No, you're not lost, but you did neglect an important detail of the problem -- only a portion (of length d) of the rope is vertical.

If length d of the rope is vertical, the mass of that piece of rope is (0.159/1.07)*d.

Otherwise, your method is essentially correct. The weight of Block B plus the weight of the rope gives the tension. We want this tension to just balance the force of (static) friction acting on Block A, which you correctly calculated as
(mu_s)(M_A)g. So find the value for d such that

[M_B + (0.159/1.07)*d]g = (mu_s)(M_A)g

Note that you can divide g out.

And at least for this problem, the coefficient of kinetic friction is not needed.