Evaluate the motion - Smooth pulley:
Block B is accelerated along the horizontal plane via mass A attached to it by a flexible inextensible massless rope passing over a smooth pulley as illustrated in Figure (a). Suppose that the coefficient of friction among block B and the plane is 0.20. Mass of block B is 2 kg and that of A is 1.20 kg. Discuss the motion.
Solution
We draw Free Body Diagram for the block under consideration as shown in Figure (b). Suppose acceleration of the block B is a to the right as illustrated. We also should note the position of N_{B} (reaction of the block B) which is illustrated acting at a distance x from the vertical line through the mass centre of block B. (Block B being in motion, its reaction has an eccentricity (= x) to balance the moments).
Now we can write down equilibrium equations for the block B.
∑ F_{x} = 0, T - 0.2 N _{B} - 2 a = 0
T = 0.2 N _{B} + 2 a -------- (1)
∑ F_{y} = 0, N _{B} = 2 × 9.81 = 19.6 N --------- (2)
Likewise, considering ∑ F_{y} = 0, for block A, we get
T = 1.2 g - 1.2 a
= 11.77 - 1.2 a
From Eqs. (1) and (3), we obtain
0.2 N_{ B} + 2 a = 11.77 - 1.2 a
0.2 × 19.6 + 2 a = 11.77 - 1.2 a ------(3)
∴ 3.2 a = 7.85
∴ a = 2.45 m / sec ^{2}
∴ T = 11.77 - 1.2 a
= 11.77 - 1.2 × 2.45 = 8.83 N.
∑ M _{C.G.} = 0 for the block B.
We get 150 T + 0.2 N _{B} × 150 + N _{B} x = 0
Substituting the values of T and N_{B}, we obtain
(150 × 8.83 + 0.2 × 19.6 × 150 /19.6)= - x
∴ x = - 97.58 cm.
Negative sign mention that the reaction N_{B} is acting on the right of the mass centre rather than to the left assumed.