Evaluate the motion - smooth pulley, Mechanical Engineering

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.


1098_Evaluate the motion - Smooth pulley.jpg



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 NB (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.

∑ Fx  = 0,     T - 0.2 N B  - 2 a = 0

                                       T = 0.2 N B  + 2 a               -------- (1)

∑ Fy  = 0,   N B  = 2 × 9.81   = 19.6 N       --------- (2)

Likewise, considering ∑ Fy  = 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 NB, we obtain

(150 × 8.83 + 0.2 × 19.6 × 150 /19.6)= - x

∴ x = - 97.58 cm.

Negative sign mention that the reaction NB is acting on the right of the mass centre rather than to the left assumed.

Posted Date: 1/29/2013 2:55:48 AM | Location : United States

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