Consider the system shown in the figure (Faure 1). Block A weighs 47.2 N and blo

Question

Consider the system shown in the figure (Faure 1). Block A weighs 47.2 N and block B weighs 28.4 N. Once block B is set into downward motion, it descends at a constant speed. Calculate the coefficient of kinetic friction between block A and the tabletop. A cat, also of weight 47.2 N, falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude? A cat, also of weight 47.2 N, falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction?

Explanation / Answer

Given weight of block A is 47.2 N, block B = 28.4N

forces acting on block A,B are T - mue_k *Mg_A = Ma----->1
                                T- Mg_B = Ma

solving for mue_k = Mg_B / Mg_A = 28.4/47.2 = 0.602 now

when cat falls sleep on block A with wieght 47.2 N the weight of block is = 2*47.2 N = 94.4 N

coefficient of kinetic friction is mue_k = 28.4/94.4 = 0.30

and tension in the string T - Mg_B = Mg*a
               T = M_B(a+g)       Mg_B = 28.4 N => M_B = 2.9 kg

                                          Mg_A = 94.4 N ==> M_A = 9.63 kg

                                              T = 2.9(a+9.8 ) N

now substituting the tension in 1

           2.9(a+9.8 )- mue_k*Mg_A = M_A*a
             2.9(a+9.8 )- 0.3*94.4 = 94.4*a  

               a = 0.001093 m/s^2

moves upwards.

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