The setup in the figure shorn a cart/block of mass mi attached by a cord under t

Question

The setup in the figure shorn a cart/block of mass mi attached by a cord under tension, T, over a pulley to a hanging mass of m2. There is friction between m1 and the table. Assume m1 and m2 move with the same acceleration, a, and that the coefficient of kinetic friction between m1 and the table is mux. Calculate the coefficient of kinetic friction m in terms of a, m1, m2 and g. The setup in the figure below is similar to that above, except that now m1 resides on an incline. There is friction between mt and the incline with the same coefficient of kinetic friction mux as in part A. Assume the acceleration of the hanging mass is downwards, and calculate the acceleration, a, in terms of 0, m1, m2, g and mux.

Explanation / Answer

A )

m2g - T = m2a ;

T - m1g - m1g = m1a ;

adding the two equations :

(m2-m1)g - (m2+m1)a =  m1g

=> =[(m2-m1)g - (m2+m1)a]/m1g

B )

m2g - T = m2a

T - m1gsin - m1gcos = m1a

adding the two equations :

we have

m2g -m1gsin - m1gcos = (m1+ m2)a

=> a = [m2g -m1gsin - m1gcos]/ (m1+ m2)

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