Perhaps there is a pulley and block B is hanging ?
call tension in line = T
T = 2.25 (g-a)
Static friction force = .2(4.5+mc)g
so
T - .2 (4.5+mc)g = (4.5+mc) a
so
2.25 (g-a) -.2(4.5+mc)g = (4.5+mc)a
Now in PART A
a = 0
2.25 g -.2(4.5+mc)g = 0
2.25 = .9 + mc
so
mc = 1.35 Kg
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PART B is the same EXCEPT use 0.15 instead of .2
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For Part C, mc = 0 but leave a (and g in the equations
Block A is 4.5 kg and Block B is 2.25 kg.
a. determine the mass of block C that must be placed on block A to keep it from sliding if the coefficient of static friction between block A and the table is 0.2
b. If the coefficient of kinematic friction between the surface and block A is 0.15, what must the mass of block C be in order to move across the surface at a constant speed?
c. If block C is lifted, what is the acceleration of the block system if the coefficient of friction between block A and the table is 0.15
1 answer
1 answer