Question
The coefficient of static friction, $${\mu _s},$$ between block $$A$$ of mass $$2\,kg$$ and the table as shown in the figure, is 0.2. What would be the maximum mass value of block $$B,$$ so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless $$\left( {g = 10\,m/{s^2}} \right)$$
The coefficient of static friction, $${\mu _s},$$ between block $$A$$ of mass $$2\,kg$$ and the table as shown in the figure, is 0.2. What would be the maximum mass value of block $$B,$$ so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless $$\left( {g = 10\,m/{s^2}} \right)$$
A.
$$2.0\,kg$$
B.
$$4.0\,kg$$
C.
$$0.2\,kg$$
D.
$$0.4\,kg$$
Answer :
$$0.4\,kg$$
Solution :
Let the mass of the block $$B$$ be $$M.$$
In equilibrium, $$T - Mg = 0 \Rightarrow T = Mg\,.......\left( {\text{i}} \right)$$
If blocks do not move, then $$T = {f_s}$$
where, $${f_s}$$ = frictional force $$ = {\mu _s}R = {\mu _s}mg$$
$$\therefore T = {\mu _s}mg\,......\left( {{\text{ii}}} \right)$$
Thus, from Eqs. (i) and (ii), we have
$$\eqalign{ & Mg = {\mu _s}mg \cr & {\text{or}}\,M = {\mu _s}m \cr & {\text{Given,}}\,{\mu _s} = 0.2,\,m = 2\,kg \cr & \therefore M = 0.2 \times 2 = 0.4\,kg \cr} $$
Let the mass of the block $$B$$ be $$M.$$
In equilibrium, $$T - Mg = 0 \Rightarrow T = Mg\,.......\left( {\text{i}} \right)$$
If blocks do not move, then $$T = {f_s}$$
where, $${f_s}$$ = frictional force $$ = {\mu _s}R = {\mu _s}mg$$
$$\therefore T = {\mu _s}mg\,......\left( {{\text{ii}}} \right)$$
Thus, from Eqs. (i) and (ii), we have
$$\eqalign{ & Mg = {\mu _s}mg \cr & {\text{or}}\,M = {\mu _s}m \cr & {\text{Given,}}\,{\mu _s} = 0.2,\,m = 2\,kg \cr & \therefore M = 0.2 \times 2 = 0.4\,kg \cr} $$
