Question

Block A of mass 2 kg is placed over block B of
mass 8 kg. The combination is placed over a rough
horizontal surface. Coefficient of friction between
B and the floor is 0.5. Coefficient of friction between
Aand B is 0.4. Ahorizontal force of 10 N is applied
on block B. The force of friction between A and B
is (g = 10 m/s2)​

Answer 1

The fig. shows two blocks A and B of masses 2 kg and 8 kg respectively according to the question.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(-25,0){\line(1,0){50}}\put(-10,0){\framebox(20,10){\sf{8\ kg}}}\put(-5,10.3){\framebox(10,10){\sf{2\ kg}}}\put(11.5,-4){ \sf{\mu_1=0.5} }\put(-25,10){ \sf{\mu_2=0.4} }\put(10,5){\vector(1,0){20}}\put(31,4){ \sf{10\ N} }\put(-1,22){ \sf{A} }\put(11.5,9){ \sf{B} }\end{picture}$

The reaction acted upon between the blocks A and B is equal to the weight of the block A, i.e.,

$\longrightarrow\sf{R=2\times10\ N}$

$\longrightarrow\sf{R=20\ N}$

Then the frictional force acting between the blocks A and B is,

$\longrightarrow\sf{f=20\times0.4\ N}$

$\longrightarrow\sf{f=8\ N}$

But this frictional force is expected. Actually there's no friction between the blocks since the system undergoes no motion!

The reaction acted upon between the block B and the floor is equal to the weight of the whole system, i.e.,

$\longrightarrow\sf{R'=(2+8)\,10\ N}$

$\longrightarrow\sf{R'=10\times10\ N}$

$\longrightarrow\sf{R'=100\ N}$

So the frictional force acting between the block B and the floor is,

$\longrightarrow\sf{f'=100\times0.5\ N}$

$\longrightarrow\sf{f'=50\ N}$

Hence the net frictional force acting on the block B is,

$\longrightarrow\sf{f_{net}=f+f'}$

$\longrightarrow\sf{f_{net}=8\ N+50\ N}$

$\longrightarrow\sf{f_{net}=58\ N}$

This frictional force is very greater than the applied force, 10 N. Hence the block B doesn't move, and hence there's no relative motion between the blocks A and B, and therefore there's no friction between them.