Question

Block $A$ of mass $3kg$ rests on another block $B$ of mass 7 $kg$. The coefficient of friction between $A$ and $B$ is $0.4$ while the coefficient of friction between $B$ and the horizontal floor on which $B$ rests is $0.55$. Find the force of friction between $A$ and $B$, when a horizontal force of $50N$ is applied on the block $B$. (Use $g=10m/s^2$ )

• A. 0
• B. $5N$
• C. $4N$
• D. $1.2N$

Answer 1

Answer: (A)

Given, mass of block $A$ and $B$ are $\left(m_A\right),\left(m_B\right)$ be $3kg$ and $7kg$, respectively.
Coefficient of friction between block $A$ and $B,B$ and floor $\left({\mu }_A\right),\left({\mu }_B\right)$ be $0.4$ and $0.55$ respectively,
Applied force $(F)$ on block $B$ be $50N$.
Let $f_{A^{\prime }}{f}_B$ be the friction forces between blocks $A,B$ and $B$, floor respectively.

Now, from block $B$ diagram,
$f_B={\mu }_B{N}_B\dots \text{ (i) }$
where, $N_B$ be the normal reaction on block $B$
$={\mu }_B\left(m_A+m_B\right)g$
$=0.55(3+7)10=55N$
Similarly, $f_A={\mu }_A{N}_A$
where, $N_A$ is normal reaction on block $A$.
$\therefore f_A={\mu }_A{m}_{A}g$
$f_A=0.4\times 3\times 10$
$=12N$
Now, since, $f_B>f$
$\therefore$ the block $B$ will not move.
Hence, $f_A=0N$