Question

Block $A$ of mass $3 kg$ 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 $50 N$ is applied on the block $B$. (Use $g =10 \,m / s ^2$ )

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

Answer 1

Answer: (A)

Given, mass of block $A$ and $B$ are $\left(m_A\right),\left(m_B\right)$ be $3 \,kg$ and $7 \,kg$, 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 $50 \,N$.
Let $f_{A^{\prime}} f_B$ be the friction forces between blocks $A, B$ and $B$, floor respectively.

Now, from block $B$ diagram,
$f_B=\mu_B N_B \ldots \text { (i) }$
where, $N_B$ be the normal reaction on block $B$
$ =\mu_B\left(m_A+m_B\right) g$
$ =0.55(3+7) 10=55 N$
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 $
$ =12 N$
Now, since, $f_B>f$
$\therefore$ the block $B$ will not move.
Hence, $f_A=0 N$