Question

A block $A$ of mass $5\text{kg}$ rests over another block $B$ of mass $3\text{kg}$ placed over a smooth horizontal surface. There is friction between $A$ and $B$. A horizontal force $F_1$ gradually increasing from zero to a maximum value is applied to $A$ so that the blocks move together without relative motion. Instead of this, another horizontal force $F_2$ gradually increasing from zero to a maximum value is applied to $B$ so that the blocks moves together without relative motion. Then -

• A. $F_1\left(\text{max}\right):F_2\left(\text{max}\right)=25:24$
• B. $F_1\left(\text{max}\right):F_2\left(\text{max}\right)=5:3$
• C. $F_2\left(\text{max}\right):F_1\left(\text{max}\right)=25:24$
• D. $F_2\left(\text{max}\right):F_1\left(\text{max}\right)=5:3$

Answer 1

The correct option is B $F_1\left(\text{max}\right):F_2\left(\text{max}\right)=5:3$

Here, we have two different cases.

Case-$\text{I}$ :
Let the coefficient of friction between the blocks be $\mu$ and their maximum acceleration be $a$.

So, we have FBD of both blocks as -


For $5\text{kg}$ block:

$N^{\prime }=5g...\left(i\right)$, and,

$F_1-f=5a$

$\Rightarrow F_1-\mu N^{\prime }=5a$

$\Rightarrow F_1-5\mu g=5a...\left(ii\right)\left[\text{From}\right(i\left)\right]$

For $3\text{kg}$ block:

$f=3a$

$\Rightarrow \mu N^{\prime }=3a$

$\Rightarrow 5\mu g=3a$

$\Rightarrow a=\frac{5\mu g}{3}...\left(iii\right)$

From equation $\left(ii\right)$ and $\left(iii\right)$,

$F_1-5\mu g=\frac{25\mu g}{3}$

$\Rightarrow F_1\left(\text{max}\right)=\frac{40\mu g}{3}...\left(A\right)$

Case-$\text{II}$ :

For $5\text{kg}$ block:

$N^{\prime }=5g...\left(iv\right)$, and,

$f=5a...\left(v\right)$

$\Rightarrow \mu N^{\prime }=5a$

$\Rightarrow 5\mu g=5a\left[\text{From}\right(iv\left)\right]$

$\Rightarrow a=\mu g...\left(vi\right)$

For $3\text{kg}$ block:

$F_2-f=3a$

$\Rightarrow F_2-5a=3a\left[\text{From}\right(v\left)\right]$

$\Rightarrow F_2=8a$

$\Rightarrow F_2\left(\text{max}\right)=8\mu g\left[\text{From}\right(vi\left)\right]...\left(B\right)$

From $\left(A\right)$ and $\left(B\right)$,

$F_1\left(\text{max}\right):F_2\left(\text{max}\right)=\frac{40\mu g}{3}:8\mu g=5:3$