Question

In the figure shown, find the maximum value of $F$ (in $N$ ) for which both the blocks move together. Force $F$ is applied on upper block. (Take $g=10\, m / s ^{2}$ )

Answer 1

$f_{1 \,s \max }=0.6 \times 30 \times 10=180\, N$
$f_{1 \,k}=0.5 \times 30 \times 10=150\, N$
$f_{2\, s \max }=0.3 \times 35 \times 10=105 \,N$
$f_{2\, k}=0.2 \times 35 \times 10=70\, N$

If the two blocks move together, they must have common acceleration.
$\therefore a=\frac{F-f_{2 k}}{30+5}=\frac{F-70}{35}\,\,\,...(i)$
and $F-f_{1}=30\, a$
$\Rightarrow F-f_{1}=30\left(\frac{F-70}{35}\right)$[From (i)]
$\Rightarrow F-30\left(\frac{F-70}{35}\right)=f_{1}$
$\Rightarrow f_{1}=F-\frac{6}{7}(F-70)$
$ f_{1}=\frac{F}{7}+60$
But $f_{1}$ is static
$ \therefore \cdot f_{1} \leq f_{1 \,s \max }$
$\frac{F}{7}+60 \leq 180$
$\frac{F}{7} \leq 120 $
$ \therefore F_{\max }=840\, N$