Question

What should be the maximum value of $M$ so that the $4\text{kg}$ block does not slip over the $5\text{kg}$ block: (Take $g=10{\text{m/s}}^2)$

• A. $12\text{kg}$
• B. $8\text{kg}$
• C. $10\text{kg}$
• D. $6\text{kg}$

Answer 1

The correct option is D $6\text{kg}$

Static friction $\left(f\right)$ will act at its maximum value, so that both blocks $4\text{kg}$ and $5\text{kg}$ can move together corresponding to maximum mass of block $M$.
Hence, system of blocks $4\text{kg},5\text{kg}$ and $M$ will move together with acceleration ${}^{\prime }{a}^{\prime }$, due to string constraint.


Net external force on system of blocks $(4\text{kg},5\text{kg},M)$ is $Mg$ downwards.
$\therefore a=\frac{F_{\text{net}}}{\text{Total mass}}=\frac{Mg}{M+5+4}$
$=\frac{Mg}{M+9}$ ... (i)

From FBD of $4\text{kg}$:


$N_1=mg=40\text{N}$
& $f_{\text{max}}=4a$ ... (ii)
$f_{\text{max}}=\mu N_1=0.4\times 40=16\text{N}$
$\therefore a=\frac{16}{4}=4{\text{m/s}}^2$

Putting in eq.(i) gives:
$4=\frac{M\times 10}{M+9}\Rightarrow 6M=36$
$\therefore M=6\text{kg}$