Question

If coefficient of friction between all surfaces is 0.4, then find the minimum force $F$ required to just keep the system from moving.
[Take $g=10{\text{m/s}}^2$]

• A. $62.5\text{N}$
• B. $150\text{N}$
• C. $135\text{N}$
• D. $50\text{N}$

Answer 1

The correct option is A $62.5\text{N}$

If tension in string connecting $15\text{kg}$ is $T$, then tension in other string connecting $15\text{kg}$ block is $2T$
[from equilibrium of pulley]


Normal force acting on each block is $F$.
[since wall is vertical]
Hence, limiting friction $f=\mu F=0.4F$

From FBD of the two blocks,
For vertical equilibrim of $25\text{kg}$ block:
$2T=250$ or $T=125\text{N}$

For vertical equilibrim of $15\text{kg}$ block:
$T+0.4F=150$
i.e $F=62.5\text{N}$ is the minimum force for which the system doesn't move.