Question

If the coefficient of friction between all surfaces is $0.4$ then find the minimum force $F$ applied to have system in equilibrium.

• A. $62.5\text{N}$
• B. $72.5\text{N}$
• C. $82.5\text{N}$
• D. $92.5\text{N}$

Answer 1

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

Force has been applied along the horizontal direction, hence the minimum value of $F$ will lead to the minimum value of normal reaction between all surfaces.
$FBD$ of $15\text{kg}$ block:


The block will tend to slip downward, therefore friction will act upwards.
Applying equilibrium condition along horizontal,
$N=F$ ...........(1)
and along vertical,
$T+f=150$
$\Rightarrow T+\mu N=150$
$\Rightarrow T+0.4N=150$
$\Rightarrow T+0.4F=150$ .........(2) [from (1)]
$FBD$ of $25\text{kg}$ block-


Applying equilibrium condition along horizontal,
$N^{{}^{\prime }}=N=F$ ...........(3)
and along vertical,
$2T+f^{{}^{\prime }}=250+f$
$\Rightarrow 2T+\mu N^{{}^{\prime }}=250+\mu N$
$\Rightarrow 2T+\mu N=250+\mu N$ [from (3)]
$\Rightarrow T=125\text{N}$
Putting this in (2), we get,
$F=62.5\text{N}$