Question

A solid cube is placed on a horizontal surface. The coefficient of friction between them is $\mu$, where $\mu <\frac{1}{2}$. A variable horizontal force is applied on the cube’s upper face, perpendicular to one edge and passing through the mid-point of edge, as shown in figure. The maximum acceleration with which it can move without toppling is

• A. $g(1-2\mu )$
• B. $g(1+2\mu )$
• C. $\frac{g}{(1-2\mu )}$
• D. $\frac{g}{(1+2\mu )}$

Answer 1

The correct option is A $g(1-2\mu )$

The free body diagram of the block:
Here,
The normal reaction shifts at the edge of cube so as to counter the couple offered by $P$ and $f$.

Balancing the vertical forces gives $N=mg$
Now, equating torques in both direction to avoid cube's toppling, we get
$f\times l/2+P\times l/2=N\times l/2$
$\Rightarrow P=N-f=mg-f=mg-\mu mg$
Therefore, $P=mg-\mu mg$

As the block moves forward with acceleration $a$,
$P-f=ma$
$mg-\mu mg-\mu mg=ma$

$\Rightarrow a=g(1-2\mu )$