Question

a horizontal force is applied at the top of an equilateral triangular block having mass m. The minimum coefficient of friction required to topple the block before translation will be

Answer 1

see diagram.

At the point C,  there is friction in the direction opposite to F.  So friction force must be more than or equal to F, so that the block does not slide forward.

Balancing forces in the horizontal and vertical directions:
       N = m g    and         f  >=  F

  f = force of friction =  μ N = μ m g  >= F

=>  $μ \ge \frac{F}{mg}$

Force F required to topple the block, to make the triangular block rotate :
       
Let BC make an angle x with the horizontal.  So angle ACH = 90 - (60+x) = 30 - x.
The arm length for Force F from pivot C is :    CH = a Cos (30-x)

the angle GCJ = 30 +x .  So Arm length for weight mg : GC * Cos(30+x)
     GC = 2/3  CE = 2/3 * √3/2 * a = a/√3
 
Hence, balancing the torques/moments of forces:
            F * a Cos (30-x) = mg * $\frac{a}{\sqrt{3}}$ * Cos (30+x)

   =>     $\frac{F}{mg} = \frac{Cos (30+x)}{\sqrt{3} Cos (30 - x)}$

 Force F needed to lift the triangular block and make it rotate / topple:
       $\frac{F}{mg} = \frac{1}{\sqrt{3}} * \frac{Cos (30+x)}{Cos (30-x)}$
           x has range:  0 <= x < 60.

For x > 60,  The weight mg and F both are in the same direction. So the block will topple.

$\frac{F}{mg}$ =  1/√3    when  x = 0
           = 1/√3 * Cos50/ Cos10  for x = 20 deg
           = 1/√3 * Cos45/Cos15  for x = 15 deg
           = 1/√3 * Cos55/Cos5    for x = 25 deg
           = 1/2    for x = 30 deg
           = 1/√3 * Cos 75/cos 15  for x = 45 deg
           = 0  for x = 60 deg.

Force F required to rotate , lift the triangular block is highest when x = 0.

This we want the minimum friction to be more than  maximum force F required.

Hence,    μ >=  F / mg
       =>  μ >= 1/√3