A tall block of mass `M=50kg` and base width `b=1 m` and height `h=3 m` is kept on rough inclined surface with coefficient of friction `mu=0.8` as shown in figure. The angle of inclination with the horizontal is `37^@`. Determine whether the block slides down or topples over.

Explanation

Here `mu=0.8, b/h=1/3, =0.33 tan theta=3/4=0.75`

Since `mugtb//h` and `mugttantheta`, therefore the block has a greater tendency to topple over. From the condition of translation equilibrium , we have
`N=Mgcos37^@=(50)(10)(4/5)` or `N=400N`
`f=Mg cos 37^@=(50)(10)(3/5)`
`=300Nltf_("mass")=muN=0.8(400)=320N`
For rotational equilibrium `Sigma tau_(c)=0`
`N_(x)=f h/2`
Thus, `x=f/N h/2=(300/400)(3/2)=9/8ltb/2`
Since the value of `x` is more than `b//2` which is not possible, therefore the block topples over.