Answer:
The value is [tex]\mu =0.4286[/tex]
Explanation:
From the question we are told that
The mass of the steel cabinet is [tex]m = 95 \ kg[/tex]
The threshold angle is [tex]\theta = 23.2^o[/tex]
Generally at the point just before the steel cabinet starts to slide
The horizontal force on the steel cabinet is equal to the frictional force that is
[tex]F_f = F_h[/tex]
Here [tex]F_h[/tex] is the horizontal force which is mathematically represented as
[tex]F_h = m * g cos (theta )[/tex]
and [tex]F_f = \mu N[/tex]
Here N is the normal force acting on the steel cabinet and this is mathematically represented as
[tex]N = mg sin (\theta )[/tex]
=> [tex] m * g cos (theta )= mg sin (\theta ) * \mu [/tex]
=> [tex]\mu = tan (\theta )[/tex]
=> [tex]\mu = tan (23.2)[/tex]
=> [tex]\mu =0.4286[/tex]
