Answer:
The coefficient of static friction between the book and the vertical back of the wagon is [tex]\mu_{s} = mg/N[/tex]
Explanation:
Since the book is pressed against the vertical side of the wagon, the weight of the book will be acting downwards. i.e. [tex]W_{b} = mg[/tex]
Let the frictional force between the book and the vertical back of the wagon be [tex]f_{r}[/tex]
For the book not to drop down from the back vertical side of the wagon,
[tex]f_{r} = mg[/tex]........................(1)
A static frictional force is given as [tex]f_{r} = \mu_{s} N[/tex]............(2)
where N= Normal Reaction
Equating (1) and (2)
[tex]mg = \mu_{s} N\\ \mu_{s} = mg/N[/tex]
