In order for the object not to slip, the component of the weight parallel to the surface must be equal to the frictional force (which acts in the opposite direction):
[tex]F_{//}= F_a[/tex]
The parallel component of the weight is:
[tex]F_{//} = mg \sin \alpha[/tex]
where m is the object mass and [tex]\alpha[/tex] is the angle of the inclined plane.
The frictional force is
[tex]F_a = \mu m g \cos \alpha[/tex]
where [tex]\mu[/tex] is the coefficient of static friction.
Equalizing the two forces, we have
[tex]mg \sin \alpha = \mu m g \cos \alpha[/tex]
from which we find
[tex]\mu = \tan \alpha[/tex]
and so, in our problem the coefficient of static friction must be
[tex]\mu=\tan 51^{\circ} =1.23[/tex]
