If friction is acting along the plane upwards
then in this case we will have
For equilibrium of 100 kg box on inclined plane we have
[tex]mgsin\theta = F_f + T[/tex]
also for other side of hanging mass we have
[tex]T = Mg = 50(9.8) = 490 N[/tex]
now we have
[tex]100(9.8)sin\theta = 100 + 490[/tex]
[tex]980sin\theta = 590[/tex]
[tex]sin\theta = 0.602[/tex]
[tex]\theta = 37 degree[/tex]
In other case we can assume that friction will act along the plane downwards
so now in that case we will have
[tex]mgsin\theta + F_f = T[/tex]
also we have
[tex]T = Mg = 50(9.8) N[/tex]
now we have
[tex]100(9.8)sin\theta + 100 = 50(9.8)[/tex]
[tex]980sin\theta + 100 = 490[/tex]
[tex]980 sin\theta = 490 - 100[/tex]
[tex]sin\theta = 0.397[/tex]
[tex]\theta = 23.45 degree[/tex]
So the range of angle will be 23.45 degree to 37 degree
