Frictioned Plane

Start by summing the forces parallel to the incline.

$\Sigma F_\parallel = F_f - mg\sin(45^{\circ}) = 0$ $F_f = mg\sin(45^{\circ})$ $\mu F_N = mg\sin(45^{\circ})$

To find the normal force, sum the forces perpendicular to the incline.

$\Sigma F_\perp = F_N - mg\cos(45^{\circ}) = 0$ $F_N = mg\cos(45^{\circ})$

Substitute this into our original equation.

$\mu mg\cos(45^{\circ}) = mg\sin(45^{\circ})$ $\mu \cos(45^{\circ}) = \sin(45^{\circ})$ $\mu = \tan(45^{\circ}) = 1$

the weight is halved : tangent force = normal force = 1/2 weight for equilibrium, coefficient *normal force = tangent force => coefficient = 1