Inclined plane

there's a formula for the coefficient of friction .....tan(theta)=coeff of friction ......where theta is the angle of the slope ........using that .....tan(45)=1=coeff of friction

Use 2nd Newton Law: F = ma, we got: Normal force = mgcos45. Friction force = coefficient of friction * normal force = mgsin 45. So, coefficient of friction = (mgsin 45)/(mgcos 45) = tan45 = 1

$µe$ = $\tan \theta$

$µe$ = $\tan 45^\circ$

$µe$ = $1$

Co-efficient of friction is given as $\mu_{s} = \tan \theta$ now $\theta = 45$

As formula: Tan theta = Meu s (coefficient of friction), now putting the value of theta in above formula

Coefficient of limiting friction = tan(\theta) Therefore Coefficient of limiting friction = tan π/4 =1.

R-mg cos45=0,mg sin45-fs=0, so co eff= sin45/cos 45

at 45 angle

mgsin(45) = k mgcos45 ( k= coefficient of friction )

k=1