a 142kg box is placed on an inclined plane. when the angle makes with the horizontal is increased to theta , the crate begins to slide down. if the coefficient of static friction between the crate anf the rampl is 0.65, what is the angle theta when the crate just begins to slide downward?

Question

a 142kg box is placed on an inclined plane. when the angle makes with the horizontal is increased to  theta , the crate begins to slide down. if the coefficient of static friction between the crate anf the rampl is 0.65, what is the angle theta when the crate just begins to slide downward?

anonymous · Answer

$$mu= \tan \theta$$

0.65 = tan theta

$$\theta=\tan^{-1} 0.65$$

anonymous · Answer

|dw:1336835441582:dw|

no matter whatever the mass may be at if u increase the inclination above the maximum value body will start to slide down..
we have
$$mu *N=mgSin \theta$$

and we have N(Normal reaction=mg)
hence 
$$mu=\tan \theta $$
$$\theta=\tan^{-1} mu$$