Two touching bars 1 and 2 are placed on an inclined plane
forming an angle α with the horizontal. The masses of
the bars are equal to m1 and m2, and the coefficients of friction between the inclined plane and these bars are equal to k1 and k2 re-
spectively, with k1 > k2 . Find:
(a) the force of interaction of the bars in the process of motion;
(b) the minimum value of the angle a at which the bars start slid-
ing down.
http://irodov.nm.ru/ebooks/irodov-1979.djvu (pg 17 fig 1.10)
23they will hav interaction while only when they both have same acceleration.
m1gsinθ + N - k1 m1gcosθ = m1a (1)
for m2
m2gsinθ= N + f2
i.e
m2gsinθ - N - k2 m2gcosθ=m2a (2)
eliminate a frm 1 and 2
N = m1m2gcosθ(k1-k2)m1+m2
Now when they are about to slide
m1gsinθ + N = k1 m1gcosθ
m2gsinθ - N = k2 m2gcosθ
(m1+m2)gsinθ =(k1m1+k2m2)gcosθ
so θ = tan-1(k1m1+k2m2m1+m2)
1For M1 both the blocks will exert a equal and opposite force? i.e., N and thts why we have taken the equ as m1gsinθ + N am i rite?
1thnku but can u explain hw u got tht angle? whn they r abt to slide thn wht is the logic?
23see when they are about to slide , it means the friction has reached its peak value by wich it can prevent sliding . and that peak value will be fsmax i.e μN