23
qwerty
·2010-10-11 12:21:17
they 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)
1
Shubh
·2010-10-16 09:39:34
For 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?
23
qwerty
·2010-10-17 05:22:30
yeah... newtons 3rd law ..
1
Shubh
·2010-10-17 06:35:43
thnku but can u explain hw u got tht angle? whn they r abt to slide thn wht is the logic?
23
qwerty
·2010-10-17 09:30:06
see 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