The figure shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2kg and the incline is μ1, and that between the block of mass 4kg and the incline is μ2. Calculate the acceleration of the 2kg block if μ1=0.30 and μ2=0.20. Ans= 2.4 m/s2
71You just need to consider the FBD of the individual blocks.
Just the trick here is to consider the reaction force by blocks on each other, obtain two equations and solve them for the Accn.
I obtained :
m1gsin30-fA-R = m1w ..... (1)
R+m2gsin30-fB = m2w ......(2)
Expressions have their usual meanings. Add the both and substitute the values.
30I had posted this question as a doubt earlier since my answer was not matching and I thought something was wrong in my approach.. After posting, I checked my calculations and the answer turned out correct! So the question remains for others to try.
Coming back to the question:
Since, the 4kg has a friction coefficient, the blocks move separately and hence acceleration can be found using the force eqn of 2kg block.
11Yeah, you can get the answer by using just one equation here [1]
anyways its a simple one
1
FRICTIONAL FORCE FOR 4kg mass-
F=0.3*4*9.8cos30°
downward force=4*9.8*sin30°
therefore acc.=9.8*(12 -210*√32)
=0.23596 m/s2 ≈0.24 m/s2
1@khyati
don't think it simple one
it is one of those prolems whih lets u know what normal reaxn actually is...Reaxn.=0 case is the most interesting and most inspiring one!