IIT JEE , AIEEE Forum - Physics , Mechanics

ramahoo

block M slides down on frictionless inclined as shown.

find the min friction co efficient so that m does not slide with respect to M

netkid07

minimum friction coefficient..................0.6

tell me if i am correct...........i will post the solution...........

Anna_LL

Hi netkid , ur ans seem ... correct. i will do and compare to ur ans . And ... where are u from netkid ? Are u from Vietnam ?

karthik2007

*Edited*

learner

karthik could you pls explain why acc. in horizontal direction is gsin@cos@ ?

karthik2007

gsin@ is along the incline, so along the horizontal which makes an angle of 37 with the incline, its component will be gsin@cos@.

@topicstarter - can u verify our answers please?

edit - ignore this, Rooney is correct. Made a mistake while considering normal reactions.

KS

See
(M+m)gsin37 = Ma FBD Of M
Now a psuedo force will be acting on m along the slope passing through the centre of mass m
Now Fbd of m relative to M would be
m( (M+m)gsin37 )cos37/M - umg = 0
So u = (M+m)sin37 )cos37/M

KS

naveen07

Is the answer 1.2?

rooney

I think its 0.75.
Acceleration of both blocks along the plane is gsin37.
Just form the equation of mass m along the plane

mgsin37 - Nsin37 + uNcos37=mgsin37

u = tan 37 = 0.75

ramahoo

hey rroney u r are correct man.
answer is 0.75 only.
can u please explain ur working to all of us

karthik2007

I think he is offline, so I'll explain it for you:

See, the acceleration of the system is gsin37.

Now, the forces acting on the block of mass m are:

1) mgsin37 down the incline
2) fcos37 (f - force due to friction) down the incline) [ As friction opposes relative slipping, and the mass m has a tendency to move backward, so f acts in the forward dirn]
3) Nsin37, up the incline

Now, the acceleration of the system is gsin37, so the force acting on the block is mgsin37

So, we have:

mgsin37 + kNcos37 - Nsin37 = mgsin37

or k = tan37= 3/4 = 0.75.

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