IIT JEE , AIEEE Forum - Physics , Mechanics

ananth910

The figure shows a block of mass m released from rest on the incline plane on a smooth wedge of mass M that is ''NOT FIXED " at its bottom to the floor.there is no friction between the block and the wedge.Find the acceleration of the block with respect to the wedge and the acceleration of the wedge.

rajvarun

force=mgsin

a=gsin

the accleratio of the wedge=Mgcos

rajat.khanduja

Assume the acceleration of the wedge to be a and then apply a pseudo force on the block = ma.

Now solve it. The acceleration of the block, now, is actually with respect to the wedge. And 'a' can also be calculated.

PS : nudge me if you have any doubts and rate me if you find this useful

tarun_bits

hey,,,if the block will accelerate..will it not become a non-inertial frame of reference..

tarun_bits

and the force contributing the acceleration of the wedge will be in the horizontal direction...not mgcos@

rajat.khanduja

@ tarun,

so what if it becomes a non-inertial frame.

All of us here know how to use the concept of pseudo force.

tarun_bits

Re:NEWTONS LAWS

tarun_bits

ya rajat,,,i was just giving that post earlier than you...i mean to say that we hav to apply pseudo force..........ans by rajvarun is wrong

waterdemon

Here's the solution:

Let the acceleration of Wedge be "a₁".
Considering the motion of the block in frame of the wedge.

The Forces acting on the block will be:
1)N - Normal Force
2)mg - Downward
3)ma₁- forward (pseudo force)

The Block is sliding down the inclined plane.
we have:

ma₁Cos@ + mgSin@ = ma
a = a₁Cos@ + gSin@ ...........(1)

also,
N + ma₁Sin@ = mgCos@........(2)

Now we will consider the motion of the Wedge frm the
ground fram.No Pseudo force is neede as the frame is
Inertial.
1)Mg =Downward.
2)N =Normal to the incline by the block
3)N' =Normal upward by the horizontal ground.

We have,
NSin@ = Ma₁
N = Ma₁/Sin@...................(3)

Putting the value from equation(3) in Equation(2)

Ma₁/Sin@ + ma₁Sin@ = mgCos@
a₁ = mgSin@Cos@
       M+mSin²@
So acceleration of Wedge = mgSin@Cos@
                                 M+mSin²@

In equation (1)
a = a₁Cos@ + gSin@
a = (mgSin@Cos@*Cos@/M+mSin²@) + gSin@

On solving the above expression we get:
Acceleration of Block:
a = (M+m)gSin@
      M+mSin²@

Hope you find it useful.
Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

kmmankad

I think this problem also be solved by applying a constraint relation on the wedge?...the relation in terms of acclerations is
tan@=acc. Block/acc.Wedge
and then if i solve,then the acc of the wedge g/(tan@+kCot@),where k is ratio of masses...................

i think we can also solve using conservation of mechanical energy instead of newtons 2nd law..??

bvsatyaram

waterdemon is perfectly right but the value of 'a' we got finally is not the absolute acceleration of block. It is the acceleration of block with reference to wedge. So, all this rel acceleration vector to the wedge's acceleration vector to get block's acceleration vector. It is essential to add the accelerations vectorially, not algebraically.

bvsatyaram

kmmankad is wrong. The expression he stated is correct only when the block is constrained to move vertically. but here that is not the case in this problem.

kmmankad

I realised tht 2.....Thnx

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