IIT JEE , AIEEE Forum - Physics , Mechanics - point out my mistake.....PLEASE ....... !!!!!! (friction)

biki

Find the relative accl^n of A with respect to B when both blocks are given the same initial velocity towards right.
a)ug/2 b)ug c)ug/4 d)3ug/4

u = coeff. of friction

MY SOLUTION (PLEASE HAVE A LOOK AND POINT OUT THE MISTAKE)

I tried it like this........ (USING HC VERMA pg-92 VOL-1 CONCEPT)
let a1 = accl^n of A      and a2 = accl^n of B
As both blocks move towards right hence the only force i,e, the frictional force on A should be towards right.
hence ma1 = umg ____ (1)
   or a1 = ug
this frictional force has an equal and opposite reaction on block B which opposes its motion along with the frictional force between the block B and the surface.
also force reqd. to move A => (2m + m)(u/2)g + umg
this is equal to 2m.(a2)
thus............    3m.(u/2)g + umg = 2m.(a2)
                 or (5/2)umg = 2m(a2)
                 or a2 = (5/4)ug
Thus relative accl^n of A with respect to B is a1 - a2
                                            or ug - (5/4)ug
                                            or - ug/4
since we considered motion along right as (+)ve. So the relative accl^n of A with respect to B is -ug/4 towards left.

BUT THE ANSWER IS GIVEN --------->>>> 3ug/4 towards left....

PLEASE PROVIDE A SOLUTION TO THE PROBLEM AND

WHAT IS MORE IMPORTANT IS THAT PLEASE POINT OUT MY MISTAKE....

savvej

ug=a1

for b,
3/2umg -umg = 2 m a2
umg/2 = 2ma2
ug/4 = a2
a1 - a2 = ug - ug/4
= 3ug/4
plz refer the fbd for b:

biki

explain nicely man....

biki

where was i wrong.................... point it out.

savvej

see,
the upper block and the lower block both move with the same initial velocity as given.
in the hcv problem the the lower surface is frictionless and also the force is applied only to the lower block.so the force in this case is simultaneouly applied on both the blocks.so the friction force umg will opppse the motion of upper block.similarly the frictional forces to be overcome by lower block is 3umg/2 whereas umg will help it to overcome it acting in the opposite direction.hence..

savvej

arey see umg is ading the motion whereas u have taken it as opposing the motion.that is ur mistake.

biki

" similarly the frictional forces to be overcome by lower block is 3umg/2 whereas umg will help it to overcome it acting in the opposite direction "

if this is true...
than why did you substract umg from 3umg/2 instead of subtracting
3umg/2 from umg

biki

we always substract the force opposing the direction of motion from the force causing the motion...
according to your diagram and explanation...umg causes the motion and 3umg/2 is opposing it. Still the body moves rightward.
So why substract umg from 3umg/2 ???
and why not 3umg/2 from umg ????

FURTHER ... HOW TO SHOW THAT RELATIVE ACCL^N OF A WITH RESPECT TO B IS TOWARDS LEFT.....??????

ramyani

I tried it like this........ (USING HC VERMA pg-92 VOL-1 CONCEPT)
let a1 = accl^n of A      and a2 = accl^n of B
As both blocks move towards right hence the only force i,e, the frictional force on A should be towards right.
hence ma1 = umg ____ (1)
   or a1 = ug

ok

Now draw the FBD for B

As Savej showed,

                        B is acted upon by following forces :

1. umg from right to left
2. gravitational force 2mg downwards
3. contact force of A upon B which is mg
Reaction R1 in the upword dirn.

R1 = 2mg + mg = 3 mg
so that Frictional force is (u /2) *3mg = 3umg /2

Now the important point is what r the dirns of 3umg /2 and umg. There I think u went wrong. Both cannot act in the same dirn.
Then

3/2umg -umg = 2 m a2
umg/2 = 2ma2
ug/4 = a2

so that

a1 - a2 = ug - ug/4
= 3ug/4

ramyani

umg is from left to right and 3umg/2 is opposing it. The body moves rightward ( becoz both the blocks r given initial vel in that dirn ) and comes to rest , retarded.

biki

@ RAMYANI....

we always substract the force opposing the direction of motion from the force causing the motion...
according to your diagram and explanation...umg causes the motion and 3umg/2 is opposing it. Still the body moves rightward.
So why substract umg from 3umg/2 ???
and why not 3umg/2 from umg ????

BUT HOW IS RELATIVE ACCL^N OF A W.R.T. B IS TOWARDS LEFT.....?????
ANYONE EXPLAIN IT PLEASE......

biki

HOW IS RELATIVE ACCL^N OF A W.R.T. B IS TOWARDS LEFT.....?????

ANYONE EXPLAIN THIS PLEASE....

AND why substract umg from 3umg/2 ???
and why not 3umg/2 from umg ????

ramyani

Hmmm !

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