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Cool goIITian

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3 Jul 2008 08:08:16 IST

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H.C.verma:friction:Q28, 30, 31 pls help

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friends, pls help to solve the problems from H.C. Verma, chapter: FRICTION, exercise problems: 28, 30, 31.

I tried hard but failed, so posting here friends and experts fo the board, pls help me!

thank u,

note: if u need me to write the problem with diagram, pls post here, i will write them all.

Thanks

mail: hi_rumki@yahoo.com

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rumki chakraborty

Cool goIITian

Joined: 10 Oct 2007

Posts: 41

3 Jul 2008 16:44:22 IST

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solved

VARUN RAJ

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3 Jul 2008 17:21:51 IST

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Q31)
IF THE MAN PULLS IT WITH UNEQUAL FORCES THEN
HE WILL FALL THEREFORE HE WILL PULL WITH EQUAL FORCES
FOR EQUILIBRIUM
2uN=400
uN=200
N=250N

VARUN RAJ

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Posts: 1825

3 Jul 2008 17:23:53 IST

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28) [ friction ]

This one is a little tricky problem. To find the solution, not only you need to identify the forces acting on the blocks but also figure out the relative motion of the two as well.

As the string is inextensible, whenever the lower end of the string (connected to M) displaces by x units to the right, the other end of the string (connected to block m) lowers down by 2x units. This is to be showed by 'constant length of the string' method as described well in HCV - 1 pg 73 , example 6 in ch. Newton's Laws of Motion

So accl^n of m (vertically downward) = 2 ( accl^n of block M in horizontal right direction )

Let accl^n of M = a

So accl^n of m = 2a.

Also, the block m is always in contact with the larger block M as far as motion in the horizontal direction is concerned. So accl^n of block m in horizontal direction is also a (same as that of M)

Now look at the figure.

The blue coloured are the forces on M and red ones act on m

Motion of m:

the forces are:

1) mg downwards

2) R ( contact force by M ) towards right

3)

₁R ( frictional force) upwards

4) T ( Tension) upward

For horizontal direction.

R = ma

In vertical direction

mg - T -

₁R = m(2a)

or T = mg - ma(2 +

₁)

Motion of M:

The forces on M are:

1) Mg downwards

2) R ( contact force by m ) towards left

₁R ( reaction of frictional force on m ) downwards

4) T ( Tension due to string on the lower end ) towards right

5) N ( contact force by ground ) upwards

₂N ( frictional force due to ground ) towards left

7) T ( Tension due to string on pulley attached to M ) towards right

8) T ( Tension due to string on pulley attached to M ) downwards

for vertical equilibrium

N = Mg + T +

₁R

or N = Mg + T +

₁ma {since R = ma}

In horizontal direction

2T - R -

₂N = Ma

Putting values of R, T and N.......

2T - ma -

₂(Mg + T +

₁ma) = Ma

or (2 -

₂)T - ma -

₂(Mg +

₁ma) = Ma

or (2 -

₂)[mg - ma(2 +

₁)] - ma -

₂(Mg +

₁ma) = Ma

or 2mg -

₂g(m + M) = 5ma + 2ma

₁ - 2ma

₂ + Ma

or a[M + m{5 + 2(

₁ -

₂)}] = g[2m -

₂(m + M)]

or a = g[2m -

₂(m + M)] / [M + m{5 + 2(

₁ -

₂)}]

VARUN RAJ

Blazing goIITian

Joined: 16 Mar 2008

Posts: 1825

3 Jul 2008 17:29:52 IST

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31)
THE WHOLE SYSTEM WILL BE MOVING WITH A VELOCITY V
BUT DUE TO FRICTION
A FORCE u(M+m)G FORCE WILL ACT ON THE PLANK IN THE OPP DIRECTION
DUE TO THIS THE SMALL BODY WILL ACC
AND A FORCE u/2mG WILL ACT ON THE SMALL BODY IN THE BACKWARDS DIRECTION AND IN THE FORWARD DIRECTION FOR THE PLANK
SO RESULTANT OPP FORCE IS
u(M+m)G-u/2mG
AND THE RESULTING OPP FORCE FOR THE SMALL BLOCK
u/2mG
AND THE DISTANCE CAN BE FOUND OUT EASILY
PLS RATE ME IF U FIND ME USEFUL
!!!!!!!!!CCHEERS!!!!!!!!

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