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Blazing goIITian

Joined:	5 Sep 2008
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5 Sep 2008 18:01:25 IST

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How 2 solve friction probs including 2-3 blocks one on top of the other??

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How 2 solve friction probs including 2-3 blocks one on top of the other??

Comments (5)

HeMaNt..... ThE DeViL....

Blazing goIITian

Joined: 22 Jul 2008

Posts: 654

5 Sep 2008 22:02:09 IST

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such problems are solved by wedge constrain,,,,,,

HeMaNt..... ThE DeViL....

Blazing goIITian

Joined: 22 Jul 2008

Posts: 654

5 Sep 2008 22:04:52 IST

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wedge constrain is the relation tat exists between the acceleration of wedge and the body over it.
to find these relations we analyse the motion of system after small time.
from the geometry of the figure find the relation between the displacement of connected body.and differentiate it to get velocity or acc

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

5 Sep 2008 23:58:45 IST

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i learnt a new method (new for me) to find accln of bodies kept over one another... i thought of sharing it with u people... read this carefully..

and i m sorry if it is already known to you...

consider the first example.. fig 1

A constant force F acts on the body m₁ placed over m₂. The coeff. of friction b/w the two bodies is . Let accln of m₁ be a₁ and accln of m₂ be a₂.

Here we apply our new method.

Let us calculate the force for which both bodies move with equal accln. So we consider a₁ = a₂= a. Now as both bodies move with the same accln (same direction), they can be considered a single system. So we get

F = (m₁ + m₂).a ________(1)

Now we go for seperate bodies.

Consider the static friction acting b/w the blocks be f.

So by definition of static friction f m₁g

Again for mass m₂ we get the force of friction towards the direction of F. So this being the only force,

f = m₂a

or a = f/m₂

or a (m₁/m₂)g

This denotes max. value of a. Now as a₁ = a₂= a

a_max = (m₁/m₂)g = (a₁)_max= (a₂)_max.

So from (1) F_max = (m₁ + m₂).(m₁/m₂)g

i,e if F (m₁ + m₂).(m₁/m₂)g , then a₁ = a₂ .

Now as F acts on m₁ , so a₁ a₂ . So if F > the above value, only a₁ will change but a₂ which arises due to friction remains same as previous value of a.

So when values are given,

a₁ = a₂= a = F/(m₁ + m₂) if F (m₁ + m₂).(m₁/m₂)g

and

a₂ = same as above ; a₁ = {(F - m₁g) / m₁} if F > (m₁ + m₂).(m₁/m₂)g

Now consider the second example .. fig 2

Here F acts on m₂. So obviously a₂ a₁.

Let us find the force for which a₂ = a₁=a. Let f = friction on upper block along right.

Here, for m₁ , a = f/m₁ (m₁g) / m₁ g

So a₁ = a₂ = a g

i,e, a_max = g = (a₁)_max= (a₂)_max.

Again taking m₁ + m₂ as a system, F_max = (m₁ + m₂)a_max.

So F_max= (m₁ + m₂)g

i,e if F (m₁ + m₂)g then a₁ = a₂.

Now as F acts on m₂ , so a₂ a₁ . So if F > the above value, only a₂ will change but a₁ which arises due to friction remains same as previous value of a.

So when values are given,

a₁ = a₂= a = F/(m₁ + m₂) if F (m₁ + m₂).g

and

a₁ = same as above ; a₂ = {(F - m₁g) / m₂} if F > (m₁ + m₂).g

__________________________________________________________________

Hope this helps ... try solving HCV friction 23rd problem of exercise in page 98 using this method... i,e first find the force for which the blocks have same accln and then carry on ...... tell me if it helps..

Thanks for reading... :)

ramyani chakrabarty

Blazing goIITian

Joined: 22 Apr 2007

Posts: 2537

6 Sep 2008 00:50:54 IST

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Take this problem.

Three blocks placed on top of one another on a table.there is a 7kg block.on top of it there is a 3kg block on which there is a 2kg block.frictional coefficient between 2kg and 3kg is 0.2,that between 3kg and 7 kg is 0.3 and the table is smooth.find their accelerations when 10N force is applied on (a)2 kg block (b)3 kg block (c) 7 kg block.[g=10m/s²]

This was solved elegently by swatygupta (689)

when force acts on 2kg block:

for f.b.d of 2kg block

f_max= (0.2)2.10= 4N

10 - f₁= 2a₁

a₁= 3 ms^-2

for 3kg block:

f₁- f₂= 3a₂;... (1)

for 7kg block:

f₂= 7a₃;... (2)

now, f₂ will assume the value such that the realtive motion between 3kg n 7kg block is minimum.. (friction is based on relative motion btw the surfaces)

since, f₂(max)= 15N;

a₂=a₃

so, solving these..

a₂=a₃= 0.4 ms^-2

now,

when force acts on 3kg block:

f₁= 2a₁... (1)

10 - (f₁ + f₂) = 3a₂.. (2)

f₂= 7a₃.. (3)

now, the friction adjusts itself in such a way that relative motion is least among all surfaces. so, the accelerations of blocks should be the same so that relative motion is zero.. (note that this case is not always valid.. (sometimes the friction values may not permit this.. lets try out here)

hence

a₁=a₂=a₃=a

so solving (1), (2), (3)

a=5/6 ms^-2

the case is exactly same for the third part.

AViK

Blazing goIITian

Joined: 5 Sep 2008

Posts: 953

6 Sep 2008 12:00:00 IST

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Thanks all of u

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