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Mechanics

Blazing goIITian

 Joined: 5 Sep 2008 Post: 953
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??

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,,,,,,

Blazing goIITian

Joined: 22 Jul 2008
Posts: 654
5 Sep 2008 22:04:52 IST
1 people liked this

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

Blazing goIITian

Joined: 5 Jun 2007
Posts: 978
5 Sep 2008 23:58:45 IST
1 people liked this

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

amax = (m1/m2)g = (a1)max = (a2)max.

So from (1) Fmax =  (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) / m g

So a1 = a2 = a  g

i,e, amax = g = (a1)max = (a2)max.

Again taking m1 + m2 as a system, Fmax = (m1 + m2)amax.

So Fmax = (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

__________________________________________________________________

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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..

Blazing goIITian

Joined: 22 Apr 2007
Posts: 2537
6 Sep 2008 00:50:54 IST
2 people liked this

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)

a)

when force acts on 2kg block:

for f.b.d of 2kg block

fmax= (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.

Blazing goIITian

Joined: 5 Sep 2008
Posts: 953
6 Sep 2008 12:00:00 IST
0 people liked this

Thanks all of u

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