IIT JEE , AIEEE Forum - Articles Category , Community shelf - FRICTION

biki

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... :)

Community Contributions - Articles by goIITians