IIT JEE , AIEEE Forum - Physics , Mechanics

shub

in a pulley system is it necessary for the acc. to b the same if the tensions in the strings is the same??consider the given system....wut will b the relation between the acc. of the bodies???

nivedh_89

it is not necessary tht if the tension is same then acceleration has 2 b the same ........if the masses r of different magnitude then acc. will b different for each of them..................

Omi

not necessay subh.... actually u answer ur question by finding the constraint equations......the procedure or it is vry simple..... the constraint equation gives u the relation between the accelarations.....

swashata4iit

In the diagram attached the free body diagram is draw at sides.

For pulley 2 the eqn of motion are

m₃a₁ + T₁ = m₃g .......................(1)
T₁ = m₂g + m₂a₁........................(2)

Solving We have

a₁ = [(m₃-m₂)/(m₃+m₂)]g

T₁= [2m₂m₃/(m₃+m₂)]g

Similarly for Pulley 1 We may consider the pulley 2 as a system of mass (m₃+m₂) attached to pulley 1
Therefore,

a₂ = [(m₃+m₂-m₁)/(m₃+m₂+m₁)]g

T₂ = [2m₁(m₂+m₃)/(m₃+m₂+m₁)]g

So we may conclude that T₁ is not equal to T₂

swashata4iit

Hey Guys PLEASE VOTE MY EFFORTS

shub

thnx for d reply swashata ......
i hav a confusion though.....

r u sure dat the acc. of m1 and m2 will be the same???
this question is from hcv.....

they hav givenm different acc. for all d 3 masses.....

shub

how exactly do we find the constraints equation???

it has something to do with the inextensibility of strings i guess....
but can any1 plzzz tell me d exact procedure...

rupin

The reason is simple
The string is inextensible hence if m3 goes with acc a1 downwards then m2 will have to go with acc a1 upwards since length of string is constant
also velocity and displacement of m2 and m3 will have same magnitude due 2 same reason

you can safely conclude T2=2T1 assuming pulley massless, net force on the pulley must be zero(F=ma)

hence you can solve

swashata4iit

The question is purely from newton's law of pully.
The acc generated on the string is due to the diff in values of the masses. Now if the string be mass less then the tension will be same thr out the string and the acc created will be same on both the masses. if the string is massive then due to change in tension the acc of both the masses will be different

swashata4iit

Sorry I wrote wrong it will be due to inextensibility of the string and masslessness of pulley. Imagine a pulley the one side of the string of which is coming down quicker than the other side which is arising slower. It will generate a stress on the string. But the string being massless and more importantly inextensible it is impossible to generate stress. Hence the acc on both the part will be same.

jatinroxx

You can get n constraint equation where n is the number of strings in the question.

Steps for making constraint equations

1.Mark a reference axis.

Here let us take the axis as a line perpendicular to the strings passing to the topmost pulley.

2. Mark the distance of the every point from the axis where there is a moving body or there is a turn in the string.

Here as the axis passes thru the top pulley, its distance 4m the axis is 0.

Next let m1 be at x1, the second pulley at x4, m2 at x2, m3 at x3.

3. Calculate the lengths of the strings in terms of the assigned coordinates.

Here,

length of first string is x1 + x4 = l1

and of second (x3-x4) + (x2 - x4) = l2

4. Next is differentiating them to get the required relations...

When u differentiate as the string is inextensible, l1 is constant, RHS becomes 0.

Now wen u differentiate somethin, you get the acceleration of the body dat was assigned that co-ordinate.

For instance, differentiating x1 wrt t once gives u velocity of m1 and twice, acc of m1.

I guess now u can form the relations.....

@swashata4iit

U dont have to solve the entire thing for getting the constraint equations and kinematic relations between the blocks.......

swashata4iit

Thanks Jatin it's very useful

shub

thnx for d help guys...

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