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a description to length of string method..

study the diagram

ANOTHER EXAMPLE ON CONSTARINT MOTION ----------->>>>

Masses of A=5Kg B=10Kg

_k betwn B and surfc is 0.1 take g=10

Detrmn: acc.n of A

Take accl^n of A and of pulley 2 = a , that of pulley 1 = a^// and that of B = a^/ ...

The strings are named as shown in MY diagram (Right hand side figure)

Let B represent the position of the block B. Let it is displaced to B^/ .

The position of 1st pulley is C. It goes to C ^/.

The 2nd pulley is in A along with the block A. It goes to A ^/.

E and F are the fixed ends.

The length of the string will remain constant.

So for first string.....

BD + DC + CE = B^/D + DC^/+C ^/ E

or BB^/ + B^/ D + DC + CE = B^/D + DC + CC^/+ C^/C + CE ... (from figure)

or BB^/ = 2CC^/

Thus displacement of block B = 2 x (displacement of pulley 1)

Diff.. twice with respect ot time..

we get accl^n of block B (a^/ ) = 2 x accl^n of pulley 1 (a^//)

or a^/ = 2 a^//__________(1)

Now for second string....

CA + AF = C^/ A^/+ A^/F

or CC^/ + C^/ A + AF = C^/ A + AA^/ + A^/A + AF

or CC^/ = 2 AA^/..

Thus displacement of pulley 1 = 2 x (displacement of pulley 2)

= 2 x (displacement of block A)

Diff. twice with respect to time..

we get accl^n of pulley 1 (a^// ) = 2 x accl^n of block A (a)

or a^// = 2a ________(2)

equation (1) and (2) implies

a^/ = 2(2a)

or a^/ = 4a.

i,e, accl^n of block B = 4 x (accl^n of block A) ___________(3)

Now let tension in string joining B with E be T

Since pulley 1 and 2 are massless, so force on them is zero ...

i,e tension in string joining C and F is = T + T = 2T

and tension in string joining the block A to the 2nd pulley = 2T + 2T = 4T

So writing force equations of A and B... respectively we get...

5g - 4T = 5a ________ (4)

and T -

_k(10g) = 10 a^/

or T -

_k(10g) = 10 (4a) ........... [by (3) ]

or T - 0.1(100) = 40a

or T = 40a + 10 _______ (5)

Put value of T from equ^n (5) in equ^n (4).... to get ---->>>

5g - 4( 40a + 10) = 5a

or 50 - 160a - 40 = 5a

or 165a = 10

or a = accl^n of block A = 10/165 = 0.06 m/s^2

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