Question

A block 'A' of mass m is tied to a fixed point C on a
horizontal table through a string passing round a
massless smooth pulley B (figure 5-W5). A force F is
applied by the experimenter to the pulley. Show that if
the pulley is displaced by a distance x, the block will be
displaced by 2x. Find the acceleration of the block and
the pulley.

plz explain detail solution ​

Answer 1

Initially length of string is, l1+l2=l (constant)

Let pulley and block are displaced by distance x1 and x2 respectively.

(l1+x1+x1)+(l2−x2)=l⇒2x1=x2

Differentiating twice we get acceleration of pulley (a1) and acceleration of block(a2) as,

2a1=a2

Now, for pulley,

F−2T=0⇒T=F2

For block,

a2=Tm=F2m

a1=F4m

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Answer 2

The acceleration of the block is  $\frac{F}{4m}$

Given:

Two pulleys A and B where block of mass is attached to pulley A

To Find:

To find the acceleration of the block and the pulley and to show that if pulley is displaced by distance x the block will be displaced by 2x.

Solution :

Consider, another diagram which is marked as A', B',C'

Calculate according to the new diagram

We have,

B'C + B'A'

B'C+B'A' = BC + BB' + BB' + A'B

Comparing, the above with the first diagram,

We have,

BC+AB = BC+ BB'+BB'+A'B

Upon, cancelling we get ,

AA' = 2BB'

AA' = 2x

Calculating , the tension of both the pulleys,

F =T+T

F = 2T

We get the value of tension,

T = $\frac{F}{2}$

Tension (T) = ma'

a'  =  $\frac{T}{m}$

a'  =  $\frac{F}{2m}$

a  =  $\frac{a'}{2}$

a =  $\frac{F}{4m}$

Hence, acceleration of the block and the pulley is given as

a' = $\frac{F}{2m}$  and  a =  $\frac{F}{4m}$

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