Helppppppppppp

The figure shows one end of a string pulled down by a constant velocitty V.Assuming the pulleys to be massless and smooth,prove that the tension in string AB is mg(1+b2v2/4gx3)

9 Answers

62
Lokesh Verma ·

let the length of string that hangs out be L
So,

L+2.√x2+b2 = c

dL/dt=v

Now we differentiate...

Then double differentiate to find accleration of point B

This will solve the question.

62
Lokesh Verma ·

Of course u need to write the FBD and pput that equation as well :)

11
Shailesh ·

yes i solved with this...

L+2.√x2+b2 = c

dL/dt + 2/(√x2+b2).2x.(dx/dt) = 0

v=-4x/(√x2+b2).(dx/dt)
dx/dt = -√x2+b2 v / 4x

differentiating,

dv/dt = d/dt{4x/(√x2+b2)} . (dx/dt) + 4x/(√x2+b2). d2x/dt2

0 = d/dt{x/(√x2+b2)} . (dx/dt) + x/(√x2+b2). d2x/dt2

d/dt{x/(√x2+b2)} .√x2+b2 v / 4x = x/(√x2+b2). d2x/dt2

d/dt{x/(√x2+b2)} .(x2+b2) v = 4x2. d2x/dt2

4 d2x/dt2 = d/dt{x/(√x2+b2)} .{1+(b/x)2} v

d/dt{x/(√x2+b2)} = {1/√x2+b2 + x/(x2+b2)3/2 } . dx/dt
= -{1/√x2+b2 + x/(x2+b2)3/2 }.√x2+b2 (v / 4x)
= -{1 + x/(x2+b2) }(v / 4x)

Thus,

4 d2x/dt2 = d/dt{x/(√x2+b2)} .{1+(b/x)2} v
= -{1 + x/(x2+b2) }/ 4x . {1+(b/x)2}

I dont know what mistake i am making :(

62
Lokesh Verma ·

Dont have the patience to check the mistake..

L+2.√x2+b2 = c
dL/dt + 2/2(√x2+b2).2x.(dx/dt) = 0

v=-2x/(√x2+b2).(dx/dt)
dx/dt = -√x2+b2 v / 2x
dx/dt = -√1+(b/x)2 v/2
differentiating,

2/v d2x/dt2 = 1/(2√1+(b/x)2 ) {2b/x} {-1/x2} dx/dt
2/v d2x/dt2 = {-b/x3}/(√1+(b/x)2 ) dx/dt

2/v d2x/dt2 = {b/x3}.v/2

d2x/dt2 = {b/x3}.v2/4

24
eureka123 ·

sir is this answer???????????

33
Abhishek Priyam ·

thats acc of block now draw FBD of block and find tension....

24
eureka123 ·

answer is still not coming.........plzzz help

33
Abhishek Priyam ·

ok i am trying...

33
Abhishek Priyam ·

I think tere some calculation mistake...

acc comes out to be :

sorry theres v2b2 in numerator

now i think answer is coming....

T will be (1/2cosθ)(m*acc+mg)

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