IIT JEE , AIEEE Forum - Physics , Mechanics

akshax

In the given diagram the two strings are moving downward with velocity v and the velocity of the mass M are to be found out.....in terms of v and

only..the two pulleys are fixed..

options

(a) 2vcos

(b) v/cos

(c) 2v/cos

(d) vcos

apurviitjee2008

is the answer a?
vcos theta+v cos theta.....

akshax

its not a..thats wat i thought but its not..

ammu_91

the angle teta is not clear

nishantsingh89

* post edited

netkid07

it's surely v/cos(theta) i.e (b)

consider the figure drawn..............

in that triangle::

x^2+L^2=y^2

diffrentiating both sides wrt time:::

dx/dt=v (we need to calculate this)

dy/dt=u (given)

dL/dt=0 (this is constant)

diff we get::

xv=uy
or
v=uy/x

v=u/cos(theta)

don't u think i deserve a rate !!

nudge if u hav doubt !

lokeshsardana

I think it is v/cos(theta)

waterdemon

See this is simple ,

The two ends from the pulley start descending with "V"

velocity.

Now since the angle between the string and the block is

Angle @.

Now see the length "L" of string decreases at the rate of

V m/s.

I have given the diagram refer it.

L²=b²+y²

2L dL/dT = 0 + 2y dy/dt

differentiation of b² = 0 as it remains constant.

now we get

dy/dt = l(dL)/ y(dt)

dy/dt = (1/Cos@).dL/dt

dy/dt = V/Cos@

Therefore.
The answer:(b)

Hope you find it useful.

Rate if useful.

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akshax

thanx all i got it...din think of drawin triangle and using calculus....

waterdemon

No problems dude...just is fine if it was helpful.

Cheers!!!!!!!!!@@@!!!!!!!!!!

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