IIT JEE , AIEEE Forum - Physics , Mechanics

avi_1214545

P and Q are goin down with uniform velocity "u"

find velocity of block in terms of "u" and "

"

also justify...

ganesha1991

is it 2ucos^2@

akshay.khare91

is the ans. usec

?????????????

if yes ?????
can give u explanation also...

hash_include

it's u sec

.
wait ill post proof now..

Werewolf

Hey , have u been taught by NM or AKG ?

varun.tinkle

let the tension in both the strings be T
the only formula im getting is
2Tcosa=mg

avi_1214545

we were taught by NM

avi_1214545

@ varun.tinkle

first read the question..

i have asked about velocity and not force

varun.tinkle

so how can u solve it withouth equating the forces

avi_1214545

ok...u solve watever method.

gimme the answer and solution/

hash_include

see the diagram..

velocity of the string = u = dx/dt
velocity of block = v = dy/dt.

applying pythyagoras theorem, we get

x^2 = L^2 + y^2

differentiate wrt 't'

2x dx/dt = 0 + 2y dy/dt
solving we get..
v = dy/dt = u (x/y)

but x/y = sec

according to figure..

so v = u sec

EDIT: enlarge the diagram to see it clearer.

newarnav

Hmmmmmm

N73

akshay.khare91

dy/dt= v

therefore

dl/dt=u

l^2= x^2+y^2

differentiating wrt to time we get

2dl/dt= 2dy/dt    (since x is constant)

lu = yv            (v= velocity of block)

v= lu/y    (since l/y = sectheta)

therefore v- sec

..

avi_1214545

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