IIT JEE , AIEEE Forum - Physics , Mechanics

savvej

In the arrangement shown in figurethe ends p and q of an inextensible string move downwards with uniform speed u.Pulleys A and B are fixed .The mass M moves upwards with a speed ------

THis is q.no.13 from hc verma pg 50 .
And the answer is u/cos theta.

please explain this question in detail step by step.I will surely rate u.

catch_arnnie

i'm not sure abt the solution but this is how i've done::

let the block of mass M moves vertically with velocity v.

we know, the string AM or BM moves with velocity u

=> resolving v in direction of AM or BM, we get, u = v cos (theta)

=> v = u / cos(theta)

plz correct me if i'm wrong anywhere..

stavan110

see figure 3-Q3....
let the distance from point A to point of mass M block(say point is K), be Z.
now construct a perpendicular to the base where the pulleys are affixed....
name the point as R...
let the distance KR be X....
let the distance RA be Y...
In triangle ARK,
X^2 + Y^2 = Z^2;
differentiating with respect to time....
2Z dZ/dt = 0 + 2X dX/dt...
therefore...
dX/dt= Z/X*dZ/dt...this can be written as....
dZ/dt/X/Z (here / stands for "whole divided by" )
or... u/cos

..........

dZ/dt = u and X/Z= cos

.....

plz...rate me if u find my answer useful.....

sudeep.kumar

as both the strings move downwards with equal velocity.. the mass will move vertically upwards.. let its velovity is v.

now, the length of the string PAM remains fixed, and so, in any given time interval, say dt, the distance moved by the end end M will be equal to the distance moved by the end P...... so, the vel. of end P = vel of end M...
but vel of end M = vcost
and vel of end P = u
so
vcost = u
or, v = usect.

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