Home
Class 12
PHYSICS
In the arrangement shown in figure the e...

In the arrangement shown in figure the ends of an in extensible string move downwards with uniform speed u. Pulleys A and B are fixed. Find the speed with which the mass Mmoves upwards.

Text Solution

Verified by Experts

Suppose the distance of point O from the ceiling is y and the distance of point O from each pulley is x and the distance between the two pulleys is l:
`x^(2) =y^(2) + l^(2)/4`

Differentiating the above equation wr.t. to "t"
`2x(dx)/(dt) =2y(dy)/(dt) + 1/4 2l(dl)/(dt)`
But, `u=(-dx)/(dt), v=(dy)/(dt)` and `(dl)/(dt)`
`therefore -xu =yv` and `v=-u(x/y) = u sec theta`
`therefore` Velocity of mass =`v=u sec theta` (upwards)
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise SHORT ANSWER TYPE QUESTIONS|6 Videos
  • MOTION IN A PLANE

    AAKASH SERIES|Exercise VERY SHORT ANSWER TYPE QUESTIONS|10 Videos
  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Circular Motion)|7 Videos
  • MOTION IN A STRAIGHT LINE

    AAKASH SERIES|Exercise very Short answer type question|15 Videos

Similar Questions

Explore conceptually related problems

In the arrangement shown in the figure if the blocks of masses m and 2m are released from the state of rest tension in the string is ( mu = coefficient of friction string is massless and inextensible pulley is frictionless )

Consider the arrangement shown in figure. Two identical wedges of same mass, one is fixed and of mass 'm' is attached to the string and a force F = 2mg is applied at the free end (Wedge B fixed, Wedge A moveable) The acceleration of the hanging block (m) is