Question

The alongside figure shows the combination of a movable pulley P₁ with a fixed pulley P₂ used for lifting up a load W.

(i) State the function of the fixed pulley P₂.
(ii) If the free end of the string moves through a distance x, find the distance by which the load W is raised.
(iii) Calculate the force to be applied at C to just raise the load W = 20 kgf, neglecting the weight of the pulley P₁ and friction.

Answer 1

(i) It is quite difficult to apply effort in the upward direction if no fixed pulley P₂ is used. The fixed pulley changes the direction of effort from upwards to downwards, making the application of the effort more convenient and easier.

(ii) As the movable pulley doubles the effort,

∴ Force, L = 2T

i.e., W = 2T

Mechanical advantage M.A. =`(2"T")/"T"=2`

and V.R. =`("Distance travelled by effort")/("Distance travelled by load")=(2"x")/"x"=2`

The distance travelled by load is half the distance moved by effort = x / 2.

(iii) Since W = 2T or 20 kgf = 2T Here, T = effort

∴ Effort applied =`20/2` = 10 kgf.