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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.
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Solution
(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.
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