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A pulley system with a velocity ratio of 4 is used to lift a load of 150 kgf through a vertical height of 20 m. the effort required is 50 kgf in downward direction. Calculate: (a) the distance moved by the effort. (b) the work done by the effort. (c) the mechanical advantage. (d) the efficiency of the pulley system, and (e) the total number of pulleys and the number of pulleys in each block. (g=10N" "kg^(-1)). |
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Answer» Solution :GIVEN: `V.R.=4,L=150kgf,d_(L)=20m`. `E=50kgf=50xx10N=500N,d_(E)=?` (a) V.R.=`("distance MOVED by effort "d_(E))/("distance moved by load "d_(L)) or 4=(d_(E))/(20)` `therefore` Distance moved by the effort `d_(E)=20xx4=80m` (b) Work done by the effort=effort `EXX`distance `d_(E)` `=500Nxx80m` `=40000J`. (c) Mechanical ADVANTAGE`=(L)/(E)=(150)/(50)=3` (d) Efficiency=`(M.A.)/(V.R.)xx100%=(3)/(4)xx100%=75%` (e) The TOTAL number of pulleys=4 The number of pulleys in each block=2 |
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