Question

a lever of length 100cm has effort 15kgf at a distance of 40 cm from the fulcrum at one end. what load can be applied at its other end.

Answer 1

so as lever= 100 cm
effort arm= 40 cm
load arm= 60 cm
effort=15kg

lever law
$load \div effort \: \\ = \: effort \: arm \: \div load \: arm$
so here x/15=40/60
that is:; 10/15=40/60

how: 15*4= 60
. so x*4=40
. now x= 40/4=10kg
solved