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Question Number 52340 by Necxx last updated on 06/Jan/19

A 90cm uniform lever has a load  of 30N suspended at 15cm from  one of its ends.If the fulcrum is at  the centre of gravity what is the  force that must be applied to its  other end to keep it in horizontal  equilibrium?

$${A}\:\mathrm{90}{cm}\:{uniform}\:{lever}\:{has}\:{a}\:{load} \\ $$$${of}\:\mathrm{30}{N}\:{suspended}\:{at}\:\mathrm{15}{cm}\:{from} \\ $$$${one}\:{of}\:{its}\:{ends}.{If}\:{the}\:{fulcrum}\:{is}\:{at} \\ $$$${the}\:{centre}\:{of}\:{gravity}\:{what}\:{is}\:{the} \\ $$$${force}\:{that}\:{must}\:{be}\:{applied}\:{to}\:{its} \\ $$$${other}\:{end}\:{to}\:{keep}\:{it}\:{in}\:{horizontal} \\ $$$${equilibrium}? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

F×45=30×30  F=20N  moment taken about fulcrum

$${F}×\mathrm{45}=\mathrm{30}×\mathrm{30} \\ $$$${F}=\mathrm{20}{N}\:\:{moment}\:{taken}\:{about}\:{fulcrum} \\ $$

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