Question

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A man can open a nut by applying a force of 75 kgf by using a lever of handle of 0.5 m. What should be length of the lever of another handle, if the force required to open the nut is 45 kgf?


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Answer 1

Explanation:

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Equating the torque in both the cases:

E1L1 = E2L2

150 × 0.4 = L2 × 60

$l2 = \frac{150 \times 0.4}{60}$

1.0 m

Answer 2

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$Given :- F1 \:=\: 75kgf\:and\:L1\:=\:0.5m$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:F2\:=\: 45kgf\:and\:L2\:=\:?$

$To\:find\:=\:L2(length\:of\: another\:handle.)$

$NOW,$

$F1\:×\:L1\:=\:F2\:×\:L2$

$75\:×\:0.5=\:45\:×\:L2$

$37.5\:=\:45\:×\:L2$

$37.5/45\:=\:L2$

$L2\:=\: 8.33\:m$

$Therefore,\:lenght\:of\:another\: handle\:=\:8.33\:m$