The handle of a nutcracker is 16 cm long and a nutis placed 2cm from its hinge.If a force of 4 kgf is applied at the end of handle to crack it, what weight if simply placed on the nut to crack it?
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5
As moment of force should be equal.
F1×L1 = F2 ×L2
⇒4×16 = F2 ×2
⇒F2 = 32 Kgf
That means, if 32 Kgf force is applied, the nut can be break.
F1×L1 = F2 ×L2
⇒4×16 = F2 ×2
⇒F2 = 32 Kgf
That means, if 32 Kgf force is applied, the nut can be break.
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2
32 Kgf force is applied, the nut can be break.
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