Question

A pair of nutcracker is 15 cm long and a nut is
placed 2.5 cm from the hinge. A force of 0.75 kgf
is applied at the end when it cracks the nut. What
weight if simply placed at
the
top
of nutcracker
(Ans. 4.5 kgf]
will crack it?​

Answer 1

Answer:

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=4.8kgf and in approx 4.5kgf

Explanation:

given = distance of nut from fulcrum (d)=2.5 , distance of effort from fulcrum (D)=16 cm ,Load =?

we know that,

load × load arm = effort × effort arm

$load \times 2.5cm = 0.75kgf \times 16cm$

$load = \frac{0.75 \times 16}{2.5}$

$load = \frac{75 \times 16 \times 10}{25 \times 100}$

$load = 4.8kgf$

$in \: approx \: it \: will \: be \: 4.5kgf$

so, the ANSWER is 4.5kgf