A walnut can be broken by applying a direct force of $60 kgf$ . If the walnut is placed in a nutcracker, the length of whose handle is $15 cm$ , and the nut is placed $2 cm$ from the pivot, calculate the minimum force required to crack the nut.

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Solution

Given:
Effort arm, $E_{A} = 15 cm$
Load arm, $L_{A} = 2 cm$
Walnut breaks at a load, $L = 60 kgf$
It is a class $II$ lever,
For, $\sum M_{n e t} = 0$ about point $F$ :
$Effort \times Effort arm = Load \times Load arm$
$Effort \times 15 = 60 \times 2$
$Effort = \frac{120}{15} = 8$
So, the minimum force required is $8 kgf .$

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A walnut can be broken by applying a direct force of 60 kgf. If the walnut is placed in a nutcracker, the length of whose handle is 15 cm, and the nut is placed 2 cm from the pivot, calculate the minimum force required to crack the nut.

Given: Effort arm, EA=15 cm Load arm, LA=2 cm Walnut breaks at a load, L=60 kgf It is a class II lever, For, ∑Mnet=0 about point F: Effort×Effort arm=Load×Load arm Effort×15=60×2 Effort=12015=8 So, the minimum force required is 8 kgf.

A walnut can be broken by applying a direct force of $60 kgf$ . If the walnut is placed in a nutcracker, the length of whose handle is $15 cm$ , and the nut is placed $2 cm$ from the pivot, calculate the minimum force required to crack the nut.