The handle of a nutcracker is 18 cm long and a nut is placed 2 cm from its hinge if a force of 36 kgf is placed on the nut it will crack .calculate the force which has to be applied at the end of the handle to crack the nut.

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Solution

Dear Student

Effort arm =18 cm = 18 /100 m
Effort E = 36 kgf

Load arm = 2 cm = 2 / 100

Load = L (Assume)
L x load arm = E x effort arm
L x load arm = E x effort arm
L x 2 / 100 = 36 x 18 / 100
L = 648 / 100 x 100 / 2 = 324 Kgf

Regards

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The handle of a nutcracker is 18 cm long and a nut is placed 2 cm from its hinge if a force of 36 kgf is placed on the nut it will crack .calculate the force which has to be applied at the end of the handle to crack the nut.

Dear Student Effort arm =18 cm = 18 /100 m Effort E = 36 kgf Load arm = 2 cm = 2 / 100 Load = L (Assume) L x load arm = E x effort arm L x load arm = E x effort arm L x 2 / 100 = 36 x 18 / 100 L = 648 / 100 x 100 / 2 = 324 Kgf Regards

Dear Student

Effort arm =18 cm = 18 /100 m
Effort E = 36 kgf

Load arm = 2 cm = 2 / 100

Load = L (Assume)
L x load arm = E x effort arm
L x load arm = E x effort arm
L x 2 / 100 = 36 x 18 / 100
L = 648 / 100 x 100 / 2 = 324 Kgf

Regards