An effort of 50 kgf is applied at the end of lever of the second order which supports a load of 750 kgf, such that the load is at a distance of 0.1 m from the hinge.Find the length of the lever?(Assume that lever is weightless)

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Solution

Dear Student,

$Effort, E = 50 kgf Load, L = 750 kgf Load arm, d = 0.1 m Effort arm, D = ? Effort \times Effort arm = Load \times Load arm 50 \times D = 750 \times 0.1 D = \frac{750 \times 0.1}{50} = 1.5 m$

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An effort of 50 kgf is applied at the end of lever of the second order which supports a load of 750 kgf, such that the load is at a distance of 0.1 m from the hinge.Find the length of the lever?(Assume that lever is weightless)

Dear Student, Effort, E=50 kgfLoad, L=750 kgfLoad arm, d=0.1 mEffort arm, D=?Effort×Effort arm=Load×Load arm50×D=750×0.1D=750×0.150=1.5 m

Dear Student,

$Effort, E = 50 kgf Load, L = 750 kgf Load arm, d = 0.1 m Effort arm, D = ? Effort \times Effort arm = Load \times Load arm 50 \times D = 750 \times 0.1 D = \frac{750 \times 0.1}{50} = 1.5 m$