Question

An effort of 50 kgf is applied at the end of a 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 the lever is weightless.]

Answer 1

Given:

Load = 750 kgf

Load arm = 0.1 m

Effort = 50 kgf

Formula:

Load × Load arm = Effort × Effort arm

750 × 0.1 = 50 × Effort arm

Effort arm = `(750 × 0.1)/50`

= 1.5 m

Since the effort is applied at the end of the lever, the lever is 1.5 m long.