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Levers

Reference data and engineering information about levers for miscellaneous applications.

levers

Overview

Engineering reference data for Levers in miscellaneous.

Key Formulas

Unit Conversion

y=xky = x \cdot k

Multiply by conversion factor.

Linear Interpolation

y=y1+(xx1)(y2y1)x2x1y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}

Estimate between two known points.

Percentage

p=partwhole×100%p = \frac{\text{part}}{\text{whole}} \times 100\%

Part as fraction of whole.

Variables

Symbol Description Unit
xx Input value
yy Output value
kk Conversion factor

Lever Classes

First-Order Lever

  • The fulcrum is positioned between the effort and the load.
  • The effort force is smaller than the load force.
  • The effort moves a greater distance than the load.
  • Functions as a force magnifier.

Second-Order Lever

  • The effort and the load are on the same side of the fulcrum, applied in opposite directions.
  • The load lies between the effort and the fulcrum.
  • The effort force is smaller than the load force.
  • The effort moves a greater distance than the load.
  • Functions as a force magnifier.

Third-Order Lever

  • The effort lies between the load and the fulcrum.
  • The effort force is greater than the load force.
  • The load moves a greater distance than the effort.
  • Functions as a distance magnifier.

Multiple Forces on a Lever

When a lever is subject to multiple load forces, the principle of moments is extended. The generic equation for the effort force (FeF_e) required to balance multiple loads (FlA,FlB,...,FlNF_{lA}, F_{lB}, ..., F_{lN}) is:

Fe=FlAdlA+FlBdlB++FlNdlNdeF_e = \frac{F_{lA} d_{lA} + F_{lB} d_{lB} + \dots + F_{lN} d_{lN}}{d_e}

Where dlA,dlB,...d_{lA}, d_{lB}, ... are the respective distances from each load force to the fulcrum.

Example: A lever is subjected to three loads: FlA=1 lbF_{lA} = 1\ \text{lb} at dlA=1 ftd_{lA} = 1\ \text{ft}, FlB=2 lbF_{lB} = 2\ \text{lb} at dlB=2 ftd_{lB} = 2\ \text{ft}, and FlC=3 lbF_{lC} = 3\ \text{lb} at dlC=3 ftd_{lC} = 3\ \text{ft}. The required effort force at an effort arm distance de=2 ftd_e = 2\ \text{ft} is:

Fe=(1 lb1 ft)+(2 lb2 ft)+(3 lb3 ft)2 ft=7 lbF_e = \frac{(1\ \text{lb} \cdot 1\ \text{ft}) + (2\ \text{lb} \cdot 2\ \text{ft}) + (3\ \text{lb} \cdot 3\ \text{ft})}{2\ \text{ft}} = 7\ \text{lb}

References