SIMPLE MACHINES / WEDGE

Why does a shallow wedge need less pushing force?

Use horizontal travel to lift a vertical load by a smaller distance.

EffortLoad

ORIGINAL IDEAL EXAMPLE

500 N load
100 N effort

For the ideal single wedge, horizontal input work equals vertical output work. A shallow working face trades travel for force.

Conceptual schematic; use the model below for parameter-linked geometry.
wedge.napkin / editable
✓ checks pass
Wedge working face
XZ / UNDEFORMED GEOMETRYFITTED TO VIEW

Wedge working face selected · related inputs are marked.

pushing force100.00 N✓ passes

BASELINE INPUTS

CALCULATED

angle11.31 deg
effort100 N
mechanicalAdvantage5
faceLength101.98 mm
centerX50 mm
centerZ10 mm
negativeAngle-11.31 deg

Edits change the baseline source.

Calculated from your source · not saved
Baseline · Baseline
pushing force100.00 N ✓ passes
steeper wedge · steeper wedge
pushing force200.00 N × fails

Calculated from the same .napkin source. Edits preview a temporary study and are not saved.

How the model works

For the ideal single wedge, horizontal input work equals vertical output work. A shallow working face trades travel for force.

effort = load × rise / length

Try the alternative

A steeper wedge increases lift per push but requires more force. This example is a lifting wedge, not a cutting or fracture model.

Choose a scenario, change an input, then inspect the equations and force check in Source.

Assumptions and limits

Friction, self-locking, contact stresses and material fracture are omitted. The geometric model shows the inclined working face rather than a solid wedge feature.