Equilibrium of a Rigid Bar

A rigid bar of negligible mass is supported on either end by two triangular blocks. On top of the bar sits a box of variable mass that can be moved along the bar. The forces , , and act at , , and .
This Demonstration shows how to use balancing of force and torque to find the forces exerted by the triangular blocks on the bar due to the weight of the box.
The sum of the forces in the or direction, and , must be zero. In the direction this means that the force exerted by the block on the bar, , must be balanced by the forces from the triangular blocks on the bar, and ; therefore , or . There are no forces acting in the direction.
The sum of the counterclockwise torques must equal the sum of the clockwise torques. In this case this means that , or .

SNAPSHOTS

  • [Snapshot]
  • [Snapshot]
  • [Snapshot]
    • Share:

Embed Interactive Demonstration New!

Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site. More details »

Files require Wolfram CDF Player or Mathematica.