Notions of force, torque and mechanical equilibriumA 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 sum of the forces in the x or y direction, Fx and Fy, must be zero. In the direction y this means that the force exerted by the block on the bar, Fa, must be balanced by the forces from the triangular blocks on the bar, Fb and Fc; therefore Fa=Fb+Fc, or Fb=Fa-Fc. There are no forces acting in the x direction. The sum of the counterclockwise torques must equal the sum of the clockwise torquesComponente Curricular::Ensino Médio::FísicaComponente Curricular::Educação Superior::Ciências Exatas e da Terra::Física