Question

A weightless rod CD,1.0m long is acted upon by horizontal forces of 2N, 2N, 4N and 8N. what force will be necessary to produce equilibrium

Answer 1

With the horizontal placement of the rod, 2N-forces acting to the left and 4N- and 8N-forces acting to the right the necessary force for equilibrium is

$$0=2+2-4-8+F,\\ F=8\text{N}.$$

This force must act to the left.