Answer to Question #84703 in Mechanics | Relativity for Godknows Ogun Jonathan

Question

A weightless rod CD, 1.0m long is acted upon by horizontal forces of 2N,2N, 4N, 8N is shown in the figure. What force will be necessary to produce equilibrium?

Answer 1

Since there is no figure shown, hope these basic principles will help to solve the real problem. Let's suppose that we have a rod with orange axis of rotation in the middle and forces applied as shown.

Since there is no figure shown, hope these basic principles will help to solve the real problem. Let's suppose that we have a rod with orange axis of rotation in the middle and forces applied as shown.

Moment of force equals force times distance (from the point where the force is applied to the axis of rotation):

$$M=F\cdot d.$$

Forces that try to move a rod upward are positive, downward - negative. For equilibrium, a sum of moments on the left must be equal to the one on the right:

$$-x\cdot \frac{L}{2}-2\cdot\frac{L}{4}+2\cdot\frac{L}{2}=-4\cdot\frac{L}{4}+8\cdot\frac{L}{2},$$

find

$$x$$

:

$$x=-5 \text{ N}.$$