Answer to Question #168418 in Physics for Andu

Question

8. The forces F1 = 17N, F2 = 25N and F3 = 23N act on a horizontal rod that rotates around the axis O. Give OA = 16cm, OB = 37cm,

 OC = 60cm and a = 30 ° a) Calculate the moment of each force in relation to the axis of rotation. b) Calculate the resultant moment acting on the rod. c) Is the rod in balance? Why? Discuss the result.

Answer 1

a) Calculate the moment of each force in relation to the axis of rotation.

As we know, the moment if lever length times perpendicular component of force. Moments clockwise are negative and counterclockwise are positive.

Moments of the forces:

$$M_1=F_1·D_{OA}=17·0.16=2.72\text{ N-m}.\\ M_2=F_2·D_{OB}=25·0.37=9.25\text{ N-m}.\\ M_3=-F_2D_{OB}=-23\text{ cos}30°·0.6=-11.95\text{ N-m}.$$

b) Calculate the resultant moment acting on the rod:

$$M=M_1+M_2+M_3=0.02\text{ N-m}.$$

c) The rod is not in balance because the resultant moment is not zero. Thus, these forces will slowly rotate the rod counterclockwise.