Question

A half metre rod is pivoted at the centre with two weights of 30 gf and 20 gf suspended at a perpendicular distance of 20 cm and 30 cm from the pivot respectively as shown below.

(i) Which of the two forces acting on the rigid rod causes clockwise moment?

(ii) Is the rod in equilibrium?

(iii) The direction of 20 kgf force is reversed. What is the magnitude of the resultant moment of the forces on the rod?​

Answer 1

Answer:

Assuming the rod is of 100 cm as given in the figure

i) Force of 20 gf

ii) Rod is in equilibrium

iii)

Explanation:

i) Pretty much self explanatory : D

ii) Anticlockwise moment= F*d =30 gf * 20 cm= 600 gfcm

   Clockwise moment= F*d =20 gf * 30 cm= 600 gfcm

Anticlockwise moment= Clockwise moment

Therefore, rod is in equilibrium

iii) New direction is upwards

    Hence moment of force = (600+600) = 1200 gf cm anticlockwise