Question

A uniform rod of weight 30gf is balanced at 40cm mark a weight of 100gf is suspended at 5 cm mark where must a weight of 80gf be suspended ro balance the metre rod

Answer 1

rods weight exerts at middle o w=30gf at 50 cm

100gf at 5 cm

so

clockwise movement=30*(50-40)gfcm=30*10 gfcm=300gfcm

Anticlockwise movement=100*(40-5)gfcm=100*35gfcm=3500gfcm

Another weight of 80 gf must be hang on left side of fulcrum .

So total clockwise movement=300+80*m=300+80m

∴clockwise movement=Anticlockwise movement

=300+80m=3500

=80m=3500-300

=80m=3200

=m=40m

So.Weight of 80gf must hang of 40 m away from fulcrum i.e;on 80 cm mark.