A uniform meter rule is balance at 30cm mark when of 50grams is hanging from its zero cm mark.calculate the weight of the rule?

m*(50-30) = 50*30 gives MASS, m, in grams (not really weight but is this math or physics?)

the centers of mass of the two segments of the rule are halfway from the balance point to the respective ends

let x = mass of the rule in grams

(50 g * 30 cm) + (.3 x * 15 cm) = .7x * 35 cm

after dividing by 5 ... 300 g-cm + .9x g-cm = 4.9x g-cm

solve for x

Although that should work fine Anonymous, you do not have to split the 100 cm ruler into sections left and right of the balance point. The weight of a rigid body acts as if it were all at the center of mass, no matter what forces are acting on it where.

I want the answer please

I need the workings

physics

clockwise moments=anti-clockwise moments

weight1 multiplied by distance1= weight2 multiplied by distance2
50-30=20
change to newtons
0.5 multiplied by 0.3= 0.2 multiplied by distance2
=0.75 newtons

Answer

0.75 N

0.75N