PHYSICS! Rotational equilibrium?

For translational equilibrium

N=Mg

19.6=(.1+.7+m)*9.8

m=1.2kg, m is unknown mass

For rotational equilibrium

torques on both sides must be equal

.04*20 + .7*35 = .06*30 + 1.2*x

x=19.5833 cm

here distances are from the point where the string is attached

true x = 40+19.5833=59.5833cm

mass is attached at 59.5833 cm mark

Start with A. You know the total force from the ceiling, and you know the mass of the stick and one weight. All you need is the second weight.

19.6N = g * (.1kg + .7kg + x)

19.6N/g - .8kg = x

2kg - .8kg = x

1.2kg = x

Now for b, the two hanging masses plus the stick must produce equal moments around the support string. And the moment is simply the mass times the distance from the support.

To the left of the support is 40cm of stick: Avg distance = 20cm, mass = .04kg

also mass 1 is 35cm left of the support (40 - 5)

To the right of the support is 60cm of stick: Avg distance = 30cm, mass = .06kg

and mass2 (1.2kg) is on that side as well.

20cm * .04kg + 35cm * .7kg = 30cm * .06kg + x * 1.2kg

.8kg cm + 24.5kg cm - 1.8kg cm = 1.2kg * x

23.5kg cm / 1.2kg = x

19.6 cm = x

That is to the right of the support at the 40cm mark so :

19.6cm + 40cm = 59.6cm