A uniform meter stick of mass 40 g is used to suspend three objects as follows: 25 g at 0m, 40 g at 0.34 m and 20 g at 0.7 m. Where could you place your finger to balance the stick in the horizontal position?

Question

A uniform meter stick of mass 40 g is used to suspend three objects as follows: 25 g at  0m, 40 g at 0.34 m and 20 g at 0.7 m. Where could you place your finger to balance the stick in the horizontal position?

anonymous · Answer

To solve the problem we must use the law of the lever
$$F _{1}D _{1}=F _{2}D _{2}$$
But here we have three forces applied to three different points, so in our case the law will look like
$$F _{1}D _{1}+F _{2}D _{2}=F _{3}D _{3}$$
F=mg
So after some manipulations we get 
$$25D _{1}+40D _{2}=20D _{3}$$
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Object I at o cm
Object II at 34 cm
Object III at 70 cm of meter stick
from the figure above you may notice that
$$D _{1}+D _{3}=70 cm$$
$$D _{3}+D _{2}=36 cm$$
Now you can find D1, D2 and D3
D1 = 32.5 cm
D2=-1.5 cm
D3=37.5 cm 
from pivot point
negative sign of D2 means that the object II with mass 40g is located to the right of pivot point.
So the answer is:
To balance the stick with these three objects you must place your finger below 32.5 cm mark of meter stick.