Question

a 0.15 kg meter stick is balanced with the pivot point at the 18 cm mark. A weight of 3.2N is hung from the shorter end. Where should the weight be positioned in order to balance the meter stick?

Answer 1

To do this there are two approaches
1) figure out the contact force from the pivot and use sum of forces. That requires multiple steps.
2) we can sum torques at the contact point of hte pivot, thereby eliminating the pivot contact force from our analysis in a proper way :) I called that point P.

We want to know where to put the 3.2 N weight wh acts downwards. The distance from point P have labeled as 'x'.
Call counterclockwise positive. The first term deals with the weight (which we can simplify and analyze as it is acting int he center because it is, presumably, a uniform stick. The second term is the 3.2N block.
\[\sum T = \] - 0.15 kg *9.81 m/s^2 * (0.5-0.18) m+ 3.2N * x
x = 0.147 m
This means that the weight is hung 0.147 from the pivot.
Alternatively you can report your answer as 0.18 m-0.147 m = 0.033 m, 3.3 cm from the left end of the stick as drawn.