# Choosing the right point when calculating moments

A body is composed of two straight pins that are joined at a right angle. They have lengths $a$ and $b$ and the mass per unit length is $\rho$. When the body is balanced on a flat surface, as shown, how large is the normal force against the ground in the right point of contact? 4 options as can be seen in the picture.

Let me point out that this is a conceptual question. When I first tried to solve this problem I decided to choose to calculate the moment around the left contact-point in order to reduce one term(left normal force). This seems like a natural way but gives a false answer($N_2= \rho g$. If I instead calculate about the vertex of the triangle and use Newton's second law I get the correct solution.(answer is D).

So how should I choose the "correct" point?

Around left contact point:

By dropping an altitude h at the right angle we get that:

Notice that: $\cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$, $\cos\beta=\frac{b}{\sqrt{a^2+b^2}}$ and $m_ag=\rho ag$ and $m_b=pbg$

$-\frac{\rho a^2g}{\sqrt{a^2+b^2}}-\frac{\rho b^2g}{\sqrt{a^2+b^2}}+N_2\sqrt{a^2+b^2}=0$. Simplifying gives $N_2=\rho g$

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Related: physics.stackexchange.com/q/56108. (Which you should probably have linked just so that it didn't look like you were trying to sneak it past...). I don't believe we have a policy on problem solving strategy questions as yet. I'm inclined (heh!) to be against, but won't take unilateral action on the matter. In any case, answerers should stick to the strategy question and not to answering the particulars of this problem. –  dmckee Mar 7 '13 at 21:59
Assuming that the last line there is meant to be the torques (moments in engineer-speak) about the point where the short piece hits the ground you've written both contributions from the masses incorrectly. They both act through their own CoG... –  dmckee Mar 7 '13 at 22:27
dmckee: Yes,that point. I am not really following: Both masses act from the midpoint on each stick, that is $\frac{a}{2}$ and $\frac{b}{2}$. Altitude h can easily be computed since $\frac{ab}{2}=\frac{h\sqrt{a^2+b^2}}{2}$. It follows that $h=\frac{ab}{\sqrt{a^2+b^2}}$ Hence the values for $\cos\alpha$ and $\cos\beta$ –  EricAm Mar 7 '13 at 22:36