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 $N_2=\rho g$
