Leaning sticks on a horizontal surface [closed]

A body is composed of two straight pins that are joined at a right angle. They have lengths α and β and the mass per unit length is ρ. 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.

We can bisect the angle with a downward altitude, call it h. Recall that the area of the triangle is:

$\frac{ab}{2}=\frac{\sqrt{a^2+b^2}h}{2}$ and we have $h=\frac{ab}{\sqrt{a^2+b^2}}$. It follows now that $sin\alpha=\frac{h}{a}=\frac{b}{\sqrt{a^2+b^2}}$.

We proceed now by calculating the moment around the left contact-point and recall that $M=F*r*sin\alpha$

$(\frac{ρag}{2})(\frac{a}{2})(\frac{b}{\sqrt{a^2+b^2}})+(\frac{ρbg}{2})(\frac{b}{2})(\frac{a}{\sqrt{a^2+b^2}})-N_2\sqrt{a^2+b^2}=0$

Simplifying and express in terms of N_2 gives:

$N_2=\frac{ρabg}{4(a^2+b^2)}(a+b)$ which is wrong.

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 Hi AdamYac - this is a site for conceptual questions about physics, not general homework help. In the future, when you have a homework-like question to ask, please narrow it down to the specific concept that is giving you trouble and ask about that. See our FAQ and homework policy for more information. – David Zaslavsky♦ Mar 7 at 1:18

closed as too localized by David Zaslavsky♦Mar 7 at 1:17

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.

Here's the setup: first use Newton's second Law in the vertical direction to get $$N_a + N_b - (m_a+m_b)g = 0$$ next balance the torques about the vertex of the triangle to get $$-aN_a\cos\alpha+\frac{1}{2}am_ag\cos\alpha+bN_b\cos\beta-\frac{1}{2}bm_bg\cos\beta=0$$ where $\alpha$ and $\beta$ are the angles between the sides of the triangle and the horizontal on the left and right sides respectively.