Leaning sticks on a horizontal surface [closed]

Question

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.

Answer 1

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.

I'll let you attempt to do the rest.