Playing with rods and sticks

Actually , problem is simpler than it looks, i really thank you for giving good angles and for connecting the rods normally that too at mid point,,

and i must say, how did you know that all those weird numbers would add up to finally give such a simple integer result, it was hard to expect, and at the end i see them cancel out so fine,

ok so heres my way, a bit hectic though, and i will just drop in the equations and the reasons, solve them yourself, , (dont worry its simple to solve them, just dont make calculation mistakes)

ok, consider first rod,, the forces on it are

1) Mg 2) normal reaction of rod acting perpendicularly upon it at midpoint 3) ground reaction

and torque is only provided by ground reaction obviously,

so let the centre of mass move right with a1 and down with a2 as acceleration and let rod rotate about it with acceleration alpha,

also note that its end touching ground has no vertical acceleration,

Now equations concerned are

$Mg-Ncos(x)-R=ma2\\ Rcos(x)\frac { l }{ 2 } =I\alpha \\ Nsin(x)=ma1\\ \\ \frac { \alpha l }{ 2 } cos(x)=a2$

Now these effectively yield two equations that we will use later (basically since i dont give a shit about ground reaction, i eliminate that, also i express angular accelration in terms of a2)

Note that R is ground reaction, N is what we want,

$mg-Ncos(x)=\frac { 13 }{ 9 } ma2\\ Nsin(x)=ma1$

Now note that i have skipped some steps and input some angle values,

Now similarly solving for second rod yields me the two effective equations

$Ncos(x)+Mg=\frac { 7 }{ 3 } Mb2\\ Nsin(x)=Mb1$

Where b1 is acceleration of com of 2nd rod (right side one) along x and b2 along y,,

Now focus, this is the real thing,, since the rods are in contact, and it has been very intelligently mentioned that the ends of the rod are smooth and curved, so at all instants i can draw a common normal at the contact point and obviously they must have similar accelerations along normal, so we have

$a2cos(x)-a1sin(x)=b1sin(x)+b2cos(x)$

Now input the values of a1 a2 b1 b2 which i have given and isolate N, and finally everything messy cancels out to yield 9g,

It should have been mentioned to take g as 10m/s^2 btw

answer is 90N,

Nice problem,

I am giving only formula for calculating $N$ in terms of ${m}_{AB} , {m}_{CD} , \theta$

$N=\frac { 3{ (m }_{ AB }{ m }_{ CD })(g)(cos\theta )(cos2\theta ) }{ { m }_{ AB }(1+{ 3cos }^{ 2 }\theta )(4{ sin }^{ 2 }\theta )+{ m }_{ BC }(1+{ 3 }sin^{ 2 }\theta )(1+2{ cos }^{ 2 }\theta ) }$

i was the 90th solver , yeah cool co-incidences . btw @Ronak Agarwal @Mvs Saketh if you r still active do try my problems waves and shm both r original !