A sphere is placed over a block as seen in the figure from below. The mass of the sphere is $10\,kg$ and the mass of the block is $4\,kg$. Assume that the block is supported over two triangular stands and is aligned horizontaly with respect to the floor. Under these conditions. Find the maximum distance in centimeters from $P$ the sphere can be placed so that the system remains in equilibrium.

The alternatives given in my book are as follows:

$\begin{array}{ll}

1.&\textrm{0.4 cm}\\

2.&\textrm{0.8 cm}\\

3.&\textrm{1.2 cm}\\

4.&\textrm{40.0 cm}\\

5.&\textrm{80.0 cm}\\

\end{array}$

How exactly should I use the equilibrium condition on this problem?. I don't know exactly how to cancel the reaction coming from both supports.

So far the only thing which I could establish (for which I'm not sure if it is correct)

Assuming $N$ is the reaction occuring in both triangular supports

$2N=40+100$

$N=70$

Assuming that the torque happens in the second support seen from the left:

$-N\cdot 3 - 100\cdot (x)+ 40 \cdot 1 = 0$

$-70 \times 3 - 140 x +40=0$

But from solving this system yields a negative quantity. What could I be missunderstanding?. Can someone help me here?. What would be the most appropiate approach here?.