This is why you always hang a Monet by THREE nails.

When two nails are in contact with the wall, it is clear that the y-component of the force by each nail is $F_y=\frac{mg}{2}$ .

Right after one nail lose contact with the wall, the y-component of the force by the other nail do not change, so the picture fall down with the acceleration: $a_0=\frac{mg-F_y}{m}=\frac{g}{2}$

image

Beside falling down, the picture begin rotating around the remaining nail.

Since the speed of the picture is 0, the angular acceleration right after one nail lose contact with the wall is $\gamma = \frac{a_0}{\frac{b}{2}}=\frac{g}{b}$ .

We have: $mg\frac{b}{2}=\gamma I=\frac{g}{b} \frac{m(a^2+b^2)}{3}$

$\frac{b^2}{2}=\frac{a^2+b^2}{3}$

$\frac{b}{a}=\sqrt{2}=1.41.$

The moment of inertia of a plane lamina (like a picture) of width $w$ and height $l$ about its center is: $I_{com} = \frac{m}{12} \left(l^2 + w^2\right)$

The parallel axis theorem allows us to shift the axis of rotation to one of the corners: $I_{corner} = \frac{m}{3} \left(l^2 + w^2\right)$

The remaining nail does not break off, so the picture may only rotate about it. There are only two external forces acting on the picture: the force of gravity and the force of the nail. Since the nail is the pivot point, however, the nail's force exerts no torque about the nail. Thus, the only force contributing to the torque on the picture about the nail is that of gravitation, and the magnitude of its torque is: $\tau_g = r_{\perp} F_g = \frac{l}{2} (m g) = m \frac{g l}{2}$

The angular acceleration of the picture is therefore as follows: $\Sigma \tau = I_{corner} \alpha$ $\frac{m}{3} \left( l^2 + w^2\right) \alpha = m \frac{g l}{2}$ $\alpha = \frac{3g}{2} \frac{l}{l^2 + w^2}$

The picture is rigid, so this is the angular acceleration of all points of it. We now want to find the linear acceleration of the center of mass , since Newton's 2nd and 3rd laws allow us to state that: $\Sigma F_{ext} = M_{tot} a_{com}$

Rigidity allows us to say: $a_{com} = r_{com} \alpha = \frac{1}{2} \sqrt{l^2 + w^2} \alpha$ $a_{com} = \frac{3 g}{4} \frac{l}{\sqrt{l^2 + w^2}}$

Note that we had to assume uniformity to state that the center of mass lay at the center of the rectangle. The center of mass is not at the same height as the corner! This means that the $a_{com}$ found above is not downward, but at an angle. It is not difficult to show (similar triangles arising from perpendiculars) that the linear acceleration of the center of mass is at an angle to the vertical whose tangent is equal to $w / l$ . Therefore, the vertical component of the linear acceleration of the center of mass is equal to the acceleration times the cosine of the above angle, or: $a_{y, com} = \frac{3 g}{4} \frac{l}{\sqrt{l^2 + w^2}} \left(\frac{l}{\sqrt{l^2 + w^2}}\right)$ $a_{y, com} = \frac{3g}{4} \frac{l^2}{l^2 + w^2}$

The vertical linear acceleration of the center of mass is equal to the quotient of the sum of the total external vertical forces and the total mass, or: $\Sigma F_{ext, y} = M_{tot} a_{y, com}$

The only external forces are that of the nail and gravitation. The vertical component of the nail's force is equal to half of that of gravitation, since the problem statement specifies that the vertical component of the nail's force is the same as what it was before the other nail fell off. Before the nail fell off, symmetry dictated that the vertical forces of both nails be half of gravitation. The nail's force is upward, however, so the total external vertical forces act as follows: $\Sigma F_{ext, y} = m g - \frac{m g}{2} = \frac{m g}{2}$

Synthesizing all this, we have: $\Sigma F_{ext, y} = M_{tot} a_{y, com}$ $\frac{m g}{2} = m \frac{3g}{4} \frac{l^2}{l^2 + w^2}$ $\frac{2}{3} = \frac{l^2}{l^2 + w^2}$ $\frac{3}{2} = \frac{l^2 + w^2}{l^2} = 1 + \frac{w^2}{l^2}$ $\frac{w^2}{l^2} = \frac{1}{2}$ $\frac{l^2}{w^2} = 2$ $\frac{l}{w} = \boxed{\sqrt{2} \approx 1.41}$

There is likely a faster way, but this was the surefire way that came to me initially. This ratio is incidentally that used in the A0, A1, ... series of paper sizes.

Let the dimensions be $a$ (width) $\times b$ (height) and its mass be $m$ , and $p = \frac{\sqrt{a^2 + b^2}}{2}$ = distance of center of mass from a nail.

Clearly, initially, $N_{y}$ = $\frac{mg}{2}$

When one nail looses, due to the torque produced by the weight of picture about stationary axis perpendicular to the wall and passing through the other nail, the picture gets some angular acceleration(say $\alpha$ ).

Using perpendicular axis theorem moment of inertia about nail = $I = m\frac{a^2 + b^2}{3}$ ,

Taking torque about above mentioned axis,

$\boxed{mg \frac{a}{2} = m\frac{a^2 + b^2}{3} \alpha}$ .... $(i)$

Now, let us talk about the acceleration of center of mass, since initial angular velocity is zero , the centripetal acceleration would be zero and the tangential acceleration would be $a_{t} = p \alpha$ .

Now, say the line joining center of mass and the nail makes an angle $\theta$ with horizontal, then , $a_{t}$ , being perpendicular to this line, would make $\theta$ with vertical. Hence the component of $a_{t}$ along vertical would be $a_{vertical} = a_{t} \cos\theta = p \alpha \cos\theta = \frac{a}{2} \alpha$ (as $p\cos\theta = \frac{a}{2}$ )

Applying Newton's law along vertical,

$mg - N_{y} = mg - \frac{mg}{2} = \boxed{\frac{mg}{2} = \frac{ma \alpha}{2}}$ .... $(ii)$

Divide $(i)$ by $(ii)$ to get :

$3a^2 = 2(a^2 + b^2) \Rightarrow \boxed{\frac{a}{b} = \sqrt{2}}$