# Proving an intuitive fact rigorously

We place a wooden stack upon two supports and place a weight of mass $$m$$ at one of its ends. Now it is almost obvious intuitively that just at the verge of the mass $$m$$ falling down, the normal force $$P_2$$ exerted by the right-most support will be $$0$$. But is there any way we can prove it mathematically or rigorously? It seems so obvious that I can't think of a rigorous way to prove this fact.

• Hint: don't forget to take into account the plank weight. It's not drawn in your example, but it should be from the plank center of mass downwards. Apr 20, 2023 at 4:10

The typical way to do something like this is to assume the system is at equilibrium (so that the net force and torque on the plank vanish) and then solve for $$P_2$$. The result will be a function of $$m$$ which is positive for small $$m$$, zero for some $$m=m_0$$, and negative for $$m>m_0$$

The interpretation of this is that for $$m>m_0$$, $$P_2$$ would have to be negative (i.e. the plank would have to adhere to the right-most support) in order for the system to be in equilibrium. Assuming that no such adhesive forces are present equates to the mathematical constraint that $$P_2\geq 0$$. Assuming that $$P_2\geq 0$$ and $$m>m_0$$ results in a contradiction, which means that it is not possible for the system to be in equilibrium and that it will begin to rotate.

The threshold point - the last point at which equilibrium is possible - corresponds to $$m=m_0$$ and $$P_2 =0$$.

• Pardon please but i am finding it hard to understand where mass $m_0$ comes from? We place only an object of mass $m$ at the end. Isn't this mass $m$ fixed? Sorry if i am being too dumb. Apr 19, 2023 at 21:57
• @madness The quantity $m_0$ will be proportional to the mass of the plank itself and depends on the location of the supports with respect to its center of mass. I'm not saying that $m$ changes dynamically, I'm saying that if you solve for $P_2$ under the assumption that the system is in equilibrium, you will end up with an expression of the form $P_2 = (\text{something...})(m_0- m)$. If $m<m_0$ then the system will be stable, but if $m>m_0$ then it will start to rotate since $P_2$ cannot be negative. If $m=m_0$ then $P_2=0$ and any increase in $m$ will cause the system to rotate. Apr 20, 2023 at 3:26

Indeed, this can be shown using the laws of conservation of momentum and conservation of angular momentum.

Assume we have a two-dimensional body held over two support points. Let's call the normal forces by the two supports $$\vec P_1$$ and $$\vec P_2$$ and let's denote the horizontal distance from each of the points to the center of mass of the body by $$x_1$$ and $$x_2$$ respectively.

If the body is in equilibrium, then the sum of forces on it is zero. Also, the sum of angular moments around the center of mass is zero. Thus we can write: $$P_1 + P_2 = mg$$ $$P_1x_1 + P_2x_2 = 0$$ On the verge of tipping, the center of mass of the body is right below support 1, so that $$x_1 = 0$$. This means from the second equation above that $$P_2 = 0$$, and then from the first equation $$P_1 = mg$$.

So if the beam has mass M, the distance from the end to the supports $$P_i$$ are $$L_i$$, and they apply forces $$F_i$$ then:

$$(m+M)g = F_1 + F_2$$

keeps the beam from falling, while to keep it from pivoting around $$P_1$$, you have to balance the torques:

$$mgL_1 = F_1(L_1-L_1) + F_2(L_2-L_1)$$

Solved those and show that as $$m\rightarrow m_{max}$$, $$F_2\rightarrow 0$$.

• Pardon please but could you please tell me what we mean by as $m$ approaches $m_\mathrm{max}$? What do we mean by maximum mass in this case? Isn't the mass $m$ fixed? Apr 19, 2023 at 21:54
• $m$ is from your drawing. If the beam is massless ($M=0$), it's flipping over. Since you said if $m$ is on the verge of $F_2=0$, then I would think the equations have a solution for $m \in [0, m_{max}]$, I just don't have time to solve them.
– JEB
Apr 19, 2023 at 22:30
• @JEB, why will $m$ flip over if beam is massless? Apr 20, 2023 at 7:52
• @a_i_r Energy wise, it cost no energy to pivot about $P_1$, so $m$ is going to seek lower $mgh$. In fact, it may be better to solve the problem by computing the energy as a function of flip angle...probably a lot easier. If the center of mass gets lower when the beam goes rip: flip.. Ofc: OP asked about $F_2=0$, so that's how I tackled it.
– JEB
Apr 20, 2023 at 21:29

When the stick is on the verge of tipping over, it is perfectly balanced on a single fulcrum point. Add more mass to the end of the stick, it will tip one way, remove mass, and it will tip the other way. Only when the stick is on "the verge of tipping" does it not tip either way. Since the balanced stick has no tendency to move one way or another, we can remove any supports at one end of the stick or the other without changing anything. When the stick is balanced, removing the P2 support has no effect whatsoever, therefore it can't be imparting any force to the stick.

Indeed, this is the entire concept of balance, that the torque induced by gravity on one side of the fulcrum exactly balances the torque induced by gravity on the other side of the fulcrum. If you need any other forces besides gravity and the fulcrum normal force, you do not have balance. We can't have the stick naturally balanced on the leftmost support while feeling support from the rightmost support, as a balanced lever only gets support from the fulcrum. It would be like using a balance scale while propping up one side with your hand and claiming the scale was balanced.

You can also look at this from a Newtons third law perspective. The only way the rightmost support can be pushing up on the stick is if the stick is pushing down on it. But that implies that if you were to remove the support, the stick would fall over to the right, contradicting the premise that the stick was on the verge of falling over to the left.