# Two standing sticks with a hinge, seems to lead to a paradox?

I have kind of a paradoxal question about two sticks connected through a hinge that rest on a frictionless floor at the bottom and frictionless walls on each side at the top.

What are the forces in the hinges if the total weight of the sticks is equal to 2*F (F the weight of each stick seperately, the hinge is considered weightless).

I worked out force and moment equilibria, but it seems to lead to a contradiction. I think this configuration should be in equilibrium, but mathematically I cant find a solution. My calculations are:

For the force of the ground on each stick we have $$F_{ground} = F$$, with $$F$$ the weight of a stick.

On the walls: $$F_{wall} = F \frac{(l-a)}{a} \cot \alpha$$, with $$l$$ the length of the stick and $$a$$ the distance from the top of the stick to the hinge point. $$\alpha$$ is the angle of the stick with the ground.

The hinge should thus exert a force of $$-F_{wall}$$ to achieve equilibrium, but in that case the moments around the contact point of stick and wall are not in equilibrium anymore, leading to a paradox.

EDIT: solved! I forgot to include the weight of the sticks in the moment equation...

• But if the potential energy decreases, then the opening angle at the bottom should increase, leading to a paradox as the walls don't permit the structure to open up its angle. I'm starting to be really confused by now. – AWally May 25 at 12:20
• You should provide a better drawing. At the moment there is nothing to support the sticks from below. And if they are not behaving like a wall it is obvious that the system is not in equilibrium because a force in x direction is excerted on them. – MachineLearner May 25 at 12:37
• @Awally, I've deleted my comment. You're right: if the sticks can't reach wider at the top (and if the floor is fixed and frictionless), the linkage has nowhere to move. I think that if you take both vertical and horizontal forces into account, you will show that the system is in equilibrium. More to the point, if you re-cast the problem so that the constraints are built into your equations, you should find success. – S. McGrew May 25 at 12:42
• you have not included torque due to weight acting at centre of mass. – Monocerotis May 25 at 12:49
• Indeed, I had a brain fart it seems. Thanks for the help! – AWally May 25 at 13:00 This is an example of why it will always be equal no matter what the weight is. Let the canvas seat take take the place of your two stick walls.

If $$G=mg$$ for the bar and if the tension force in the strings is equal to $$F$$ then we can realize that forces in the $$x$$-direction nullify. In $$y$$-direction we get

$$G = 2F \sin \alpha$$

You should be able to get the angle $$\alpha$$ by using $$\cos \alpha = \dfrac{w/2}{l}$$. In which $$l$$ is the lenght of the upper strings and $$w$$ is the distance between the two blue sticks.

If you set up the torque equilibrium at the hinch which is connecting all the strings then you will get $$0=0$$, because all forces can be moved into this point.

• The blue 'sticks' are actually walls. Maybe it is not so clear from the figure, sorry. So the system consists of the two sticks connected through the hinge. The top of each stick rests on the blue wall on each side, the bottom on the floor. No friction exists on the floor or the walls. The hinge is weightless and eacht stick weighs F. – AWally May 25 at 12:28
• I thought of the sticks as walls. What is the problem with my solution. It gives you a unique solution but you will need to provide the length of the connecting strings and the distance between the left and right wall. – MachineLearner May 25 at 12:32
• I updated the description. I hope this will clear things up. – AWally May 25 at 12:35

The horizontal (normal) forces at the top are equal but opposite. The weight is supported by the normal forces at the bottom. The sum of the torques about the hinge is zero. The system is in equilibrium.

• But the force at the hinge for one beam should be equal to $-F_{wall}$. If the equilibrium of the moments around the top of the stick are then calculated, the system is not in equilibrium, by definition. – AWally May 25 at 12:39
• @AWally remember there is weight acting at the centre of mass that produces a counterclockwise torque – Monocerotis May 25 at 12:42
• Yes, but there is still no equilibrium possible in that case. – AWally May 25 at 12:46
• Ow ok, I'm sorry. I know what you mean! – AWally May 25 at 12:49