When the center weight of weight 2W moves a distance h from lowest to highest position, the other two outer weights (each of weight W) move down through an equal distance h. So the gravitational potential energy at the upper and lower extremes of motion is the same, whatever happens in between, if no additional energy is supplied to the system. The only energy you could get out of this device is the small energy you supplied with that initial upward "mysterious force" on the blue mass. That alone kills any hope of perpetual motion with this device.

This energy argument will likely not satisfy the inventor, for most who design perpetual motion and free energy devices do not accept the principle of energy conservation. Analysis of forces and torques give us a better idea of what going on in this interesting mechanical device. We have demolished this perpetual motion idea, but further analysis is instructive.

## The system center of gravity and equilibirum positions.

Look at the center of gravity of the system in all positions. At the extremes of motion the system center of gravity is on the dotted line, halfway between the two axles.When all the weights are on the dotted line, the center of gravity is zero also. There are three positions where the center of gravity is on the dotted line. Are any of these positions stable? What about points elsewhere?

If the blue weight is at the top position, and then nudged a bit downward, the system center of gravity moves lower. So this position is not stable. If the blue weight is at the bottom position, and then moved a bit upward, the center of gravity moves upward . The blue weight wants to stay at the bottom, but only because it has reached the limit the slotted beams allow. If they were longer, the blue weight would move lower. If the weights are all on the dotted line, any disturbance in either direction causes the system to move away from that position. There is also a potition with the blue weight below the dotted line where the equilibrium is unstable, and a nudge in either direction cause the system to move away from that point.

So there is one position of stable equilibrium and it's not at the extremes of allowed motion as the inventor claimed. There is also one position of unstable equilibrium, At both of these positions the net force and net torque on the system are both zero.

Where are these equilibrium positions?

## Force and Torque analysisto find positions of static equilibrium.

When the system is in static equilibrium, the net force on each part of the sytem is zero, and the sum of torques about any axis is also zero. Let's concentrate on the blue weight. It slides in slots in the rotating beams. The system is symmetric, and can be further simplified. Break it into two parts, each having a rotating beam with two equal masses. The blue mass moves only up and down, so we can consider just the left hand part of the system with half of the blue weight riding in a fixed vertical slotted bar. (This is a sneaky application of Galileo's principle of superposition.) This greatly simplifies the analysis, but you may use our method on the entire original system if you enjoy messy mathematics. We guarantee you'll get the same answer.

I call this device the "taptoy" for reasons that will be revealed. The figure shows the geometry of the system, with the vertical bar at distance L from the axle, and the red weight moving in an arc of radius R. Both weights have size W. (When we split the system, we split the blue weight in half as well.)

Now we focus on forces on the blue weight. Here lies a trap for the unwary. If we draw forces on the diagram it isn't always clear which forces act on what. We should focus our attention on forces acting on just one part of the system. It is wise to draw a separate diagram showing only forces acting on that part of the system. The forces acting on the blue weight, are, the force due to gravity, W, the force due to the tilted bar, B (perpendicular to the rotating bar, and the force due to the vertical bar, H (which is horizontal). That's all.

The free body diagram shows these three forces, and for static equilibrium, they must sum to zero. Looking at the verticdal components, we conclude that B cosβ = W. From horizontal components, H = B sinβ but we won't need that equation.

The only forces acting on the blue weight are the gravitational foarce, W, the force due to the vertical bar, H, and the force due to the tilted bar, B. They are shown as black arrows. We also show the vertical and horizontal components of B in violet color.

Now, from Newton's third law, we know that if the tilted bar exerts a force B on the blue weight, the blue weight exerts a force −B on the bar. That is the only force on that end of the bar! Do not be tempted to include anything else, for the vertical bar does not exert a force on the tilted bar, it only exerts a force on the blue weight. That's the subtle trap we warned you about. This force, B, on the bar acts downward and to the right, It provides the torque that opposes the torque due to the red weight on the left end of the bar.

The blue weight exerts force B on the tilted bar (normal to the bar), with lever arm given by L/cosβ and torque −BL/cosβ, by the usual convention that clockwise torques are negative. The torque that the red weight causes at the other end of the bar is +WR/cosβ. For force equilibrium, these must be equal. Therefore BL = WR. Substituting for B we get LW/cosβ = WR, so L/R = cosβ. This determines the angle β at which the tilted bar is in stable equilibrium. When L = R/2, β = 60°. When L = R/3, β = 70.53.

This turned out to be an elementary physics homework exercise. We found the one position of stable equilibrium. Are there others? No. But there's a position of unstable equilibrium with the blue mass below the dotted line. At that position the forces and torques on all parts of the system are zero, but the slightest disturbance in either direction causes the system to move away from that position. Use our method above to find that position. Hint: the force B now is aimed upward and to the right, so H points to the left.

The original flipping mechanism (Fig. 1) also showed a geometry with L = R/2. This is a special case, and one position of stable equilibrium (the one we just found) happens to have the red and blue weights equidistant from the axle. This leads the unwary student to think that that condition is always true. It is not, as we will show.

One should always test results in the real world. It isn't difficult to make a small model of our simplified device, and allow motion beyond the stable equilibrium points. Steel construction sets are handy for prototype devices. European-made metric sets (Eitech, for example) have nice slotted strips and girders, ideal for this device. The figure shows my model.

Of course this model has considerable friction, even after smoothing the inner faces of the slots with emery paper. The movable beam is balanced. The left weight is fixed to it. The right hand weight is free to slide in its slot and in the slot of the fixed vertical strip. Here friction is useful, If we tap the base of the device repeatedly, it settles to its equilibrium position and stays there. That's why I call this the taptoy. Or we can sit the device on a vibrating platform. The rotating arm settles to the same equilibrium position whether the right hand weight is above or below it. And, no surprise, that equilibrium position is when the weights are equidistant from the axle.

Now we can test whether this result is general.

## Generalze?

Our inventor chose to diagram a special case, where L = R/2 and the slotted strips only allowed motion to the positions of unstable equilibrium, not beyond. He incorrectly asserted that one of them was stable. He never mentioned, or found, the true position of stable equilibrium.

His special case also had the result that at stable equilibrium the weights are equidistant from the axle. This tempts one to naively suppose that this is true for all such mechanisms. But one should never generalize from just one special case. Is it true if the vertical slotted strip were tilted? No, it isn't, but this left as an exercise for the reader. We can test this with our prototype model. Just tilt the entire model, which is equivalent to just tilting the vertical strip. When this is done the system settles to an equilibrium position that is not when the weights are equidistant from the axle. Depending on the direction of tilt, equilibrium can be with the right weight farther or nearer the axle than the left weight.

An improved model is shown in Fig. 4. It has L/R = 5.00 cm / 1.25 cm = 4. It more clearly demonstrates the position of unstable equilibriuim, and allows exploration of the full range of motion along the vertical slot. There is, of course, a position of stable equilibrium when the right hand weight is at −∞ but our model is inadequate to demonstrate that.

Latest revision April 2, 2018.