Page History

Part A

What are the forces acting on the Yardstick before the hands start moving?

Solution

A Force Diagram of the yardstick looks like this. gravity (interaction) pulls downward at the center of mass with force mg (where m is the yardstick's mass) and is resisted by the normal forces F₁ and F₂ of the hands

We now write the equations of Newton's 2nd Law for the center of mass and of torque balance for the system. Since the system is at rest, we can put the axis wherever we choose. The simplest point is at the center of mass , since that removes one term from the expression for torque (single-axis). The torque must be zero, since the yardstick is not rotating.

Part B

Now imagine that the hands start moving slowly toward the center. There is frictional force preventing easy sliding of the yardstick on the hands, with coefficient of friction μ , but eventually the stick must slide against the hands. What happens?

The yardstick will try to remain stationary, with the skin on the hands moving against the underlying tissue until it can go no farther. We know that the force of friction will prevent motion until the force exceeds F_max given by

\begin{large}\[ F_{\rm max} = \mu N \]\end{large}

where N is the normal force.

Solution

We again sketch the situation, this time adding friction.

The magnitude of the sum of the horizontal forces is

\begin{large} \[ F_{\rm total} = F_{f1} - F_{f2} = \mu F_{1} - \mu F_{2}\]\end{large}

Substituting from Part A for the normal forces F₁ and F₂ gives

\begin{large}\[ F_{\rm total} = \mu mg \frac{b - a}{a + b} \]\end{large}

If a > b , then the normal force F₂ is greater than the normal force F₁, and hence the frictional force due to the hand a distance b from the center of mass is greater. Therefore that hand will remain fixed relative to the yardstick and the other hand will slide, bringing it closer to the center of mass of the yardstick.

As it does so, the distance a will change, and the distribution of normal forces will change – that on the left hand will begin to increase, and that on the right hand will decrease by the same amount. The two normal forces will become equal when the new distance between the left hand and the center of mass equals b. At this point, the normal forces will be equal, and so will the frictional forces.

As you continue to press your hands towards each other, the forces will eventually cause the yardstick to slip against your hand. In an ideal world, with the same normal force on each hand and the same coefficient of friction, the forces on both hands will be equal and you would expect both hands to move toward the center. Inevitably, however, all things will not be equal – the coefficient of friction will vary from place to place on the yardstick, for instance, and one side will move first. But, by the same logic as above, the hand closer to the center of mass will have more normal force on it, and hence the frictional force will increase and will eventually stop the hand from moving.

We then have the re-appearance of the initial state, with one hand closer to the center of mass than the other, and consequently having more normal force, so it is the hand farther from the center of mass that now moves until the hands are again equidistant from the center of mass.

The hands and the yardstick proceed in this way, with one hand moving a little bit closer, then the stick stopping and the alternate hand catching up, until the hands meet, at which point the center of mass will be very nearly overhead, to within the range of one of these motions.

Part C

How can we make the hands meet at some other point than the center of the yardstick?

Solution