Value and direction of the force applied by a rubberband on an object

Question

Is there a model that would allow me to compute the value and the direction of the force F applied to the object in this case: I have to fixed points: $ A = (-l/2,0) $ and $B=(l/2,0)$. A rubberband of initial length $l$ is tied between $A$ and $B$. A hockey puck $H$ is catching the rubberband; the setup is like a bow under tension. Now I would like to compute the force applied on the puck in order to compute its trajectory.

Answer 1

The situation looks like this. Note that to avoid annoying factors of two I've made $l$ the distance from the ends of the rubber band to the middle, i.e. the distance from A to B is $2l$. My $l$ is half the size of yours.

In this diagram you've pulled the puck back a distance $d$. The force on the puck is $F$. Suppose the tension in the rubber band is $T$ then the force on the puck is:

$$F = 2 T sin(\theta)$$

So this gives you the force provided you know the tension in the rubber band, $T$. Assuming the rubber band obeys Hookes law, and assume the tension when the rubber band is straight is $T_0$, then the tension $T$ is given by:

$$T = A\left(\frac{l}{cos(\theta)} - l\right) + T_0$$

where $A$ is the spring constant. This expression may look a bit opaque, but $l/cos(\theta)$ is the length of the band from the point B to the puck, so $l/cos(\theta) - l$ is the amount of extra stretch. Multiply this by the spring constant, $A$, and you get the extra tension due to pulling the band back. Then add on the initial tension $T_0$ and you get the total tension.

The $sin$ and $cos$ make life a bit difficult because you have to calculate $\theta$. You can do this by noting that $tan(\theta) = d/l$ or if the distance $d$ is small compared to $l$ you can use the approximations:

$$sin(\theta) = \frac{d}{l}$$ $$cos(\theta) = 1$$

to make life easier. If you do this you get the very simple approximate expression:

$$ F = 2T_0 \frac{d}{l} $$