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

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.

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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:

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

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Is the general case significantly more complicated ? I did not specify that the puck was on the perpendicular bisector of $[AB]$. I think that this implies to introduce the friction between the rubberband and the puck. I'd like to use this in a casual game where the player must find an angle and a distance to reach a goal. Does this also imply a spin around the vertical axis ? – alecail Apr 28 '12 at 20:48
The physics is actually pretty simple. You have a tension in the rubber band, which is simply given by the extension. The force on the puck is simply the vector sum of the tension in the string either side of the puck. If the puck isn't in the middle it just adds more variables. If the puck isn't in the middle the force won't be normal to $AB$, and there will be a torque on the puck so if you want to include the puck spinning also need to consider the geometry of the puck and it's moment of inertia. – John Rennie Apr 29 '12 at 6:24