# Force Transfer Between two bodies linked by a rope [closed]

When two bodies, free of movement, are linked by a physical constraint (a maximum distance between them) how do their masses and velocities influence each other? Which formula with theses parameters gives the force applied by one of the two on the other?

For example :
I have two box (grey and red) that are next to each other attached by a rope of maximum length L, the grey one jump following the yellow curve C and i'm trying to get the equation to find the force applied to the red box when the distance between the boxes reach L.
V is the velocity when the gray box reach the red circle.
It would probably split into the radial force and the tangential force depending on the angle. I guess the answer would probably work for every position on the red circle.

I tried to read about pendulum but there was only the gravity taken into consideration, and it kinda felt like the tension was vertical. There might be an answer for me in one ticket but i just couldn't find it.

• Welcome on Physics SE and thank you for the precise sketch :) Is there gravity in your simulation? Is the rope flexible (i.e., can it change length)? Where does the grey box continue to? Commented Aug 12, 2016 at 10:36
• Thanks you. Yes there is also gravity. The rope is not flexible. The grey box continue following Ft (or -Ft if the angle between Ft and v > 90°) Commented Aug 12, 2016 at 11:14
• if there is gravity, is the red box fixed? Commented Aug 12, 2016 at 11:46
• not always, in this i've imagine the redbox on a fixed floor but eventually it could be in the air/fall too or be in movement with an other velocity for example (1,1). The gravity on the redbox is already applied by the engine but yes if they fall in the same direction and the same speed there should be no additional force on the redbox. (i've just realised o_O) Commented Aug 12, 2016 at 12:48
• Are you saying that once grey box reaches the red circle it is then confined to move only along that circle?
– Deep
Commented Aug 13, 2016 at 4:39

The string is inextensible, so it becomes taut instantaneously. The momentum of A changes instantaneously, hence in this model the tension in the string becomes infinite. Instead of dealing with force, we have to deal with an impulse, causing a change in momentum for particles A and B.

The traditional assumption has been that the final momentum is shared equally by the particles attached to the ends of the string. This is equivalent to a completely inelastic collision. However, this does not agree with the observed behaviour of real strings.

These authors suggest that it is more realistic to regard the string as elastic but with a very large spring constant $k$. The result is that kinetic energy is conserved as well as momentum along the line of the string - ie we treat this as an elastic collision.

Instead of calculating kinetic energies it is easier to apply a coefficient of restitution $e=1$ to the collision, so that (along the string) the relative velocity of separation equals the relative velocity of approach.

For real strings neither model (completely inelastic or completely elastic) is completely accurate. The correct model is somewhere in between and depends on the material, construction and history of the particular string. Typically $e \approx 0.65$ according to Roper & Hartley p 18.

It appears that particle A is a projectile, falling under gravity in a parabola. When the string becomes taut its distance from B is $L$. You can work out the position and velocity of A when this happens. Then apply conservation of momentum and the restitutions law to the (partially) elastic collision along the line of the string. B is "jerked" towards A; the radial component of the velocity of A is reduced, but the tangential component is not affected. After this "collision" A again moves as a projectile, until the string becomes taut a second time.

If particle B is not free to move but is fixed in position then it has an effective mass much greater than that of A. The result of the collision in this case is that the radial velocity of A is reversed, like that of an elastic ball bouncing back after colliding with the ground.

well, I am not too sure about the physicallity of the whole situation anymore, but the force $F_r$ (so the radial part of the force $m_{grey} \vec{a}_{grey}$ needed to change the velocity of the grey box) will get transferred through the rope directly onto the red box. The direction will just be along the connecting line of those two boxes.

If an object is following a path, and some instant the tangent vector is $\hat{e}$ and the normal vector $\hat{n}$ then the velocity and accelerations are decomposed as

\begin{align} \vec{v} &= v\, \hat{e} \\ \vec{a} &= \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} \end{align}

Where $v$ and $\dot{v}$ are the scalar speed and acceleration along the path and $r$ is the radius of curvature of the path

Finally you add-up all the forces vectorially to get the equations of motion

$$F_t \hat{e} + F_r \hat{n} + m \vec{g} + T \frac{\vec{\ell}}{\|\vec{\ell}\|} = m \vec{a} = m \dot{v}\, \hat{e} + \frac{m v^2}{r} \hat{n}$$

where $\vec{\ell}$ is the vector separation of the rope and $T$ is the tension