# What does it really mean to have infinite tension?

Let an object of mass $$m$$ lie on a wire. The wire is much smaller in mass than the object and so its mass can be negligible. As the object lies on the wire, the wire is pulled down to a position where the object is no longer accelerating and is in equilibrium.

We can calculate the tension in the wire,

considering forces in the $$x$$ direction, $$F_{net_{x}} = m\vec{a_x}$$ $$T_{R_x}=T_{R_x}$$ $$T_{R}\cos\theta=T_{L}\cos\theta$$ $$T_{R}=T_{L}=T$$

considering forces in the $$y$$ direction, $$F_{net_{y}} = m\vec{a_y}$$ $$T_{R_y}+T_{L_y}-mg = 0$$ $$T_{R}\sin\theta+T_{L}\sin\theta = mg$$ $$T \cdot \ 2\sin\theta = mg$$  Finally, the tension force in the wire of an object of mass $$m$$ lying at equilibrium. $$\boxed{T = \frac{mg}{2\sin\theta}}$$

I have two follow up questions to this,

Firstly, we can see that for $$\theta = 0$$ and $$\theta = \pi$$ the tension is infinite, in a mathematical perspective this seems fine but in a physics point of view what does it actually mean to have infinite tension?

Secondly, lets suppose the object is stuck to the wire and is oscillating up and down along with the wire, because the angle is changing at every instant so does the tension in the wire. How can one derive an equation that describes the tension of the wire at any instance in time while the object-wire system is oscillating up and down at some constant rate?

• but in a physics point of view what does it actually mean to have infinite tension? It means it simply cannot exist: $\theta$ will aways at most be slightly smaller that $\pi$. Infinite forces cannot exist.
– Gert
Apr 17 at 14:11
• How can one derive an equation that describes the tension of the wire at any instance in time while the object-wire system is oscillating up and down at some constant rate? That equation is called the wave Equation for a taut string or wire. It's the basis for the math treatment of plucked string instruments.
– Gert
Apr 17 at 14:13

It means that $$\theta = 0$$ will never happen.

In real life, there is always going to be some sag in the system and $$\theta >0$$. It is a physical impossibility to have infinite tension and this is to be avoided both practically, and mathematically.

You statement that $$1 \over 0$$ is fine mathematically isn't accurate either.

For the second part, you need an estimate of the vertical stiffness of the wire given a small pertubation from the equilibrium position. The original equation isn't valid because it pertains to static equilibrium only.

The get the full dynamic model of the system you will need some assumptions. For example the wire will need to be elastic and the pertubation small enough to use small angle approximations. Even so, the resulting equation is non-linear and has 2DOF as the mass can sway left and right as well as up and down due to the elasticity of the wire. This considers the tension waves going back and forth along the wire.

If you restrict the interaction between the wire and mass to a vertical load only you will get a function of resotring force (applied on mass) as a function of displacement and this is also non-linear.

Further assumptions are needed to linearize the above and to get it in the form of $$F = -k \,x$$ where $$F$$ is the resoring form and $$x$$ the vertical deflection from equilibrium. The solution would be harmonic motion with frequency $$\omega = \sqrt{ \tfrac{k}{m} }$$ and the form $$x=X \sin(\omega t)$$.