# Interesting inertia problem

Consider the following.

A car is accelerating with acceleration $$a$$. A string is attached to the roof of the car and to the bottom of the string, an object of mass $$m$$ is attached. Given $$\theta$$, the angle between the vertical and the string (which is not $$90^\circ$$ due to inertia of the object).

How to derive an expression for the acceleration $$a$$ of the car given $$\theta$$ and $$m$$?

And when does $$\theta$$ remain constant?

I found a similar question, but the answers to that post were too low quality (as is also evident by the fact that the user didn't accept any of those as solutions); so don't flag this post as a duplicate of that.

• This is a common question with many solutions explained on the internet eg Physics problem solving - hanging ball in an accelerating truck. Commented Mar 7, 2021 at 8:33
• Draw FBD of object in frame of car(Non-inertial) and also include Pseudo Force in it. Balance forces in horizontal and vertical direction. Commented Mar 7, 2021 at 8:37
• And what about the second question? Commented Mar 7, 2021 at 8:37
• Two equations are $Tsin\theta=ma$ and $Tcos\theta=mg$. Commented Mar 7, 2021 at 8:41

So if we consider the x-component of the tension $$Tsin(\theta)$$ and given the car moves with acceleration $$a$$ then

$$ma - Tsin(\theta) =0$$

and so

$$a=\frac{Tsin(\theta )}{m}$$

for a mass $$m$$. Remember that the mass experiences an inertial force and so $$a$$ is the acceleration of the car and mass for an observer inside the car. We can write this in terms of theta as

$$\theta=sin^{-1}(\frac{ma}{T})$$ This tells us that $$\theta$$ will continue to increase if $$a$$ increases, and if the rope does not break, it appears that $$\theta$$ can take on any value from $$0\le \theta \le 90$$ degrees.

It could snap at some point $$\theta$$ but to calculate this we would need a maximum value for $$T$$. If $$\theta =90$$ degrees then you’d have the condition $$T-ma=0$$

• Why not write $$\theta=arctan\Big(\frac{a}{g}\Big)$$ by balancing forces in vertical direction? Commented Mar 7, 2021 at 8:39
• @joseph What is the expression for tension here? Commented Mar 7, 2021 at 8:47
• We could have @user1000 but I wanted to keep the Tension symbol. Commented Mar 7, 2021 at 8:56
• $T=mgcos\theta$ is wrong It should be $Tcos\theta=mg$. Commented Mar 7, 2021 at 8:58
• @josephh On the other hand, why does a rod behave differently in this situation?(string replaced by rod) Commented Mar 7, 2021 at 13:03

In summary, when $$\theta$$ is constant we have

$$T \sin \theta = ma \\ T \cos \theta =mg \\ \displaystyle \Rightarrow \tan \theta = \frac a g \text{ ; } T = m \sqrt{a^2+g^2}$$