# Hanging a bob in an inertial frame

Suppose a bob is hanging from the ceiling of a car by a thread. The car is moving with a constant acceleration $a$ towards right. The thread is making an angle $x$ with the vertical. If we see from the frame of car we call say that the pseudo force balances the horizontal component of tension. Then how can we explain why the bob is fixed from the ground frame as there is no pseudo force in this case?

• In the ground frame there is a real force F=ma. – safesphere Sep 16 '17 at 18:27
• Hanging Bob will result in his death regardless of whether the frame is inertial or not. – Emilio Pisanty Sep 16 '17 at 20:55
• 'Death'? I don't follow ? – Hardik Raj Sep 17 '17 at 3:10
• @Hardik A bob is an object. When you just say "Bob", it seems like you mean a person named "Bob", and "hanging" is a method of execution. – JMac Sep 18 '17 at 11:28

From the perspective of an observer from both frames one thing which is common is that the bob is inclined to the vertical making an angle $x$ from the pivoted point.

In the car frame, you are absolutely correct about the pseudo force. I am sure things are pretty clear for you in the non-inertial frame. So, most of my answer will cover calculations made for the bob's motion from the inertial frame.

Let $a_{b,g}$ and $a_{b,c}$ be the acceleration of the bob in the ground frame and car frame respectively.

$\therefore \vec a_{b,g}= \vec a_{b,c}+\vec a_c$

Since, $|\vec a_{b,c}|=0$,

$\vec a_{b,g}= \vec a_c$ $\tag 1$

From the ground, you will observe to bob to be accelerating (not fixed), with an acceleration equal to the acceleration of the car, $'a'$. One might ask why is the bob accelerating. The answer to this question comes into light when you realize that the bob is not completely vertical. The force that is imparting acceleration to the bob is the tension in the string.

Free body diagram of the bob : In the horizontal :

$ma_x=\Sigma F_x$

$ma=T\sin{x}$ $,(\because \Sigma F_x = T\sin{x}$)$\tag 2$

In the vertical :

$ma_y=\Sigma F_y$

but $a_y=0$,

$\therefore \Sigma F_y=0$

$mg-T\cos x=0 \implies mg=T\cos x$ $\tag 3$

Normally one would not be able to form equations of motion using Newton's laws for a body in a non-inertial frame. To make it possible all we have to do is to add a pseudo force in the FBD of the body. This allows us to use Newton's laws in non-intertial frames.

In the non-inertial frame, we just have to balance the pseudo force and the horizontal component of tension in order to satisfy the observation we make from the car's frame that the bob is stationary. Same goes for the vertical axis.

In the car frame there is a pseudo force to explain why the bob hangs at rest at an angle to the vertical. The sum of forces is then zero. In the ground frame the bob is accelerating; this is explained by an unbalanced force in the direction of acceleration.