Explain the stationary position of the bob from ground frame of reference

Question

Ignore the friction in the problem.The string used is mass less.

Ok, we usually see these kinda of situation in the pendulum toys in our cars.When we accelerate or cars in the forward direction we usually find the toys(ones hanging via string) accelerate in the backward direction,and more interestingly if we keep our cars moving with same acceleration(or nearly the same),Then we find that the toy,that had accelerated in the opposite direction as that of the car,will be hovering in its position,just like the bob in my diagram(the non-dotted one).

Now replace the car with the box,and the toy with the bob(because,I was not able to draw a car,or a toy...... obviously.)

So initially(represented by dotted lines),the bob was as it is shown in the picture,after acceleration the bob is accelerated in backward direction as shown.So imagining the acceleration of the box to be constant,then the bob will remain at the position(non dotted one).So we can say that the forces on the bob are summing up to zero,because of which it is remaining stationary.

Now,if we look from inside of the box i.e in an non-inertial frame of reference,then clearly,we can add the required pseudo force and explain the stationary condition of the bob.[MY QUESTION]But what if one looks from the ground(inertial frame of reference),would he be able to explain the stationary condition of the bob,because as far as I know pseudo forces are just added in non-inertial frames to make newton's law valid.....Because form ground frame only Tsin(theta) is acting on the bob,so how is it getting balanced...

Answer 1

From your question, I gather that you have found the velocity $V(t)$ of the box that guarantees that, in the frame of reference of the box, bob is immobile. You have found that the acceleration $\mathrm{d}V/\mathrm{d}t$ was a nonzero constant.

Thus, in the frame of reference of ground and along $x$ direction, the bob accelerates also by $\mathrm{d}V/\mathrm{d}t$ and its rate of change of momentum is exactly equal to the unbalanced force you refer to, in accordance to Newton's second law.