enter image description here

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. enter image description here

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...

  • $\begingroup$ Try to make your questions concise. $\endgroup$ – Joce Sep 17 '20 at 15:58
  • $\begingroup$ The bob is not immobile in the ground frame of reference, it is accelerating due to the force $T \sin\theta$ $\endgroup$ – Wolphram jonny Sep 17 '20 at 20:59
  • $\begingroup$ @Wolphramjonny,yes you are absolutely right with your statement,can draw the motion of the bob ,when seen by man from ground frame of reference $\endgroup$ – PATRICK Sep 18 '20 at 9:22

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.

  • $\begingroup$ Please tell me what in the elements of my answer is not in agreement with your question. $\endgroup$ – Joce Sep 17 '20 at 15:59
  • $\begingroup$ @PATRICK, in the ground frame the object is accelerating. But your question says "it is remaining stationary". These are not compatible. It is only stationary in the box frame, not in the ground frame. $\endgroup$ – BowlOfRed Sep 17 '20 at 16:50
  • $\begingroup$ Actually I have thought about it,for quiet a while and you are absolutely right.My problem was that I was not able to imagine the motion of the pendulum toy from ground frame of reference,besides it was so easy,I don't know why I had this problem but anyways,got throught it anyways...... Thank ypu for your time and contribution everyone.. $\endgroup$ – PATRICK Sep 18 '20 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.