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Question

A train is travelling West to East along a straight horizontal track. A child suspends a pendulum from the roof of one of the carriages and notices that the pendulum is inclined at an angle of 4◦ to the vertical. Calculate the acceleration of the train.

Where I am at so far:

Let the mass of the bob be m.

I don't know where to start, I don't even know if my diagram is correct.enter image description here

T is the tension in the string.

SFR
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    General relativity nitpick: What's "the vertical" if not exactly the direction a pendulum hangs? – Arthur Aug 06 '19 at 11:50

3 Answers3

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You know two things:

  1. The vertical component of $T$ must equal the weight of the bob, since the bob is not accelerating vertically. So:

$T \cos(4^o)=mg$

  1. The horizontal component of $T$ must be the force required to accelerate the bob horizontally at the same acceleration $a$ as the train (since the bob is stationary with respect to the train). So:

$T \sin(4^o)=ma$

You can eliminate $T$ and $m$ from these two equations and find an expression for $a$ in terms of $g$.

gandalf61
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  • So does that mean my diagram is incorrect? The orange force that I drew is not meant to be there? – SFR Aug 06 '19 at 11:11
  • @CubbyKushi Yes. You can remove the orange arrow. Because the train is accelerating, the bob is not in equilibrium so you do not need a force to balance the horizontal component of $T$. It is the horizontal component of $T$ that is accelerating the bob so that it stays stationary w.r.t. the train. – gandalf61 Aug 06 '19 at 11:26
  • Thanks so much for your help! – SFR Aug 06 '19 at 11:29
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    Strictly speaking, the orange arrow is correct in a frame of reference attached to the train car. That frame of reference is a non-inertial frame and in that frame there is a so-called "fictitious" force that pulls on everything toward the left. But non-inertial frames are a more advanced topic and you don't need them for this problem. It is better to use an inertial frame, where we all understand the rules. – David K Aug 06 '19 at 11:30
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Start with balance of forces projected: $$ \begin{align} ma = T \cos(90^\circ - 4^\circ)\\ mg = T \sin(90^\circ - 4^\circ) \end{align} $$

guest
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Suppose the pendulum hangs at an angle of $\theta$. Then, we can equate horizontal and vertical forces to give $ma = T \sin \theta$ and $mg = T \cos \theta$. This yields $m = \frac{T}{a} \sin \theta$, so $\left ( \frac{T}{a} \sin \theta \right )g = T \cos \theta$, so $\boxed{a = g \tan \theta}$.

Sharky Kesa
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