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A lift can move in $x$ axis and also in $y$ axis. A bob of mass m is suspended with inextensible thread inside the ceiling of elevator. Determine tension $T$ in the string when elevator moves in $x$ direction. Analysing the motion w.r.t. lift (considering that lift moves with acceleration $a$) the forces on the bob are tension upwards, $mg$ downwards and $ma$ left ( pseudo force). Equating forces in $y$ direction as there is no motion along $y$, I end up getting $T=mg$. Where have I done wrong? The next question in same paragraph is what is the tension when lift moves down with constant velocity. The only downward force on bob is $mg$ and upward force is $T$. Pseudo force is $0$ as the lift is moving with constant velocity. Again I am getting $T=mg$. Where I have gone wrong?

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Your free body diagram holds only for $t=0$.

Observe that at $t=0$, due to the force $ma$, there exists a torque on the bob, $\tau=ma\ell$ where $\ell$ is the length of the string. This causes the bob to rotate. If $a$ is directed to the left for the elevator, the inertial pseudo-force on the bob will be directed to the right, causing the bob to swing anticlockwise.

As long as $ma$ isn't too great, the bob will tend to reach an equilibrium, namely where the torque becomes zero. Let $\theta$ be measured from the vertical. Then the torque becomes zero precisely when: $$mg\ell\sin{\theta}=ma\ell\cos{\theta}$$ Which can be shown with geometry. This reduces to: $$\theta=\arctan{\frac{a}{g}}$$ Intuitively, observe that this makes sense. The larger the lateral acceleration, the larger we can expect the angle of deflection to be.

Since at this point, the bob is in equilibrium, the tension in the string must be equal to the magnitude of the vector sum of weight and lateral force. In other words: $$T=m\sqrt{a^2+g^2}$$ We use the Pythagorean theorem since the lateral force and weight are always orthogonal.

For the second part, as far as I know, since the velocity is constant, $T=mg$. There are no torques at play here. The problem is entirely one-dimensional.

Please comment if anything seems off!

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  • $\begingroup$ I cant understand first part last step $\endgroup$ – dhanesh vijay Jun 7 '19 at 12:13
  • $\begingroup$ Which part exactly are you referring to? $\endgroup$ – Andrew Paul Jun 7 '19 at 15:58

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