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A door is hinged at one end and is free to rotate about a vertical axis (figure). Does its weight cause any torque about this axis? Give reason for your answer.
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Let us consider the given diagram, where weight of the door acts along negative $y$-axis.
We know that a force can produce torque only along a direction normal to itself as $\vec{\tau}=\vec{r} \times \vec{F}$, here axis of rotation of door is along $y$-axis and the door is in the $x y$-plane and force $F$ can be applied along $\pm z$-axis, the torque produced by gravity can only be $\pm z$ direction, never about an axis passing through $y$-direction.
Hence, the weight of door will not produce any torque or door cannot rotate about $y$-axis.
We know that a force can produce torque only along a direction normal to itself as $\vec{\tau}=\vec{r} \times \vec{F}$, here axis of rotation of door is along $y$-axis and the door is in the $x y$-plane and force $F$ can be applied along $\pm z$-axis, the torque produced by gravity can only be $\pm z$ direction, never about an axis passing through $y$-direction.
Hence, the weight of door will not produce any torque or door cannot rotate about $y$-axis.
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