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A door is hinged at one end and is free to rotate about a vertical axis. Does its weight cause any torque about this axis? Give reason for your answer.

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 In \overrightarrow{\tau }=\overrightarrow{r }\times \overrightarrow{F } the direction of \tau is perpendicular to the plane of  \overrightarrow{r } and \overrightarrow{F } . So a force can produce torque only along the axis in the direction normal to force.

The weight of the door acts along –y-axis. The door is in X-Y plane. So it can rotate the door in the axis along the z-axis and not along the y-axis.

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