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March 6, 2015

March 6, 2015

Posted by **Anonymous** on Sunday, December 9, 2007 at 3:33pm.

- physics -
**drwls**, Sunday, December 9, 2007 at 4:11pmAssuming the hands are rodlike (which is not true), the torque exerted by a hand is half the length, times the mass, times g, times the sine of the angle that the hand makes with vertical. The direction of the torque must also be considered. It is positive (clockwise) from the 12 to the 6 positions, then the negative.

(a) At 3 PM, only the hour hand exerts a torque because the minute hand is straight up. That torque is

(60 kg)*1.35 m*9.8 m/s^2*1.00 = 794 M*m

(b) This is a trickier one. You have to consider both hands, and the hour hand is between numbers.

(c) Both hands are on a vertical axis of symmetry. Their weight acts along a line that passes through the axis of rotation. There is therefore no torque.

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