Dover+Lab+19+-+Torque+and+Rotational+Equilibrium

=Case III: Unknown Mass--the Balance Principle= Mass E at 93.8 cm (.492-.1)(9.8m) = (.938-.492)(.2*9.8) Mass of E = 227.55 grams Percent Error = (227.55-206.14)/(206.14) Percent Error = 10.39%

Mass R at 79.3 cm (.392)(9.8m) = (.793-.492)(.2*9.8) Mass of R = 153.57 grams Percent Error = (153.57 - 132.40)/(132.40) Percent Error = 15.99% =Case IV: Meter Stick with One Mass= Find the mass of the meter stick using one mass

(.297)(.1*9.8) = (.492 - .2)(9.8m) Mass of meter stick = 153.09 Percent Error = (153.30 - 153.09)/(153.30) Percent Error = 0.14%

=Questions= 1) How can a large force produce little or no torque, and a small force produce a large torque? A torque is produced by the resultant of the radius from the pivot point, the force, and the angle at which the force is applied. Therefore, a large force with a very small radius can produce little to no torque, and a small force can produce a very large torque.

2) In order to keep the meter stick and hanging masses from falling to the table, the support must exert an upward force on the meter stick, whose magnitude is equal to the sum of all the downward forces. Why do we not include this "support force" in our torque calculations? We do not include this force in the torque calculations because the force is applied directly in the middle (at the pivot point) and thus would not have an effect on either side. Also the angle at which the force is would result in the torque being 0.

3) A uniform meter stick is at rotational equilibrium when 220 g is suspended at 5.0cm, 120 g is suspended at 90cm, and the support stand is placed at the 40 cm mark. What is the mass of the meter stick? (.35)(.220) = (.1m) + (.50 * 120) Mass of meter stick = 170 grams