EWeaver+Lab+19

=Torque and Rotational Equilibrium=

Case III:
(m1)(x1) = (m2)(x2)

Mass D: (m1)(.33) = (200)(.45) m1=272.727g Actual mass: 271.02g

Mass B: (m1)(.33) = (200)(.42) m1 = 254.54g Actual mass: 257.51

Mass R: (m1)(.33) = (200)(.235) m1 = 142.42g Actual mass: 132.40g

Mass E: (m1)(.33) = (200)(.37) m1 = 224.24g Actual mass: 206.14g

Mass F: (m1)(.33) = (200)(.325) m1 = 196.97 Actual mass: 196.7

Percent Error:
Mass D: (272.727-271.02)/(271.02) x100 = .63%

Mass B: (254.54-257.71)/(257.71) x100 = 1.23%

Mass R: (142.42-132.40)/(132.40) x100 = 7.57%

Mass E: (224.24-206.14)/(206.14) x100 = 8.78%

Mass F: (196.97-196.7)/(196.7) x100 = .137%

Case IV
(m1)(x1) = (m2)(x2) (100)(.239) = (m2)(.261) m2 = 91.57g Mass of meter stick is equal to 96.43g Percent Error: .05%

Questions
1) A large force can produce little or no torque while a small force can produce a large torque. This is because Torque = Lever arm * Force. Therefore, if there is a large force, to make it a small torque, the lever arm is decreased. For a torque to be large from a small force, the lever arm has to be greater.

2) The downward and upward forces acting on the meter stick cancel each other out, to balance the masses. The forces directly balance each other out, so force of the meter stick is irrelevant. Any change in mass at the fulcrum would not change the balance of the masses.

3) (m1)(x1) = (m2)(x2) + (m3)(x3) (220)(.35) = (m2)(100) + (.5)(120) Mass of the meter stick = 170g