Pfautz+lab+19

Torque and Rotational Equilibrium

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

Mass D: (m)=(200) M=276g

Mass B: M=200 M=267g

Mass F: M=200 M=208g

Mass E: M=200 M=218

Mass R: M=200 M=147

Percent Error: (Actual - expected) / expected D= .1.8% B= 3.68% F= 5.7% E= 5.7% R= 11.03%

Case IV New Support position: 24.3 (9.8)(100)(.243)=m(.257)(9.8) m=94.6g, .0946kg

Actual mass: .0921kg Percent Error: 2.7%

Discussion Questions:

1) A large force produces little torque and a small force can produce a large torque because T=rF which is both the length of the arm and the force. When the lever arm is very long and a small force is applied, a large torque results. The opposite is also true.

2) The support force is not included in torque calculations because the upward force is canceled out by the downward force. The meter stick is balanced so there is no acceleration; therefore, no net force exerted by either the upward or downward forces.

3) (m1)(x1) = (m2)(x2) + (m3)(x3) (.22)(.35)=(.01)(m)+(.5)(.12) m=.17kg