Roth+Torque+and+Rotational+Equilibrium

Lab 19 Torque and Rotational Equilibrium Mass E: Left side of fulcrum: m= x kg r=.392 m g=9.8 m/s^2 theta= 90 degrees Right side of fulcrum: m= .2 kg r= .446 m g=9.8m/s^2 theta= 90 degrees The meter stick is balanced, so the net torque is zero, therefore: rFsintheta = rFsintheta .392(x)(9.8)sin90 = .446(.2)(9.8) sin 90 3.8416x = ,87416 X = .228 kg Accepted Value = .20614 kg Percent Error = l (actual – experimental) / actual l x 100= l (.20614 - .228) / .20614 l x 100 = 10.6% error
 * Case III**

Mass B Left side of fulcrum; m=x kg r= .301 m g=9.8 m/s^2 theta = 90 degrees Right side of fulcrum: m=.2 kg r= .446 m g=9.8m/s^2 theta= 90 degrees rFsintheta = rFsintheta .301(x)(9.8)sin90 = .446(.2)(9.8) sin 90 2.95x = .87 X= .296 kg Accepted Value = .13240 kg Percent Error = l (actual – experimental) / actual l x 100= l (.25751 - .296) / .25751 l x 100 = 14.95 % error


 * Case IV**

New support position = .298 m when m= .1g hangs off of one end of the meterstick Original equilibrium position = .492 m rFsintheta= rFsintheta .298(.1)(9.8)sin90 = (.492-.298) (x) (9.8) .292= 1.90x .1536 kg = x Actual mass of meter stick= .1532 kg Percent Error = l (actual – experimental) / actual l x 100= l (.1532 - .1536) / .1532 l x 100 = 0.261 % error


 * Analysis:**
 * 1) A large force can produce little or no torque if the radius, or distance from the axis of rotation, is small since the torque = rFsintheta. Likewise, a small force can produce a large torque if the radius is larger.
 * 2) The total downward force of the system would be the weight of the system. The total upward force on the meter stick is equal to that of the weight because the force must be enough to hold the system in place. The downward force, or weight, would be negative because gravity is negative. The upward force would be positive because the force is directed upward. If you add a positive force to a negative force of the same value, the net force is zero. Therefore, the support force is not included in calculations.
 * 3) When a meter stick is at rotational equilibrium when 220 g is suspended at 5 cm, 120 g is suspended at 90, and the support stand is at 40cm, the mass of the meter stick can be calculated in the following equation: rFsintheta = rFsintheta + rFsintheta. When the numbers are plugged in, then the equation becomes (40 - .05)(.220)(9.8)sin90 = (.90-.40)(.120)(9.8)sin90 + (.50-.40)(x)(9.8)sintheta. When the equation is solved, then x, the mass of the meter stick is 0.17 kg.