- Statics Board
- Mass and Hanger Set
- Balance Arm and Protractors
Gravity is a universal force; every bit of matter in the universe is attracted to every other bit of matter. Therefore, when the Balance Arm is supported by the pivot, every bit of matter in the Balance Arm is attracted to every bit of matter in the Earth.
Fortunately, the sum of all these individual gravitational forces produces a single resultant. This resultant acts as if it were pulling between the center of the Earth and the center of mass of the Balance Arm. The magnitude of the force is the same as if all the matter of the Earth were located at the center of the Earth, and all the matter of the Balance Arm were located at the center of mass of the Balance Arm.
An object thrown so that it rotates tends to rotate about its center of mass, and the center of mass follows a parabolic path. An object whose center of mass is above a support tends to remain in rotational equilibrium (balanced on the support). In this experiment, you will use your knowledge of torque to understand and locate the center of mass of an object.
- Mass of the beam: 80.6g = 0.81g
- Mass of protractor #1: 0.01kg
- Mass of protractor #2: 0.01kg
- The total mass of the beam plus protractors plus mass hangers (5 grams each):
=>Mass of system: 0.83kg + 0.01kg = 0.84kg
*** We made sure that the beam is balanced and level, and then tightened the thumbscrews to hold the protractors and beam in place.
- Why would the Balance Arm necessarily rotate if the resultant of the gravitational forces and the force provided by the pivot were not concurrent forces?
Answer: The total force would not equal 0 which will cause the rotation; the net force would be 0.
- What is the relationship between the sum of the clockwise torques about the center of the mass and the sum of the counterclockwise torques about the center of mass? Explain.
Answer: Equal in magnitude but opposite in direction
Next step: Add 50 grams to one mass hanger and 100 grams to the other mass hanger. Loosen the thumbscrew on the beam and slide the beam through the pivot until the beam and masses are balanced.
- The pivot is still supporting everything (beam, protractors, mass hangers, and hanging masses), but at the new center of mass of the system –the pivot point.
- Calculate the three torques, t1, t2, and t3 provided by the three forces F1, F2, and F3 acting about the new pivot point position.
- Are the clockwise and counterclockwise torques balanced?
Answer: Unfortunately, our two trials of the experiment didn’t show balanced torque –they were slightly unbalanced. They should be balanced, but we are guessing that this was caused because of lack of precision.