PASCO ME-9502 Statics System - Manual (Page 30)

S t a t i c s S y s t e m E x p . 5 A : C e n t e r o f M a s s
®
26 012-12876B
Level the Beam and Mark the Center of Mass
Loosen the thumbscrew and adjust the beam so that the indicator marks on the pivot are aligned with the zero mark
on the beam. If necessary, adjust the positions of the two protractors until the bubble in the bubble level is midway
between the two lines on the level. Once the beam is balanced and level, tighten the thumbscrews to hold the pro-
tractors and beam in place.
Put a pencil mark on the beam to indicate the center of mass of the system. All of the mass of the system (beam,
protractors, mass hangers) acts as if it is concentrated at the center of mass.
Once the beam is in balance, the force at the pivot point must be the equilibriant of the total gravitational force act-
ing on the beam. Since the beam does not rotate, the gravitational force and its equilibriant must be concurrent
forces.
Experiment
1. Why would the Balance Arm necessarily rotate if the resultant of the gravitational forces and the force pro-
vided by the pivot were not concurrent forces?
• Think of the Balance Arm as a collection of many small hanging masses. Each hanging mass is pulled down
by gravity and therefore produces a torque about the pivot point of the Balance Arm.
2. 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.
• 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. Tighten the thumbscrew.
• 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.
3. Calculate the three torques,

1
,

2
,
and

3
provided by the three forces
F
1
, F
2
, and F
3
acting about the new
pivot point position. Be sure to indi-
cate whether each torque is clock-
wise (cw) or counterclockwise (ccw).
4. Are the clockwise and counterclockwise torques balanced?
• Remove the 50 gram mass from the left hand mass hanger, but leave the hanger and protractor in place.
(Removing the mass effectively removes F
1
). Reposition the beam in the pivot until the torque provided by F
3
balances the torque provided by F
2
and the beam is level again.
Position Mass (kg) Force (F = mg) Distance (m) Torque (  = F d)
1 0.050
2 0.100
3
Pivot
Center of mass
of the beam
F
2
F
3 F
1
Figure 5.2: Torques and the Center of Mass