Static Equilibrium of Parallel Forces
(Center of Mass)
The objective is to verify that
1) in static equilibrium of a rigid object, the sum of moments of external forces acting on it about any point is zero, and
2) the weight of a uniform solid beam acts at its geometric center that is also its center of gravity.
A meter sick, 3 meter stick pivots, 3 meter stick clamps, 3 weight hangers, slotted weights, and a mass scale
If weights are hung from a meter stick placed on a pivot such that balance is achieved, the sum of clockwise moments equals the sum of counterclockwise moments. The moment of each force is equal to the product of that force and its perpendicular distance from the pivot.
In Fig. 1, W3X3 is the only clockwise moment about the pivot. The counterclockwise moments about the pivot are W1X1 and W2X2. The weight of the meter stick does not create any moment about the pivot because its weight vector passes through it. For the figure shown, the equations for static equilibrium are:
ΣFx = 0 , ΣFy = 0 , and ΣT = 0.
If the center of the meter stick (the 50-cm line) is not at the pivot but some distance XCG off the pivot (see Fig. 2), then the torque due to the weight of the meter stick must be included. The weight of the meter stick can be considered to be acting at its center of gravity. The center of gravity (C.G.) is the point which the entire weight of the meter stick can be assumed to be concentrated at. The center of gravity is on the 50-cm mark on the meter stick. Using Fig. 2, the equation for the balance of moments is
ΣTcw = ΣTccw.
W2X2 + W3X3 + WBeam XC.G. = W1X1.
where WC.G. is the weight of the meter stick that presumably acts at its center of gravity C.G. or the 50-cm line.
Note: XCG is the distance from the beam's center of gravity (the midpoint of the ruler) to the pivot.
In this experiment it will be verified that the sum of the clockwise moments equals the sum of the counterclockwise moments when the applied external forces acting on a rigid object are in static equilibrium. The static equilibrium will then be used to show that the weight of the meter stick acts at its geometric center. This will further verify that for a uniform beam the geometric center and center of mass coincide.
A) Meter Stick Pivoted at its Center of Gravity
1) Measure the weight of the meter stick and record its value in the data section. This may be used as Waccepted in Step B-4.
2) Arrange a table similar to Table 1, as shown under the data section. It must include each weight in gram-force (gf) including the weights of its clamp and hanger, its position relative to the pivot (in cm), and the calculated torque in (gf-cm) for each weight.
3) Set up the meter stick so that it pivots about its center of gravity (the 50.0-cm line). It could be slightly off due to imperfections. Record the actual cm value.
4) Attach (or hang) approximately 50.0g at the 10.0-cm mark, 150.0g at the 30.0-cm mark, and 250.0g somewhere on the other side of the pivot so that balance is achieved. In order to do this, you need to first slide 3 stick hangers over the meter stick (beam), and then hang 3 regular hangers from them. The aforementioned weights should then be placed on the regular weight hangers. It is better to use a 50.0-gram regular weight hanger for each. Next, calculate the exact values of W1, W2, and W3 and record them in Table 1. For example, W1 = 50.0gf + 50.0gf + the weight of its stick hanger. Each stick hanger should be weighed separately. Their weights might be slightly different.
5) Determine the distance of each weight from the pivot and record its value in Table 1. Calculate the sum of clockwise (-) moments and the sum of counterclockwise (+) moments and record their values. The sums might by slightly different. Calculate their percent difference.
6) Repeat the steps 4 and 5 above but using the values: 150g at 20.0cm, 250g at 40.0cm, and 350g on the other side of the pivot somewhere so that the meter stick is in equilibrium or balance is reached. Calculate the total cw moment and the total ccw moment again and find the percent difference.
7) Repeat steps 4 and 5 again with values: 300g at 10.0cm, 50.0g at 70.0cm, and 250g somewhere on the same side where the 50.0g weight is so that balance is achieved. This is a somewhat different arrangement compared to the previous two cases, but it is still true that the total cw torque must equal the total ccw torque. Calculate both totals as well as the corresponding percent difference.
B) Meter Stick Not Pivoted at Its Center of Gravity
In this part, pivot the meter stick at its 30.0cm mark. The center of gravity of the stick is now approximately 20 cm from the pivot (Fig. 2). Because the center of gravity is off the pivot, the weight of the meter stick causes an extra torque. This torque is equal to the weight of the meter stick (Wbeam) times the 20cm distance that its center of gravity has from the pivot.
1) Attach 50.0g at 60.0cm, 50.0g at 80.0cm, and on the other side of the pivot, 350g where balance can be reached.
2) Compute the distance each weight has from the pivot and record.
3) Assuming that the weight of the meter stick is unknown (call it Wmeasured), calculate and record the torque of each weight including the torque of the meter stick but in terms of Wmeasured. Apply the static equilibrium equations to solve for Wmeasured. Record its value in Table 1 in the space provided.
Repeat Steps 2 and 3 above for the new values: 30.0g at 40.0cm, 50.0g at 70.0cm, and 450g somewhere on the other side of the pivot where balance is reached. Apply the static equilibrium equations to solve for Wmeasured again.
4) Find the average value of Wmeasured and calculate the percent error on it using the accepted value Waccepted you found under A-1.
g = 9.81 m/s2.
Measured: Waccepted = ...........
|%diff. on ΣT||RA=
|Part A||Meter Stick Pivoted at its Center of Gravity|
|Part B||Meter Stick Not
Pivoted at its Center of Gravity
In this case, treat Wbeam as unknown.
|%error on Wbeam|
Comparison of the results:
Provide the %error and % difference formulas used and their calculated amounts.
State your conclusions of the experiment.
Provide a discussion if necessary.
1) In the last column of Table 1, why is RA = W1+W2+W3+Wbeam ?
2) What does a low %error on Wbeam in Part B of the experiment indicate?