LR - Torques, Equilibrium, and Center of Gravity PDF

Preview

Summary

lab report...

Description

Steeven Imbaquingo 06/27/17 PHY 215 Prof. Kibrewossen Tesfagiorgis Torques, Equilibrium, and Center of Gravity Introduction: Forces makes a difference when it applies on the object. However, when we have two extended solid objects it becomes a rotational motion, which is deliberated a rigid body. A rigid body is when the two objects’ distances are fixed and continue to be uniform. Therefore, a rigid body applies static equilibrium like bridges (RFK bridge) and beams that you see in construction. Furthermore, when it is considered a static equilibrium it means that the objects or particles are slightly balanced. Objectives: Our objective in this lab is to explain mechanical equilibrium and how it is applied to rigid bodies. Also, to comprehend the difference between center of mass and center of gravity. Finally, describe how a laboratory beam balance measure mass. Procedure: Apparatus with Support Point at Center of Gravity: 1. A general experimental setup is illustrated in where the masses or weights are suspended by knife-edge clamp weight hangers. The hooked masses may also be suspended from small loops of string, which can be slid easily along the meterstick. The string allows the position of a mass to be read easily and may be held in place by a small piece of masking tape. (a) Determine the mass of the meterstick (without any clamps), and record it in the Laboratory Report. (b) Weights may be suspended by loops of string or clamps with weight hangers. The string method is simpler; however, if you choose or are instructed to use weight hangers, weigh the three clamps together on a laboratory balance and compute the average mass of a clamp. Record in the Laboratory Report. 2. With a knife-edge clamp on the meterstick near its center, place the meterstick (without any suspended weights) on the support stand. Make certain that the knife edges are on the support stand. (The tightening screw head on the clamp will be down.) Adjust the meterstick through the clamp until the stick is balanced on the stand. Tighten the clamp screw, and record in Data Table 1 the meterstick reading or the distance of the balancing point xo from the zero end of the meterstick.

3. Case 1: Two known masses (a) With the meterstick on the support stand at xo, suspend a mass m1 = 100 g at the 15cm position on the meterstick—that is,15 cm from the zero end of the meterstick.

(b) Set up the conditions for static equilibrium by adjusting the moment arm of a mass m2 = 200 g suspended on the side of the meterstick opposite m1. Record the masses and moment arms in Data Table 1. If clamps are used instead of string, do not forget to add the masses of the clamps. Remember the moment arms are the distances from the pivot point to the masses. (c) Compute the torques, and find the percent difference in the computed values (that is, compare the clockwise torque with the counterclockwise torque). 4. Case 2: Three known masses Case (a) (i) With the meterstick on the support stand at xo, suspend m1 = 100 g at the 30-cm position and m2 = 200 g at the 70-cm position. Suspend m3 = 50 g, and adjust the moment arm of this mass so that the meterstick is in static equilibrium. Record the data in Data Table 1. (ii) Compute the torques and compare as in Procedure 3. Case (b) (i) Calculate theoretically the lever arm (r3) for the mass m3 = 50 g for the system to be in equilibrium if m1 = 100 g is at the 20-cm position and m2 = 200 g is at the 60-cm position. (Remember to add the masses of the hanger clamps if used.) Record this value in the data table. (ii) Check your results experimentally, and compute he percent error of the experimental value of r3, taking the previously calculated value as the accepted value. 5. Case 3: Unknown mass—The balance principle: A balance (scale) essentially uses the method of moments to compare an unknown mass with a known mass. Some balances have constant and equal lever arms, and others do not. This procedure will illustrate the balance principle. (a) With the meterstick on the support stand at x0, suspend the unknown mass (m1) near one end of the meterstick (for example, at the 10-cm position). Suspend from the other side of the meterstick an appropriate known countermass m2 (for example, 200 g) and adjust its position until the meterstick is “in balance” or equilibrium. Record the value of the known mass and the moment arms in Data Table 1. (b) Remove the unknown mass, and determine its mass on a laboratory balance. (c) Compute the value of the unknown mass by the method of moments, and compare it with the measured value by calculating the percent error.

Apparatus Supported at Different Pivot Points:

In the previous cases, the mass of the meterstick was not explicitly taken into account since the fulcrum or the position of the support was at the meterstick’s center of gravity or center of mass. In effect, the torques due to the mass of the meterstick on either side of the support position canceled each other. The centers of gravity of the lengths of the stick on either side of the support are equidistant from the support (for example, at the 25-cm and 75-cm positions for a uniform stick) and have equal masses and moment arms. For the following cases, the meterstick will not be supported at its center-of-gravity position (x0) but at some other pivot points (designated in general by x’0. In these cases, the mass of the meterstick needs to be taken into account. To illustrate this very vividly, let’s start off with a case with only one suspended mass. 7. Case 5: Meterstick with one mass: Suspend a mass m1 5 100 g at or near the zero end of the meterstick. Record the mass position x1 in Data Table 2. If a string loop is used, a piece of tape to hold the string in position helps. Move the meterstick in the support clamp until the system is in equilibrium. (This case is analogous to the solitary seesaw—sitting on one side of a balanced seesaw with no one on the other side.) Record the support position x’0 in Data Table 2. Since the meterstick is in balance (static equilibrium), the point of support must be at the center of gravity of the system; that is, the torques (clockwise and counterclockwise) on either side of the meterstick must be equal. But where is the mass or force on the side of the meterstick opposite the suspended mass? The balancing torque must be due to the mass of length L2 of the meterstick. To investigate this: (a) Using the total mass m of the meterstick (measured previously) as m2, with a moment arm r2, compute the counterclockwise and clockwise torques, and compare them by computing the percent difference. Record it in Data Table 2. (b) Now the masses of the lengths of meterstick will be taken into account. Compute the average linear mass density of the meterstick, and record it in the data table. If we assume that the mass of the meterstick is uniformly distributed, the center of mass (or center of gravity) of the length of meterstick L2 on the opposite side of the support from m1 is at its center position. Compute the mass m2 of this length of stick and record. Also, record the center position of L2, where this mass is considered concentrated (x2), and find the length of the lever arm r2. It should be evident that r2 = L2/2. Compute the torque due to m2, and record it as Tcw. From the linear mass density, compute the m3 of the portion of the meterstick remaining to the left of the pivot. Calculate the torque due to this portion of the meterstick, add it to the torque due to mass m1 to find the total counterclockwise torque, and record it as Tcc. Compare the torque differences with those found in Case 5(a).

Data Set:

Data Table 1: Mass of the meterstick: 90.09g Total mass of clamps: 45g Average mass of one clamp (mc): 15g Balance position of meterstick, xo: 51cm Diagram Values (add mc to masses) Case 1

Case 2(a)

Case 2(b)

Case 3

m 1=115 g x 1=15 cmm2=215 g x 2=69.5 cm m 1=115 g x 1=30 cm m 2=215 g x 2=70 cm m 3=65 g x 3=23.3 cm m 1=115 g x 1=20 cmm2=215 g x 2=60 cm m 3=50 g x 3=79.3 cm x 1=10 cm m 2=200 g

x 2=67 . 3 cm

Moment (lever) arms r 1=. 36 m r 2=.18 5 m r 1=. 21 m r 2=. 19 m r 3=.27 7 m

Results T cc=4140 N∗m T cw=3977.5 N∗m Percent Diff.= 4.0% T cc=4 215.5 N∗m T cw=4085 N∗m Percent Diff.= 3.1%

r 1=31 c m r 2=9 c m

r 3=32.6 cm(calc .) r 3=28.3 cm ( measured ) Percent Error = 13%

r 1=41 c m r 2=16 3 c m

m 1=79 g ( calc .) m 1=70 g (measured) Percent Error = 11%

Data Table 2: m =¿ 1.05g/m L Results

Linear mass density of meterstick, μ= Diagram

Values (add mc to masses)

Case 5(a)

m 1=115 g x 1=0.01 cmm2=105.09 g x 2=77.6 cm x ' 0=22. 4 cm m 1=115 g x 1=0.01 cmm2=72.2 g x 2=77.6 cm m 3=20.2 g x 3=11.2 cm x ' 0=22. 4 cm

Case 5(b)

Moment (lever) arms r 1=.22 39 m r 2=.38 8 m

r 1=.2239 m r 2=.388 m r 3=.11 2 m

T cc=25 .75 N∗m T cw=40 . 77 N∗m Percent Diff.= 45% T cc=28.011 N∗m T cw=28.014 N∗m Percent Diff.= 0.01%

Calculations: Data Table 1: Case 1  Our center of gravity is 51cm, so we subtracted with x to get r.  Σ T cc =Σ T cw Σ T cc=( 115 g∗. 36 m )=4140 N∗m Σ T cw= ( 215 g∗.185 m )=3977.5 N∗m Percent difference: 4.0% Case 2(a) Σ T cc=( 115 g∗. 21 m )+(27.7∗65 cm)=4140 N∗m Σ T cw= ( 215 g∗. 19 m)=3977.5 N∗m Percent difference: 4.0% Case 2(b) Σ T cc=Σ T cw

( 31 cm∗115 g) = ( 225 g∗9 cm ) + ( 50 g ) r 3

r 3=32.6 cm

Percent error: 13% Case 3 Σ T cc=Σ T cw

( 10 cm )m 1= ( 16.3 cm∗200 )

m 1=79 g

Percent error: 11% Conclusion: In conclusion, we got an understanding how rotational motion or static equilibrium expresses through the calculations we did in the lab. These calculation shows us the equilibrium or where it is balanced when it is a rest. Furthermore, we learned how to calculate the sum of all the torques counterclockwise and clockwise. Therefore, torques and static equilibrium are important when two objects are fixed in distances to have a balance, just like bridges because they are slightly moving but we don’t feel it (beams in construction too)....