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Abstract . Torque is a measure of how much a force is acting on an object causes that object to rotateand he second condition of equilibrium is simplified as the net torque which is acting on the body is zero.This could be achieved by equating the sum of all counter clockwise torques to the sum of all clockwisetorques. The objectives of the study are to analyze the systems that are in equilibrium using the secondcondition and to distinguish some of the second condition of equilibriums use and significance. In thethree parts of experiment which finds the weight of the pan, force exerted and weight of the beamrespectively, we noticed how torque is affected by the forces acting on the system and their radial

distance from the axis and also, how the rotational equilibrium is applied. We have come up to theconclusion when second condition of equilibrium is satisfied, there is no angular acceleration and body willnot be moving and will be in rotational equilibrium.

Introduction Torque is a measure of how much a force isacting on an object causes that object torotate. It is also called as the moment of force.The object rotates about an axis, which we callthe pivot point. As the force applied increases,torque also increases. Therefore, torque isdirectly proportional with the force applied on theobject. Furthermore, we must note that the forceapplied on the object should be perpendicular toits axis of rotation.

An example would be pushing a door to open it.The force of your push (F) causes the door torotate about its hinges (the pivot point). Howhard you need to push depends on the distanceyou are from the hinges (r). The closer you are tothe hinges (the smaller r is), the harder it is topush. This is what happens when you try to push

open a door on the wrong side. The torque youcreated on the door is smaller than it would havebeen had you pushed the correct side (away fromits hinges).

Equilibrium simply implies a state of balance. Itsfirst condition of equilibrium states that thevector sum of all forces acting on it must be zero.It is when a body at rest or moving with uniformvelocity has zero acceleration.

Meanwhile, the second condition of equilibriumstates that the net torque acting on the bodyshould be zero for angular acceleration to bezero. Thus for a body in equilibrium, the vectorsum of all the torques acting on it about anyarbitrary axis should be zero.

By convention, the sum of all counter clockwisetorques is equated to the sum of all clockwisetorques.

In the first part of the experiment, we are tryingto determine the weight of the pans using thesecond condition of equilibrium. By equating theforces that will make the system rotate clockwiseto the forces that will make the system rotatecounter clockwise, we can derive the formulaneeded:

= ( ) = ( )

( ) ( ) ( )

( )

For the second part of the experiment, we aretrying to determine the force needed to be in

equilibrium using the second condition ofequilibrium. By equating the forces that will makethe system rotate clockwise to the forces that willmake the system rotate counter clockwise, wecan derive the formula needed:

= ( ) (( )( ) ( )

( )( )( )

For the third part of the experiment, we aretrying to determine the weight of the beam usingthe second condition of equilibrium. By equatingthe forces that will make the system rotateclockwise to the forces that will make the systemrotate counter clockwise, we can derive theformula needed:

= ( ) ( )

( )( )( )

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Methodology For this experimentwhich is abouttorque: the secondcondition ofequilibrium, wewere given a set ofa model balance, a

set of weights, 1piece of meter stick,a protractor, 2pieces of weightpans, a springbalance and adigital weighingscale.

In the first part of the experiment which isdetermining the weights of the pans, we set upthe model balance in a levelled table top andmake sure that the axis of rotation is passingthrough the center of gravity of the beam. Wethen hang the two pans on the beam. We place a10 gram (W 1 ) weight on the first pan and try tomake a state of equilibrium to the system. Wecan know if the state of equilibrium is achieved ifthe beam is perfectly horizontal even thoughthere are unequal masses on the pans and wecan do this by placing the pans on farther andcloser to the axis of rotation. If the state ofequilibrium is now achieved, we now measure thedistance of the two pans from the axis of rotationand record there as L 1 and L 2 . After measuring,

we now reset the system and place a 5 gram(W 2) weight on the second pan making sure thatwe remove the 10 gram weight on the earlierpart of the experiment. We again try to create astate of equilibrium to the system and record thedistances of the pan as L 3 and L 4 . After recording,we again reset the system and make the secondtrial with W 1 as 15 grams and W 2 as 25 gramsand the third trial with W 1 as 30 grams and W 2 as20 grams with the same procedures as the firsttrial. Using the values got, we can compute forthe weight of the pan using the formulasprovided.

For the secondpart of theexperimentwhich isdetermining theforce needed tobe inequilibrium, we

use the sameset up as thefirst part of theexperiment butnow, we willmake use ofthe springbalance. First,we place 50grams (W 1) onthe first pan onthe left side ofthe beam anduse the spring balance to achieve the state ofequilibrium. We must make sure that the springbalance makes an acute angle or an angle that isless than 90 with the beam that is in horizontalposition. After achieving the state of equilibrium,we record the reading of the spring balance asFmeasured . We also measure the angle of inclinationof the spring balance from the beam and thedistance of the pan and the spring balance fromthe axis of rotation and mark it as L 1 and L 2respectively. After which, we can now computefor the force exerted on the system. For the

second trial of the second part of the experiment,we still use the same set up of the system butthis time, we now place the spring balance on theright side of the beam. Afterwards, the sameprocedure applies as the first trial of thisexperiment. Using the values got, we cancompute for the force applied to the system.

For the third part of the experiment which isdetermining the weight of the beam, we nowplace the axis of rotation in the other holeprovided. We now hang the first pan on the rightside of the beam and place 50 grams (W 1) on itand adjust its position until the state ofequilibrium is acquired. We then measure thedistance of the pan from the new axis of rotation(L1) and the distance of the former axis ofrotation and the new axis of rotation (L 2). Afterwhich, we now use 60 grams and 70 grams as W 1for the second and third trials and the sameprocedures applies as the first trial. Using thevalues got, we can compute for the weight of thebeam using the formulas provided.

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Results and DiscussionIn conducting the experiment, we had a bit oftrouble in making the system in the state ofequilibrium since the beam is very unstableespecially when weights are added. When ourgroup have finally achieved the state ofequilibrium, the problem facing us is measuringthe distances of the pans from the axis ofrotation without compromising the state ofequilibrium of the beam. The group also had a bitof trouble in measuring the angle that the springbalance makes with the beam in the second partof the experiment. Finally, the last part of theexperiment is the trickiest of the all parts sincethe axis of rotation is not on the center, ratherit s just close to the center but the group hasmanaged to still put it in its state of equilibrium.

In the first part of the experiment, wedetermined the weight of the pans using thesecond condition of equilibrium. The following is a

sample computation of weight of the pans for thefirst trial.

( )( )( ) ( )( )( )( )( ) ( ) ( )

( )

The table below shows the weights added on thepans and the distances of the pans from the axisof rotation.

T W (g)L1

(cm)L2

(cm)L3

(cm)L4

(cm)P1(g)

P2(g)

1W 1 = 10

12.5 17.8 23.2 19.5 25.53 24.96W 2 = 5

2W 1 = 15

13 20.7 15.5 7.6 24.38 24.73W 2 = 25

3W 1 = 30

10.7 23.5 23.9 13.4 25.34 25.19W 2 = 20

ave. W of P1 25.08g %diff of P1 1.12%

ave. W of P2 24.96g %diff of P2 0.64%

With the tabulated results of the experiment, wecan notice that when one side of the beam with apan has added weights, the other side of thebeam with a pan without added weights tend tobe father away to balance the system. This isimportant since to balance the system, the forcesmaking the system to rotate clockwise needs tobe equal to the forces that makes the system to

rotate counter clockwise and since torque iscomputed by multiplying the force and itsdistance from the axis of rotation, the only wayfor the other side of the beam to be equal to theother since that has a greater force, it needs agreater distance.

In the second part of the experiment, wedetermined the force applied needed to be inequilibrium using the second condition ofequilibrium. The following is a samplecomputation of force applied needed to be inequilibrium for the first trial.

( )( )( )

( )( )( ( ))

The table below shows the force applied neededto be in equilibrium and the distances of the pansfrom the axis of rotation.

TrialL1

(cm)L2

(cm)W1 + P1

(g)Fcomputed

(g)Fmeasured

(g)%diff(%)

1 20.7 9.4 74.8 195.83 200 2.11

2 18.8 9 74.8 203.97 200 1.97

With the tabulated results of the experiment, wecan notice that in the second trial, the side withthe pan and the weights tend to be farther thanthe spring balance on the other side. This isbecause the pan with weights tends to have lessforce compared to the force that the springbalance applies, so to make the system inequilibrium, the pan with weights needs to havea greater distance to make the systemequilibrium. With regards to the first trial whenboth pan and spring balance is on the left side,the spring balance just cancels the force exertedby the pan with weights which makes thesystem, in equilibrium.

In the third part of the experiment, wedetermined the weight of the beam using thesecond condition of equilibrium. The following is asample computation of weight of the beam forthe first trial.

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( )( )( )

( )( )( )

The table below shows the weights added on thepans and the distances of the pans from the axisof rotation.

TrialL1

(cm)L2

(cm)W1 + P1

(g)Wcomputed

(g)Wmeasured

(g)1 14.5 8 50 135.58

136.82 12.8 8 60 135.46

3 11.9 8 70 141.57

ave. W of beam 137.54g

%diff 0.54%

We can notice that the pan with weights is inequilibrium with the weight of the beam itself.

With regards to the tabulated results, we cannotice that as we increase the weight added tothe pan, the shorter the distance needed to makethe system in equilibrium. This is notable sinceagain, the weight of the beam itself (clockwisemotion) which is placed on its center of gravityneeds to be equal in magnitude to the weight ofthe pan with weights (counter clockwise motion)to have equilibrium.

ConclusionThe objectives of the study are to analyze thesystems that are in equilibrium using the secondcondition and to distinguish some of the secondcondition of equilibriums use and significance. Inthis experiment, we saw how torque is affectedby the forces acting on the system and theirradial distance from the axis and also, how theequilibrium is applied.

In this experiment, second condition ofequilibrium or rotational equilibrium has beenapplied in all parts of the experiment. We knowthat torque is a measure of how much a force isacting on an object causes that object to rotate

and that it is the product of the force and itsdistance from its axis. In analyzing systems inrotational equilibrium, we need to equate thesum of all torques that will make the systemrotate clockwise to sum of all the torques thatwill make the system rotate counter clockwise.We must take note that the torque is directlyproportional (the greater the force, the greaterthe torque) to the force applied and that everyforce applied in the system may or may not haveequal distance from its axis since when the force

is great, it will need less distance from the axisand vice versa for it to be a rotationalequilibrium. Therefore, when second condition ofequilibrium is satisfied, there is no angularacceleration and body will not be moving and willbe in rotational equilibrium.

Acknowledgments

I would like to thank my group mates for a jobwell done in performing this experiment.Although this is our first time to work togethersince this is the first time we became classmates,we still work well and I thank them for that. Iwould not have done this experiment by myself

I would also like to thank Facebook and Tumblrfor keeping me entertained while writing thislaboratory report. Because of all the amusingblogs and games, it kept me awake while I amwriting this laboratory report.

I would also like to thank my brother Joash, forpatiently waiting for his turn to use the computerwhile I was writing this report.

And especially, I would like to thank AlmightyGod for giving me the knowledge, and thewisdom I need to write this laboratory report.

References [1] http://en.wikipedia.org/wiki/Torque[2] http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

[3] http://auto.howstuffworks.com/auto-parts/towing/towing-capacity/information/fpte4.htm[4] http://www.blurtit.com/q700116.html[5] Young, H., Freedman, R., Ford, L., UniversityPhysics with Modern Physics, 12th Edition, 2008

http://en.wikipedia.org/wiki/Torquehttp://www.blurtit.com/q700116.htmlhttp://www.blurtit.com/q700116.htmlhttp://en.wikipedia.org/wiki/Torque