IntroductionTheoreticalExperimentalConclusionIntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 1 of 16Go BackFull ScreenCloseQuitThe Force of a Falling ChainLinda Lindsley and Gina GiaconeMay 19, 2000AbstractThe purp ose of the experiment is to determine the magnitude of theforce exerted on a force sensor by a falling chain. A mathematical modelof the force was used to find the theoretical values. Then, the experimen-tal values were determined using Science Workshop and compared to thetheoretical values.IntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 2 of 16Go BackFull ScreenCloseQuit1. IntroductionImagine a long chain with a length L and mass M, made up of many links. Oneend is attached to a force sensor, and the other is held directly above the forcesensor so that the chain exerts no force on the sensor to begin with, see Figure 1.Upon release of the chain, there is a force (which exceeds the weight force)exerted on the sensor, until the last link falls and the chain comes to a rest.The question is: what is that force exerted on the sensor by the chain as afunction of x (the distance that the top of the chain has fallen) during the timethe chain is falling. Through a series of derivations, by using the laws of physics,the total force that the chain exerts on the sensor is found.IntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 3 of 16Go BackFull ScreenCloseQuitFigure 1: The set-upIntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 4 of 16Go BackFull ScreenCloseQuitFigure 2: Diagram of Chain FallingIntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 5 of 16Go BackFull ScreenCloseQuit2. TheoreticalThe force exerted by the chain on the sensor is a combination of two forces. Thefirst force, F1, is the impulse force (momentum) of each link as it is stopped bythe link above it. The second force, F2, is the weight force of the links alreadyhanging from the hook. The total force Ftexerted on the sensor is:Ft= F1+ F2(1)Looking at the process of the falling chain, we see that the chain is made upof links that have individual mass elements dm, which are associated with theirlength increments dx, see Figure 3.This ratio is comparable to the ration of thetotal mass to the total length of the chain. Using this relationship we come upwith the equation:dmdx=ML(2)After a little rearrangement we are left with:dm =MLdx (3)Before continuing further on the derivation of the equations for the totalforce the chain exerts on the sensor, we must understand the concept of centerof mass. The center of mass is defined to be, “the point at which we assumeall the mass of an object is concentrated in order to determine its motion inresponse to external forces.”[2] We find the center of mass by adding up all ofthe mass increments times each corresponding x coordinate and dividing thissum by the total mass of the chain (which stays constant).Xc=PmixiM(4)IntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 6 of 16Go BackFull ScreenCloseQuitFigure 3: This diagram shows the relationship between x and dxIntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 7 of 16Go BackFull ScreenCloseQuitBy letting the number of elements approach infinity we can find the exact valuefor the center of mass. By replacing this limit with an integral we come up with:Xc=1MZL0xdm (5)The relationship between position and velocity are put into play by using thefact that the velocity of the center of mass is equal to the derivative of theposition of the center of mass with respect to t. This allows us to state therelationship that the sum of the mass increments times the velocities of thosemass increments, divided by the total mass of the chain is equal to the velocityof the center of mass.Vc=dXcdt=1MXmividxidt=PmiviM(6)Knowing that momentum is defined in physics to be equal to the mass timesthe volume, we rearranged Equation 6 to show that:MVc=Xmivi=Xpi= P (7)We also know that there will be an acceleration of the center of mass whichis equal to the derivative of the velocity of the center of mass with respect totime. This tells us that the acceleration of the center of mass is equal to the sumof the masses times the accelerations of each of these mass increments, dividedby the entire mass of the chain. Giving us the equation:Ac=dVcdt=1MXmidvidt=1MXmiai(8)IntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 8 of 16Go BackFull ScreenCloseQuitRearranging Equation 8 we find that the total mass of the chain times theacceleration of the center of mass is equal to the sum of all the mass incrementstimes the accelerations of each of these mass increments. This leaves us withNewton’s second law which states that the sum of the forces F1is equal to thesum of the mass increments times the accelerations of the mass increments.MAc=Xmiai=XFi= F1(9)A force on a particle can include both external and internal forces. TakingNewton’s third law into consideration, where the force of particle A on particle Bis equal to the force of particle B on particle A, we can see that when summingall of the internal forces, that they cancel in pairs and we are left only withexternal forces. We then use Newton’s second law and Newton’s third law toshow that the sum of the external forces is equal to the total mass of the chaintimes the acceleration of the center of mass. Then by taking the derivative ofthis with respect to t(knowing that the mass of the system is changing and thevelo city is constant), and knowing that momentum is equal to the velocity timesthe mass, we can show that:F1=dPdt= vdmdt(10)In order to find F1in terms of variables that we know, we need to substituteEquation 3, into Equation 10. We then arrive at the conclusion that:F1= vML(dxdt) =MLv2(11)The only variable left in Equation 11 that we do not know is the velocity. Byusing one of the kinematic equations we can find a substitution for the velocity.IntroductionTheoreticalExperimentalConclusionHome PageTitle PageContentsJJ IIJ IPage 9 of 16Go BackFull ScreenCloseQuitv2f= v20+ 2a∆x (12)In order to simplify Equation 12, we have known information that can be of use.a = g, (acceleration due to gravity)∆x = 2xv0= 0(13)Equation 13 is true because as x amount of chain has fallen at the bottom ofthe sensor, two times that