Physics I Labs 1 The fact that the steel pellets sink in ...

1 Physics I Labs1 archimedes PrincipleGoals In this exercise, you will be studying archimedes prin-ciple, specifically exploiting it to measure the specificgravity of a few different Triple-beam balance, two solid metal cylinders, piece ofwood or cork, unknown liquid, vernier caliperIntroductionWhen an object is submerged into a fluid, it experiencesan upward force called the buoyant force (or buoyancy). Thisforce is due to the difference in pressure of the fluid on thetop and on the bottom of the submerged object. To showthis, consider a cube with area A and height h.

2 Then thebuoyant force Fbequals the difference in the force of pressureon the bottom of the cube versus the top of the (Pbottom Ptop)A= [(Patm+ gh2) (Patm gh1)]A= g(h2 h1)A= ghA= gVHere is the density of the fluid andV(=hA) is the volumeof the submerged object that is equal to the volume of thedisplaced liquid. This equation is just another way of statingArchimedes Principlewhich states thatthe upward forceon a partially or completely submerged object is equal to theweight of the fluid that it that this statement should match your experiencesin the real world.

3 For example, if an object is less densethan the fluid it is immersed in ( such as wood placedin water), then its weight will be less than the weight of thewater it displaces when submerged, and thus, the net force onthe object will be upward, , the object will float. (Whenthe wood is floating on the surface, the weight of fluid dis-placed is equal to the weight of the wood.) However, if anobject is instead denser than the fluid it is immersed in ( steel pellets placed in water), then its weight will be greaterthan the weight of the water it displaces and the net force onthe object will be downward, , the object fact that the steel pellets sink in water does not meanthat there is no buoyant force; it simply means that the buoy-ant force is less than the weight of the water displaced.

4 Inother words, a submerged object will have an apparent weight less than its true weight when the buoyant force isnot present. In this lab, you will be submerging an objectand comparing its mass when not submerged, m, to the ap-parent mass, msub, when submerged. A force diagram for asubmerged object is shown Vgbuoyant waterrF =Vggravity objectrNotice that this means the apparent weight of the submergedobject ismsubg=Fgravity Fbuoyant=mobjectg mwatergmsubg= objectV g waterV g(1)and the weight of the object outside the water ismsubg=Fgravity=mobjectmg= objectV g(2)Based on the above equations, it can be shown that the spe-cific gravity of the object (the ratio of its density versus thatof water, water= 1000 kg/m3) is.

5 Object water=mm msub(3)This method will work if the object is denser than water (andthus sinks) and you will use it to measure the specific gravityof two metal blocks. However, in addition to this, you willalso explore ways of using archimedes Principle to measurespecific gravities of solids less dense that water (and thusfloat) and of liquids (which would mix with water).Part I: Weight of the Fluid Displaced Tie a string to one of the two metal I Labs3 Next fill a graduated cylinder with about 200 ml ofwater andrecord the volume of water in the lower the metal object into the reading on the graduated cylinder with the metalobject submerged in water.

6 Compute thevolume ( V) of water displaced by theobjectand use water V g to calculate the weight ofthe fluid displaced. You will need this value in Part II. Repeat the above procedure for the second metal objectand for the wood block. For each object use a separatepiece of string to tie the all your mea-surements and calculate the weight of the fluid displacedfor each : Wood will float on water. In thiscase you are measuring the volume displaced with woodfloating on II: Buoyant Force and Specific Gravity of Ob-jects Denser than Water Hang one of the two metal objects from the hook onthe underside of the balance and determine its the mass of the metal object in air, Next partially fill a beaker and arrange to have the ob-ject hang completely submerged in water but not touch-ing the bottom of the and record thebalance reading for the submerged mass, msub.

7 Compute thedifference in the weight of the mass in airand in water. This difference is the buoyant force. Isit equal (within the uncertainty of the measurements)to the weight of the fluid displaced you obtained in theprevious part? (See statement on archimedes princi-ple.) Next, compute the specific gravity of the metal objectusing equation the specific gravity of the metalobject and use this to compute the density of the metalobject. Calculate the volume, V, of the metal object (using aVernier caliper to determine its length l and diameterd, and calculatingV= d24l).

8 Use this volume and themass, m, of the metal object to determine its densitydirectly using metal=mVCompare this to your density obtained using archimedes Principle in the previous step. Repeat the above procedure for the second metal all your measurements and both compu-tations of the buoyant force and density of the III: Buoyant Force and Specific Gravity of Ob-jects Less Dense than WaterWood has a density lower than water (which is why itfloats). In order to measure the specific gravity of wood, wewill have to forcibly sink the wood by tying a denser mass we do this, we can compute the specific gravity of wood byusing the relationship wood water=mm1 m2(4)where m is the mass of the wooden block, m1is the apparentmass of the wood plus a piece of metal when the metal issubmerged, and m2is the apparent mass of the wood andmetal when both are submerged.

9 If you are concerned thatequation 4 seems to have popped out of thin air, don t worry,you will get a chance to derive this equation in the later partof this lab. Hang the wooden block from the hook on the undersideof the balance andrecord its mass. Next partially a beaker with water and arrange to havethe wood block hang from the balance and floating and record the balance reading for thefloating wood block, msub. Compute thedifference in the weight of wood in airand when floating on water. The difference is due tothe buoyant force.

10 Is it to equal weight of the fluiddisplaced you found in Part I? Continue by hanging a heavier metal object to the woodenblock and arranging to have only the metal object com-pletely submerged in the the apparentmass of the two objects when the metal weight is sub-merged, and call it m1. Now arrange to have both the wood and the metal ob-ject completely submerged. Once again, record the ap-parent mass of both objects when both are submergedand call the mass I Labs5 Use equation 4 tocompute and record the specific gravityof the wood. Use this result to determine the density ofwood and record this as well.