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UPTHRUST IN FLUIDS, ARCHIMEDES’, PRINCIPLE AND FLOATATION, , , , , , Syllabus :, , Buoyancy, Archimedes’ principle, floatation, relationship with density; relative density, determination of relative, density of a solid., , Scope : Buoyancy, upthrust (F,); de 1; different cases, F, > = or < weight W of the body immersed;, characteristic properties of upthrust; 2s’ principle; explanation of cases where bodies with density, f > = or < the density p’ of the fluid in which it is immersed. R.D. and Archimedes’ principle. Experimental, determinations of R.D. of a solid and liquid denser than water, Floatation; principle of floatation; relation, between the density of a floating body, density of the liquid in which it is floating and the fraction of volume, of the body immersed; (p,/p, = V,/V,); apparent weight of floating object; application to ship, submarine,, , , , , , iceberg, balloons, etc. Simple numerical problems involving Archimedes’ principle and floatation., , , , , , , , (A) UPTHRUST AND ARCHIMEDES’ PRINCIPLE, , 5.1 BUOYANCY AND UPTHRUST, , When a body is partially or wholly immersed, in a liquid, an upward force acts on it. This, upward force is known as upthrust or buoyant, force. It is denoted by the symbol F,. Thus, , The upward force exerted on a body by the fluid in, which it is submerged, is called the upthrust or, buoyant force., , , , , , , , , , The property of liquid to exert an upward, force on a body immersed in it, is called, buoyancy. This property can be demonstrated by, the following experiments., , Exp. 1. Pushing an empty can into water : Take an, empty can. Close its mouth with an airtight stopper. Put, it in a tub filled with water. It floats with a large portion, of it above the surface of water and only a small portion, of it below the surface of water., , If we push the can into water, we feel an upward, force which opposes the push and we find it difficult to, push the can further into water. It is also noticed that as, the can is pushed more and more into water, more and, more force is needed to push the can further into water,, till it is completely immersed. When the can is fully, inside water, a constant force is still needed to keep it, stationary in that position. Now if the can is released at, this position, it is noticed that the can bounces back to, the surface and starts floating again., , Exp. 2. Pushing a cork into water : If a piece of cork is, placed on the surface of water in a tub, it floats with, , nearly 2th of its volume inside water. If the cork is, , pushed into water and then released, it again comes to, the surface of water and floats. If the cork is kept, , immersed, our fingers experience some upward force., The behaviour of cork is similar to that of the empty can., , Explanation : When the can or cork is put in, the tub of water, two forces act on it : (i) its weight, (i.e., the force due to gravity) W which pulls it, downwards, and (ii) the upthrust F, due to water, which pushes the can or cork upwards. It floats in, the position when the two forces become equal in, magnitude (i.e., W = F,). Now as the can or cork, is pushed more and more inside water, the upthrust, F, exerted by water on it increases and becomes, maximum (= F 5) when it is completely immersed, in water. So when it is released, the upthrust F, e, exerted by water on it being greater than its weight, W (or force due to gravity), it rises up. To keep the, can or cork immersed, an external downward force, (= F3, — W) is needed to balance the net upward, force., , Note : Like liquids, gases also have the, property of buoyancy, i.e., a body immersed (or, placed) in a gas also experiences an upthrust. All, objects including ourselves, are also acted upon, by a buoyant force due to air, but we do not feel it, because it is negligibly small as compared to our, own weight. On the other hand, a balloon filled, with hydrogen (or any gas less denser than air), rises up because the upthrust (or buoyant force) on, balloon due to the surrounding air is more than the, weight of balloon filled with the gas., , Condition for a body to float or sink in, a fluid : When a body is immersed in a fluid,, two forces act on the body : (i) the weight W of, , 98

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the body which acts vertically downwards and, (ii) the upthrust F, which acts vertically, upwards. We have noticed that the upthrust, depends on the submerged portion of the body., It increases as the submerged portion of body, inside the fluid increases and becomes maximum, (= F3) when the body is completely immersed, inside the fluid. Fig. 5.1 shows a body held, completely immersed in a fluid with two forces, W and F;, acting on it., , F, (Max. upthrust), , , , , , , , , , , , Fig. 5.1 Forces acting on a body held immersed, inside a liquid, , Depending upon the density of the fluid, the, maximum buoyant force F’, can be greater than,, equal to or less than the weight W of the given, body. Whether the body will float or sink in a, fluid, depends on the relative magnitudes of forces, W and F;, (buoyant force when the body is fully, immersed)., @ If F4, > W or F’, = W, the body will float, (it will not sink). If F’, > W, the body will, float partly immersed with only that much, part of it inside liquid, the upthrust F, due, to which becomes equal to the weight W of, body (i.e., F, = W). But if F’, = W, the, body will float with whole of it immersed, inside the liquid. Thus for a floating body,, net force acting downwards (i.e., apparent, weight) is zero., If F;, < W, the body will sink due to the net, force (W— F') acting on the body downwards., If m is the mass of body, it will go down into, the liquid with an acceleration a such that, ma = W-F,, or a=(W- F;)/m. Here we, have ignored the viscous force of the liquid., Unit of upthrust : The upthrust, being a, force, is measured in newton (N) or kgf., 5.2 CHARACTERISTIC PROPERTIES OF, , UPTHRUST, , The upthrust has the following three, characteristic properties :, , Gi), , 99, , (i) Larger the volume of body submerged in a, fluid, greater is the upthrust., , (ii) For same volume inside the fluid more the, density of fluid, greater is the upthrust., , (iii) The upthrust acts on the body in upward, direction at the centre of buoyancy i.e., the, centre of gravity of the displaced fluid., , (i) Larger the volume of body submerged in, , a fluid, greater is the upthrust, , In the experiment of pushing an empty can or, cork into water as described above, it is, experienced that the upthrust on the body due, to water increases as more and more volume of it, is immersed into water, till it is completely, immersed., , Similarly, when a bunch of feathers and a, pebble of same mass are allowed to fall in air,, the pebble falls faster than the bunch of feathers., The reason is that upthrust due to air on pebble, is less than that on the bunch of feathers because, the volume of pebble is less than that of the bunch, of feathers of same mass. However in vacuum,, both the bunch of feathers and pebble will fall, together because there will be no upthrust., , (ii) For same volume inside the fluid more the, density of fluid, greater is the upthrust, If we place a piece of cork A into water and, another identical cork B into glycerine (or mercury),, we notice that the volume of cork B immersed in, glycerine (or mercury) is smaller as compared to, the volume of cork A immersed in water. The, reason is that the density of glycerine (or mercury), is more than that of water. Now if we want to, immerse cork B in glycerine to the same extent as, cork A in water, then an additional force is needed, on cork B, to immerse it to the same level as cork, A. This shows that for same volume of a body inside, the liquid, a denser liquid exerts a greater upthrust., (iii) The upthrust acts on the body in upward, direction at the centre of buoyancy (i.e., the, centre of gravity of the displaced liquid), For a uniform body completely immersed, inside a liquid, the centre of buoyancy coincides, with the centre of gravity of the body (Fig. 5.1)., But if a body floats in a liquid with its part, submerged (Fig. 5.2), the centre of buoyancy B is, at the centre of gravity of the displaced liquid, (i.e., at the centre of gravity of the immersed part

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Fig. 5.2 A body floating with part of it submerged, , of the body) which lies below the centre of gravity, G of the entire body. The weight of the body, W acts downwards at G, while upthrust F, acts, upwards at B such that W = F,., , 5.3, REASON FOR UPTHRUST, , We have read that a liquid contained in a vessel, exerts pressure at all points and in all directions., The pressure at a point in liquid is same in all, directions (upwards, downwards and sideways). It, increases with depth inside the liquid. When a body,, say a block of area of cross section A, is immersed, ina liquid (Fig. 5.3), the pressure P, exerted upwards, on the lower face of block (which is at a greater, depth) is more than the pressure P, exerted, downwards on the upper face of block (which is at, a lesser depth). Thus there is a difference in pressure, (= P, — P,) between the lower and upper faces of, block. Since force = pressure x area, the difference, in pressures due to liquid on the two faces of, block causes a net upward force (i.e., upthrust), =(P,—P,)A on the body. However, the thrust on the, side walls of body get neutralised as they are equal, in magnitude and opposite in directions., , , , , , Fig. 5.3 A block immersed in a liquid, , Note : If a lamina (thin sheet) is immersed in a, liquid, the pressure on its both surfaces will be, nearly same, so the liquid will exert negligible, upthrust on it, causing it to sink into the liquid due, to its own weight., , 5.4 UPTHRUST IS EQUAL TO THE, WEIGHT OF DISPLACED LIQUID, (Mathematical proof), , When a body is immersed in a liquid, upthrust, on it due to liquid is equal to the weight of the, , , , eT 100 7, , liquid displaced by the submerged part of the, body., , Proof : Consider a cylindrical body PQRS, of cross-sectional area A immersed in a liquid, of density p as shown in Fig. 5.4. Let the upper, surface PQ of body be at a depth h, while its, lower surface RS be at a depth h, below the free, surface of liquid., , , , Fig. 5.4 Calculation for upthrust, , At depth h,, pressure on the upper surface PQ, , P, =hipg, .. Downward thrust on the upper surface PQ, F, = pressure x area = hpg A wei), At depth h., pressure on the lower surface RS, P, =hpg, .. Upward thrust on the lower surface RS, F, =hpgaA seni), , The horizontal thrust at various points on the, vertical sides of body get balanced because liquid, pressure is same at all points at the same depth., , From above eqns. (i) and (ii), it is clear that, F,, > F, because h, > h, and therefore, the body, will experience a net upward force., , Resultant upward thrust (or buoyant force), on the body, , F, = F,-F,, = h,pgA —h,pgA, = A(h, — h,) pg, , But A (h,—h,) = V, the volume of the body, , submerged in liquid., wan(5.1), Since a solid when immersed in a liquid,, , displaces liquid equal to the volume of its, submerged part, therefore, , , , Vpg = Volume of solid immersed x density of, liquid x acceleration due to gravity.

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or Vpg = Volume of liquid displaced x density, of liquid x acceleration due to gravity., = mass of liquid displaced x acceleration, due to gravity., = Weight of the liquid displaced by the, submerged part of the body., , a ae, , +(5.2), Note : (1) If the body is completely immersed, in a liquid, the volume of liquid displaced will be, equal to its own volume and upthrust then will be, maximum (= F'5)., (2) Although the above result is derived for a, cylindrical body, but it is equally true for a body of, any shape and size., , Factors affecting the upthrust, , From the above discussion, it is clear that, the magnitude of upthrust on a body due to a, liquid (or fluid) depends on the following two, factors :, , (i) volume of the body submerged in liquid (or, fluid), and, , (ii) density of the liquid (or fluid) in which the, body is submerged., , Effect of upthrust : The effect of upthrust is, that the weight of body immersed in a liquid, appears to be less than its actual weight. This, can be demonstrated by the following experiment., Experiment : Lifting of a bucket full of water from a, well. Take an empty bucket and tie a long rope to it. If the, bucket is immersed in water of a well keeping one end of, rope in hand and the bucket is pulled when it is deep inside, water, we notice that it is easy to pull the bucket as long as, it is inside water, but as soon it starts coming out of the, water surface, it appears to become heavy and now more, force is needed to lift it., , This experiment shows that the bucket of water, appears lighter when it is immersed in water than its actual, weight (in air)., , Similarly, when pulling a fish out of water,, it appears lighter inside water as compared to, when it is out of water., , Similarly, a body weighed by a sensitive spring, balance, will weigh slightly less in air than in, vacuum due to upthrust of air on the body., , ==, , , , , , ee 101 2, , 5.5 ARCHIMEDES’ PRINCIPLE, , When a body is immersed in a liquid, it, occupies the space, which was earlier occupied by, the liquid i.e., it displaces the liquid. The volume, of liquid displaced by the body is equal to the, volume of the submerged part of the body so the, body experiences an upthrust equal to the weight, of the liquid displaced by it., , It is the upthrust due to which a body, immersed in a liquid appears to be of weight, less than its real weight. The apparent loss in, weight is equal to the upthrust on the body. This, is called the Archimedes’ principle. Thus, , This principle applies not only to liquids, but, it applies equally well to: gases also., , 5.6 EXPERIMENTAL VERIFICATION OF, ARCHIMEDES’ PRINCIPLE, , Archimedes’ principle can be verified by, , either of the following experiments., Expt. (1) : Take two, cylinders A and B of the, same volume. The cylinder A, is solid and the cylinder B, is hollow. Suspend the two, cylinders from the left, arm of a physical balance, keeping the solid cylinder A, below the hollow cylinder B., Then balance the beam by, keeping weights on right, arm of the balance. In this, situation, both cylinders A, and B are in air., , The solid cylinder A is, now completely immersed, into water contained in a, beaker D placed on a bench, Cas shown in Fig. 5.5, taking care that the cylinder A does, not touch the sides and bottom of the beaker. It is observed, that the solid cylinder A loses weight i.e., the left arm of, the balance rises up. Obviously the loss in weight is due to, upthrust (or buoyant force) of water on the cylinder A., , Now pour water gently in the hollow cylinder B till it is, completely filled. It is observed that the beam balances again., , Thus, it is clear that the buoyant force acting, on solid cylinder A is equal to the weight of water, , SEE RG, , , , , , , , , , Fig. 5.5 Verification of, Archimedes’ principle

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filled in the hollow cylinder B. Since the cylinders, A and B both have equal volume, so the weight of, water in the hollow cylinder B is just equal to the, weight of water displaced by the cylinder A. Hence, the buoyant force acting on the cylinder A is equal, to the weight of water displaced by it. Thus, it, verifies the Archimedes’ principle., , Expt. (2) : Take a solid (say, a metallic piece). Suspend, it by a thin thread from the hook of a spring balance, [Fig. 5.6(a)]. Note its weight., , , , , , MEASURING CYLINDER, Fig. 5.6 Verification of Archimedes’ principle, , Now take a eureka can and fill it with water up to its, spout. Arrange a measuring cylinder below the spout of, the eureka can., , Now immerse the solid gently into water of the eureka, can. The water displaced by it gets collected in the, measuring cylinder [Fig. 5.6 (b)]. When water stops, dripping through the spout, note the weight of the solid, and the volume of water collected in the measuring, cylinder,, , In Fig. 5.6, the solid weighs 300 gf in air and, 200 gf when it is completely immersed in water., , The volume of water collected in the measuring, cylinder is 100 ml i.e., 100 cm?., , Loss in weight = 300 gf — 200 gf = 100 gf scse(i),, Volume of water displaced = Volume of solid, = 100 cm?, Since density of water = 1 gcm™, . Weight of water displaced = 100 gf weee(ii), , From eqns. (i) and (ii), Weight of water displaced = Upthrust or loss in weight., Thus the weight of water displaced by solid is, , equal to the loss in weight of the solid. This verifies, Archimedes’ principle., , 5.7 SOLID BODIES WITH DENSITY (p), GREATER THAN DENSITY OF, LIQUID (p,) SINK WHILE WITH, DENSITY (p) LESS THAN DENSITY OF, LIQUID (p,) FLOAT, , Let a body of volume V and density p be, immersed completely in a liquid of density p,., The weight of the body acting downwards will, be W = Vpg and the maximum upthrust on the, body acting upwards will be F’, = Vp,g., Following three cases may arise :, , (i) If W > F’, or Vpg > Vp,g or p > p,, the, body will sink due to net force (W — F’,) acting, downwards., , (ii) If W = F’, or Vpg = Vp,g or p = p,,, the body will float and the net force on the body, is zero., , (iii) If W < F’, or Vpg < Vp,g or p < p,,, the body will float due to net force (F R - W), acting upwards and only that much volume v of, the body will submerg inside the liquid due to, which upthrust F;, (= vp,g) balances the weight, W. The net force on the body is zero in this, situation also., , Thus a body of density p sinks in a liquid, of density p, if p > p,, while it floats if p = p,, or p < p,. This can be demonstrated by the, following experiments., , Expt. (1) : Take an iron nail and a piece of, cork both of same mass. First place the iron nail, on the surface of water contained in a cup. The, nail sinks. It implies that the force of gravity (or, weight). on iron nail pulling it downwards is, greater than the upthrust of water on nail pushing, it upwards. Now place the piece of cork on the, surface of water. The cork floats. It means that, upthrust on cork, when fully immersed is more, than that on nail because the density of water is, more than the density of cork, while the density, of water is less than that of iron nail., , Expt. (2) : Take few solid bodies of different, materials of known density and place them on, the surface of water. It is observed that if the, , 102