Exam Question 9: Hydrostatics. March 6, Applied Mathematics: Lecture 8. Brendan Williamson. Introduction. Density, Weight and Volume

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1 Exam Question 9: Hydrostatics March 6, 2017

2 This lecture is on hydrostatics, which is question 9 of the exam paper. Most of the situations we will study will relate to objects partly or fully submerged in water. As a result we will talk a lot about the density, volume and surface area of objects. We will also discuss pressure, thrust, buoyancy and. We will use the idea of moments and force, but no other prior knowledge is required.

3 Density and Weight The density of an object is its mass per unit volume. In other words, if an object or substance has mass m and volume V then its density ρ = m V and is measured in kg/m3. As 1 gram of water has volume 1 cm 3, 1 m 3 of water has mass 1,000 kg and so the density of water is 1,000 kg/m 3. We then define the relative density of an object or substance to be the ratio of its density to that of water. So an object with relative density of 0.87 has 87% the density of water, or a density of 870 kg/m 3. Sometimes relative density is referred to as specific gravity. The weight of an object is the force it exerts due to gravity, and is given by W = mg = V ρg.

4 The volume of an object is the amount of physical space it takes up. The volume of a cube is given by length breadth height. The volume of a sphere of radius r is given by 4 3 πr 3. The volume of a cylinder of radius r and height h is given by πr 2 h. The volume of a cone of base radius r and height h is given by 1 3 πr 2 h. The volume of a frustum, shown below, of base radius R and top radius r, and height h is given by 1 3 πh(r2 + Rr + r 2 ).

5 Density of objects Example: A solid frustum of height 5cm, base radius 3cm and top radius 10cm has mass 2kg. Find its relative density. First of all we should convert our units to standard units, so that the volume of the frustum is 1 3 π(0.05)( (0.1) ) = m 3. Therefore its density is 2/ = 2748 kg/m 3, so that its relative density is

6 Pressure When we apply a force to a certain area, the effect of that force depends on the size of the area in question. For example, pushing the palm of your hand against a piece of cardboard will have very little effect compared to pushing a knife into the same cardboard with the same force. The pressure applied to a surface is defined as the force per unit area. More mathematically, Pressure = Force Area. When underwater, the pressure exerted on you comes from the weight of the water above you. Therefore if we go twice as deep, there is twice as much water above you and so as your surface area doesn t change the pressure doubles. More generally, the pressure at a certain depth of any liquid is proportional to the depth. As a result we often talk about pressure at a point. Imagine a flat disc of any shape of area A lying horizontally a distance h under the surface of a liquid of density ρ. The liquid directly above it is a prism of volume Ah, and so the weight of the liquid is Ahρg. Therefore the pressure exerted on the disc is Ahρg/A = hρg. As this is true regardless of the shape we say that ρgh is the pressure at a point of depth h.

7 The corresponding force causing this pressure is called thrust. The thrust on the base of a container containing a liquid is given by the pressure at any point on the base multiplied by the surface area of the base, regardless of the shape of the container. Example: A frustum of height 5 metres, base radius 3 metres and top radius 2 metres is filled with water. Find the thrust on the base of the frustum. As the pressure at the base of the frustum is ρgh = 5000g and the area of the base of the frustum is 9π, the thrust on the base is 45, 000πg N. It may seem strange that this is true. If the water was instead solid, then the thrust exerted on the base would just be the weight of the block, which is less than our answer. However, because it is liquid it fills up the whole space, which means it pushes against the side of the frustum and this causes extra pressure. To further see that the thrust is the same as it would be in the case of a cylinder consider the following diagram.

8 It should be clear that the height of the water is the same on both sides. If there was more pressure on the right side then the water on the right would successfully push down, pushing the other column of water up and leaving a difference in height. Therefore the pressure must be the same.

9 A U-tube shown below contains 10cm of water on one side and 11cm of oil on the other side. Find the relative density of the oil. Similar to before, as we are in equilibrium we know that the pressure at a point on the left hand side is the same as on the right. Therefore if ρ is the density of the oil, ρg(0.11) = 1000g(0.1) ρ 909.1kg/m 3.

10 : Examples Example: Oil of relative density 0.8 is poured into one side of a U-tube that already contains water so that its height is 10cm. What is the difference in height of the two branches? The diagram is similar to before. In this case if the difference in height between the lowest point of the oil and the water in the second branch is x, then 800(0.1)g = 1000xg x = 0.08 m. Example: Consider the example given by the diagram. If we now pour oil of relative density 0.8 into the side that has water, what will be the height of this oil when both sides are of equal height? Similar to before, the pressure 11cm down from the level of the oil on the left side will be the same as the pressure at the same height on the right. The pressure on the left is easy to calculate, it is (909.1)0.11g = 100g. On the right we can calculate it in two, equivalent ways. Say the height of the oil is x m. Then the x 0.11 and it is of height density of the mixture is 800 x m, so the pressure is 800xg (0.11 x)g. Equivalently the pressure from the oil plus the pressure from the water is 800xg (0.11 x)g.

11 : Examples Setting these pressures equal to each other gives us 800xg (0.11 x)g = 100g x = 0.05m, so the height of the liquid is 5 cm.

12 principle states that if an object is fully or partially submerged in a liquid, an upward force, known as buoyancy or upthrust acts on the object, and the magnitude of this force is equal to the weight of the liquid displaced. Informally the reason why there is some sort of upthrust is that if you have an object in a liquid, the thrust on the parts of the object that are deeper are greater than those near the top, so that on aggregate there is an upwards force. More formally, consider a hollow object with infinitely thin walls full of the same liquid as it is submerged in. If fully submerged it will not rise or fall, and the only forces on it are buoyancy and gravity, which therefore must be equal. If it is partly filled then it will be in equilibrium when the liquid level inside the object lines up with the liquid level in the larger container. Example: A sphere of relative density 0.8 and radius 2m is attached to a piece of string which is attached to the bottom of a container. The container is filled with water so that the sphere floats but is tethered to the base of the container by the string. What is the tension in the string?

13 : Example Below is a visualisation of the system. There are three forces acting on the sphere, buoyancy, gravity and tension. The buoyancy acts upwards and is of magnitude 4 3 π23 (1000)g = 32,000 3 πg. Gravity acts downwards and is of magnitude 4 3 π23 (800)g = 25,600 3 πg. Tension acts downwards and is of unknown magnitude T. Therefore 32, 000 πg = 3 25, 600 πg + T πg = T.

14 : The Rectangular Prism We can actually see in the case of simple shapes like the rectangular prism. Imagine a rectangular prism submerged underwater so that its top face is 3m underwater. The face has dimensions 2 5m and its length is 4m. The thrust on the upper face is the pressure, (1000)g(3) = 3000g, multiplied by the area, 2(5) = 10, so it is equal to 30, 000g and acts downwards. Similarly the thrust on the lower face is 70, 000g. The horizontal thrusts are easily seen to cancel out and so the resultant of these thrusts is an upwards force of 40, 000g = 1000g(2 5 4), the weight of the water displaced.

and This method can be used to find the thrust on complicated surfaces. Consider the following situation similar to the sphere in the first example. NOTE: HL Only.

15 and This method can be used to find the thrust on complicated surfaces. Consider the following situation similar to the sphere in the first example. NOTE: HL Only. What is the thrust on the lower surface if the sphere is again of radius 2m and the upper surface is 4m below the surface of the water? We know that the thrust on the upper surface is 1000g(4)(π2 2 ) = 16, 000πg. The weight of the water displaced is 2 3 π g = 16,000 3 πg, so that the thrust on the lower surface is 16,000 3 πg + 16, 000πg = 64,000 3 πg.

16 in different liquids If an object has volume V it has a buoyancy of B W = 1000Vg when placed in water. When placed in a liquid of relative density s it has buoyancy B L = 1000sVg = sb W, so that s = B L B W. This equation can help you relate buoyancy and relative density of liquids but gives no information about the object. Also, if an object is placed in a liquid with the same relative density as it then B L = W so s = W B W. An object has buoyancy 10N in water and 7N in oil. What is the relative density of the oil? Answer= What would the buoyancy of the object be if it was submerged in a liquid of relative density 1.1? Answer= 10(1.1) = 11N. The object appears to weigh 4N when submerged in the oil. What is its real weight? Answer= = 11N.

A cube of side 2m has relative density 0.8 and is placed in water. How high is its top face above the water? The following is a visualisation of the situation.

17 A cube of side 2m has relative density 0.8 and is placed in water. How high is its top face above the water? The following is a visualisation of the situation. Say the object is submerged so that its bottom face is x metres below water. Then the buoyancy is equal to x(2)(2)1000g = 4000gx. The weight of the object is 8(800)g = 6400g, and as the object is assumed to be in equilibrium 4000gx = 6400g, so that x = 1.6. It is true in general that if an object with relative density s < 1 is placed in water that 100s% of it will be underwater in equilibrium.

18 : Example We have already seen items that are forced underwater with string, but instead now consider a block of mass M placed on top of the object, so that now 90% of the object is underwater (notice the use of block and object for differentation). Find M. If 90% of the object is underwater then the buoyancy is 0.9(8000g) = 7200g. The weight of the object is 6400g and the force from the block is Mg. As the net force is 0 we have So the block weighs 800kg. 6400g + Mg = 7200g M = 800.

19 : Example An upright cylinder is submerged in water, and some of it floats above the surface. Then oil of relative density 0.8 is poured on top of the water until the cylinder is just covered. If 40% of the cylinder is in the oil, what is the relative density of the cylinder? Below is a picture of the problem.

20 : Example The easiest way to solve this problem is to imagine the cylinder as two separate objects on top of each other, one contained entirely in the water and the other contained entirely in the oil. In this case each piece of the cylinder has three forces acting on it, buoyancy, gravity, and the normal reaction between the objects, whose magnitude we will denote as R. If the density of the cylinder is s and its volume is V, then 0.4V (800)g + R = 0.4Vsg 0.6V (1000)g = 0.6Vsg + R. Finding R in both equations and setting them equal to each other, we get 0.4Vsg 320Vg = 600Vg 0.6Vsg s = 920. Notice that s = 0.4(800) + 0.6(1000), and that we never used the fact that the object was a cylinder.

21 Rods floating in water (HL Consider the following situation, drawn below. A uniform rod of length 1m is attached at one end to the edge of a container containing water, and half of the rod is submerged in the water. Find the relative density of the rod. We solve this problem using moments. Consider a one-dimensional object (like this rod) and some reference point a on the rod. If we apply a force of magnitude F perpendicular to the rod a distance d from a that would cause the rod to rotate anti-clockwise, we say that the anti-clockwise moment of this force is of magnitude Fd. The resultant of all the clockwise and anti-clockwise moments acting on an object decides whether the objects rotates around a. In all cases here the object will not rotate, so the clockwise and anti-clockwise moments will be equal.

22 Rods floating in water The following diagram shows all of the forces acting on the rod. As the forces are not perpendicular to the rod we would ordinarily need to resolve them. If the rod is inclined at an angle θ to the horizontal then the clockwise moments about a are 1 4W cos θ and the anti-clockwise moments are 3 4R cos θ. Setting these equal we get W = 3R. As the rod also does not move up or down the upwards and downwards forces are equal so we get R + B = W. But B = W 2s where s is the relative density of the rod so we get W 3 + W 2s = W, so s = 3 4.

Rods floating in water Now consider a rod of relative density 2 attached at either end to different strings of different lengths which are attached to the ceiling, so that the rod is not level.

23 Rods floating in water Now consider a rod of relative density 2 attached at either end to different strings of different lengths which are attached to the ceiling, so that the rod is not level. It is immersed so that its lower 40% is immersed in water and the upper 60% is immersed in oil of relative density 0. What is the tension of each string in terms of W? The following is a visualisation of the system.

24 Rods floating in water The following are all of the forces acting on the rod. The numbers in between are a ratio of the given lengths. From the equation B = W s we get B 1 = 2W 5 2 = W 5, and from the equation B L = s L B W we get B 2 = 0.8 3W 5 2 = 3W 10. Therefore T 1 + T 2 + W 2 = W so that T 1 + T 2 = W 2. Also, if we take moments around the point where B 1 is centered, we get 2T W 5 = 53W T 2.

25 Rods floating in water Solving these simultaneous equations in T 1, T 2 gives us T 1 = T 2 = W 4.

26 Boards floating in water Now consider a uniform rectangular board shown below, with one corner attached to a string attached to the ceiling, two corners on the surface of the water and with ab bc.

27 Boards floating in water If the liquid is water, what is the relative density of the board and the tension in the string (in terms of W, the weight of the board)? Similar to before we will use the fact that the board is not moving, so the vertical forces add to 0. We will also look at moments about the line ac to exploit the fact that the board is also not rotating. The important thing to note here is that the centre of gravity of each section of the board is not halfway between a (or c) and the centre point, but two-thirds the way to the centre point, as is always the case with centres of gravity (centroids) of triangles. So this gives us the following forces.

28 Boards floating in water First of all B = W 2 s = W 2s, so equating vertical forces gives us T + W 2s = W. If we look at moments about the point where buoyancy is centered, we get 2( W 2 ) = 4T, so that T = W 4 and so s = 2 3.

. In an elevator accelerating upward (A) both the elevator accelerating upward (B) the first is equations are valid

. In an elevator accelerating upward (A) both the elevator accelerating upward (B) the first is equations are valid IIT JEE Achiever 2014 Ist Year Physics-2: Worksheet-1 Date: 2014-06-26 Hydrostatics 1. A liquid can easily change its shape but a solid cannot because (A) the density of a liquid is smaller than that of