EQUILIBRIUM OBJECTIVES PRE-LECTURE

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1 27 FE3 EQUILIBRIUM Aims OBJECTIVES In this chapter you will learn the concepts and principles needed to understand mechanical equilibrium. You should be able to demonstrate your understanding by analysing simple examples of equilibrium. You will also learn about the interactions between fluids and solid bodies as well as the concepts - buoyant force and pressure - used to describe the interactions. Minimum learning goals When you have finished studying this chapter you should be able to do all of the following. 1. Explain, interpret and use the terms translational motion, rotational motion, rigid body, equilibrium, stable equilibrium, unstable equilibrium, neutral equilibrium, axis, torque [moment of a force], centre of gravity, buoyancy, buoyant force, Archimedes' principle, pressure, pascal, density, barometer. 2. State and apply the relation between force and torque. 3. State the conditions for equilibrium and apply them to simple problems. 4. Describe and explain how the centre of gravity of a body can be located. 5. Describe and explain the forces acting on a body which is wholly or partly immersed in fluid. Solve simple problems involving buoyancy. Introduction PRE-LECTURE This chapter deals mainly with the equilibrium of rigid bodies. The conclusions about rigid bodies can also be applied to some examples of non-rigid bodies, such as bodies of fluid at rest. We start with two simple examples of objects in equilibrium: an object at rest and one moving with constant velocity. All the examples and principles discussed in this chapter are restricted to systems in which all the forces can be described in two dimensions - a plane. The extension to general three-dimensional systems uses the same concepts but is mathematically much more complex. 3-1 TRANSLATION AND ROTATION The simple descriptions of motion that we have used so far implicitly treated the motion of only one point in a body. That is a good kind of description provided that all points in the body follow similar, parallel, paths. That kind of motion is called pure translational motion. As well as translational motion, a body can also execute rotational motion (like that of a spinning wheel) and vibration (like that of a shaking jelly). A body which cannot vibrate noticeably is said to be rigid; its shape and size do not change significantly when it is acted on by a system of forces. The most general kind of motion of a rigid body is, therefore, a combination of translation and rotation; the flight of a boomerang is a good example.

2 FE3: Equilibrium EQUILIBRIUM OF FORCES Q3.1 A block is at rest on a table. Figure 3.1 An object at rest. What are the forces? a) On figure 3.1 draw in all the forces acting on the block. The block is not accelerating, so the net force on it must be zero. What does this tell you about the vertical forces and any horizontal forces that may be present? b) Suppose that a horizontal force is applied in an attempt to push the block along the table, but the block does not move. What is opposing the horizontal force? Q3.2. Block moving at constant velocity A block is being pulled across the table at constant velocity (figure 3.2). pull Figure 3.2 An object moving with constant velocity Draw in the other forces. A frictional force opposes the motion. Draw in this force and the other forces on the block. Again, the block is not accelerating so the net force on it must be zero. What does this tell you about the vertical and horizontal forces? The meaning of equilibrium The examples in questions 3.1 and 3.2 are simple illustrations of equilibrium. In both cases the velocity of the block is constant, i.e. its acceleration is zero, and the total force on it is zero. However it is not enough that the forces balance in order to have equilibrium. This guarantees only that there is no change of translational motion, i.e. that the motion of the body as a whole does not change. The definition of equilibrium needs to be extended to include the requirement that the rotational motion of the body also remains constant. Thus, for example, a body which is completely at rest is in equilibrium only if it does not start to move or rotate. As another example, a wheel rotating about a fixed axle, is defined to be in equilibrium only if its rotational speed does not change. In order to consider this rotational aspect of equilibrium we need the concept of torque. We shall then see that for equilibrium, torques, as well as forces, must balance.

3 FE3: Equilibrium 29 LECTURE 3-3 TORQUE Example 3.1. Wheel on a fixed axle Consider a wheel which can rotate about its axle. The axle remains in a fixed position. An object is hung by a string from the rim of the wheel as shown in figure 3.3. F A W F Figure 3.3 An unbalanced wheel When the object is released, the wheel starts to rotate so, clearly, it is not in equilibrium. The forces acting on the wheel are shown. W is the wheel's weight, F A is the force exerted by the axle on the wheel and F is the force exerted by the string attached to the falling object, which causes the wheel to rotate. Note that these forces are not all acting through the same point. In this example, rotation occurs about an axis - a line in space - which is perpendicular to the plane of the forces involved. In such cases we define torque as follows. The torque of a force F about a specified axis is defined as τ = Fx... (3.1) where x is the perpendicular distance from the axis to the force's line of action. Torque is also known as the moment of a force. Note that we can, in principle, define many torques for each force, one for every possible choice of axis. It is not necessary for the axis to be a possible axis of rotation. Example continued x F Figure 3.4 Torque on a pivoted wheel In this example, we consider torques about the wheel's axis of rotation, its axle. Only the force F has a non-zero torque about the axle so there is a net torque, due to F, acting on the wheel. This net torque produces an angular acceleration, i.e. it changes the angular speed of rotation about the axle.

4 FE3: Equilibrium 30 Demonstrations Consider two wheels which have the same shape and the same total mass. One has a dense metal rim, the other has a dense metal axle. Identical loads are hung from their rims. This demonstration shows that the angular acceleration depends on the distribution of mass in the object. metal rim metal axle Figure 3.5 Effect of mass distribution on angular acceleration Identical loads were attached to the wheels and then released. The angular acceleration was greater for the wheel whose mass is more concentrated near the axle. Other demonstrations show rotational motion when the axis is not fixed. 3-4 EQUILIBRIUM OF TORQUES Example continued The wheel has no translational motion so the total force on it must be zero. F A Forces W F Torques Torque Fx Figure 3.6 Forces and torque on the pivoted wheel The vertically upward component of the total force = F A - W - F = 0, but the net torque, Fx, is unbalanced. Note that this torque produces a clockwise rotation. Example 3.2: Equilibrium of the wheel To bring our wheel back into equilibrium, a torque of the same magnitude but in the opposite (anti-clockwise) sense would have to be provided. Another object with the same mass could be attached to the opposite side of the wheel in order to achieve this. See figure 3.7. (Note that the supporting force F A will take on a new value.) Forces F A Torques -Fx Fx F W F Figure 3.7 Forces and torques at equilibrium

5 FE3: Equilibrium CONDITIONS FOR EQUILIBRIUM In general, then, the conditions for equilibrium of an object which is free to rotate about a fixed axis are: (i) total force acting on the object = 0; (ii) total torque about the axis = 0. Note that, since force is a vector quantity, the calculation of the net force must take account of directions. This can be done using the method of components introduced in chapter FE2. Torque, as defined here, is a scalar quantity whose values need to be associated with either clockwise or anti-clockwise rotation. We can assign positive values to one of these two senses, and negative values to the other. (The usual convention makes anti-clockwise values positive.) Example continued When we apply these conditions to the wheel with two objects hanging from its rim, we can choose vertically down as a component direction for the forces (all horizontal force components are equal to zero). (i) F + W + F - F A = 0. (F A takes on a new value when the second object is attached.) The condition for balancing the torques is satisfied because the two torques have the same magnitudes but opposite senses: (ii) F x - F x = 0. It is obviously inconvenient to have to draw two diagrams for each example - one showing forces and one showing torques. Henceforth both forces and torques will be shown on the same diagram. Remember, however, that forces and torques are quite different entities and must be combined separately. 3-6 CENTRE OF GRAVITY Example 3.3. Centre of gravity of a flat object A flat object is pivoted at the point P. See figure 3.8. Imagine that the object is divided up into little pieces. The weight of each piece provides a torque about the pivot. The object will be in equilibrium only if the torques due to all these pieces add up to zero. P Figure 3.8 Distribution of gravitational forces on an object It is obviously inconvenient to have to consider the weights of all the little pieces of the object separately. Fortunately, their total can be represented by the total weight, W, acting through a point called the centre of gravity.

6 FE3: Equilibrium 32 Torque Equilibrium P P Centre of gravity W W Figure 3.9 Centre of gravity and the total weight force Then W provides a torque about the pivot (equal to the sum of the many small torques). The object will be in equilibrium if its centre of gravity lies vertically below the fixed pivot point. In this position the length of the perpendicular from the pivot to the line of action of the weight is zero, and so is that from the pivot to the supporting force at the pivot. Therefore the torques of the weight and the supporting force about the pivot are both zero. Example 3.4. Locating the centre of gravity To locate the centre of gravity of a flat object, first mark a vertical line showing the line of action of the weight. The centre of gravity must be somewhere on this line. Then choose a different pivot point and repeat the process. The centre of gravity must be also be somewhere on the new line; so it must be at the intersection of the lines. A wheel is symmetric about its axis and the centre of gravity is at the centre of the wheel. This is easily verified. 3-7 EQUILIBRIUM OF A SYSTEM OF OBJECTS Example 3.5: Two children balancing on a seesaw. This example is analysed in terms of forces and torques acting on a system which consists of the two children and the plank. The forces acting on this system are the weights of the two children, the weight of the seesaw's plank and a vertical supporting force at the pivot. The weights of the children, W 1 and W 2 act at distances x 1 and x 2 from the pivot, as shown on the figure. These forces give torques W 1 x 1 (anticlockwise) and W 2 x 2 (clockwise) respectively about the pivot. Suppose that the centre of gravity of the plank is directly above the pivot so that the weight W S of the seesaw plank acts downwards at the pivot. N is the upward supporting force exerted by the pivot on the plank. Each of these two forces gives zero torque about the pivot. N x 1 x 2 W 1 x 1 W 2 x 2 W 1 W S W 2 Figure 3.10 Balancing on a seesaw For equilibrium the following conditions must be satisfied. (i) Total force acting on the system = 0. Taking force components in the vertically downward direction:

7 FE3: Equilibrium 33 W 1 + W 2 + W S - N = 0. (ii)total torque about an axis through the pivot = 0. Taking clockwise as the positive sense: W 2 x 2 - W 1 x 1 = 0. This analysis is essentially the same as that for a beam balance. 3-8 EQUILIBRIUM OF A FREE OBJECT A free object is one that is not pivoted. This is a more general situation than the case of a fixed axis. Demonstration push moves forward spins anticlockwise push moves forward spins clockwise push moves forward no spin Figure 3.11 Motion of a free object The centre of gravity is shown as a heavy dot. A net force not acting through the centre of gravity of a rigid body will cause translational acceleration of the object as well as change in its rotational motion. The resulting motion can be described as a combination of translational motion of the centre of gravity and rotational motion about the centre of gravity. 3-9 GENERAL CONDITIONS FOR EQUILIBRIUM The general conditions for equilibrium are as follows (i) The total force must be zero (as before). (ii) The total torque about any axis must be zero. In many cases it is convenient to consider torques about axes through the centre of gravity BUOYANCY When a solid object is wholly or partly immersed in a fluid, the fluid molecules are continually striking the submerged surface of the object. The forces due to these impacts (which are sometimes called pressure forces) can be combined into a single force, the buoyant force.

8 FE3: Equilibrium 34 Figure 3.12 Forces exerted by a fluid Note that, for clarity, we show only the forces exerted by the surrounding fluid in this and the following diagram. We want to find the magnitude of the buoyant force and the point through which it acts. Buoyant force on a completely submerged object Figure 3.13 Buoyant force on a submerged object To work out how big this buoyant force is, and where it acts, we use the trick of thinking about a 'block' of fluid, which has exactly the same shape and size as the solid object. This imaginary portion of fluid is often called the 'displaced fluid'. Figure 3.14 Fluid "displaced" by the submerged object The pressure forces on this imaginary displaced fluid are exactly the same as the pressure forces on the solid object. So their total effect, the buoyant force, will be the same, irrespective of what is inside the broken outline. The buoyant forces on the solid object and the 'displaced fluid' are identical. Now the displaced fluid must be in equilibrium. Since the only other force on it is its weight, which acts through its centre of gravity, the buoyant force must be equal to its weight, and it must act vertically upward through its centre of gravity. Hence the buoyant force on the submerged block must be equal to the weight of the displaced fluid and it must act vertically up through the centre of gravity of the 'displaced' fluid body.

9 FE3: Equilibrium 35 Buoyant force Weight of displaced fluid Figure 3.15 Forces on the displaced fluid This conclusion is known as Archimedes' principle. Buoyant force on a partly submerged object Air Buoyant force Liquid Figure 3.16 Buoyant force on a partly submerged object Note that there is also a gravitational force (weight). In this case the object is immersed in two fluids one of which is the air. The diagram shows only the total force exerted by these fluids on the object. Consider an imaginary block composed of the two fluids 'displaced' by the object. Displaced air Displaced liquid Figure 3.17 Fluids displaced by the partly submerged object This fluid block is in equilibrium and, just as before, buoyant force = total weight of the displaced fluids, and acts vertically upward through the centre of gravity of the displaced fluids. Note that, for an object floating in a liquid, the buoyant force due to the displaced air is usually negligibly small.

10 Example 3.6: A heavy object supported in water by a string FE3: Equilibrium 36 String Force exerted by string Water Buoyant force Weight Figure 3.18 A submerged object in equilibrium The force exerted by the string adjusts to balance the weight and the buoyant force. Here all the forces on the submerged object are shown. For equilibrium of this object: force exerted by string = weight - buoyant force. We might call the weight of the block minus the buoyant force the "effective gravitational force". This situation is demonstrated using a spring balance. If the buoyant force were greater than the weight, a force would have to be applied to hold the object down. This happens, for example with a helium-filled balloon. Example 3.7: Floating object Air Buoyant force Liquid Weight Figure 3.19 A floating object in equilibrium The displaced liquid adjusts so that the forces balance. For equilibrium, taking components in the direction vertically down: weight - buoyant force = 0. The total torque will be zero if the buoyant force and the weight act along the same line. This means that the centres of gravity of the floating object and the displaced fluids must lie in the same vertical line MOMENT OF INERTIA POST-LECTURE The physical quantity which describes the distribution of matter about the axis of rotation is the moment of inertia. Objects with their masses concentrated about the axis of rotation (or axle) have smaller moments of inertia about that axis. More precisely, if we divide the object up into small pieces each with mass Δm and at some distance r from the axle, then the moment of inertia is I = Σ Δm r 2 where the sum is taken over all the small pieces.

11 FE3: Equilibrium 37 Axle Δ m r Figure 3.20 Defining the moment of inertia For objects with a fixed axis of rotation, total torque about the axis equals moment of inertia about the axis times angular acceleration. This is the reason why the wheel with the metal axle in the lecture demonstration had the larger angular acceleration QUESTIONS AND PROBLEMS ON EQUILIBRIUM Q3.3 a) A tall block sitting on a table is being pulled. F Figure 3.21 Tipping an object If the pull is applied near the top, the object will tip over instead of sliding along the table. Why? Hint: to get the block moving, the force F must be greater than the maximum frictional force acting at the bottom on the block. F cannot be any smaller than this. Now consider torques. What point is the block going to rotate about? How could you pull the block along the table without tipping it? b) Pulling trees down with a tractor can be a dangerous occupation. Which of the methods shown below is the less dangerous way to tie the rope to the tractor? Why? Equilibrium of pivoted objects Figure 3.22 Uprooting a tree. Which way is safer? Q3.4 Two children are balancing on a seesaw. Their weights are 200 N and 300 N. The smaller child is 1.80 m from the centre. Suppose that the weight of the seesaw is 500 N and the centre of gravity is directly above the pivot. Draw all the forces acting on the seesaw. a) Calculate the magnitude of the force that the pivot exerts on the seesaw. b) How far from the centre is the larger child at equilibrium? Q3.5 Explain why a beam balance gives the same value for the mass of an object on the Moon as on the Earth STABLE, UNSTABLE AND NEUTRAL EQUILIBRIUM We sometimes distinguish between stable and unstable equilibrium. For example, consider a cone.

12 FE3: Equilibrium 38 Stable Unstable Neutral Stable Unstable Neutral Figure 3.23 Stable and unstable equilibrium In both cases the cone is in equilibrium because the total force is zero and the total torque is zero. But the first case is stable, a slight displacement has no effect, while the second case is unstable, a slight displacement causes the cone to tip over. When the cone is lying on its side it is in neutral equilibrium. Q3.6 What forces are acting and where are they acting in each case? What torques are responsible for restoring the cone to its original position or otherwise? Centre of Gravity Q3.7 a) Does the centre of gravity always lie within an object? If not, give examples. b) Suggest a way of locating the centre of gravity of a "lumpy" object (not a flat object) FLUIDS Pressure When a solid object is immersed in a fluid, the force exerted on the object by the fluid is distributed over the contact surface. For a complete description we need to look at the force acting on each small part of the surface. We can define the average pressure on a flat surface to be the component, δf n, of the force perpendicular to the surface divided by the area, δa, of the surface. The limit of this quotient as we take smaller and smaller pieces of the contact surface (and hence also smaller and smaller forces) is the pressure, P, at a point on the surface: δf P = lim n.... (3.2) δa 0 δa Provided that the body and the fluid are not moving, the force on each small part of the contact surface is perpendicular to the surface (see figure 3.12) so the interaction can be described completely in terms of pressure. (On the other hand, if there is relative motion between fluid and solid object, the force has components parallel to the surface, not described by the pressure.) This idea of pressure can be used also to describe what goes on inside the fluid; just imagine the fluid divided into two parts as in the argument about buoyancy. Wherever we draw the fluid boundary, we can define a pressure exerted by one part of the fluid on the other part. So we can say that pressure exists within the fluid. The following are important statements about pressure. Pressure is a scalar quantity - it has no direction. The pressure within a uniform stationary fluid is the same at all points in the same horizontal plane. The SI unit of pressure is the pascal, symbol Pa; 1 Pa = 1 N.m -2.

13 Density FE3: Equilibrium 39 Pressure variations within a fluid are affected by its density. The density, ρ, of a uniform substance is defined as the quotient: mass volume: m ρ = V.... (3.3) The barometer h S S' Figure 3.24 A mercury barometer A tube, closed at one end, is filled with mercury and is then inverted over a container of mercury as shown. If the tube is sufficiently long, the mercury falls from the top leaving an evacuated region there. By considering the equilibrium of a part of the surface SS' of the mercury outside the tube, we can show that the pressure in the mercury just under that surface is equal to the atmospheric pressure. So too is the pressure at the same level inside the tube. We consider the equilibrium of a body consisting of the column of mercury in the tube which is above the level SS'. The forces on this column of mercury are its weight, W, downward and the upward force, F, exerted by the mercury below the column. The magnitude of the upward force must be equal to atmospheric pressure multiplied by the cross-sectional area, A, of the tube: F = PA. The weight of the mercury column is equal to the product of its density, its volume and g. Since the volume is equal to the product of the column's height and cross-sectional area, W = ρhag. These two forces, must have equal magnitudes, so P = ρgh....(3.4) The density of mercury is Pa (101 kpa). Q3.8 a) Estimate the height of a column of mercury in a barometer. kg.m -3 and normal atmospheric pressure is b) The density of water is kg.m -3. Estimate the height of the column of water in a water barometer QUESTIONS ON BUOYANT FORCES Q3.9 Density of air = 1.29 kg.m -3. Density of helium = 0.18 kg.m -3. A balloon is filled with helium. Its volume is 1.00 m 3. What downward force must be applied to stop the balloon rising up into the sky? Q3.10 Estimate the buoyant force exerted on you by the atmosphere. Q3.11 An ice cube is floating in a glass of water. The water and the ice are at about 0 C. Density of ice at 0 C = 917 kg.m -3. Density of water at 0 C = 1000 kg.m -3. a) What fraction of the ice cube is submerged?

14 FE3: Equilibrium 40 b) Explain what happens to the level of the water as the ice melts. Q3.12 Steel is about eight times as dense as water. How can ships float? Q3.13 The density of water varies with temperature as shown in the graph below. The curve has a maximum at about 4 C Density/kg.m Figure Temperature/ C How the density of water changes with temperature a) When a beaker of water is heated from below, why does the warm water at the bottom rise? b) When a lake ices over, why does freezing occur only at the top of the lake? Is there any biological significance in that?

15 Interlude 3: Masses 41 INTERLUDE 3 - THE RANGE OF MASSES IN THE UNIVERSE mass/kilograms Mass of the Universe Mass of our Galaxy Mass of the Sun Mass of the Earth Mass of the Moon Ocean liner Person 1 Sugar cube Postage stamp Red blood corpuscle Protein molecule Oxygen molecule Proton Electron