Upthrust and Archimedes Principle

Transcription

1 1 Upthrust and Archimedes Principle Objects immersed in fluids, experience a force which tends to push them towards the surface of the liquid. This force is called upthrust and it depends on the density of the fluid and the volume of the fluid. Archimedes Principal When an object is totally immersed in a fluid, the upthrust is equal but opposite to the weight of fluid displaced. Deriving Archimedes Principle Let us consider a cylinder of height (h) and cross sectional area (A). It is immersed in a fluid of density (ρ) and is at a depth (h) below the surface as shown in the diagram below. Therefore, the force experienced at the top of the cylinder can be calculated by using the formula. F = P A F x = hρg A Hence, the force experienced by the bottom of the cylinder can be calculated by F y = Hρg A Hence the resultant force is F y F x Resulatant Force = HρgA hρga Resulatant Force = ρga(h h)

2 2 From looking at the diagram above we realize that (H h) is the height of the cylinder. Therefore Resulatant Force = ρga h But we know that A h = volume so we can say that Resulatant Force = ρgv (NB. volume of the cylinder is also the volume of the water displaced) We also know that m = pv and can there state that Resulatant Force = ρgv = mg We can therefore conclude that the upthrust is equal to the weight of the water displaced. Flotation When an object floats in a liquid, the upthrust is equal to the weight of the object. Hence the weight of the object is equal to the weight of the liquid displaced. The principal of flotation stated that a floating body displaced its own weight of fluid. Example 1 A glass stopper of weighed in air and them immersed wholly in water and then reweighed. The readings obtained are 2.4 N in air and 2.0 N in water. Given that the density of water is 1000 kg/m 3, calculate the density of the stopper. Example 2 A sphere of radius 1 m is held below the surface of the sea by a string attached to the bottom of the sea bed. The density of the sea water is 1000 kg/m 3. Calculate: a) the volume of the sphere b) the mass of the sphere c) mass of the water displaced d) tension in the string the initial acceleration if the string breaks

3 3 Motion of Particles through Fluids When objects move through fluids, (example: a ball through a liquid), they experience a retarding force due to friction between fluid molecules and surface of the object. Additional forces associated with the object push away the fluid in order to advance. When the molecules of the fluid are pushed away they gain momentum. This momentum change required a force which is provided by the object. According to Newton s 3 rd Law, there is an equal but opposite force acting on the object. This force is called the drag-force. When the velocity of the object is low, the drag-force (FD) is approximately proportional to the velocity of the object (v). In other words F D v or F D = kv. However, when the velocity of the object is high, the drag-force (FD) is approximately proportional to the velocity squared of the object (v 2 ). In other words F D v 2 or F D = kv 2. When an object is dropped in a fluid (example: a ball in a liquid) the forces acting on it are: (i) (ii) (iii) the weight of the object (W) the upthrust of the liquid (U) the frictional force of the liquid (F) When the object is first dropped into the liquid, its weight is greater than (F + U). The resultant force is downwards and it produces acceleration so that the velocity increases. Over a period of time, the frictional force increases so that (F + U) is now equal to the weight of the object (W). At this point, the resultant force is now zero and the object moves at a constant velocity (terminal velocity). Example An object is released from a plane and falls to the ground. The mass of the object is 2.5 kg and the drag force obeys the equation F D = kv 2 where k is 0.4 Nms -2. Taking g as 10m/s 2, calculate the terminal velocity of the object.

4 4 Moments (Torque) The moment (or torque) of a force is defined as the product of force by perpendicular distance from the pivot to the line of action of force. moment (torque) = force distance T = F d Example Calculate the torque for the situation below m 10 N Principle of Moments D2 D3 D1 F 3 F1 F2 sum of the anticlockwise moments = sum of the clockwise moments F 3 D 3 = F 1 D 1 + F 2 D 2

5 5 When the system is in equilibrium, the net force is zero, hence the total upward forces is equal to the total downward forces. We can calculate the upward reaction by using the equation R = F 1 + F 2 + F 3 Example 1 A hanged trap door of mass 15 kg and length 1 m is opened with a force (F) at 45 0 as shown in the diagram below. Calculate (i) (ii) the value of F the horizontal force on the hinge F m W Example 2 A uniform plank AB is 6 m long and has a weight 300 N is supported horizontally by two vertical ropes at A and B. A weight of 150 N rests at C where AC is 2 m. Find the tension in each rope. T A T B 2 m C A 3 m B 150 N W

6 6 Couples When two forces are equal in magnitude and acting in the opposite direction on an object we call it a couple. We can calculate the torque or the moment of a couple by multiplying once of the force by the distance between the two. Total Torque about O = (F AO) + (F OB) = F (AO + OB) = F (AB) = F d (NB: rque of a couple = one force separation of forces )

7 7 Equilibrium Review: Equilibrium under the Action of Concurrent Forces Centre of Gravity and Centre of Mass The centre of gravity of a body is a single point at which the entire weight of the body can be considered to be acting. The centre of mass is the point at which the total mass of the body is concentrated. Types of Equilibrium There are three types of equilibrium: 1. Stable Equilibrium If a body returns to its equilibrium position after it has been displaced slightly. 2. Unstable Equilibrium If a body does not return to its equilibrium position after it has been displaced slightly. 3. Neutral Equilibrium If a body is slightly displaced and it stays in the displaced position.

8 8 Let us consider a heavy beam (example: a ladder) with one end leaning against a smooth vertical wall and the other end resting on a rough horizontal floor as shown in the diagram below. A R W is weight acting vertically downwards R is normal reaction on the wall FV is normal reaction for θ FH is the frictional force F V W Ѳ F H B If the beam is in equilibrium, we can make the following statements (i) (ii) There is no movement in any direction, therefore the resultant force is zero and the total upward forces must equal to the total downward forces. F V = W There is no movement from left to right F H = R (iii) Because the ladder is in equilibrium, the total torque or moment about any point is zero. Example 1 A uniform ladder of weight 400 N leans in equilibrium against a frictionless wall. The foot of the ladder rests on a rough ground and makes an angle of 53 0 to the horizontal. Find: a) the magnitude and the direction of the force acting on the top of the ladder by the wall. b) the horizontal and vertical components of the force exerted by the ground by the ladder. c) the magnitude and the direction of the resultant force exerted by the ground on the foot of the ladder.

9 9 Example 2 A uniform ladder which is 5 m long has a mass of 20 kg and leans with it its upper end against a smooth vertical wall and its lower end on the rough ground. The bottom of the ladder is 3 m from the wall. Calculate the frictional force between the ladder and the ground given that g = 10 ms -2.

10 10 Gravitational Potential Energy (GPE) Suppose a constant force (F) is used to move a mass (m) a distance (d) above the ground. The work done (W) by the force is given by: Therefore W = mgh GPE = mgh In a system in which the only forces acting are associated with the potential energy, the sum of the kinetic and the potential energy is constant. ie. KE + PE = constant Power The power of a machine is the rate at which it does work power = work done time Let us look at the power equation P = W t We also know that W = Fs Hence P = Fs t Also v = s t Therefore we can state that P = Fv

11 11 Example 1 A small block slides down a smooth track as shown in the diagram below. If the speed of the block is 2 m/s at A, calculate (i) (ii) the speed at B the speed at A Example 2 A block of mass 3 kg is pulled up a smooth plane inclined at 30 0 to the horizontal by a force of 25 N parallel to the plane. Calculate the velocity of the block.