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Chapter 4 ((F= ma (F,Force (N, newtons), m, mass (kg kilograms), a,…

- - - - i.e. on different planets
        
        9.81 on earth
        
        1.6 on moon
      - use a newton meter to determine weight
- - - - the gravitational force acting on an object through its centre of mass
    - - The force that arises when two surfaces rub against each other
    - - The resistive force on an object travelling through a fluid (e.g. air and water); the same is fraction
        
        Magnitude of drag force depends on
        
        Speed of object(most important)
        
        drag force is directly proportional to speed^2
        
        Shape of object
        
        modern vehicles have smooth streamlined shapes to reduce air resistance exerted against them whilst travelling
        
        this increases the top speed and reduces the amount of fuel used
        
        Cross sectional area (most important)
        
        large CSA results in greater drag
        
        Roughness/texture of object
        
        density
        
        Drag experienced is often known as air resistance
        
        Terminal velocity
        
        During a vertical fall through air or another fluid, the wieght of the object remains constant bu the drag force increases as the speed increases.
        
        At the instant the object starts to fall there is no drag force on the object. The total force is equal to the wieght. The acceleration of the object is g, the acceleration of free fall
        
        As the object falls, its speed increases and this in turn increases the magnitude of the opposing drag force. The resultant (net) force on the object decreases and the instantaneous acceleration of the object becomes less than g.
        
        3.Eventually the object reaches terminal velocity which is when the drag force on the object is equal and opposite to its weight. At terminal velocity, the object has zero acceleration and therefore a constant speed.
        
        page 55 for diagram
        
        velocity-time graph for an object falling through air and the corresponding free-body diagrams at three different times.
        weight is mg.
        D is drag.
        instantaneous acceleration is a.
        
        Investigating motion in a fluid
        
        To investigate the motion of an object falling affected by a drag force by using a motion sensor connected to a data-logger or a laptop.
        
        The falling object is attached ti a light polystyrene ball by a thin thread passed over a pulley.
        
        The object is then dropped through a cylinder of liquid such as water or glycerol, pulling the polystyrene ball vertically upwards. The motion of this ball is identical to that of the object falling through the fluid.
        
        You can generate and analyse velocity-tim and acceleration-time graphs with this arrangements
        
        page 56
    - - the force within a stretched cable or rope
    - - an upward buoyancy force acting on an object when it is in a fluid
    - - force arising when one object rests against another obect
      - normal means 'at a right anlge to'
  - - - Fy = N = mgcosθ
- - - - It is possible to draw a triangle of forces for the forces passing through P as all three coplanar forces pass through point P in space.
        
        This method simplifies a complex problem
- - - - Density (kgm^-3)
      - mass(kg)
      - Volume (m^-3)
      - ρ = m/V
    - - Need to know mass, which can be measured directly using a digital balance.
      - Need to know volume
        
        For liquids, simply use a measuring cylinder to determine the volume,
        
        For a regular-shaped solid, calculate the volume using measurements taken from ruler, digital calipers, or a micrometer.
        
        For an irregular solid, the volume can be determined by displacement of a liquid. Measure the volume of a liquid in a measuring cylindar before the object is added, and then after the object is added.
        
        The difference is the volume of the object
  - - - P = F/A
      - Pressure(Nm^-2 or Pa) = Normal force(N) / Cross-sectional area (m^-2)
        
        Pressure is measured in Nm^-2 or Pa where 1 Pa = 1 Nm^-2
    - - Tiny flutuations in this pressure are respinsible for the variety of weather patterns that we so on Earth
  - - - They exert pressure on surfaces because of the constant bombardment by their molecules.
    - - At sea level the atmospheric pressure about 101 kPa
      - At the top of Ben Nevis britains highest mountain, the pressure is only 87 kPa
    - - p = hρg
        
        This equation show that
        
        Pressure does not depend on the cross sectional area
        
        Pressure is directly proportional to height
        
        therefore water pressure increases with depth
        
        Denser liquids will exert greater pressure
      - Pressure p(Nm^-2 or Pa)
      - h height(m)
      - density ρ (kgm^-3)
      - acceleration of freefall g (9.81ms^-2)
    - - is the resultant upwards force of submerging an object in a fluid
      - If you push a small piece of wood into water then let go, the wood will immediately pop out of the water, bob up and down on the surface, then remain afloat
        
        The buoyent force on the submerged wood can be explained in terms of the pressure differences at its upper and lower surfaces
        
        Force at the top surface = hρgA
        
        force at the bottom surface = (h + x)ρgA
        
        Resultant upward force = (h+x)ρgA - hρgA = xρgA
        
        Upthrust (N)= Axρg
      - Archimedes' Principle
        
        The upthrust exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
        
        An object will sink if the upthrust is less than the wiehgt of the object. For a floating object, the object must equal the weight of the object. This in turn means that the weight of a floating object must be equal to the weight of the fluid it displaes.
      - Icebergs
        
        about 9/10 of an iceberg lies underwater as at 0° the density of ice is about 9000kgm^3, whereas the density of water is about 1000kgm^-3
        
        A cube of ice with sides 1m will therefore have a weight of (900x9.81) = 8.83 x 10^3N.
        
        When it floats, it will displace water of weight 8.83 x 10^3 Which has a volume of (8.83 x 10^3) / (1000x9.81) = 0.9m^3
        
        so therefore the 1m^3 ice cube will sink until it has displaces 0.9m^3 of water, which is 9/10ths of the ice cube that is underwater
- - - - This point is often called the centre of gravity, which on the earth, coincides with the centre of mass.
        
        If an object's centre of gravity falls outside of the base of the object, it will topple as a moment (turning force) is introduced.
      - Finding the centre of gravity of a regular or irregular object can be done with this practical
        
        Make small holes along the edges of the object made from care, insert a pin through one of the holes and hold it firmly in a clamp.
        
        Allow the object to swing freely, it will come to rest with its centre of gravity vertically blow the pin
        
        Hang a plumb line from the pin and draw a line along the vertical string of the plumb-line. Repeat this process for other holes. The centre of gravity will be the point of intersection of the lines.
        
        Check if the position of the centre of gravity ismarked correctly now by trying to balance the card on your finger at the point.