Forces In Equilibrium

Vectors and Scalars (7.1)
1. Vectors
  1. Vector Quantities have magnitude and direction
  2. Examples of Vectors:
    1. Force
    2. Displacement
    3. Velocity
    4. Momentem
2. Scalars
  1. Examples of scalars:
    1. Mass
    2. Tempreture
    3. Length
    4. Speed
  2. Scalar Quantities only have magnitude
3. Vertical and Horizontal Componets
  2. How to find the Horizontal Component of a force:
  3. How to find the Vertical Component of a force:
  4. If no right angle present, draw scale diagram
4. Resultant Forces
  2. Using Pythagoras
  3. Using Trigonometry
  4. Resolving force to perpendicular forces
Balanced Forces (7.2)
1. 2 Forces
  1. When a object (e.g. block) is placed on a surface (e.g. floor), the support force from the floor is equal to the weight, force on block due to gravity
2. 3 Forces
  1. When an object (e.g. block) is placed on a slope, 3 forces are present.
    1. To resolve the three forces parallel and perpendicular to the slope:
      1. Horizontal: F = W sin(θ)
      2. Vertical: S = W cos(θ)
      3. W = √F +√S
  2. When a suspended weight is being supported by two strings
    2. To Resolve T1, T2 using:
      1. Horizontal
        T1 x Sin(θ1) = T2 x Sin(θ2)
      2. Vertical
        T1 x Cos(θ1) + T2 x Cos(θ2) = W
The Principle Of Moments (7.3)
1. Turning Effects
  1. To calculate the moment of the force
    1. Moment = Force x distance
      1. Moment = Fd
  3. The greater the distance (d), The greater the moment
2. The Principle Of Moments
  1. =
    1. Sum of the clockwise moments
      1. Sum of the anticlockwise moments
  2. W1 provides an anti-clockwise moment from the pivot
    1. W2 provides a clockwise moment from the pivot
    2. For equilibrium applying the priciple of moments
      1. W1 x d1 = W2 x d2
3. Centre of Mass
  1. The centre of mass of a body is the point through which a single force on the body has no turning effect
    1. Centre of mass test can be done by doing the experiment in the picture above
4. Calculating the weight of a meter ruler
  1. weight W1 provides the anti-clockwise moment
    1. The weight of the ruler W0 (unknown) provides the clockwise moment (located at the C.O.M of the uniform ruler)
      1. All of this is calculated from the pivot
  2. Therefore applying the principle of moments
    1. W1 x d1 = W0 x d0
More on Moments (7.4)
1. Single Support Forces
  1. When an object in equilibrium is supported at one point only, the support force on the object is equal and opposite to the total downwards force
    1. E.g. with a uniform rule balanced on a support, Support force S, Weight force W
      1. S = W1 + W2 + W0
        where W0 is the weight of the rule
        This is taken from the support force and as the distance equals 0 from the pivot, the force has a Zero moment
  2. Fig 1
2. Two Support Forces
  1. consider a uniform beam support on two pillars X & Y at distance D apart
    1. The weight of the beam is shared between two pillars
  2. If the C.O.M of the beam is midway between the pillars, the weight of the beam is shared equally between the two pillars
    1. If the centre of mass of the beam is at dx distance from pillar X and distance dy from pillar Y, then taking moments about X,
      2. In addition
        If the centre of mass of the beam is at dy distance from pillar Y and distance dx from pillar X, then taking moments about X,
    2. Fig 2
3. Couples
  1. A couple is a pair of equal and opposite forces acting on a body, but not along the same line
    1. The moment of a couple is sometimes referred as torque
      1. The moment/torque of a couple = Force x Perpendicular distance (moment = Fd)
        As seen below the total moment is
        Fx + F(d-x) = Fx +Fd - Fx = Fd
Stability (7.5)
1. Stable & Unstable Equilibrium
  1. Stable
    1. If a body in stable equilibrium is displaced then released, it resturns to its equilibrium position
      1. This is because the C.O.M is directly below the point of suppoert
        Figure 1
  2. Unstable
    1. If a body in unstable equilibrium is displaced then released, it does not resturn to its equilibrium position
      1. This is because the C.O.M is not directly below the point of suppoert but above so when diplaced the the C.O.M is not above the support point therefore acting against the equilibrium forces
        Figure 2
2. Tilting and Toppling
  1. Tilting
    1. This is when an object at rest is acted by a force raising it up a side
      1. Figure 3
        The entire support from the floor acts at pount P
        The clockwise moment of F about P = Fd
        where d is the perpendicular distance from the line of action
        The anti-clockwise moment of W about P = Wb/2
        where b is the width of the base
        Therefor, for tilting to occur Fd > Wb/2
  2. toppling
    1. A tilted object will topple over if it is tilted too far. This happens if the line of action passes closer to the pivot
3. On a Slope
  1. A tall object will topple over if the slope is too great
    1. This will happen if the line of action of the weight passing through the centre of mass lies outside the wheelbase of the vehicle (as seen in figure 5)
      1. Fig 5
        as the line of action is inside the wheelbase, the vehicle will not topple over
        For equilibrium, resolving forces parallel and perpendicular on the slop gives
        Parallel
        F = Wsin(θ)
        Perpendicular
        Sx + Sy = Wcos(θ)
        Note that Sx is greater tha Sy because X is lower than Y
  2. consider the forces acting on the vehicle on a slope at rest. the sideways friction F, support weights S, and gravity of vehicle act as shown in figure 5.
    1. Fig 5

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