Mechanics 2 (4762)


      d1 - 4 Friction:

      • An object moving over a surface will have at least three forces acting on it; weight, normal reaction force and friction
      • μ\mu is the coefficient of friction. It has no units
      • If an object is on a surface with coefficient of friction μ\mu and being pulled by a force FF,
        - if it's still stationary, FμRF \leq \mu R
        - if it's on the point of sliding, F=μRF = \mu R
        - if it's accelerating, F>μRF \gt \mu R
      • You can use Newton's laws to help solve these problems. If there are multiple tension forces acting on an object, remember to create simultaneous equations for each system

      d7 - 8 Moments:

      • moment=Fd\mathrm{moment} = Fd where FF is a force and dd is the distance between the point you are taking the moment about and dd. FF must be acting perpendicular
      • Therefore, the units of a moment are N m\mathrm{N}\ \mathrm{m}
      • To check that an object is in equilibrium, you need to confirm that
        - the resultant force is 00
        - the resultant moment about any point is 00
           - (only one point needs to be checked)
      • If a direction of force is not perpendicular to the object it's acting on, you'll need to resolve it in the perpendicular direction

      d9 Sliding and toppling:

      • Draw a vertical line from the centre of mass of the object. If it leaves the shape before reaching the edge in contact with the surface, the object will topple
      • If this is not the case, then the object will slide if FμRF \geq \mu R
      • So to find if an object will topple or slide first, find the minimum angle for each to occur. The one with the smallest angle will be what would actually happen

      d10 Light frameworks:

      • It's best to assume that rods are under tension. If it is actually a thrust, you'll get a negative value
      • First take moments about a suitable point (to exclude unnecessary forces)
      • Then find internal forces by considering the equilibria at joints

      d11 Tension and compression:

      • Tension is a pulling force
      • Compression is a pushing force

      Centre of mass

        G1 - 2 Centre of mass of a system of particles:

        • In the xx direction, (imi)x¯=imixi(\sum\limits_{i}{m_i}) \overline{x} = \sum\limits_{i}{m_i x_i}. The same applies for yy and zz
        • i(mi)\sum\limits_{i}(m_i) is the total mass of the system
        • x¯\overline{x} is the xx-coordinate of the centre of mass, from the origin
        • i(mixi)\sum\limits_{i}(m_i x_i) is the sum of masses multiplied by their distance from the origin


        • We have a 1 kg1\ \mathrm{kg} mass at (32)\pmatrix{3 \\ 2} and a 2 kg2 \ \mathrm{kg} mass at (37)\pmatrix{3 \\ 7}
        • You can instantly tell that the xx-coordinate of the centre of mass must be 33, but we'll work it out anyway
        • Using the above equation, (1 kg+2 kg)×(x¯y¯)=1 kg(32)+2 kg(37)(1\ \mathrm{kg} + 2\ \mathrm{kg}) \times \pmatrix{\overline{x} \\ \overline{y}} = 1\ \mathrm{kg} \pmatrix{3 \\ 2} + 2\ \mathrm{kg} \pmatrix{3 \\ 7}
        • We can simplify this to (3x¯3y¯)=(32)+(614)\pmatrix{3\overline{x} \\ 3\overline{y}} = \pmatrix{3 \\ 2} + \pmatrix{6 \\ 14}
        • So x¯=3+63=3\overline{x} = \frac{3 + 6}{3} = 3 and y¯=7+23=3\overline{y} = \frac{7 + 2}{3} = 3
        • Therefore, the centre of mass of the two particles is (33)\pmatrix{3 \\ 3} from the origin

        G3 Centre of mass of a composite body:

        • Above, we dealt with particles only. However, the same process can be used to find the centre of mass of any shape (a composite body is something made up from smaller things)
        • Treat each piece of the composite body as a particle. It's position vector can be treated as its centre of mass
        • So you just need to divide the body or shape into smaller pieces with known centres of mass and mass. Then use the above equation and method
        • If you need to know the centre of mass of a shape, they are given in the formula booklet on page 5, or page 67 of the textbook
        • You can also use subtraction in the centre of mass equation. For example, if you are finding the centre of mass of a cube with a small piece removed, use (imi)x¯=(mcube×centre of masscube)(mremoved piece×centre of massremoved piece)(\sum\limits_{i}{m_i}) \overline{x} = (m_{cube} \times \mathrm{centre\ of\ mass}_{\mathrm{cube}}) - (m_{\mathrm{removed\ piece}} \times \mathrm{centre\ o f\ mass}_{\mathrm{removed\ piece}})

        Work, energy and power:

        w1 Work done:

        • If a force is used to move an object, it has done work on the object
        • Work is a scalar quantity
        • W=FdW = Fd where
          - WW is the work done by a constant force FF, measured in J\mathrm{J}
          - FF is the force (N\mathrm{N})
          - dd is the distance moved in the direction of the force (m\mathrm{m})

        w2 Kinetic energy:

        • Kinetic energy is the energy of a body due to its movement
        • It is a vector quantity
        • Ek=12mv2E_{\mathrm{k}} = \frac{1}{2}mv^2 where
          - EkE_{\mathrm{k}} is the kinetic energy of the object (J\mathrm{J})
          - mm is the mass of the object (kg\mathrm{kg})
          - vv is the velocity of the object (m s1\mathrm{m}\ \mathrm{s}^{-1})

        w3 Mechanical energy:

        • There are two types of mechanical energy, kinetic energy and potential energy
        • The total mechanical energy is always conserved if only gravity does work (i.e. for a falling object, kinetic energy will increase at the same rate of the decrease of gravitational potential energy)

        w4 The work-energy principle:

        • The total work done by all forces acting on a body is equal to the increase in kinetic energy
        • So for an accelerating/decelerating object, it's often useful to use the relationship Fd=12mv212mu2Fd = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 (where uu is the initial velocity and vv is the final velocity)

        w5 Conservative and dissipative forces:

        • A conservative force is a force where mechanical energy is conserved - e.g. gravitational forces
        • A dissipative force is one which dissipates (i.e. converts) mechanical energy - e.g. friction and drag forces

        w6, w7 Gravitational potential energy:

        • ΔGPE=mgh\Delta \mathrm{GPE} = mgh where
          - ΔGPE\Delta \mathrm{GPE} (a.k.a ΔPE\Delta \mathrm{PE} or ΔEgrav\Delta E_{\mathrm{grav}}) is the change in gravitational potential energy in J\mathrm{J}
          - mm is the mass of the object in kg\mathrm{kg}
          - gg is the gravitational constant (9.8 N kg19.8\ \mathrm{N}\ \mathrm{kg}^{-1})
          - hh is the change in height in m\mathrm{m}

        w8, w9 Power:

        • Power is the rate of work being done (i.e. the work done per second)
        • Therefore, P=P = Wt\frac{W}{t}
        • Power is measured in watts (W\mathrm{W})
        • You can also use P=FvP = Fv to calculate power from force and velocity

        Momentum and impulse

          i1 Impulse:

          • Impulse is the change in momentum over a period of time
          • impulse=F×t\mathrm{impulse} = F \times t and impulse=mvmu\mathrm{impulse} = mv - mu
          • It has the units N s\mathrm{N}\ \mathrm{s} or kg m s1\mathrm{kg}\ \mathrm{m}\ \mathrm{s}^{-1}

          i2 Momentum:

          • The symbol for momentum is pp
          • p=mvp = mv
          • Its units are the same as impulse; N s\mathrm{N}\ \mathrm{s} or kg m s1\mathrm{kg}\ \mathrm{m}\ \mathrm{s}^{-1}

          i3 The impulse-momentum equation:

          • The above formulae can be combined to Ft=mvmu=mΔvFt = mv - mu = m\Delta v

          i4 Conservation of momentum:

          • Without external forces, momentum is always conserved in a system

          i5, i6 The conservation of momentum equation for collisions between two particles:

          • The total momentum before a collision == the total momentum after collision
          • mAuA+mBuB=mVvA+mBvB\Rightarrow m_A u_A + m_B u_B = m_V v_A + m_B v_B

          i7, i8 Newton's experimental law:

          • The coefficient of restitution is a constant (for two particular surfaces) between 0 and 1. It has the symbol ee
          • e=e = speed of separationspeed of approach\mathrm{speed\ of\ separation} \over \mathrm{speed\ of\ approach}
          • So a perfectly elastic collision will have e=1e = 1 and a perfectly inelastic collision will have e=0e = 0
          • This can be combined with the conservation of momentum equation with simultaneous equations to solve some problems

          i9 Mechanical energy in impacts:

          • In impacts, momentum is always conserved, but mechanical energy is not (unless e=1e = 1, which is impossible in practice) - some is transferred to other types of energy

          i10 Oblique impacts - vertical component:

          • When an object hits a smooth plane, an impulse acts perpendicular to the plane

          i11 Oblique impacts - horizontal component:

          • When an object bounces from a smooth plane, its horizontal velocity remains constant

          i12 Oblique impacts - velocity direction and magnitude:

          • The direction is reversed (horizontally mirrored)
          • The magnitude of the perpendicular component of the velocity is multiplied by the coefficient of restitution

          i13 Oblique impacts - loss of kinetic energy:

          • Use Ek=12mv2E_\mathrm{k} = \frac{1}{2} mv^2 to calculate the loss of kinetic energy in an impact (along with the coefficient of restitution)