Mechanics 2 (4762)

Force

d1 - 4 Friction:

Test yourself
• An object moving over a surface will have at least three forces acting on it; weight, normal reaction force and friction
• $\mu$ is the coefficient of friction. It has no units
• If an object is on a surface with coefficient of friction $\mu$ and being pulled by a force $F$,
  - if it's still stationary, $F \leq \mu R$
  - if it's on the point of sliding, $F = \mu R$
  - if it's accelerating, $F \gt \mu R$
• 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:

Test yourself
• $\mathrm{moment} = Fd$ where $F$ is a force and $d$ is the distance between the point you are taking the moment about and $d$. $F$ must be acting perpendicular
• Therefore, the units of a moment are $\mathrm{N}\ \mathrm{m}$
• To check that an object is in equilibrium, you need to confirm that
  - the resultant force is $0$
  - the resultant moment about any point is $0$
     - (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:

Test yourself
• 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 $F \geq \mu R$
• 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:

Test yourself
• 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:

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

Centre of mass

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

Test yourself
• In the $x$ direction, $(\sum\limits_{i}{m_i}) \overline{x} = \sum\limits_{i}{m_i x_i}$. The same applies for $y$ and $z$
• $\sum\limits_{i}(m_i)$ is the total mass of the system
• $\overline{x}$ is the $x$-coordinate of the centre of mass, from the origin
• $\sum\limits_{i}(m_i x_i)$ is the sum of masses multiplied by their distance from the origin

Example:

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

G3 Centre of mass of a composite body:

Test yourself
• 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 $(\sum\limits_{i}{m_i}) \overline{x} = (m_{cube} \times \mathrm{centre\ of\ mass}_{\mathrm{cube}}) - (m_{\mathrm{removed\ piece}} \times \mathrm{centre\ o f\ mass}_{\mathrm{removed\ piece}})$

Work, energy and power:

w1 Work done:

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

w2 Kinetic energy:

Test yourself
• Kinetic energy is the energy of a body due to its movement
• It is a vector quantity
• $E_{\mathrm{k}} = \frac{1}{2}mv^2$ where
  - $E_{\mathrm{k}}$ is the kinetic energy of the object ($\mathrm{J}$)
  - $m$ is the mass of the object ($\mathrm{kg}$)
  - $v$ is the velocity of the object ($\mathrm{m}\ \mathrm{s}^{-1}$)

w3 Mechanical energy:

Test yourself
• 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:

Test yourself
• 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 $Fd = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ (where $u$ is the initial velocity and $v$ is the final velocity)

w5 Conservative and dissipative forces:

Test yourself
• 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:

Test yourself
• $\Delta \mathrm{GPE} = mgh$ where
  - $\Delta \mathrm{GPE}$ (a.k.a $\Delta \mathrm{PE}$ or $\Delta E_{\mathrm{grav}}$) is the change in gravitational potential energy in $\mathrm{J}$
  - $m$ is the mass of the object in $\mathrm{kg}$
  - $g$ is the gravitational constant ($9.8\ \mathrm{N}\ \mathrm{kg}^{-1}$)
  - $h$ is the change in height in $\mathrm{m}$

w8, w9 Power:

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

Momentum and impulse

i1 Impulse:

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

i2 Momentum:

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

i3 The impulse-momentum equation:

Test yourself
• The above formulae can be combined to $Ft = mv - mu = m\Delta v$

i4 Conservation of momentum:

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

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

Test yourself
• The total momentum before a collision $=$ the total momentum after collision
• $\Rightarrow m_A u_A + m_B u_B = m_V v_A + m_B v_B$

i7, i8 Newton's experimental law:

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

i9 Mechanical energy in impacts:

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

i10 Oblique impacts - vertical component:

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

i11 Oblique impacts - horizontal component:

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

i12 Oblique impacts - velocity direction and magnitude:

Test yourself
• 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:

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