# Force

## d1 - 4 Friction:

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

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

• 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$$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:

• 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 $x$$x$ direction, $(\sum\limits_{i}{m_i}) \overline{x} = \sum\limits_{i}{m_i x_i}$$(\sum\limits_{i}{m_i}) \overline{x} = \sum\limits_{i}{m_i x_i}$. The same applies for $y$$y$ and $z$$z$
• $\sum\limits_{i}(m_i)$$\sum\limits_{i}(m_i)$ is the total mass of the system
• $\overline{x}$$\overline{x}$ is the $x$$x$-coordinate of the centre of mass, from the origin
• $\sum\limits_{i}(m_i x_i)$$\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}$$1\ \mathrm{kg}$ mass at $\pmatrix{3 \\ 2}$$\pmatrix{3 \\ 2}$ and a $2 \ \mathrm{kg}$$2 \ \mathrm{kg}$ mass at $\pmatrix{3 \\ 7}$$\pmatrix{3 \\ 7}$
• You can instantly tell that the $x$$x$-coordinate of the centre of mass must be $3$$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}$$(1\ \mathrm{kg} + 2\ \mathrm{kg}) \times \pmatrix{\overline{x} \\ \overline{y}} = 1\ \mathrm{kg} \pmatrix{3 \\ 2} + 2\ \mathrm{kg} \pmatrix{3 \\ 7}$
• We can simplify this to $\pmatrix{3\overline{x} \\ 3\overline{y}} = \pmatrix{3 \\ 2} + \pmatrix{6 \\ 14}$$\pmatrix{3\overline{x} \\ 3\overline{y}} = \pmatrix{3 \\ 2} + \pmatrix{6 \\ 14}$
• So $\overline{x} = \frac{3 + 6}{3} = 3$$\overline{x} = \frac{3 + 6}{3} = 3$ and $\overline{y} = \frac{7 + 2}{3} = 3$$\overline{y} = \frac{7 + 2}{3} = 3$
• Therefore, the centre of mass of the two particles is $\pmatrix{3 \\ 3}$$\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 $(\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}})$$(\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 = Fd$$W = Fd$ where
- $W$$W$ is the work done by a constant force $F$$F$, measured in $\mathrm{J}$$\mathrm{J}$
- $F$$F$ is the force ($\mathrm{N}$$\mathrm{N}$)
- $d$$d$ is the distance moved in the direction of the force ($\mathrm{m}$$\mathrm{m}$)

## w2 Kinetic energy:

• 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$$E_{\mathrm{k}} = \frac{1}{2}mv^2$ where
- $E_{\mathrm{k}}$$E_{\mathrm{k}}$ is the kinetic energy of the object ($\mathrm{J}$$\mathrm{J}$)
- $m$$m$ is the mass of the object ($\mathrm{kg}$$\mathrm{kg}$)
- $v$$v$ is the velocity of the object ($\mathrm{m}\ \mathrm{s}^{-1}$$\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 = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$Fd = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ (where $u$$u$ is the initial velocity and $v$$v$ 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:

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

## w8, w9 Power:

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

# Momentum and impulse

## i1 Impulse:

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

## i2 Momentum:

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

## i3 The impulse-momentum equation:

• The above formulae can be combined to $Ft = mv - mu = m\Delta v$$Ft = 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 $=$$=$ the total momentum after collision
• $\Rightarrow m_A u_A + m_B u_B = m_V v_A + m_B v_B$$\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 $e$$e$
• $e =$$e =$ $\mathrm{speed\ of\ separation} \over \mathrm{speed\ of\ approach}$$\mathrm{speed\ of\ separation} \over \mathrm{speed\ of\ approach}$
• So a perfectly elastic collision will have $e = 1$$e = 1$ and a perfectly inelastic collision will have $e = 0$$e = 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 = 1$$e = 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 $E_\mathrm{k} = \frac{1}{2} mv^2$$E_\mathrm{k} = \frac{1}{2} mv^2$ to calculate the loss of kinetic energy in an impact (along with the coefficient of restitution)