# Mechanics 2 (4762)

# 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

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**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

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**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

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**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

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**d11** Tension and compression:

**Tension**is a pulling force**Compression**is a pushing force

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# 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

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## 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}})

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# 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} )

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**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} )

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**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)

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**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)

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**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

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**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}

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**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

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# 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}

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**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}

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**i3** The impulse-momentum equation:

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

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**i4** Conservation of momentum:

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

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**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

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**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

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**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

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**i10** Oblique impacts - vertical component:

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

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**i11** Oblique impacts - horizontal component:

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

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**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

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**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)

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