Further mechanics: A-level Maths OCR revision notes

3.02

Projectiles

Modelling a projectile and resolving the initial velocity 3.02

Definitions

Projectile: a body moving freely under gravity alone, modelled as a particle with air resistance neglected and $g = 9.8 m s^{- 2}$ taken as constant.

Key results

A launch speed $u$ at angle $θ$ above the horizontal resolves into $u_{x} = u cos θ$ and $u_{y} = u sin θ$ .
Horizontally no force is assumed to act, so $a_{x} = 0$ , the horizontal velocity stays at $u_{x}$ for the whole flight, and the horizontal displacement is simply $x = u_{x} t$ .
Vertically the acceleration is $a_{y} = - g$ with upward positive, and every suvat equation applies to the vertical components alone.

Notes

Horizontal and vertical motion share only the time $t$ ; finding $t$ from one direction and substituting it into the other is the engine of nearly every projectile question.
A body projected horizontally is the case $θ = 0$ , so $u_{y} = 0$ and its vertical motion is identical to that of a body simply dropped from rest at the same height.
The modelling assumptions matter: neglecting air resistance overestimates range and landing speed for a light or fast-moving object such as a shuttlecock.
Resolve the launch velocity into components as the very first step of any projectile question — trying to work with the launch speed and angle directly, without splitting them, is the most common source of a stuck solution.

Maximum height and time of flight 3.02

Key results

At the greatest height the vertical component of velocity is zero; the horizontal component is unchanged, so the speed there is $u_{x}$ , not zero.
Time to the greatest height: $t = \frac{u _{y}}{g}$ .
Greatest height above the point of projection: $h_{m a x} = \frac{u _{y}^{2}}{2 g}$ .
Over level ground the vertical motion is symmetric, so the total time of flight is $T = \frac{2 u _{y}}{g}$ and the landing speed equals the launch speed.

Method

Resolve the launch velocity into $u_{x} = u cos θ$ and $u_{y} = u sin θ$ .
Work entirely in the vertical direction, with $a = - g$ , to find the time to the top or the total time of flight.
Use $v_{y}^{2} = u_{y}^{2} - 2 g h$ with $v_{y} = 0$ for the greatest height.
Return to the horizontal direction, where the velocity is constant, only when a horizontal distance is required.

Notes

$h_{m a x}$ is measured from the point of projection; for a projectile launched from a height $h_{0}$ above the ground, the greatest height above the ground is $h_{0} + \frac{u _{y}^{2}}{2 g}$ .
Time to greatest height and total time of flight are related but different quantities — over level ground the second is exactly double the first, but that relationship breaks as soon as launch and landing heights differ.

Range and the equation of the trajectory 3.02

Key results

Range over level ground: $R = u_{x} T = \frac{2 u ^{2} sin θ cos θ}{g} = \frac{u ^{2} sin 2 θ}{g}$ .
Since $sin 2 θ$ is greatest when $2 θ = 90°$ , the maximum range over level ground occurs at $θ = 45°$ and equals $\frac{u ^{2}}{g}$ .
Any range short of that maximum is achieved by two complementary launch angles, $θ$ and $90° - θ$ ; for instance $30°$ and $60°$ give the same range from the same speed.
Eliminating $t$ between $x = u t cos θ$ and $y = u t sin θ - \frac{1}{2} g t^{2}$ gives the trajectory $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}$ , showing the path is a parabola.

Notes

The trajectory equation is the fastest route to 'does the projectile clear the obstacle', since it gives the height directly from the horizontal distance without ever finding the time.
The range formula $\frac{u ^{2} sin 2 θ}{g}$ is valid only for launch and landing at the same height; from a cliff or a raised platform it does not apply and the flight time must be found from a quadratic instead.
The two complementary angles $θ$ and $90° - θ$ giving equal range have very different flight profiles — one is a low, fast, flat trajectory and the other a high, slow, lofted one — worth sketching to distinguish them in context.

Projection from a height 3.02

Key results

The landing speed comes from combining components: $v_{x} = u_{x}$ and $v_{y} = u_{y} - g t$ , so $v = v_{x}^{2} + v_{y}^{2}$ , with the angle below the horizontal equal to $arctan (\frac{∣ v _{y} ∣}{v _{x}})$ .
The vertical landing speed can also be found without the time from $v_{y}^{2} = u_{y}^{2} + 2 g h_{0}$ .

Method

Take upward as positive with the origin at the point of projection, so ground level is at $y = - h_{0}$ .
Write the vertical equation $- h_{0} = u_{y} t - \frac{1}{2} g t^{2}$ and rearrange to $\frac{1}{2} g t^{2} - u_{y} t - h_{0} = 0$ .
Solve this quadratic for $t$ and discard the negative root.
Substitute the positive root into $x = u_{x} t$ to find the horizontal distance travelled.

Notes

The shortcuts $T = \frac{2 u _{y}}{g}$ and 'landing speed equals launch speed' hold only for level ground; from a height the flight lasts longer and the landing speed is greater.
The discriminant of the flight-time quadratic is always positive for $h_{0} > 0$ , which simply reflects the fact that a projectile launched above the ground must eventually reach it.
Setting up the origin and sign convention explicitly at the start (upward positive, ground at $y = - h_{0}$ ) prevents the sign errors that a projection-from-a-height question is specifically designed to test.

Worked examples

Worked example

A stone is thrown horizontally at $15 m s^{- 1}$ from the top of a cliff $45 m$ above the sea.

Taking $g = 9.8 m s^{- 2}$ , find the time before it hits the sea, the horizontal distance travelled, and the speed and direction with which it enters the water.

Show worked solution

Horizontally, $u_{x} = 15 m s^{- 1}$ throughout; vertically, $u_{y} = 0$ and the stone falls $45 m$ .

Using:

s = \frac{1}{2} g t^{2}

downward: $45 = 4.9 t^{2}$ , so:

t^{2} = \frac{45}{4.9} = 9.1837

and $t = 3.03 s$ (3 s.f.).

Horizontal distance:

x = u_{x} t = 15 \times 3.0305 = 45.5 m

(3 s.f.).

Vertical speed on entry, from:

v_{y}^{2} = 0 + 2 (9.8) (45) = 882

v_{y} = 29.70 m s^{- 1}

Speed:

v = 1 5^{2} + 882 = 1107 = 33.3 m s^{- 1}

(3 s.f.).

Direction:

arctan (\frac{29.70}{15}) = 63.2°

below the horizontal (3 s.f.).

Worked example

A ball is projected at $25 m s^{- 1}$ at $30°$ above the horizontal from level ground.

Taking $g = 9.8 m s^{- 2}$ , find the maximum height, the time of flight, and the horizontal range.

Show worked solution

u_{x} = 25 cos 30° = 21.65 m s^{- 1}

u_{y} = 25 sin 30° = 12.5 m s^{- 1}

Maximum height:

h = \frac{u _{y}^{2}}{2 g} = \frac{12. 5 ^{2}}{2 ( 9.8 )} = \frac{156.25}{19.6} = 7.97 m

(3 s.f.).

Time of flight:

T = \frac{2 u _{y}}{g} = \frac{25}{9.8} = 2.55 s

(3 s.f.).

Range:

R = u_{x} \times T = 21.65 \times 2.551 = 55.2 m

(3 s.f.).

Worked example

A projectile is launched at $30 m s^{- 1}$ at $45°$ above the horizontal from level ground.

A vertical wall of height $25 m$ stands $50 m$ away.

Determine whether the projectile clears the wall.

Show worked solution

Use the trajectory equation:

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}

with $θ = 45°$ , so $tan 45° = 1$ and:

cos^{2} 45° = \frac{1}{2}

The denominator is:

2 (3 0^{2}) (\frac{1}{2}) = 900

At $x = 50 m$ :

y = 50 - \frac{9.8 ( 5 0 ^{2} )}{900} = 50 - \frac{24500}{900} = 50 - 27.22 = 22.8 m

(3 s.f.).

Since:

22.8 m < 25 m

the projectile does not clear the wall; it strikes it about $2.2 m$ below the top. (As a check on the scale of the answer, the full range at $45°$ would be:

\frac{u ^{2}}{g} = \frac{900}{9.8} = 91.8 m

so at $x = 50 m$ the projectile is just past its highest point and descending, consistent with $y$ being below the peak height of $22.96 m$ .)

Worked example

A ball is projected at $20 m s^{- 1}$ at $30°$ above the horizontal from a point $2 m$ above level ground.

Taking $g = 9.8 m s^{- 2}$ , find the greatest height above the ground, the time of flight, and the horizontal distance travelled before it lands.

Show worked solution

u_{x} = 20 cos 30° = 17.32 m s^{- 1}

u_{y} = 20 sin 30° = 10 m s^{- 1}

Greatest height above the launch point:

\frac{u _{y}^{2}}{2 g} = \frac{100}{19.6} = 5.102 m

so above the ground the greatest height is:

2 + 5.102 = 7.10 m

(3 s.f.).

For the flight time, take upward as positive with the origin at the launch point, so the ground is at $y = - 2 m$ :

- 2 = 10 t - 4.9 t^{2}

giving:

4.9 t^{2} - 10 t - 2 = 0

Then:

t = \frac{10 \pm 100 + 4 ( 4.9 ) ( 2 )}{9.8} = \frac{10 \pm 139.2}{9.8} = \frac{10 \pm 11.798}{9.8}

The negative root is rejected, leaving:

t = \frac{21.798}{9.8} = 2.22 s

(3 s.f.).

Horizontal distance:

x = 17.32 \times 2.2243 = 38.5 m

(3 s.f.).

Note this exceeds the $35.4 m$ that the level-ground formula:

\frac{u ^{2} sin 2 θ}{g} = \frac{400 sin 60°}{9.8}

would give, as expected for a launch from above ground level.

3.04

Moments

Moments of a force 3.04

Definitions

Moment of a force about a point: $M = F d$ , the product of the magnitude of the force and the perpendicular distance from the point to the force's line of action, measured in newton metres ( $N m$ ).

Key results

A moment measures turning effect and carries a sense, clockwise or anticlockwise, which must either be stated or fixed by a sign convention.
A force $F$ acting at distance $L$ along a rod, at angle $θ$ to the rod, has moment $F L sin θ$ about the end, since $L sin θ$ is the perpendicular distance to its line of action.
A force whose line of action passes through the chosen point has zero moment about that point.

Notes

The weight of a uniform rod or beam acts at its midpoint; for a non-uniform body the weight acts at the centre of mass, whose position is often the unknown being sought.
The perpendicular distance, not the distance along the rod, is what enters $M = F d$ — for a force at an angle these differ, and confusing them is a standard error.
State a consistent sense (clockwise or anticlockwise) as positive at the start, exactly as with a sign convention for displacement — mixing conventions partway through a moments equation is a common source of error.

Equilibrium of a rigid body 3.04

Key results

A rigid body is in equilibrium only when both conditions hold: the resultant force is zero, and the resultant moment about any point is zero.
Balanced forces alone are not sufficient — two equal and opposite forces with different lines of action form a couple, producing rotation with zero resultant force.
If the resultant force is zero and the moments balance about one point, they balance about every point, so the pivot for taking moments may be chosen freely.
Moment of a force about a point: $F \times d$ , where $d$ is the perpendicular distance from the point to the line of action (unit N m).
A force $F$ applied at distance $ℓ$ along a rod, at angle $θ$ to the rod, has moment $F ℓ sin θ$ about the end.
Conditions for equilibrium: $\sum F_{x} = 0$ , $\sum F_{y} = 0$ and $\sum M_{P} = 0$ about any point $P$ .

Method

Draw the body with every force marked at its point of application, together with all the relevant distances.
Resolve and set the total force to zero in two perpendicular directions.
Take moments about a point through which an unknown force acts, so that unknown drops out of the moment equation.
Write sum of clockwise moments equals sum of anticlockwise moments, solve for the remaining unknown, and verify by taking moments about a different point.

Notes

Choosing the pivot well is the single biggest labour saver in this topic: a good choice can reduce a pair of simultaneous equations to a single line of arithmetic.
A body can satisfy 'resultant force zero' while still rotating under a couple — always check both conditions rather than assuming zero net force alone guarantees equilibrium.

Supports, reactions and tilting 3.04

Key results

A rod resting on two supports has a vertical reaction at each; one force equation and one moment equation are enough to determine both.
As a load moves towards one support, the reaction at the far support decreases; the body is on the point of tilting when that reaction reaches zero.
At the point of tilting the body is effectively pivoted at the remaining support, so taking moments about that support gives the critical position immediately.
Example: a uniform 6 m beam $A B$ of mass 20 kg rests on supports $C$ and $D$ , 1 m from $A$ and 2 m from $B$ , with 5 kg at $B$ . Moments about $C$ : $3 R_{D} = 20 g \times 2 + 5 g \times 5$ , so $R_{D} = \frac{65 g}{3} = 212$ N; resolving, $R_{C} = 25 g - R_{D} = 32.7$ N.
On the point of tilting about $D$ , $R_{C} = 0$ : $20 g \times 1 = m g \times 2$ , so the greatest load at $B$ is $m = 10$ kg.

Notes

A reaction that comes out negative means the support would have to pull down rather than push up, which a simple support cannot do — physically, the body has already tilted.
For a non-uniform rod, the unknowns are typically the weight and the distance of the centre of mass from one end; two moment equations about different points, or one moment equation plus resolving vertically, determine both.
Taking moments about the support whose reaction is unknown or about to vanish (the 'about to tip' support) is usually the fastest route to the tilting condition, since it removes one unknown from the equation immediately.

Worked examples

Worked example

A uniform beam of length $4 m$ and weight $60 N$ is pivoted at one end and held horizontal by a vertical force $F$ applied at the other end.

Find $F$ .

Show worked solution

The beam's weight acts at its midpoint, $2 m$ from the pivot.

Taking moments about the pivot: the clockwise moment of the weight is:

60 \times 2 = 120 N m

balanced by the anticlockwise moment of $F$ at the far end, $4 m$ from the pivot: $F \times 4 = 120$ , so $F = 30 N$ .

Worked example

A uniform plank $A B$ of length $6 m$ and weight $200 N$ rests horizontally on two supports, one $1 m$ from $A$ and the other $1 m$ from $B$ .

A person of weight $600 N$ stands $2 m$ from $A$ .

Find the reaction at each support.

Show worked solution

Measure distances $x$ from $A$ : the supports are at $x = 1$ and $x = 5$ , the plank's weight acts at its midpoint $x = 3$ , and the person is at $x = 2$ .

Let the reactions be $R_{1}$ at $x = 1$ and $R_{2}$ at $x = 5$ .

Resolving vertically:

R_{1} + R_{2} = 200 + 600 = 800 N

Taking moments about the first support (which eliminates $R_{1}$ ), clockwise moments are the plank's weight at distance $2 m$ and the person at distance $1 m$ , and the anticlockwise moment is $R_{2}$ at distance $4 m$ :

4 R_{2} = 200 (2) + 600 (1) = 400 + 600 = 1000 N m

so $R_{2} = 250 N$ and hence:

R_{1} = 800 - 250 = 550 N

Check by taking moments about the second support:

4 R_{1} = 200 (2) + 600 (3) = 400 + 1800 = 2200 N m

giving $R_{1} = 550 N$ as before.

The nearer support carries the greater share, as expected.

3.03

Friction

Friction and the coefficient of friction 3.03

Definitions

Friction: a force resisting relative sliding between two surfaces in contact, acting parallel to the surfaces and opposing the direction of actual or impending relative motion.
Coefficient of friction $μ$ : the dimensionless constant for a given pair of surfaces relating the maximum available friction to the normal reaction.

Key results

At all times $F \leq μ R$ , where $R$ is the normal reaction; the body remains stationary as long as the friction required for equilibrium does not exceed $μ R$ .
Limiting equilibrium is the borderline case in which the body is on the point of sliding and $F = μ R$ exactly.
Once sliding, friction acts at its maximum value $F = μ R$ , directed opposite to the motion.
A smooth surface is simply the case $μ = 0$ .

Notes

The most common error in the topic is writing $F = μ R$ for a body safely at rest; short of limiting equilibrium, friction takes only whatever smaller value equilibrium demands.
The normal reaction is not always $m g$ : any applied force with a vertical component changes $R$ , and on a slope $R$ involves $cos θ$ , so $R$ must be found by resolving rather than assumed.
$μ$ is a property of the pair of surfaces in contact, not of either object alone — the same block has a different $μ$ against ice than against carpet.

Limiting equilibrium and motion with friction 3.03

Key results

A body on a rough plane inclined at $θ$ is on the point of sliding when $m g sin θ = μ m g cos θ$ , that is when $tan θ = μ$ ; this critical angle is called the angle of friction.
The mass cancels in that condition, so whether a body slips on a given slope depends only on $θ$ and $μ$ , not on how heavy it is.
While sliding down a rough slope, $ma = m g sin θ - μ m g cos θ$ , so $a = g (sin θ - μ cos θ)$ .
Pulling with tension $T$ at angle $α$ above the horizontal reduces the normal reaction to $R = m g - T sin α$ , so it both supplies a horizontal component $T cos α$ and reduces the friction opposing it.

Method

Assume the body is in equilibrium and find the friction force that would be required to maintain it.
Resolve perpendicular to the surface to find $R$ , and hence the maximum available friction $μ R$ .
Compare the two: if the required friction is at most $μ R$ the body stays at rest with friction equal to the required value; if it exceeds $μ R$ the body slides.
If it slides, apply $F = ma$ along the surface with friction fixed at $μ R$ and directed opposite to the motion.

Notes

Because friction opposes relative motion, its direction reverses when the motion reverses — a body projected up a rough slope decelerates faster going up ( $g sin θ + μg cos θ$ ) than it accelerates coming back down ( $g sin θ - μg cos θ$ ).
Always check first whether the body actually moves at all — a body can be in limiting equilibrium at a particular angle and yet remain perfectly at rest at a shallower one, since $tan θ < μ$ there.

Worked examples

Worked example

A block of mass $4 kg$ is placed on a rough plane inclined at $20°$ to the horizontal, with coefficient of friction $μ = 0.3$ between block and plane.

Taking $g = 9.8 m s^{- 2}$ , determine whether the block remains at rest, and find its acceleration if it does not.

Show worked solution

Weight:

= 4 (9.8) = 39.2 N

Resolving perpendicular to the plane:

R = 39.2 cos 20° = 39.2 (0.93969) = 36.84 N

so the maximum available friction is:

μ R = 0.3 (36.84) = 11.05 N

Resolving parallel to the plane, the component of weight down the slope is:

39.2 sin 20° = 39.2 (0.34202) = 13.41 N

Since:

13.41 N > 11.05 N

the friction required for equilibrium exceeds the maximum available, so the block slides down.

Applying $F = ma$ down the slope with friction at its maximum:

4 a = 13.41 - 11.05 = 2.356 N

so:

a = 0.589 m s^{- 2}

(3 s.f.), directed down the slope.

This agrees with the criterion $tan θ > μ$ : the critical angle is:

arctan (0.3) = 16.7°

and:

20° > 16.7°

Worked example

A box of mass $20 kg$ rests on rough horizontal ground with $μ = 0.25$ .

It is pulled by a rope inclined at $30°$ above the horizontal.

Taking $g = 9.8 m s^{- 2}$ , find the tension for which the box is on the point of moving.

Show worked solution

Let the tension be $T$ .

Resolving vertically, the rope's upward component reduces the normal reaction:

R + T sin 30° = 20 (9.8) = 196

so:

R = 196 - 0.5 T

On the point of moving, friction is limiting, $F = μ R$ , and the horizontal forces balance:

T cos 30° = 0.25 (196 - 0.5 T)

So:

0.86603 T = 49 - 0.125 T

giving:

0.99103 T = 49

and $T = 49.4 N$ (3 s.f.).

Check: with $T = 49.444 N$ ,

R = 196 - 24.72 = 171.3 N

so $μ R = 42.82 N$ , and:

T cos 30° = 42.82 N

which balances exactly.

Note that pulling at an angle needs less tension than pulling horizontally, where $T$ would have to reach:

μ m g = 0.25 (196) = 49 N

against an undiminished reaction of $196 N$ — here the vertical component lightens the contact and so lowers the friction to be overcome.