The Resultant Decides Everything
A car of mass $1000\ \text{kg}$ rolls along a level road with a driving force and a resistance of $400\ \text{N}$. The two horizontal forces oppose each other, so the resultant is the driving force minus the resistance — and the mass decides how quickly that resultant changes the car's speed. Drag the driving-force slider and watch the resultant and the acceleration move together.
resultant $F = D - 400$ = 200 N · acceleration $a = \dfrac{F}{m}$ = 0.20 m s⁻² · speeding up
Driving force against resistance: a resultant of zero means constant speed, not no motion. The car is speeding up when $D > 400\ \text{N}$ and slowing down when $D < 400\ \text{N}$.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. Two forces act along the same line in opposite directions: $9\ \text{N}$ one way and $4\ \text{N}$ the other. Their resultant is:
R2. The resultant force on a body of mass $m$ and acceleration $a$ is:
R3. The newton is equivalent to:
For motion along a line, the forces decide the acceleration through Newton's second law:
$F = ma$
where $F$ is the resultant force — the single force that replaces all the real ones. Choose a positive direction, give every force in that direction a $+$ sign and the others a $-$ sign, then add. Driving forces and resistances are not special; only the resultant matters.
Add the forces along the line to get $F$, then divide by the mass: $a = \dfrac{F}{m}$.
The constant acceleration goes into the week's formulae to give a speed, a distance or a time.
A car of mass $1000\ \text{kg}$ has a driving force of $600\ \text{N}$ and a resistance of $400\ \text{N}$. Resultant $= 600 - 400 = 200\ \text{N}$, so $a = \dfrac{200}{1000} = 0.2\ \text{m s}^{-2}$. From rest, after $10\ \text{s}$: $v = at = 2\ \text{m s}^{-1}$ and $s = \dfrac{1}{2}at^{2} = 10\ \text{m}$.
| Situation | What it tells you |
|---|---|
| moving at constant speed | the resultant force is zero — the resistance exactly balances the driving force |
| braking (no driving force) | the resistance alone is the resultant, so the acceleration is negative |
| falling freely | the weight alone is the resultant |
Mini-example: a car of mass $1200\ \text{kg}$ braking with a force of $2400\ \text{N}$ has $a = -\dfrac{2400}{1200} = -2\ \text{m s}^{-2}$. From $20\ \text{m s}^{-1}$ it stops in $t = \dfrac{20}{2} = 10\ \text{s}$, covering $\dfrac{1}{2}(20 + 0)(10) = 100\ \text{m}$.
Groups of 4: move the mass slider and the force slider and read the two live tables. The mass table keeps the force fixed and doubles and triples the mass; the force table keeps the mass fixed and doubles and triples the force. Describe the two patterns to each other, decide what they say about a = F/m, then answer the three challenge questions.
Work alone. Four pieces of student working about forces and motion are shown. Decide for each one whether it is correct or contains an error, then read the explanation. Two are sound and two hide a classic slip — check whether every force has the right sign, and whether the division goes the right way round.
The engine of a car of mass $1.2\ \text{tonnes}$ can produce a driving force of $2000\ \text{N}$. Ignoring any resistance forces, find (i) the car's resulting acceleration and (ii) the time taken for the car to go from rest to $27\ \text{m s}^{-1}$.
Convert tonnes to kilograms first. With no resistance the driving force is the resultant, so $a = \dfrac{F}{m}$. Then use $v = u + at$ with $u = 0$ to find the time.
$1.2\ \text{tonnes} = 1200\ \text{kg}$. With no resistance the resultant is the driving force:
$a = \dfrac{2000}{1200} = 1.67\ \text{m s}^{-2}$ (3 s.f.).
$v = u + at$ with $u = 0$: $27 = 1.667t$, so $t = 16.2\ \text{s}$ (3 s.f.).
An ice skater of mass $65\ \text{kg}$ is initially moving with speed $2\ \text{m s}^{-1}$ and glides to a halt over a distance of $10\ \text{m}$. Assuming that the force of resistance is constant, find (i) the size of the resistance force, (ii) the distance he would travel gliding to rest from an initial speed of $6\ \text{m s}^{-1}$ and (iii) the force he would need to apply to maintain a steady speed of $10\ \text{m s}^{-1}$.
Work backwards: the kinematics gives the deceleration from the two speeds and the distance ($v^{2} = u^{2} + 2as$), and then $F = ma$ turns that deceleration into the resistance. For part (iii), steady speed means the resultant is zero.
$v^{2} = u^{2} + 2as$ with $v = 0$, $u = 2$ and $s = 10$: $0 = 4 + 20a$, so $a = -0.2\ \text{m s}^{-2}$.
$F = ma = 65 \times 0.2 = 13\ \text{N}$ opposing the motion.
Same deceleration $0.2\ \text{m s}^{-2}$: $0 = 36 + 2(-0.2)s$, so $s = 90\ \text{m}$ — nine times the speed, nine times the distance, because $s \propto u^{2}$.
Steady speed means zero resultant, so the skater must push with exactly the resistance: $13\ \text{N}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. For a body moving along a straight line, the equation of motion is:
Q2. A car moves at a steady speed with a driving force of $800\ \text{N}$. The resistance is:
Q3. A resultant force of $30\ \text{N}$ acts on a body of mass $6\ \text{kg}$. The acceleration is:
Q4. A body moves along a straight line at constant speed. The resultant force on it is:
Work each on paper, then reveal the solution to check your working.
Calculate the resultant force in newtons required to produce each of these accelerations: (i) a car of mass $400\ \text{kg}$ accelerating at $2\ \text{m s}^{-2}$; (ii) a blue whale of mass $177\ \text{tonnes}$ accelerating at $0.5\ \text{m s}^{-2}$; (iii) a freight train of mass $42000\ \text{tonnes}$ braking with deceleration $0.02\ \text{m s}^{-2}$.
Reveal Solution
(i) $F = ma = 400 \times 2 = 800\ \text{N}$. (ii) $177\ \text{tonnes} = 177000\ \text{kg}$, so $F = 177000 \times 0.5 = 88500\ \text{N}$ (88.5 kN). (iii) $42000\ \text{tonnes} = 4.2 \times 10^{7}\ \text{kg}$, so $F = 4.2 \times 10^{7} \times 0.02 = 8.4 \times 10^{5}\ \text{N}$ (840 kN), acting against the motion.
A resultant force of $100\ \text{N}$ is applied to a body. Calculate the mass of the body when its acceleration is (i) $0.5\ \text{m s}^{-2}$, (ii) $2\ \text{m s}^{-2}$, (iii) $0.01\ \text{m s}^{-2}$ and (iv) $10g$ (take $g = 9.8\ \text{m s}^{-2}$).
Reveal Solution
$m = \dfrac{F}{a}$: (i) $\dfrac{100}{0.5} = 200\ \text{kg}$; (ii) $\dfrac{100}{2} = 50\ \text{kg}$; (iii) $\dfrac{100}{0.01} = 10000\ \text{kg}$; (iv) $10g = 98\ \text{m s}^{-2}$, so $\dfrac{100}{98} = 1.02\ \text{kg}$.
A man pushes a car of mass $400\ \text{kg}$ on level ground with a force of $200\ \text{N}$. The car is initially at rest and the man maintains this force until the car reaches a speed of $5\ \text{m s}^{-1}$. Ignoring any resistance forces, find (i) the acceleration of the car and (ii) the distance the car travels while the man is pushing.
Reveal Solution
(i) $a = \dfrac{F}{m} = \dfrac{200}{400} = 0.5\ \text{m s}^{-2}$. (ii) $v^{2} = u^{2} + 2as$ with $u = 0$: $25 = 2(0.5)s$, so $s = 25\ \text{m}$.
A top sprinter of mass $65\ \text{kg}$ starting from rest reaches a speed of $10\ \text{m s}^{-1}$ in $2\ \text{s}$. (i) Calculate the force required to produce this acceleration, assuming it is uniform. (ii) Compare this with the force exerted by a weight lifter holding a mass of $180\ \text{kg}$ above the ground (take $g = 9.8\ \text{m s}^{-2}$).
Reveal Solution
(i) $a = \dfrac{10 - 0}{2} = 5\ \text{m s}^{-2}$, so $F = ma = 65 \times 5 = 325\ \text{N}$. (ii) The lifter supports a weight of $180 \times 9.8 = 1764\ \text{N}$, about $5.4$ times the sprinter's driving force — holding a heavy mass still takes far more force than sprinting.
Pat and Nicholas control a canal barge by ropes attached to each end, horizontal and along the line of motion. The mass of the barge is $12\ \text{tonnes}$ and the total resistance to forward motion is $250\ \text{N}$ at all times. Initially Pat pulls the barge forwards from rest with a force of $400\ \text{N}$ and Nicholas leaves his rope slack. (i) Write down the equation of motion for the barge and calculate its acceleration. Pat continues to pull with the same force until the barge has moved $10\ \text{m}$. (ii) What is the speed of the barge at this time, and how long did Pat pull? Pat now lets her rope go slack and Nicholas brings the barge to rest by pulling with a constant force of $150\ \text{N}$. (iii) Calculate (a) how long it takes the barge to come to rest, (b) the total distance travelled from when it first moved and (c) the total time taken.
Reveal Solution
(i) $400 - 250 = 12000a$, so $a = \dfrac{150}{12000} = 0.0125\ \text{m s}^{-2}$.
(ii) $v^{2} = 2as = 2(0.0125)(10) = 0.25$, so $v = 0.5\ \text{m s}^{-1}$; the time is $\dfrac{0.5}{0.0125} = 40\ \text{s}$.
(iii)(a) Braking: resultant $= -(150 + 250) = -400\ \text{N}$, so $a = -\dfrac{400}{12000} = -\dfrac{1}{30}\ \text{m s}^{-2}$, and the barge stops after $0.5 \times 30 = 15\ \text{s}$. (b) Braking distance $= \dfrac{1}{2}(0.5 + 0)(15) = 3.75\ \text{m}$, giving a total of $10 + 3.75 = 13.75\ \text{m}$. (c) Total time $= 40 + 15 = 55\ \text{s}$.
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