Force and acceleration · Kinematics & dynamics

F = ma

Newton’s second law: the resultant force on a body equals its mass times its acceleration.

F = ma \qquad (\text{N} = \text{kg} \times \text{m s}^{-2})

When several forces act, find their resultant first (see resultants of forces). A force acting for a time $t$ changes the momentum: $Ft = m(v - u)$ .

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q18 (a)

Forces $F_1 = (3\text{ N}, 210^\circ)$ and $F_2 = (4\text{ N}, 120^\circ)$ act on a particle of mass $7\text{ kg}$ which is at rest. Calculate the: (i) acceleration of the particle; (ii) velocity of the particle after 3 seconds.

The resultant force
- The forces are at right angles, so ${F = \sqrt{3^2 + 4^2} = 5}$ N.
Think first. The bearings differ by 90°.
Acceleration
- ${a = \frac{F}{m} = \frac57 \approx 0.71}$ m/s².
Velocity after 3 s
- ${v = u + at = 0 + \frac57 \times 3 = \frac{15}{7} \approx 2.14}$ m/s.
Think first. It starts at rest.

More: force and acceleration

Lifts

A body standing in a lift has two forces on it: its weight $mg$ down and the reaction $R$ of the floor up. Take the direction of the acceleration as positive:

In a lift accelerating upwardsR − mg = ma, so R = m(g + a)

Accelerating downwards, ${mg - R = ma}$ , so ${R = m(g - a)}$ . At a steady speed, ${R = mg}$ .

The reaction in a liftSet the mass and the acceleration

600 Nreaction R500 Nweight mg

mass (kg) = 50acceleration (up +) = 2 m/s²

R − mg = ma, so R = 50(10 + 2) = 600 N. The acceleration is upwards, so the floor pushes harder than the weight: the body feels heavier.

Worked example · NECO 2023

NECO 2023 · Paper 1 · Q40

A 50 kg bag of rice is placed in a lift which moves with an upward acceleration of $8\text{ ms}^{-2}$ . What is the reaction between the floor of the lift and the rice? [Take $g = 10\text{ ms}^{-2}$ ]

Resultant force upwards
- ${R - mg = ma}$ , with ${m = 50}$ , ${g = 10}$ and ${a = 8}$ .
Think first. The acceleration is upwards: which force is bigger?
Solve
- ${R = 50(10 + 8) = 900}$ N: option C.

More: lifts

WAEC 2016 · Paper 2 · Q14A block of mass $0.5\text{ kg}$ rests on the floor of a lift which is moving upwards with an acceleration of $2\text{ m s}^{-2}$ . …

Rough floors and smooth planes

On a rough floor, friction $F = \mu R$ acts against the motion, so the resultant force is the push minus the friction:

Pushed along a rough floorP − μR = ma, with R = mg

On a smooth plane there is no friction, and the part of the weight down the slope, ${mg\sin\theta}$ , is the resultant force: ${a = g\sin\theta}$ (see the slope figure in equilibrium).

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q15 (a)

A body of mass $4\text{ kg}$ placed at the top of a smooth plane inclined at an angle of $35^\circ$ to the horizontal slides from rest down the plane. If the plane is $300\text{ m}$ long, calculate, correct to two significant figures, the speed of the body when it has travelled half the length of the plane. $[\text{Take } g = 10\text{ m s}^{-2}]$

The acceleration
- Only ${mg\sin 35^\circ}$ acts down the plane, so ${a = g\sin 35^\circ = 5.736}$ m/s².
Think first. The plane is smooth. What makes the body speed up?
Half-way down
- ${v^2 = 0 + 2(5.736)(150) \approx 1720.7}$ .
- ${v \approx 41}$ m/s (2 s.f.).
Think first. Half of 300 m, from rest.

More: friction

Connected particles over a pulley

Two masses joined by a light inextensible string over a smooth pulley move together: they have the same acceleration $a$ , and the tension $T$ is the same all along the string. Write $F = ma$ for each mass on its own, taking the way it moves as positive. With $m_1 > m_2$ , $m_1$ goes down and $m_2$ goes up:

m_1 g - T = m_1 a \qquad T - m_2 g = m_2 a

Adding the two equations removes $T$ :

a = \frac{(m_1 - m_2)g}{m_1 + m_2}

Then put $a$ back into either equation to find $T$ .

Two masses over a pulleym₁g − T = m₁a and T − m₂g = m₂a

A block on a table, pulled by a hanging massm₂g − T = m₂a and T − F = m₁a

For a block of mass $m_1$ on a table pulled by a hanging mass $m_2$ , the block’s weight is held up by the table ( $R = m_1 g$ ), so only the tension, and friction ${F = \mu R}$ if the table is rough, act along its motion. On a smooth table ${F = 0}$ and ${a = \frac{m_2 g}{m_1 + m_2}}$ .

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q8

The diagram shows a light inextensible string that passes over a smooth pulley and carries masses of $6\text{ kg}$ and $5\text{ kg}$ at its ends. When the system is released from rest: $[\text{Take } g = 10\text{ m s}^{-2}]$

Not to scale.

find the acceleration of each mass;

calculate, correct to two decimal places, the distance moved in 2 seconds.

Weights
- Weights: ${6 \times 10 = 60}$ N and ${5 \times 10 = 50}$ N.
- The 6 kg mass is heavier, so it goes down and the 5 kg mass goes up.
Think first. Which way does the system move?
An equation for each mass
- The 6 kg mass, down is positive: ${60 - T = 6a}$ .
- The 5 kg mass, up is positive: ${T - 50 = 5a}$ .
Think first. Resultant force = mass × acceleration, for each one.
Remove T
- Add them: ${10 = 11a}$ .
- Divide by 11: ${a = \frac{10}{11} \approx 0.91}$ m/s².
Think first. What happens if you add the two equations?
Distance in 2 seconds
- ${s = ut + \frac12 at^2}$ with ${u = 0}$ and ${t = 2}$ .
- ${s = \frac12 \times \frac{10}{11} \times 4 = \frac{20}{11} \approx 1.82}$ m.
Think first. It starts from rest.

Your turn

WAEC 2022 · Paper 2 · Q14 (b)

(b)
A load of mass $120\text{ kg}$ is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards: (i) at a constant velocity; (ii) with an acceleration of $3\text{ m s}^{-2}$ . $[\text{Take } g = 10\text{ m s}^{-2}]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(b)(i)

At constant velocity there is no acceleration: $R - 120 \times 10 = 0$ , so $R = 1200$ N.

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