The Fundamental Principle of Dynamics
Two trucks with identical engines leave a toll booth on the Mexico City–Puebla highway. One is empty and the other is loaded to the maximum. Both start with the same engine force, but the empty one picks up speed quickly while the loaded one takes much longer. The equation that explains the difference, $\Sigma\vec F = m\vec a$, is the same one used to calculate how a car brakes or how a rocket takes off.
ΣF = m · a: the fundamental principle
If you push an empty grocery cart, it accelerates easily. If you fill it with jugs of water, you have to push much harder to get the same change in speed. How exactly do acceleration, mass, and force depend on each other?
Two ideas from Newton's second law:
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If the total force on an object is not zero, its state of motion changes: it speeds up, slows down, or turns.
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For the same object, a larger force produces a larger change. For the same force, the greater the mass, the smaller the change.
The second law predicts the motion of macroscopic objects (cars, balls, rockets) very accurately. It does not work at the atomic scale or when speeds approach the speed of light.
Calculate the net force required for a toy cart of $2\ \text{kg}$ to accelerate at $3\ \text{m/s}^2$.
Calculate the net force required for a toy cart of $2\ \text{kg}$ to accelerate at $3\ \text{m/s}^2$.
A scooter together with its rider accelerates at $2\ \text{m/s}^2$ when a net force of $100\ \text{N}$ acts on the combination. Calculate the total mass.
A scooter together with its rider accelerates at $2\ \text{m/s}^2$ when a net force of $100\ \text{N}$ acts on the combination. Calculate the total mass.
A cyclist moving forwards applies the brakes. In which direction does the net force on the bicycle point while braking: forwards or backwards? What about its…
A cyclist moving forwards applies the brakes. In which direction does the net force on the bicycle point while braking: forwards or backwards? What about its acceleration? Explain the relationship between the direction of the net force and that of the acceleration.
The newton: a derived unit
When you hold an apple of about $100\ \text{g}$ in your hand, you feel a downward push (about $1\ \text{N}$). How is the unit we use to measure any force defined?
A net force of $20\ \text{N}$ acts on a box of $5\ \text{kg}$. What acceleration does the box acquire?
A net force of $20\ \text{N}$ acts on a box of $5\ \text{kg}$. What acceleration does the box acquire?
Using Newton's second law, show that $1\ \text{N}$ is equivalent to $1\ \text{kg} \times \text{m/s}^2$ and give a simple numerical example to illustrate it.
Using Newton's second law, show that $1\ \text{N}$ is equivalent to $1\ \text{kg} \times \text{m/s}^2$ and give a simple numerical example to illustrate it.
A motorcycle with its rider has a mass of $200\ \text{kg}$ and travels at $72\ \text{km/h}$. a) Express the speed in $\text{m/s}$. b) Calculate the net braking force…
A motorcycle with its rider has a mass of $200\ \text{kg}$ and travels at $72\ \text{km/h}$.
- a) Express the speed in $\text{m/s}$.
- b) Calculate the net braking force required to stop in $4\ \text{s}$.
The net force: the vector sum of all forces
Rarely does only one force act on a body. Before applying $\Sigma\vec F = m\vec a$ you have to calculate the net force: the sum, as vectors, of all the forces acting on the body, including friction if there is any. Every force has magnitude and direction.
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Forces in the same direction: the magnitudes add.
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Forces in opposite directions: the magnitudes subtract, and the result points toward the larger one.
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Perpendicular forces: the magnitude of the net force comes from the Pythagorean theorem, $\Sigma F = \sqrt{F_1^2 + F_2^2}$, and its direction is that of the diagonal of the parallelogram.
For the direction, the angle $\theta$ of the net force relative to one of the forces satisfies $\tan\theta = F_2/F_1$. Since $\vec a = \Sigma\vec F/m$ and $m$ is a positive number, $\vec a$ points exactly the same way as $\Sigma\vec F$.
Two forces of $10\ \text{N}$ and $14\ \text{N}$ push a box of $4\ \text{kg}$ in the same direction, as shown in the figure. Calculate the net force on the box and its…
Two forces of $10\ \text{N}$ and $14\ \text{N}$ push a box of $4\ \text{kg}$ in the same direction, as shown in the figure.
Calculate the net force on the box and its acceleration.
A person pushes a toolbox of $8\ \text{kg}$ with a force of $60\ \text{N}$, and the friction with the ground is $20\ \text{N}$. a) Calculate the net force on the box. b)…
A person pushes a toolbox of $8\ \text{kg}$ with a force of $60\ \text{N}$, and the friction with the ground is $20\ \text{N}$.
- a) Calculate the net force on the box.
- b) Calculate its acceleration.
Three forces act on a crate of $10\ \text{kg}$ along the same line: $50\ \text{N}$ to the right, $30\ \text{N}$ to the right and $20\ \text{N}$ to the left. a) Calculate…
Three forces act on a crate of $10\ \text{kg}$ along the same line: $50\ \text{N}$ to the right, $30\ \text{N}$ to the right and $20\ \text{N}$ to the left.
- a) Calculate the net force.
- b) Calculate the acceleration of the crate.
- c) What additional force would need to be applied so that the crate moves at constant velocity? Reason your answer.
A worker pushes a trolley of $20\ \text{kg}$ with a force of $100\ \text{N}$ and the friction is $20\ \text{N}$. a) Calculate the acceleration of the cart. b) If the…
A worker pushes a trolley of $20\ \text{kg}$ with a force of $100\ \text{N}$ and the friction is $20\ \text{N}$.
- a) Calculate the acceleration of the cart.
- b) If the worker doubles their push to $200\ \text{N}$ and the friction does not change, does the acceleration double as well? Calculate it and explain why this happens.
Two perpendicular forces of $6\ \text{N}$ and $8\ \text{N}$ act simultaneously on a wooden block of $5\ \text{kg}$. a) Calculate the magnitude of the net force on the…
Two perpendicular forces of $6\ \text{N}$ and $8\ \text{N}$ act simultaneously on a wooden block of $5\ \text{kg}$.
- a) Calculate the magnitude of the net force on the block.
- b) Calculate the acceleration of the block.
- c) State how you would determine the direction of this acceleration and justify why it coincides with that of the net force.
To pull a sledge of $30\ \text{kg}$ across the snow, a force of $90\ \text{N}$ is applied to it and the friction is $30\ \text{N}$. A student calculates the acceleration…
To pull a sledge of $30\ \text{kg}$ across the snow, a force of $90\ \text{N}$ is applied to it and the friction is $30\ \text{N}$. A student calculates the acceleration like this: $a = 90 / 30 = 3\ \text{m/s}^2$.
- a) Identify the mistake made.
- b) Calculate the correct acceleration of the sledge.
- c) Calculate the total force with which it would need to be pulled for the acceleration to actually be $3\ \text{m/s}^2$.
Acceleration and braking: the second law with kinematics
The second law gives the acceleration; kinematics tells what happens to the velocity. If the net force is constant, the acceleration is too, and the velocity changes like this: $v = v_0 + a\,t$ (or $v = a\,t$ if it starts from rest).
The sign matters: if you take the direction of motion as positive, a net force that slows the body points in the opposite direction, and that is why the acceleration comes out negative.
Solution to "your turn": $72/3.6 = 20\ \text{m/s}$; $a = -20/4 = -5\ \text{m/s}^2$; $\Sigma F = 200\times 5 = 1{,}000\ \text{N}$ against the motion.
A car of $1000\ \text{kg}$ brakes with a net force of $5000\ \text{N}$, as shown in the figure. a) Calculate its acceleration and interpret the sign of the result. b) If…
A car of $1000\ \text{kg}$ brakes with a net force of $5000\ \text{N}$, as shown in the figure.
- a) Calculate its acceleration and interpret the sign of the result.
- b) If it was travelling at $20\ \text{m/s}$, calculate how long it takes to come to a stop (use $v = v_0 + a \times t$).
A full supermarket trolley has a mass of $15\ \text{kg}$ and a net force of $30\ \text{N}$ is applied to it. a) Calculate its acceleration. b) If it starts from rest,…
A full supermarket trolley has a mass of $15\ \text{kg}$ and a net force of $30\ \text{N}$ is applied to it.
- a) Calculate its acceleration.
- b) If it starts from rest, what speed does it reach after $3\ \text{s}$?
The engine of a sounding rocket of $3000\ \text{kg}$ produces a net thrust of $12000\ \text{N}$. a) Calculate the acceleration of the rocket. b) If the thrust remains…
The engine of a sounding rocket of $3000\ \text{kg}$ produces a net thrust of $12000\ \text{N}$.
- a) Calculate the acceleration of the rocket.
- b) If the thrust remains constant, what speed does the rocket gain during the first $10\ \text{s}$?
A car of $800\ \text{kg}$ travels at $20\ \text{m/s}$ and needs to stop in $5\ \text{s}$. a) Calculate the braking acceleration. b) Calculate the required net braking…
A car of $800\ \text{kg}$ travels at $20\ \text{m/s}$ and needs to stop in $5\ \text{s}$.
- a) Calculate the braking acceleration.
- b) Calculate the required net braking force.
Constant velocity and balanced forces
A body moving does not mean there is a net force: if it travels at constant velocity (or is at rest), its acceleration is zero and, by the second law, so is the net force.
An aeroplane flies in a straight line at a constant velocity of $250\ \text{m/s}$. What is the net force on the aeroplane? What is its acceleration? Relate this to…
An aeroplane flies in a straight line at a constant velocity of $250\ \text{m/s}$. What is the net force on the aeroplane? What is its acceleration? Relate this to Newton's second law of motion.
A sledge slides along horizontal ice and we stop pushing it. A classmate claims: "since there is no longer any force pushing it, it will stop straight away". a)…
A sledge slides along horizontal ice and we stop pushing it. A classmate claims: "since there is no longer any force pushing it, it will stop straight away".
- a) According to Newton's second law, how would the sledge move if friction were exactly zero? What would the net force on it be then?
- b) On real ice the sledge eventually comes to a stop: which force is responsible and in which direction does it act?
- c) Explain why the classmate's statement confuses "there is no applied force" with "there is no net force".
Mass and acceleration: inverse proportionality
Let's go back to the two trucks from the beginning: same force, different mass. The figure shows them starting from rest.
The same force of $50\ \text{N}$ is applied first to a ball of $0.5\ \text{kg}$ and then to a medicine ball of $5\ \text{kg}$, as shown in the figure. Calculate the…
The same force of $50\ \text{N}$ is applied first to a ball of $0.5\ \text{kg}$ and then to a medicine ball of $5\ \text{kg}$, as shown in the figure.
Calculate the acceleration of each one and explain which accelerates more and why.
A pushchair of $6\ \text{kg}$ and a shopping trolley of $12\ \text{kg}$ experience the same net force of $12\ \text{N}$. a) Calculate the acceleration of each one. b)…
A pushchair of $6\ \text{kg}$ and a shopping trolley of $12\ \text{kg}$ experience the same net force of $12\ \text{N}$.
- a) Calculate the acceleration of each one.
- b) Compare the masses and accelerations: what relationship do you find between them?
A car of $1200\ \text{kg}$ tows a caravan of $400\ \text{kg}$ and the engine exerts a net force of $4800\ \text{N}$ on the combination, as shown in the diagram. a)…
A car of $1200\ \text{kg}$ tows a caravan of $400\ \text{kg}$ and the engine exerts a net force of $4800\ \text{N}$ on the combination, as shown in the diagram.
- a) Calculate the acceleration of the car-caravan combination.
- b) Calculate the force that the towbar exerts on the caravan, keeping in mind that it is the only force accelerating it.
In a laboratory, the acceleration of a trolley is measured when different forces are applied to it: with $2\ \text{N}$ it accelerates at $0.5\ \text{m/s}^2$; with…
In a laboratory, the acceleration of a trolley is measured when different forces are applied to it: with $2\ \text{N}$ it accelerates at $0.5\ \text{m/s}^2$; with $4\ \text{N}$, at $1\ \text{m/s}^2$; and with $6\ \text{N}$, at $1.5\ \text{m/s}^2$.
- a) Plot force against acceleration in your exercise book and describe the graph you obtain.
- b) Deduce the mass of the trolley from the data.
- c) Predict the acceleration of the trolley if a force of $10\ \text{N}$ were applied to it.
Two vans travel at the same speed and brake with the same net braking force, but one is loaded and has twice the mass of the other. a) Compare the braking accelerations…
Two vans travel at the same speed and brake with the same net braking force, but one is loaded and has twice the mass of the other.
- a) Compare the braking accelerations of the two vans.
- b) Compare the times they take to come to a stop.
- c) Draw a conclusion regarding vehicle load and road safety.
Two lorries with identical engines, producing the same net force of $8000\ \text{N}$, start from rest: one is empty, with a mass of $4000\ \text{kg}$, and the other is…
Two lorries with identical engines, producing the same net force of $8000\ \text{N}$, start from rest: one is empty, with a mass of $4000\ \text{kg}$, and the other is loaded, with a mass of $8000\ \text{kg}$.
- a) Calculate the acceleration of each lorry.
- b) Calculate the time each lorry takes to reach $20\ \text{m/s}$.
- c) Using these numbers, explain why a loaded lorry needs more distance to join a motorway.
Back to the beginning: the loaded truck has more mass, so with the same engine force it accelerates less. That is why it takes longer to pick up speed.