A high-speed train at 300 km/h in a straight line needs no net force to keep going; a laboratory trolley accelerates in exact proportion to the force it receives; a hurley that strikes a sliotar in Croke Park receives, in turn, a force from the sliotar. These three ideas, formalised by Newton, are the framework that solves any dynamics problem, from a crate being dragged to a system of pulleys.
First law: the principle of inertia
If you push a disc across a wooden table it stops almost at once, on ice it travels tens of metres, and on an air table it hardly loses any speed: do bodies stop because they need a force to keep moving, or because something is slowing them down?
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A high-speed train travels along a straight section at a constant speed of $300\ \text{km/h}$. a) What is the acceleration of the train on that section? b) Applying…
A high-speed train travels along a straight section at a constant speed of $300\ \text{km/h}$.
a) What is the acceleration of the train on that section?
b) Applying Newton's first law, what is the net (resultant) force acting on the train?
c) If the thrust of the engines is $5 \times 10^{4}\ \text{N}$, what is the total friction force opposing forward motion? Justify your answer.
To model a real situation with Newton's laws, we simplify it: treat the object as a single particle, list the forces, and state the assumptions.
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In the laboratory, a net force of $24\ \text{N}$ is applied to a dynamics trolley of $3\ \text{kg}$ moving along a horizontal track. a) Write the expression for Newton's…
In the laboratory, a net force of $24\ \text{N}$ is applied to a dynamics trolley of $3\ \text{kg}$ moving along a horizontal track.
a) Write the expression for Newton's second law, $\sum \vec{F} = m \cdot \vec{a}$, and make acceleration the subject.
b) Calculate the value of the acceleration of the trolley.
The motor of an electric scooter exerts a net force of $40\ \text{N}$. The diagram compares the two situations. a) Calculate the acceleration when only the scooter…
The motor of an electric scooter exerts a net force of $40\ \text{N}$. The diagram compares the two situations.
a) Calculate the acceleration when only the scooter travels, having a mass of $20\ \text{kg}$.
b) Calculate the acceleration when it carries a backpack and the total mass is $40\ \text{kg}$.
c) Explain, using $\vec{a} = \frac{\vec{F}}{m}$, why doubling the mass halves the acceleration.
Two horizontal forces along the same line act on a crate of $4\ \text{kg}$ on the loading bay: $30\ \text{N}$ to the right and $18\ \text{N}$ to the left. The diagram…
Two horizontal forces along the same line act on a crate of $4\ \text{kg}$ on the loading bay: $30\ \text{N}$ to the right and $18\ \text{N}$ to the left. The diagram shows the crate with both forces.
a) Calculate the net force on the crate, stating its direction.
b) Calculate the acceleration of the crate.
c) What additional force would need to be applied for the crate to move at a constant speed? Justify your answer using Newton's first law.
A bat hits a baseball, exerting a forward force of $200\ \text{N}$ on it. The diagram shows the instant of impact. a) State Newton's third law (action and reaction). b)…
A bat hits a baseball, exerting a forward force of $200\ \text{N}$ on it. The diagram shows the instant of impact.
a) State Newton's third law (action and reaction).
b) What force does the ball exert on the bat? State its magnitude, direction and sense.
c) Explain why, despite both forces having the same magnitude, the ball flies off while the bat barely changes its motion.
The free-body diagram: weight, normal reaction and friction
To solve any dynamics problem you must first draw all the forces on the body, identifying what exerts each one.
The normal reaction is written $R$ and the weight $W = mg$. In some exercises they appear as $N$ and $P$: they are the same forces.
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A crate of $5\ \text{kg}$ rests on the horizontal floor of a workshop. The coefficient of friction between the crate and the floor is $\mu = 0.4$ (take…
A crate of $5\ \text{kg}$ rests on the horizontal floor of a workshop. The coefficient of friction between the crate and the floor is $\mu = 0.4$ (take $g = 10\ \text{m/s}^2$).
a) Calculate the normal force that the floor exerts on the crate.
b) Calculate the friction force when the crate slides, $F_r = \mu \times N$.
c) If the crate is pushed with a horizontal force of $20\ \text{N}$, will it move at a constant speed? Give reasons for your answer.
A worker pushes a workshop trolley of $8\ \text{kg}$ with a horizontal force of $52\ \text{N}$. The coefficient of friction with the floor is $\mu = 0.5$ (take…
A worker pushes a workshop trolley of $8\ \text{kg}$ with a horizontal force of $52\ \text{N}$. The coefficient of friction with the floor is $\mu = 0.5$ (take $g = 10\ \text{m/s}^2$). The diagram shows the forces acting on the trolley.
a) Calculate the normal force and the friction force.
b) Calculate the net horizontal force on the trolley.
To move a removal crate of $20\ \text{kg}$ along the floor, the coefficient of static friction is $\mu_e = 0.5$ and the coefficient of dynamic friction is $\mu_d = 0.4$…
To move a removal crate of $20\ \text{kg}$ along the floor, the coefficient of static friction is $\mu_e = 0.5$ and the coefficient of dynamic friction is $\mu_d = 0.4$ (take $g = 10\ \text{m/s}^2$).
a) Calculate the minimum horizontal force required to set the crate in motion.
b) Calculate the horizontal force required to keep it moving at a constant speed.
c) In light of the results, explain why it takes more effort to start moving the crate than to keep it moving.
A rope can only pull, and it pulls along its own length: that force is the tension $T$. In a lift, the floor pushes on you with a normal reaction that is not always equal to your weight.
When a rope is pulled at an angle $\theta$ above the horizontal, split the pull into $F\cos\theta$ (horizontal, moves the body) and $F\sin\theta$ (vertical, helps to lift it). The normal reaction becomes $N = mg - F\sin\theta$, so the friction is smaller.
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A lamp of $4\ \text{kg}$ hangs at rest from a cable attached to the ceiling (take $g = 10\ \text{m/s}^2$). The diagram shows the lamp hanging from the cable. a) Draw in…
A lamp of $4\ \text{kg}$ hangs at rest from a cable attached to the ceiling (take $g = 10\ \text{m/s}^2$). The diagram shows the lamp hanging from the cable.
a) Draw in your notebook the forces acting on the lamp and state which body exerts each one.
b) Calculate the tension in the cable.
c) If the cable can only withstand a maximum tension of $100\ \text{N}$, what is the maximum mass of the lamp that can be hung?
You can do it on paper and send a photo with your answer.
A person of $60\ \text{kg}$ weighs themselves on scales inside a lift (take $g = 10\ \text{m/s}^2$). The diagram shows the person on the scales. a) What do the scales…
A person of $60\ \text{kg}$ weighs themselves on scales inside a lift (take $g = 10\ \text{m/s}^2$). The diagram shows the person on the scales.
a) What do the scales read with the lift at rest?
b) The lift starts moving upwards with an acceleration of $1.5\ \text{m/s}^2$. Apply Newton's second law vertically, $N - P = m \times a$, and calculate the reading on the scales.
c) Has the person's real weight changed? Explain which quantity has changed.
On an inclined plane the weight splits into a component parallel to the plane (which moves or slows the body) and a perpendicular component (which determines the normal reaction and, with it, the friction).
Constant velocity and motion up a slope
What happens to friction when a body moves up a slope? It still opposes the motion, so it now points down the slope, together with $P_x$.
From net force to motion
Newton's second law gives you the acceleration; the equations of motion then tell you how fast, how far and how long. A constant net force means a constant acceleration, so you can use these equations.
Remember that real resistive forces such as air resistance are not constant: they grow with speed. A falling object (a skydiver over the Wicklow Mountains, say) speeds up until the air resistance equals its weight. Then $F_{net} = 0$, $a = 0$ and it falls at a constant terminal velocity. Here the constant-acceleration equations no longer apply, because $F_{net}$ changes with speed.
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A remote-controlled car of $2\ \text{kg}$ moves in a straight line. Its velocity changes as follows: from $t = 0$ to $t = 4\ \text{s}$ it goes from $0$ to…
A remote-controlled car of $2\ \text{kg}$ moves in a straight line. Its velocity changes as follows: from $t = 0$ to $t = 4\ \text{s}$ it goes from $0$ to $8\ \text{m/s}$ uniformly; from $t = 4\ \text{s}$ to $t = 10\ \text{s}$ it maintains a velocity of $8\ \text{m/s}$; from $t = 10\ \text{s}$ to $t = 14\ \text{s}$ it brakes uniformly until it stops.
a) Draw the velocity-time graph in your notebook.
b) Calculate the acceleration of each of the three sections.
c) Applying Newton's second law, calculate the net force on the car in each section and state its direction.
d) In the second section the net force is zero: explain whether this means that the engine exerts no force.
You can do it on paper and send a photo with your answer.
Two identical buckets hanging from the ends of a rope that passes over a pulley stay still, but as soon as you add a nut to one of them, both start to accelerate together: how does that difference in weight travel through the rope to set both masses moving?
When two bodies are joined by a massless, inextensible rope passing over a single ideal fixed pulley, they share the same acceleration in magnitude, and the tension is the same throughout the rope.
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Two masses, $m_1 = 3\ \text{kg}$ and $m_2 = 5\ \text{kg}$, hang from an ideal pulley in the laboratory, connected by an inextensible string (take…
Two masses, $m_1 = 3\ \text{kg}$ and $m_2 = 5\ \text{kg}$, hang from an ideal pulley in the laboratory, connected by an inextensible string (take $g = 10\ \text{m/s}^2$). As the string is inextensible, both masses move with the same acceleration $a$, and the tension $T$ is the same throughout the string. The diagram shows the pulley set-up.
a) Draw in your notebook the forces acting on each mass.
b) Write Newton's second law for each mass: $T - m_1 \times g = m_1 \times a$ for the one going up and $m_2 \times g - T = m_2 \times a$ for the one going down.
c) Add the two equations to eliminate $T$ and calculate the acceleration of the system.
You can do it on paper and send a photo with your answer.
In an Atwood machine in the laboratory, masses $m_1 = 2\ \text{kg}$ and $m_2 = 6\ \text{kg}$ hang from an inextensible string passing over an ideal pulley (take…
In an Atwood machine in the laboratory, masses $m_1 = 2\ \text{kg}$ and $m_2 = 6\ \text{kg}$ hang from an inextensible string passing over an ideal pulley (take $g = 10\ \text{m/s}^2$).
a) Draw in your notebook the forces acting on each mass.
b) Set up the system of two equations with the unknowns $a$ and $T$, and solve it: calculate the acceleration of the system and the tension in the string.
c) Verify that the tension is greater than the weight of $m_1$ and less than the weight of $m_2$, and explain why this must be the case while the system is accelerating.
You can do it on paper and send a photo with your answer.
Connected bodies on a table, in a line and on a slope
Whatever the set-up, the method is the same: bodies joined by a taut, inextensible string share the same magnitude of acceleration. Write $F_{net} = ma$ for each body, taking the direction of motion as positive. For the whole system, only the external forces count.
If one body is on a slope, its equation along the slope replaces the table equation: moving up, $T - P_x - F_r = ma$, with $P_x$, $N$ and $F_r$ found as in the slope sections.
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A toy locomotive of $2\ \text{kg}$ pulls two wagons, of $2\ \text{kg}$ and $4\ \text{kg}$, connected by couplings, exerting a tractive force of $24\ \text{N}$ (consider…
A toy locomotive of $2\ \text{kg}$ pulls two wagons, of $2\ \text{kg}$ and $4\ \text{kg}$, connected by couplings, exerting a tractive force of $24\ \text{N}$ (consider friction negligible). The train consists of, from front to back: locomotive, $2\ \text{kg}$ wagon, and $4\ \text{kg}$ wagon.
a) Calculate the acceleration of the system.
b) Applying the fundamental principle only to the $4\ \text{kg}$ wagon, which is at the back, calculate the tension in the coupling between the two wagons.
c) Calculate the tension in the coupling between the locomotive and the first wagon and explain why it is greater than the one in the previous section.
A block of $6\ \text{kg}$ is on a horizontal table, connected by a string passing over an ideal pulley to another hanging block of $4\ \text{kg}$. The coefficient of…
A block of $6\ \text{kg}$ is on a horizontal table, connected by a string passing over an ideal pulley to another hanging block of $4\ \text{kg}$. The coefficient of friction between the first block and the table is $\mu = 0.25$ (take $g = 10\ \text{m/s}^2$). The diagram shows the set-up.
a) Draw in your notebook the forces acting on each block.
b) Calculate the friction force on the block on the table.
c) Set up Newton's second law for each block, taking into account that both have the same acceleration $a$ and that the tension $T$ is the same throughout the string, and calculate the value of $a$.
d) Calculate the tension in the string.
You can do it on paper and send a photo with your answer.
Back to the start: the train needs no net force at constant velocity ($F_{net} = 0$), the trolley accelerates by $a = F_{net}/m$, and the sliotar pushes back on the hurley with an equal and opposite force.
Let's practice! 5 medium exercises in this section