Newton's laws of dynamics
A high-speed AVE 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 bat that hits a ball receives, in turn, a force from the ball. 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?
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 $5 \cdot 10^{4}\ \text{N}$, what is the total friction force opposing forward motion? Justify your answer.
Second law: the fundamental principle of dynamics
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.
Third law: action and reaction
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.
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.
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.
A worker leaves a container of $10\ \text{kg}$ on a loading ramp that forms $30^\circ$ with the horizontal (take $g = 10\ \text{m/s}^2$, $\sen 30^\circ = 0,5$ and…
A worker leaves a container of $10\ \text{kg}$ on a loading ramp that forms $30^\circ$ with the horizontal (take $g = 10\ \text{m/s}^2$, $\sen 30^\circ = 0,5$ and $\cos 30^\circ \approx 0,87$). The diagram shows the container on the ramp.
- a) Calculate the weight of the container.
- b) Calculate the component of the weight parallel to the ramp, $P_x = P \cdot \sen 30^\circ$.
- c) Calculate the perpendicular component, $P_y = P \cdot \cos 30^\circ$, and state which force from the plane balances it.
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, $F_r = \mu \cdot N$.
- 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.
- c) Calculate its acceleration.
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.
Dynamics on an inclined plane
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).
In a laboratory experiment, a trolley of $2\ \text{kg}$ is released down an inclined plane of $30^\circ$, with negligible friction (take $g = 10\ \text{m/s}^2$,…
In a laboratory experiment, a trolley of $2\ \text{kg}$ is released down an inclined plane of $30^\circ$, with negligible friction (take $g = 10\ \text{m/s}^2$, $\sen 30^\circ = 0,5$ and $\cos 30^\circ \approx 0,87$).
- a) Calculate the components of the weight parallel and perpendicular to the plane.
- b) Calculate the acceleration of the trolley.
- c) If it starts from rest, calculate the time it takes to travel $10\ \text{m}$ along the plane and its final velocity (use $x = \frac{1}{2} \cdot a \cdot t^2$ and $v = a \cdot t$).
A rescue sledge of $40\ \text{kg}$ goes down a snowy slope of $37^\circ$ with a coefficient of friction of $\mu = 0,5$ (take $g = 10\ \text{m/s}^2$,…
A rescue sledge of $40\ \text{kg}$ goes down a snowy slope of $37^\circ$ with a coefficient of friction of $\mu = 0,5$ (take $g = 10\ \text{m/s}^2$, $\sen 37^\circ \approx 0,6$ and $\cos 37^\circ \approx 0,8$). The diagram shows the sledge on the slope with the forces acting on it.
- a) Calculate the component of the weight parallel to the slope.
- b) Calculate the normal force and the friction force.
- c) Calculate the acceleration of the sledge and verify that it does not depend on its mass: would a loaded sledge go down with the same acceleration as an empty one? Give reasons why.
A parcel of $4\ \text{kg}$ slides from the top of a warehouse ramp that forms $37^\circ$ with the horizontal. The coefficient of friction is $\mu = 0,25$ (take…
A parcel of $4\ \text{kg}$ slides from the top of a warehouse ramp that forms $37^\circ$ with the horizontal. The coefficient of friction is $\mu = 0,25$ (take $g = 10\ \text{m/s}^2$, $\sen 37^\circ \approx 0,6$ and $\cos 37^\circ \approx 0,8$).
- a) Draw in your notebook the forces acting on the parcel and calculate the magnitude of each.
- b) Calculate the acceleration with which it moves down.
- c) If the ramp is $8\ \text{m}$ long and the parcel starts from rest, calculate the speed with which it reaches the bottom and the time it takes (use $v^2 = 2 \cdot a \cdot x$ and $v = a \cdot t$).
You can do it on paper and send a photo with your answer.
To determine the coefficient of friction between a board and a block, the board is tilted gradually and it is observed that, when it forms $37^\circ$ with the…
To determine the coefficient of friction between a board and a block, the board is tilted gradually and it is observed that, when it forms $37^\circ$ with the horizontal, the block moves down at a constant speed (take $g = 10\ \text{m/s}^2$, $\sen 37^\circ \approx 0,6$ and $\cos 37^\circ \approx 0,8$).
- a) Draw in your notebook the forces on the block and explain why the net force is zero.
- b) Write the equations of equilibrium parallel to the plane and perpendicular to it.
- c) Deduce that under these conditions $\mu = \frac{\sen \alpha}{\cos \alpha} = \tan \alpha$ and calculate the value of $\mu$.
- d) Predict, showing the relevant calculation, the acceleration the block would have on that same ramp if the coefficient of friction were $\mu = 0,5$.
You can do it on paper and send a photo with your answer.
Systems of connected bodies: pulleys and strings
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.
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: $T - m_1 \cdot g = m_1 \cdot a$ for the one going up and $m_2 \cdot g - T = m_2 \cdot a$ 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.
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.
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.
A container of $5\ \text{kg}$ is on a ramp of $37^\circ$ in a cargo port, connected by a cable passing over an ideal pulley to a counterweight of $5\ \text{kg}$ hanging…
A container of $5\ \text{kg}$ is on a ramp of $37^\circ$ in a cargo port, connected by a cable passing over an ideal pulley to a counterweight of $5\ \text{kg}$ hanging vertically. The coefficient of friction between the container and the ramp is $\mu = 0,2$ (take $g = 10\ \text{m/s}^2$, $\sen 37^\circ \approx 0,6$ and $\cos 37^\circ \approx 0,8$). The diagram shows the ramp set-up with the counterweight.
- a) Draw in your notebook the forces acting on the container and on the counterweight.
- b) Calculate the component of the container's weight parallel to the ramp and the friction force, and reason in which direction the system moves.
- c) Calculate the acceleration of the system.
- d) Calculate the tension in the cable.
You can do it on paper and send a photo with your answer.