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Statics and dynamics · Dynamics

The Laws of Dynamics

A high-speed train at 300 km/h in a straight line needs no net force to keep going; a lab cart accelerates in exact proportion to the force it receives; the bat that hits a baseball receives a force from the ball in return. Those three ideas from Newton solve everything from a crate dragged across a corner store floor to a system of pulleys. How do you use them together?

First Law: The Principle of Inertia

If you push a disk across a wooden table, it stops quickly; on ice it travels much farther; on an air table it barely loses any speed. Do objects stop because they need a force to keep moving, or because something is slowing them down?

Here we need to distinguish two ideas. The applied force is a specific force that someone or something exerts (the push of an engine). The net force is the vector sum of all the external forces on the body, including friction. External forces come from other bodies; internal forces are the ones exerted between the parts of the system you are studying and, by the third law, they cancel in pairs inside it, so they do not change the motion of its center of mass.

Train in uniform rectilinear motion: the forward traction force is balanced by the backward resistance. Train motor thrust friction constant v → net F = 0
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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 $300\ \text{km/h}$.

  1. a) What is the acceleration of the train on that section?
  2. b) Applying Newton's first law, what is the net (resultant) force acting on the train?
  3. 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.
Solve

Second Law: The Fundamental Principle

Why? With a cart and masses in the lab you can see it: when you double the force, the acceleration doubles, and when you double the mass, the acceleration is cut in half.

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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.

  1. a) Write the expression for Newton's second law, $\sum \vec{F} = m \cdot \vec{a}$, and make acceleration the subject.
  2. b) Calculate the value of the acceleration of the trolley.
Solve
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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.

20 kg F = 40 N a = ? 40 kg F = 40 N a = ?
  1. a) Calculate the acceleration when only the scooter travels, having a mass of $20\ \text{kg}$.
  2. b) Calculate the acceleration when it carries a backpack and the total mass is $40\ \text{kg}$.
  3. c) Explain, using $\vec{a} = \frac{\vec{F}}{m}$, why doubling the mass halves the acceleration.
Solve
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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.

4 kg 30 N 18 N F net = ?
  1. a) Calculate the net force on the crate, stating its direction.
  2. b) Calculate the acceleration of the crate.
  3. 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.
Solve

Third Law: Action and Reaction

A bat hits a ball: the bat pushes the ball to the right and the ball pushes the bat with the same force to the left. bat ball F bat→ball F ball→bat Same magnitude, opposite direction, different bodies
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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.

bat ball F = 200 N on the bat = ?
  1. a) State Newton's third law (action and reaction).
  2. b) What force does the ball exert on the bat? State its magnitude, direction and sense.
  3. c) Explain why, despite both forces having the same magnitude, the ball flies off while the bat barely changes its motion.
Solve
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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.

T = ? 4 kg
  1. a) Draw in your notebook the forces acting on the lamp and state which body exerts each one.
  2. b) Calculate the tension in the cable.
  3. 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.

Solve

The Free-Body Diagram: Weight, Normal Force, Friction, and Tension

To solve a dynamics problem, first draw all the forces on the body and think about who exerts each one: the weight ($P = m\cdot g$, toward the center of the Earth), the normal force $N$ (perpendicular to the surface), the friction $F_f$ (parallel to the surface, against the sliding), and the tension $T$ of ropes and cables.

Free-body diagram of a crate on the floor: weight and normal force are vertical, applied force F points to the right and friction points to the left. crate F Ff N P
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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.

30° 10 kg P = ? Pₓ = ? Py = ?
  1. a) Calculate the weight of the container.
  2. b) Calculate the component of the weight parallel to the ramp, $P_x = P \times \sen 30^\circ$.
  3. c) Calculate the perpendicular component, $P_y = P \times \cos 30^\circ$, and state which force from the plane balances it.
Solve
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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 $g = 10\ \text{m/s}^2$).

  1. a) Calculate the normal force that the floor exerts on the crate.
  2. b) Calculate the friction force when the crate slides, $F_r = \mu \times N$.
  3. 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.
Solve
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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.

8 kg F = 52 N Fr = ? N = ? P = ? a = ?
  1. a) Calculate the normal force and the friction force.
  2. b) Calculate the net horizontal force on the trolley.
  3. c) Calculate its acceleration.
Solve
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To pull a garden cart of $10\ \text{kg}$, a rope is pulled with a force of $100\ \text{N}$ at an angle of $37^\circ$ to the horizontal. The coefficient of friction with…

To pull a garden cart of $10\ \text{kg}$, a rope is pulled with a force of $100\ \text{N}$ at an angle of $37^\circ$ to the horizontal. The coefficient of friction with the ground is $\mu = 0.25$ (take $g = 10\ \text{m/s}^2$, $\sen 37^\circ \approx 0.6$ and $\cos 37^\circ \approx 0.8$). The diagram shows the cart and the rope.

10 kg F = 100 N Fₓ = ? Fy = ? 37° N = ? P Fr = ?
  1. a) Resolve the force of the rope into its horizontal and vertical components.
  2. b) Taking into account that the rope "lifts" the cart slightly, calculate the normal force and the friction force.
  3. c) Calculate the net horizontal force and the acceleration of the cart.
Solve
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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.

60 kg N = ? P = m·g a = 1.5 m/s²
  1. a) What do the scales read with the lift at rest?
  2. 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.
  3. c) Has the person's real weight changed? Explain which quantity has changed.
Solve
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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 $g = 10\ \text{m/s}^2$).

  1. a) Calculate the minimum horizontal force required to set the crate in motion.
  2. b) Calculate the horizontal force required to keep it moving at a constant speed.
  3. c) In light of the results, explain why it takes more effort to start moving the crate than to keep it moving.
Solve

Systems of Several Connected Bodies

When several bodies are joined by couplings or inextensible ropes, they all have the same acceleration. You apply $\sum F = m\cdot a$ to the whole system to find $a$, and to a single isolated body to find the tension acting on it.

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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.

Diagram of locomotive and two wagons with tractive forces and tensionsa = ?Wagon 24 kgT₂ = ?Wagon 12 kgT₁ = ?Locomotive2 kgF = 24 N
  1. a) Calculate the acceleration of the system.
  2. 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.
  3. 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.
Solve

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 another perpendicular to it (which sets the normal force and, with it, the friction).

Block on an inclined plane at an angle α: the weight P splits into Px parallel to the plane and Py perpendicular to the plane. α P Px Py

Method: draw the forces, choose one axis parallel to the plane and another perpendicular to it, write $\sum F = m\cdot a$ along each axis (along the perpendicular one, $a = 0$), and solve.

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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 $\cos 30^\circ \approx 0.87$).

  1. a) Calculate the components of the weight parallel and perpendicular to the plane.
  2. b) Calculate the acceleration of the trolley.
  3. c) If it starts from rest, calculate the time it takes to travel $10\ \text{m}$ along the plane and its final velocity (use $x = \frac{1}{2} \times a \times t^2$ and $v = a \times t$).
Solve
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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.

37° 40 kg P Pₓ = ? Py = ? N = ? Fr = ?
  1. a) Calculate the component of the weight parallel to the slope.
  2. b) Calculate the normal force and the friction force.
  3. 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.
Solve
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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 $\cos 37^\circ \approx 0.8$).

  1. a) Draw in your notebook the forces acting on the parcel and calculate the magnitude of each.
  2. b) Calculate the acceleration with which it moves down.
  3. 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 $v^2 = 2 \times a \times x$ and $v = a \times t$).

You can do it on paper and send a photo with your answer.

Solve
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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 $\cos 37^\circ \approx 0.8$).

  1. a) Draw in your notebook the forces on the block and explain why the net force is zero.
  2. b) Write the equations of equilibrium parallel to the plane and perpendicular to it.
  3. c) Deduce that under these conditions $\mu = \frac{\sen \alpha}{\cos \alpha} = \tan \alpha$ and calculate the value of $\mu$.
  4. 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.

Solve
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A trolley of $2\ \text{kg}$ is launched upwards along an inclined plane of $37^\circ$ with an initial velocity of $8\ \text{m/s}$. The coefficient of dynamic friction is…

A trolley of $2\ \text{kg}$ is launched upwards along an inclined plane of $37^\circ$ with an initial velocity of $8\ \text{m/s}$. The coefficient of dynamic friction is $\mu_d = 0.25$ and the static coefficient is $\mu_e = 0.5$ (take $g = 10\ \text{m/s}^2$, $\sen 37^\circ \approx 0.6$ and $\cos 37^\circ \approx 0.8$). The diagram shows the trolley at the instant of launching.

37° 2 kg v₀ = 8 m/s a = ?
  1. a) Draw in your notebook the forces on the trolley while it moves up (note: friction opposes the motion) and calculate its acceleration.
  2. b) Calculate the distance it travels until it stops and the time it takes (use $v^2 = v_0^2 - 2 \times a \times x$ and $v = v_0 - a \times t$).
  3. c) Once stopped at the top, does the trolley slide back down? Compare the component of the weight parallel to the plane with the maximum static friction force and give reasons for your answer.

You can do it on paper and send a photo with your answer.

Solve
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A student calculates the acceleration of a block of $3\ \text{kg}$ sliding down a frictionless inclined plane of $30^\circ$ and writes: "since the only force is the…

A student calculates the acceleration of a block of $3\ \text{kg}$ sliding down a frictionless inclined plane of $30^\circ$ and writes: "since the only force is the weight, $a = \frac{F}{m} = \frac{m \times g}{m} = g = 10\ \text{m/s}^2$" (take $g = 10\ \text{m/s}^2$ and $\sen 30^\circ = 0.5$).

  1. a) Identify the two errors in the reasoning: which force has been omitted and which component of the weight actually accelerates the block.
  2. b) Calculate the correct acceleration of the block.
  3. c) Calculate the actual net force on the block and compare it with its weight.
Solve

Pulleys and Ropes

Two identical buckets hanging from a pulley stay still; if you add a nut to one, both accelerate together. How does that difference in weight travel through the rope?

With a rope of negligible mass, inextensible, over an ideal pulley, the two bodies have the same acceleration in magnitude and the tension is the same throughout the rope.

Atwood machine: two masses hanging from a rope over a pulley, with tension T upward and weight downward on each one. m₁ T m₁·g m₂ T m₂·g

If one block is on a table or a ramp and the other hangs, the procedure is the same: one equation per body, each along its direction of motion, including the friction on the block that slides.

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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.

m₁ = 3 kg m₂ = 5 kg a = ?
  1. a) Draw in your notebook the forces acting on each mass.
  2. 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.
  3. 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.

Solve
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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.

6 kg μ = 0.25 4 kg a = ? T = ?
  1. a) Draw in your notebook the forces acting on each block.
  2. b) Calculate the friction force on the block on the table.
  3. 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 $a$.
  4. d) Calculate the tension in the string.

You can do it on paper and send a photo with your answer.

Solve
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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 $g = 10\ \text{m/s}^2$).

  1. a) Draw in your notebook the forces acting on each mass.
  2. 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.
  3. 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.

Solve
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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.

37° μ = 0.2 5 kg 5 kg T = ? a = ?
  1. a) Draw in your notebook the forces acting on the container and on the counterweight.
  2. 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.
  3. c) Calculate the acceleration of the system.
  4. d) Calculate the tension in the cable.

You can do it on paper and send a photo with your answer.

Solve

Net Force and Motion: Reading a Graph

The second law connects the v-t graph with forces: the slope of v-t is the acceleration, and the net force is $m\cdot a$. Segment with increasing velocity: net force forward; constant velocity: zero net force; decreasing velocity: net force backward.

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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.

  1. a) Draw the velocity-time graph in your notebook.
  2. b) Calculate the acceleration of each of the three sections.
  3. c) Applying Newton's second law, calculate the net force on the car in each section and state its direction.
  4. 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.

Solve

Back to the beginning: the train, the cart, and the bat all follow the same rule: you add up the forces on one body and apply $\sum \vec{F} = m\cdot\vec{a}$.

Summary

See all the exercises in this section