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Statics and dynamics · Dynamics

The Laws of Dynamics

A high-speed train traveling in a straight line at 300 km/h needs no net force to keep going that way; a lab cart accelerates in exact proportion to the force it receives; and a bat that hits a baseball receives, in turn, a force from the ball. These three ideas, formulated by Newton, are the basis for solving any dynamics problem, from a crate dragged across a workshop floor to a system of pulleys. At the end of the page we will come back to the train.

First Law: The Principle of Inertia

If you push a puck across a wooden table, it stops almost instantly. On ice it travels tens of meters, and on an air table it barely loses any speed. Do objects stop because they need a force to keep moving, or because something slows them down?

The answer is the second one: something slows them down (friction). The less friction there is, the longer they keep their motion.

Train in uniform rectilinear motion: the forward traction force is balanced by the backward resistance force. Train motor thrust total friction constant v → net force = 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: ΣF = m·a

If you push an empty shopping cart, it accelerates easily. If you push it when it is full, with the same force, it accelerates less. And if you push harder, it accelerates more. The second law sums up these two observations.

This means that if the net force is not zero, the state of motion changes; and that, to achieve the same change, a larger mass requires a larger force. The second law predicts the motion of macroscopic objects (cars, trains, crates) with great accuracy. It does not work at the atomic scale or near the speed of light.

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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

When the bat hits the ball, the ball shoots off. But does the bat also "feel" something? Yes: forces always come in pairs. When two objects interact, each one exerts a force on the other, and that is why energy can be transferred from one to the other.

A bat hits a ball: the bat exerts a force on the ball toward the right and the ball exerts on the bat another force of the same magnitude and opposite sense. bat ball F bat→ball F ball→bat Same magnitude, same direction, opposite sense, 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

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

To solve a dynamics problem, first draw a free-body diagram: the body isolated with all the forces acting on it, identifying what exerts each one. Some are contact forces (normal, friction, tension) and others act at a distance (weight).

Free-body diagram of a crate on a horizontal surface: weight and normal force vertical, applied force F to the right and friction Fr to the left. crate F Fr N P rough surface

Here $P$ is the weight, $N$ the normal force, $\mu$ the coefficient of friction, and $f$ the friction force (the diagram uses $F_r$).

Force Applied at an Angle

When you pull a cart with a rope at an angle, the force $F$ splits into a horizontal part and a vertical part.

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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 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

Inclined Plane

On a ramp, the weight splits into a component parallel to the plane (which moves or slows the body) and a perpendicular one (which determines the normal force and, with it, the friction).

Block on an inclined plane at an angle α: the weight P, vertical, splits into Pₓ, parallel to the plane and pointing down, and Pᵧ, perpendicular to the plane and pointing into it. α P Pₓ Pᵧ

Does It Stay Still or Slide Back Down?

If a body moving up a ramp stops for an instant, what happens next? Compare $P_x = mg\sin\alpha$ with the maximum static friction $\mu_s N = \mu_s\, mg\cos\alpha$. Dividing by $mg\cos\alpha$, the comparison is equivalent to comparing $\tan\alpha$ with $\mu_s$.

For example, with $\alpha = 37^\circ$ we have $\tan 37^\circ \approx 0.75$. If $\mu_s = 0.5$, then $0.75 > 0.5$: the body does not stay stopped, it slides down again.

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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

Systems of Connected Bodies: Pulleys and Ropes

Two identical buckets hang from a rope that passes over a pulley and stay still. If you add a nut to one of them, both start accelerating at the same time. How does that difference in weight travel through the rope?

When two bodies are connected by a massless, inextensible rope that passes over an ideal pulley, they have the same acceleration (in magnitude) and the tension is the same throughout the rope.

Atwood machine: mass m₂ goes down and m₁ goes up with constant acceleration while the pulley turns; the forces T and m·g accompany each mass. After a pause, everything returns to the start. m₁ T m₁·g m₂ T m₂·g

If one block is on a table with friction and the other hangs, the method is the same: one equation per body (friction appears only in the equation for the block on the table) and the system is solved. For a chain system, like a locomotive with train cars, the common acceleration comes from $\Sigma F = (\text{total mass})\,a$, and the tension in each coupling is obtained by applying the second law to the cars behind it.

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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 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
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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

Applications of the Second Law: Elevator, Graphs, and Error Analysis

The second law also explains what a scale reads inside an elevator. The scale measures the normal force $N$, not the weight. In the vertical direction, with the positive sense upward: $$N - P = m\,a$$ If the elevator starts moving upward ($a>0$), $N > P$ and you feel heavier; if it starts moving downward ($a<0$), $N < P$. The real weight, $P = m g$, does not change: what changes is the reading of the normal force.

Another way to apply it is by reading a velocity-time graph. The slope of each segment is the acceleration, and from that $\Sigma F = m a$ gives the net force in that segment. A horizontal segment means $a = 0$ and, therefore, $\Sigma F = 0$ (first law). If the cart accelerates from $0$ to $8\ \text{m/s}$ in $4\ \text{s}$, $a = 2\ \text{m/s}^2$.

Finally, on an inclined plane the body can move up while slowing down: then friction points down the ramp, just like $P_x$, and the (braking) acceleration is $a = g(\sin\alpha + \mu_d\cos\alpha)$ in magnitude. When it stops, the condition $\tan\alpha$ versus $\mu_s$ from the previous section applies. If the body slides down at constant velocity, $\Sigma F = 0$ and from $P\sin\alpha = \mu P\cos\alpha$ we get $\mu = \tan\alpha$.

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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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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
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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

Back to the beginning: the train at 300 km/h does not accelerate because the thrust of its engines is balanced by friction (first law); the cart accelerates as $a = \Sigma F/m$ (second law); and the bat receives from the ball an equal and opposite force (third law).

Summary

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