Free-body diagrams and the combined effect of forces
Why can a sled press on the ground beneath it differently when it is pulled upward at an angle by a rope than when it simply stands still?
Prior knowledge
- Newton's second law $\Sigma\overline{F} = m\overline{a}$ from the chapter Newton's second law
- The sine and cosine of a right triangle
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
Two children take turns moving a sled on ice. One pulls on a short rope obliquely upward, and the other pushes the sled from behind with a handle that points obliquely downward. Both use an equal force. Friction on ice is small, but the runners of the sled press on the ice in different ways.
Which forces act on the sled in each case, and does the sled press on the ice with the same force as when it stands still?
Optional Review: a sliding puck
Predict
Make your predictions before investigating 2 questions
Prediction 1: the normal force in an oblique pull
Prediction 2: angle and acceleration
Investigate
In the simulation a sled on ice is pulled by a rope or pushed from a handle. On the left is the situation diagram and on the right the free-body diagram of the sled, with the lengths of the arrows to scale.
Simulation
How to use
Choose Situation (pull or push) and adjust the force $F$ and the angle $\alpha$. The free-body diagram updates immediately. Press Start and the sled starts to move; press Reset and it stops at the start. The Show the components of the force F box shows the components of the force $F$ with dashed lines. Speed slows down or speeds up the motion without changing the readings. The readings are under the picture.
Assumptions of the model
The values are model values calculated by the simulation. The ice is horizontal and friction is ignored, so only the weight $\overline{G}$, the normal force $\overline{N}$ from the ice and the force $\overline{F}$ of the rope or handle act on the sled. The rope's force is the rope's tension, usually written $\overline{T}$; here it and the handle's push share the symbol $\overline{F}$, because the simulation switches between pulling and pushing. The force is so small that the sled does not lift off the ice. There is no air drag. The blue arrow of the velocity is in the situation diagram, not in the free-body diagram, because velocity is not a force.
How the calculation works
The components of the force are $F_x = F\cos\alpha$ and $F_y = F\sin\alpha$. In a pull $F_y$ points up and in a push down. Vertically the sled does not accelerate, so $N + F_y - G = 0$ in a pull and $N - F_y - G = 0$ in a push, with $G = mg$ and $g = 9.81\,\text{m/s}^2$. Horizontally $F_x = ma$, so $a = \dfrac{F\cos\alpha}{m}$.
Experiment 1: the pulling angle changes
Test the predictions. The starting state chooses a pull, a force $F = 100\,\text{N}$ and a mass of $20\,\text{kg}$, so the weight of the sled is $196\,\text{N}$. Change only the angle $\alpha$. Record a measurement at least at the angles $0^\circ$, $30^\circ$ and $60^\circ$, and each time look at how the arrows of the free-body diagram change.
Initial state: situation: pull with a rope, force $F$ = $100\,\text{N}$, mass $m$ = $20\,\text{kg}$, angle $\alpha$ = $0^\circ$. Kept constant: situation, force $F$, mass $m$. Varied: angle $\alpha$.
| angle $\alpha$ | $N$ $(\text{N})$ | $F_x$ $(\text{N})$ | $a$ $(\text{m/s}^2)$ |
|---|
At the angle $30^\circ$ the normal force is smaller than the weight $196\,\text{N}$.
How large is the vertical component $F_y$ of the rope force then? Calculate it from the normal force in the table and the weight. Answer in newtons.
Hint 1: a guiding question
The sled does not accelerate vertically, so the sum of the vertical forces is zero.
Hint 2: what is the connection
$N$ and $F_y$ act up and $G$ down: $N + F_y - G = 0$.
Hint 3: an intermediate step
$F_y = G - N = 196\,\text{N} - N$, where $N$ is the normal force on the row $30^\circ$ of the table.
Show the solution
Given
from the table at the angle $\alpha = 30^\circ$ the normal force $N = 146\,\text{N}$; weight $G = 196\,\text{N}$ ($m = 20\,\text{kg}$); force $F = 100\,\text{N}$
Find
the vertical component $F_y$ of the rope force in $\text{N}$
Principle
The sled stays on the surface of the ice, so it does not accelerate vertically. By Newton's second law the sum of the vertical forces is then zero.
Free-body diagram
The sled as a rectangle, $y$-axis up. The normal force $N$ and the vertical component $F_y$ of the rope force act up, the weight $G$ down. The horizontal component $F_x$ points in the direction of motion, but it does not enter the vertical equation.
Formula
$N + F_y - G = 0$, so $F_y = G - N$
Conversions
All forces are in newtons, so no conversions are needed.
Substitution
Answer
$50\,\text{N}$
To the nearest newton, like the readings. The readings $196\,\text{N}$ and $146\,\text{N}$ are rounded to whole newtons, so their difference is reliable to about one newton. In a difference, the number of decimal places decides, not the number of significant figures.
Reasonableness
With trigonometry
the same result. The vertical component is smaller than the force $100\,\text{N}$, as a component must be.
Optional Experiment 2: pushing from a handle
Change pulling to pushing. The starting state chooses a push, a force $F = 100\,\text{N}$ and a mass of $20\,\text{kg}$. Change the angle and record the normal force at at least two angles.
Initial state: situation: push from a handle, force $F$ = $100\,\text{N}$, mass $m$ = $20\,\text{kg}$, angle $\alpha$ = $0^\circ$. Kept constant: situation, force $F$, mass $m$. Varied: angle $\alpha$.
| angle $\alpha$ | $N$ $(\text{N})$ | $a$ $(\text{m/s}^2)$ |
|---|
Conclusion
Using your measurements and the free-body diagram, explain why the normal force and the acceleration decrease when the pulling angle grows and the force stays the same.
Model answer
The rope force $F = 100\,\text{N}$ is split into the components $F_x = F\cos\alpha$ and $F_y = F\sin\alpha$. As the angle grows, the vertical component grows and supports a larger part of the weight, so the normal force $N = G - F\sin\alpha$ from the ice decreases: in the model values $196\,\text{N}$ in a horizontal pull and $146\,\text{N}$ at an angle of $30^\circ$. At the same time the horizontal component decreases, and since there is no friction, only it accelerates the sled: $a = \dfrac{F\cos\alpha}{m}$ is $5.0\,\text{m/s}^2$ at $0^\circ$ and $2.5\,\text{m/s}^2$ at $60^\circ$. So a force affects the motion not only by its magnitude but also by its direction.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Optional Measurement with a spring balance
Made-up example data
A student pulls a wooden block on a table with a spring balance whose string makes an angle of $30^\circ$ with the horizontal. The mass of the block is $0.60\,\text{kg}$. When the block moves at constant velocity, the balance shows the readings $2.1\,\text{N}$, $1.9\,\text{N}$, $2.0\,\text{N}$, $2.0\,\text{N}$ and $2.0\,\text{N}$ in five pulls, on average $2.0\,\text{N}$. Draw the free-body diagram of the block in your notebook before you answer.
How large is the normal force from the table on the block when the force of the balance is $2.0\,\text{N}$? Answer in newtons.
Hint 1: a guiding question
Draw the free-body diagram: weight, normal force, friction and the force of the balance. Which of them are vertical or have a vertical component?
Hint 2: what is the connection
The block does not accelerate vertically: $N + F\sin\alpha - mg = 0$.
Hint 3: an intermediate step
Show the solution
Given
$m = 0.60\,\text{kg}$, force of the balance $F = 2.0\,\text{N}$ (the mean of five readings), angle $\alpha = 30^\circ$; the block moves at constant velocity; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ from the table in $\text{N}$
Principle
The block moves uniformly and stays on the table, so its acceleration is zero. By Newton's second law the sum of the forces is zero in both directions; the vertical direction gives the normal force.
Free-body diagram
The block as a rectangle, $x$ in the direction of motion and $y$ up. The weight $\overline{G}$ down, the normal force $\overline{N}$ up, friction $\overline{F}_\mu$ backward and the force $\overline{F}$ of the balance obliquely up and forward. Vertically the forces are $N$, $F\sin\alpha$ and $G$.
Formula
$N + F\sin\alpha - mg = 0$, so $N = mg - F\sin\alpha$
Conversions
The mass is in kilograms and the force in newtons, so no conversions are needed.
Substitution
Answer
$4.9\,\text{N}$
Two significant figures like the mass $0.60\,\text{kg}$ and the force $2.0\,\text{N}$.
Reasonableness
The normal force is a little smaller than the weight $5.9\,\text{N}$, because the string lifts the block with a force of $1.0\,\text{N}$. Horizontally friction is $F\cos\alpha \approx 1.7\,\text{N}$, smaller than the force of the balance, as it should be in an oblique pull.
Theory summary
Key idea
Usually several forces act on an object at the same time. The Earth pulls it down, the surface supports it, a rope pulls and the ground resists the motion. Each force describes one interaction between the object and some other object. The motion is decided by the combined effect of these forces, the net force $\Sigma\overline{F}$, which is the vector sum of the forces. In the chapter Newton's second law you learn that the net force determines the acceleration: $\Sigma\overline{F} = m\overline{a}$.
To include every force and to keep anything extra out of the calculation, the forces are first drawn in a free-body diagram. It is a different picture from the situation diagram. The situation diagram shows the whole situation: the surface, the rope, the person pulling and the direction of motion. The free-body diagram shows only the chosen object, simplified to a rectangle, and the forces acting on it, as arrows. The direction of an arrow is the direction of the force, and the lengths of the arrows tell which force is larger.
In the figure, a box is pulled along by a rope that points obliquely upward. The force of the rope points both forward and upward. The upward part helps to hold the box up, so the floor does not have to support it as much as when the box stands still. That is why the normal force $N$ is smaller than the weight $G$. The velocity is not drawn in the free-body diagram, because velocity is not a force.
Drawing a free-body diagram
A free-body diagram is always drawn with the same eight steps. They are worth going through every time, even when the situation looks easy.
- Choose the object. Decide whose forces you draw. Only forces that act on this object go into the diagram. Forces that the object itself exerts on others do not belong in it.
- Find the interactions. List everything that acts on the object: the Earth (weight), every surface the object touches (normal force and friction), every rope (tension), a push or a pull, and air or water (drag and buoyancy). Every force has a cause outside the object. Motion alone does not cause a force.
- Draw the object and the forces. Draw the object simplified to a rectangle. Draw one arrow for each force, starting at the point where the force acts. The arrow of a pushing force may also end at that point.
- The weight $\overline{G}$ points straight down.
- The normal force $\overline{N}$ is perpendicular to the contact surface and pushes the object away from the surface. It does not always point up: a ceiling pushes down and a wall pushes sideways.
- The friction force $\overline{F}_\mu$ is parallel to the contact surface. Its direction comes from the direction in which the object slides, or tends to slide, relative to the surface. It does not always oppose the velocity. The friction arrow starts at a corner of the contact surface, outside the object, or ends there. It is not drawn along a side of the object from the middle of the side.
- The tension $\overline{T}$ acts along the rope and pulls the object toward the rope.
- Air drag is drawn only when the situation and its assumptions call for it.
- Name the forces. Give each force its own symbol. Two similar forces are told apart with subscripts, for example $\overline{T}_1$ and $\overline{T}_2$.
- Think about the magnitudes. At rest and in uniform motion the net force is zero, so the arrows cancel each other. In accelerating motion the net force points in the direction of the acceleration, which is not always the direction of the velocity. Do not invent exact lengths if the data does not fix them.
- Only forces. Velocity, acceleration and displacement are not forces. If they are needed, they are drawn beside the object as thin arrows in different colours: in this course forces are red, velocity is blue and acceleration is orange. The net force is not one more force.
- Components are not new forces. When a force is split into components, the components together replace the force in the calculation. They are drawn with a dashed line so that they are not mistaken for forces.
- Equation of motion. Only now are the axes chosen and Newton's second law written component by component from the forces and signs of the diagram.
The figure shows four situations in which the directions of the forces often surprise people. On a slope the normal force is perpendicular to the slope, not vertical, and friction holds the box in place along the slope. On a helium balloon pressed against the ceiling the ceiling pushes downward.
In a free-body diagram the object is always a simplified rectangle, and each force starts from its point of action: the weight from the centre, the normal force and friction from the contact surface, and the tension from where the rope is attached. The arrow of a pushing force may also end at the point where the push acts. In the situation diagram objects are drawn as they look, but in the free-body diagram only the rectangle and the forces are drawn.
The marking instructions of the matriculation examinations give task-specific points for the free-body diagram. For example, in the spring 2025 helium-balloon task a missing, extra or wrongly directed force cost all the points of the diagram, and so did drawing velocity or acceleration as a force on the balloon. In the autumn 2025 examination a missing name cost one point. The exact scoring varies from task to task, but every examination requires the same things: all the forces, no invented forces, correct directions and unambiguous names.
Equation of motion by components
Newton's second law concerns the net force:
$$\Sigma\overline{F} = m\overline{a}$$
where $\Sigma\overline{F}$ is the vector sum of the forces acting on the object (N), $m$ is the mass of the object (kg) and $\overline{a}$ is its acceleration (m/s²). When the forces are not along one line, two perpendicular axes are chosen and the law is written separately for each axis direction:
$$\begin{aligned} \Sigma F_x &= m a_x \\ \Sigma F_y &= m a_y \end{aligned}$$
where $\Sigma F_x$ and $\Sigma F_y$ are the sums of the components of the forces (N) and $a_x$ and $a_y$ are the components of the acceleration (m/s²). A component pointing along the axis is positive and one pointing against it is negative. It is wise to choose the axes so that one of them points in the direction of the acceleration.
The components of an oblique force $F$ come from a right triangle, when $\alpha$ is the angle between the force and the $x$-axis:
$$\begin{aligned} F_x &= F\cos\alpha \\ F_y &= F\sin\alpha \end{aligned}$$
where $F$ is the magnitude of the force (N), $F_x$ and $F_y$ are its components (N) and $\alpha$ is the angle. The components are not new forces. They replace the force $F$ in the calculation.
Common misconception
A moving object does not need a force in the direction of its motion. Only the weight, the normal force and friction act on a sliding box, and there is no "force of motion": there is no object that would cause it. Another common mistake is to think that the normal force is always equal to the weight. The normal force is exactly as large as the situation requires: in an oblique pull it is smaller than the weight, and in an oblique push or in an elevator accelerating upward it is larger. A third mistake is to draw the components as forces in addition to the force itself, which puts the same force into the calculation twice.
Summary
- A free-body diagram contains only the chosen object, as a rectangle, and the forces acting on it. The situation diagram and the free-body diagram are different pictures.
- Every force has a cause: the Earth, a surface, a rope, a pusher or the air. Velocity and acceleration are not forces, and they are drawn in different colours from the forces.
- The normal force is perpendicular to the surface, friction is parallel to the surface and tension is along the rope.
- The net force is the vector sum of the forces, and $\Sigma\overline{F} = m\overline{a}$ is written by components: $\Sigma F_x = m a_x$ and $\Sigma F_y = m a_y$.
- The components of an oblique force are $F\cos\alpha$ and $F\sin\alpha$. They replace the force in the calculation and are not new forces.
Optional Return to your predictions
Compare your predictions with the measurements. What did you learn about the size of the normal force and the importance of the direction of a force? Also return to the sled at the beginning: how does the normal force from the ice differ in a pull and in a push, and why?
Would you change your prediction?
Apply
Worked example
Problem
A sled of mass $12\,\text{kg}$ is pulled on horizontal snow by a rope that makes an angle of $35^\circ$ with the horizontal. The tension in the rope is $80\,\text{N}$ and the friction force from the snow is $30\,\text{N}$. What is the acceleration of the sled, and what is the normal force from the snow?
Step 1/8
Given and find. $m = 12\,\text{kg}$, $T = 80\,\text{N}$, $\alpha = 35^\circ$, $F_\mu = 30\,\text{N}$ and $g = 9.81\,\text{m/s}^2$. We want the acceleration $a$ and the normal force $N$.
Step 2/8
Principle. Choose the sled as the object. The Earth (weight), the snow (normal force and friction) and the rope (tension) act on it. Newton's second law is written in two directions: horizontally the sled accelerates, vertically it stays on the surface of the snow, so $a_y = 0$.
Step 3/8
Free-body diagram. The sled is drawn as a rectangle, and the forces start from their points of action. The axes are chosen so that $x$ points in the direction of motion and $y$ points up. The force of the rope is split into the components $T_x = T\cos\alpha$ and $T_y = T\sin\alpha$, drawn with dashed lines.
Step 4/8
Rearranged formula. The equations of motion are read from the diagram, and the asked quantities are solved:
$$\begin{aligned} &x\text{:}\ \ T\cos\alpha - F_\mu = m a \\ &y\text{:}\ \ N + T\sin\alpha - mg = 0 \\ &a = \frac{T\cos\alpha - F_\mu}{m} \\ &N = mg - T\sin\alpha \end{aligned}$$
Step 5/8
Unit conversions. None are needed: all quantities are in SI units, and $\tfrac{\text{N}}{\text{kg}} = \tfrac{\text{m}}{\text{s}^2}$.
Step 6/8
Substitution. Substitute the values with their units:
$$\begin{aligned} a &= \frac{T\cos\alpha - F_\mu}{m} \\ \phantom{a} &= \frac{80\,\text{N}\cdot\cos 35^\circ - 30\,\text{N}}{12\,\text{kg}} \\ \phantom{a} &\approx 2.96\,\text{m/s}^2 \\[4pt] N &= mg - T\sin\alpha \\ \phantom{N} &= 12\,\text{kg}\cdot 9.81\,\text{m/s}^2 \\ &\quad\ - 80\,\text{N}\cdot\sin 35^\circ \\ \phantom{N} &\approx 71.8\,\text{N} \end{aligned}$$
Step 7/8
Result. The acceleration is $a \approx 3.0\,\text{m/s}^2$ and the normal force is $N \approx 72\,\text{N}$, given with two significant figures like the data $80\,\text{N}$, $30\,\text{N}$ and $12\,\text{kg}$.
Step 8/8
Reasonableness. The weight of the sled is $mg \approx 118\,\text{N}$. The normal force $72\,\text{N}$ is smaller, because the vertical component of the rope force, $T_y \approx 46\,\text{N}$, supports part of the weight, and $72\,\text{N} + 46\,\text{N} \approx 118\,\text{N}$. Without friction the acceleration would be $65.5\,\text{N} : 12\,\text{kg} \approx 5.5\,\text{m/s}^2$, so friction reduces it, as it should.
Find the error
Choose the correct free-body diagram
Apply to a new situation
A fisher lifts a fish of $2.0\,\text{kg}$ with a line straight up so that the fish accelerates upward at $1.5\,\text{m/s}^2$.
How large is the tension in the line? Air drag is ignored. Answer in newtons.
Hint 1: a guiding question
Draw the free-body diagram of the fish. Which two forces act on the fish, and which one is larger when the fish accelerates up?
Hint 2: what is the connection
Choose the $y$-axis up. The equation of motion is $T - mg = ma$.
Hint 3: an intermediate step
Show the solution
Given
$m = 2.0\,\text{kg}$, acceleration upward $a = 1.5\,\text{m/s}^2$, $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ in the line in $\text{N}$
Principle
The fish accelerates, so the sum of the forces acting on it is not zero. By Newton's second law the net force is $ma$ in the direction of the acceleration, that is upward.
Free-body diagram
The fish as a rectangle, $y$-axis up. The tension $\overline{T}$ of the line up and the weight $\overline{G}$ down. $\overline{T}$ is longer, because the net force points up. The acceleration is drawn separately, not as a force.
Formula
$T - mg = ma$, so $T = m(g + a)$
Conversions
The mass is in kilograms and the accelerations in $\text{m/s}^2$, so no conversions are needed.
Substitution
Answer
$23\,\text{N}$
Two significant figures like the mass $2.0\,\text{kg}$ and the acceleration $1.5\,\text{m/s}^2$.
Reasonableness
The weight of the fish is about $20\,\text{N}$. An accelerating lift requires more, and $23\,\text{N}$ is a little larger. The excess of $3\,\text{N}$ is $ma = 2.0\,\text{kg}\cdot 1.5\,\text{m/s}^2$. If the fish were lowered with the same acceleration, the line would need only $m(g - a) \approx 17\,\text{N}$.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A ball has been thrown straight up, and it is just at the highest point of its path. Which forces act on the ball when air drag is ignored?
Answer
Only the weight $\overline{G}$, which points down. The thrower's hand is no longer touching the ball, so there is no upward force. At the highest point the velocity is zero for a moment, but the acceleration is still $g$ downward.
- Why are the components of a force drawn with a dashed line in the free-body diagram, and what happens if both the force and its components are used in the calculation?
Answer
The components are not new forces, but together they replace the force. The dashed line sets them apart from the forces. If both the force and its components are included in the calculation, the same force is counted twice and the result is wrong.
- A box is pulled on a horizontal floor by a rope that makes an angle $\alpha$ with the horizontal. The tension in the rope is $T$, the friction force $F_\mu$ and the mass of the box $m$. Write the vertical equation of motion and solve the normal force $N$ from it.
Answer
Vertically the box does not accelerate, so $N + T\sin\alpha - mg = 0$ and $N = mg - T\sin\alpha$. Friction is horizontal, so it does not belong in the vertical equation.
1. Concept
2. Calculation
A shop sign of mass $8.0\,\text{kg}$ hangs from two equally long cables. Each cable makes an angle of $50^\circ$ with the horizontal, and the sign is at rest.
How large is the tension in each cable? Answer in newtons.
Hint 1: a guiding question
Draw the free-body diagram of the sign: the weight and two tensions $\overline{T}_1$ and $\overline{T}_2$. The situation is symmetric, so $T_1 = T_2 = T$.
Hint 2: what is the connection
The horizontal components cancel each other. Vertically $2T\sin 50^\circ - mg = 0$.
Hint 3: an intermediate step
$T = \dfrac{mg}{2\sin 50^\circ}$.
Show the solution
Given
$m = 8.0\,\text{kg}$, angle of the cables to the horizontal $\alpha = 50^\circ$, the sign is at rest; $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ in one cable in $\text{N}$
Principle
The sign is at rest, so by Newton's second law the sum of the forces is zero in both directions. The situation is symmetric, so the tensions in the cables are equal.
Free-body diagram
The sign as a rectangle, $x$ to the right and $y$ up. The weight $\overline{G}$ down, the tension $\overline{T}_1$ obliquely up and to the left and $\overline{T}_2$ obliquely up and to the right, both $50^\circ$ from the horizontal. Their horizontal components $T\cos 50^\circ$ cancel each other, and the vertical components $T\sin 50^\circ$ point up.
Formula
$y$: $2T\sin\alpha - mg = 0$, so $T = \dfrac{mg}{2\sin\alpha}$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
Answer
$51\,\text{N}$
Two significant figures like the mass $8.0\,\text{kg}$.
Reasonableness
If the cables were vertical, each would carry half of the weight, $39\,\text{N}$. Oblique cables need more, because only part of the force is vertical, so $51\,\text{N} > 39\,\text{N}$ is reasonable. The flatter the cables, the larger the tension needed.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Draw it yourself
A box of mass $12\,\text{kg}$ is pushed on a horizontal floor from a handle that points obliquely downward at an angle of $30^\circ$. The pushing force is $60\,\text{N}$, and the box slides at constant velocity. There is friction between the floor and the box. Draw the free-body diagram of the box on paper: the box as a rectangle and each force from its point of action. Then write the equations of motion in the horizontal and vertical directions and calculate the friction force and the normal force.
Model diagram
The Earth, the floor and the handle act on the box. The gravitational pull of the Earth, that is the weight $\overline{G}$, points down from the centre. The floor acts from the bottom surface with two forces: the normal force $\overline{N}$ points perpendicularly up and the friction force $\overline{F}_\mu$ backward, because the box slides forward. The force $\overline{F}$ of the handle points obliquely down and forward. There is no extra force in the direction of motion. The box slides uniformly, so the net force is zero. With $x$ pointing forward and $y$ up: $x$: $F\cos 30^\circ - F_\mu = 0$, so $F_\mu = 60\,\text{N}\cdot\cos 30^\circ \approx 52\,\text{N}$. $y$: $N - mg - F\sin 30^\circ = 0$, so $N = 12\,\text{kg}\cdot 9.81\,\text{m/s}^2 + 60\,\text{N}\cdot\sin 30^\circ \approx 148\,\text{N}$. The normal force is larger than the weight $118\,\text{N}$, because the vertical component of the force of the handle presses the box against the floor.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Free-body diagram
- A drawing that shows every force acting on the object studied as an arrow.
- Normal force
- The force a surface exerts on an object, perpendicular to the surface.
- Tension
- The force a taut string or rope exerts on an object, along the string.
- Equation of motion
- Newton's second law applied: the sum of the forces on an object is linked to its acceleration.
- Component of a force
- The part of a force in a chosen direction of the coordinate system.
- Component form
- The equation of motion is written separately, for example in the horizontal and vertical directions.
- Condition of equilibrium
- In translational equilibrium the vector sum of the forces on an object is zero.$$\Sigma\overline{F} = \overline{0}$$
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 6-1Core
A lamp of mass $1.5\,\text{kg}$ hangs at rest from a vertical cord. Draw the free-body diagram of the lamp and calculate the tension in the cord.
Answer
$14.7\,\text{N} \approx 15\,\text{N}$
Show the solution
Given
$m = 1.5\,\text{kg}$, the lamp is at rest; $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ of the cord in $\text{N}$
Principle
The lamp is at rest, so by Newton's second law the sum of the forces acting on it is zero.
Free-body diagram
The lamp as a rectangle, $y$-axis up. The Earth and the cord act on the lamp: the weight $\overline{G}$ down and the tension $\overline{T}$ of the cord up along the cord. The arrows are equally long.
Formula
$T - mg = 0$, so $T = mg$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$T$$= 1.5\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 14.715\,\text{N}$Answer
$14.7\,\text{N} \approx 15\,\text{N}$
Two significant figures like the mass $1.5\,\text{kg}$.
Reasonableness
The weight of one kilogram is about $10\,\text{N}$, so the weight of a lamp of one and a half kilograms is about $15\,\text{N}$. The cord supports the whole weight, because there are no other forces acting upward.
- Exercise 6-2Core
In the free-body diagram of the figure one square corresponds to a force of $10\,\text{N}$. Four forces act on the object.
How large is the net force acting on the object?
Answer
$15\,\text{N}$
Show the solution
Given
forces read from the diagram: $F_1 = 40\,\text{N}$ to the right, $F_2 = 25\,\text{N}$ to the left, $F_3 = 30\,\text{N}$ up and $F_4 = 30\,\text{N}$ down
Find
the magnitude of the net force $\Sigma F$ in $\text{N}$
Principle
The net force is the vector sum of the forces. Calculate the sums of the components separately in each axis direction, with $x$ pointing to the right and $y$ up.
Reading the graph
The lengths of the arrows are read from the squares: $F_1$ is four squares, $F_2$ two and a half squares, and $F_3$ and $F_4$ three squares each. One square is $10\,\text{N}$.
Formula
$\Sigma F_x = F_1 - F_2$ and $\Sigma F_y = F_3 - F_4$. When $\Sigma F_y = 0$, the magnitude of the net force is $\Sigma F = |\Sigma F_x|$
Conversions
The squares are converted to newtons: $4\cdot 10\,\text{N} = 40\,\text{N}$ and $2.5\cdot 10\,\text{N} = 25\,\text{N}$.
Substitution
$\Sigma F_y$$= 30\,\text{N} - 30\,\text{N}$$= 0$$\Sigma F$$= 40\,\text{N} - 25\,\text{N}$$= 15\,\text{N}$Answer
$15\,\text{N}$
To the right, with two significant figures; the grid is read to the accuracy of half a square, that is $5\,\text{N}$.
Reasonableness
The vertical forces are equally long and cancel each other. Horizontally the arrow pointing right is one and a half squares longer, that is $15\,\text{N}$, so the net force points to the right. The net force is not the sum of the magnitudes of the forces, $125\,\text{N}$.
- Exercise 6-3
A student presses a book of $0.50\,\text{kg}$ against a vertical wall with a horizontal force of $20\,\text{N}$. The book stays at rest. Draw the free-body diagram of the book.
How large is the friction force from the wall on the book?
Answer
$4.9\,\text{N}$
Show the solution
Given
$m = 0.50\,\text{kg}$, the force of the hand $F = 20\,\text{N}$ horizontally toward the wall, the book is at rest; $g = 9.81\,\text{m/s}^2$
Find
the friction force $F_\mu$ from the wall in $\text{N}$
Principle
The book is at rest, so the sum of the forces is zero both horizontally and vertically. The wall is vertical, so its normal force is horizontal and its friction force is vertical.
Free-body diagram
The book as a rectangle, $x$ toward the wall and $y$ up. The force $\overline{F}$ of the hand toward the wall and the normal force $\overline{N}$ of the wall away from the wall, both horizontal. The weight $\overline{G}$ down and the friction force $\overline{F}_\mu$ of the wall up, because the book tends to slide down.
Formula
$y$: $F_\mu - mg = 0$, so $F_\mu = mg$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$F_\mu$$= 0.50\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 4.905\,\text{N}$Answer
$4.9\,\text{N}$
Two significant figures like the mass $0.50\,\text{kg}$.
Reasonableness
Friction supports the whole weight of the book, because it is the only force acting upward. The force of the hand, $20\,\text{N}$, does not appear directly in the answer: it determines the normal force $N = 20\,\text{N}$, but friction is exactly as large as staying at rest requires. Here the normal force does not point up but horizontally.
- Exercise 6-4
A person of mass $65\,\text{kg}$ stands in an elevator that moves upward at constant velocity. Draw the free-body diagram of the person.
How large is the normal force from the floor of the elevator on the person?
Answer
$640\,\text{N}$
Show the solution
Given
$m = 65\,\text{kg}$, the elevator moves at constant velocity; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ of the floor in $\text{N}$
Principle
In uniform motion the acceleration is zero, so by Newton's second law the sum of the forces is zero even though the person is moving.
Free-body diagram
The person as a rectangle, $y$-axis up. The weight $\overline{G}$ down and the normal force $\overline{N}$ of the floor up, equally long. The velocity is not drawn as a force.
Formula
$N - mg = 0$, so $N = mg$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$N$$= 65\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 637.65\,\text{N}$Answer
$640\,\text{N}$
Or $6.4\cdot 10^{2}\,\text{N}$, two significant figures like the mass $65\,\text{kg}$.
Reasonableness
The normal force is the same as the weight of the person, just as if the elevator were standing still. Motion at constant velocity does not require a net force. A person standing on a scale would read their ordinary weight.
- Exercise 6-5
A box is pulled with a rope whose tension is $250\,\text{N}$ and which makes an angle of $40^\circ$ with the horizontal. Draw the force of the rope and, with dashed lines, its components.
How large is the horizontal component of the tension?
Answer
$190\,\text{N}$
Show the solution
Given
$T = 250\,\text{N}$, angle to the horizontal $\alpha = 40^\circ$
Find
the horizontal component $T_x$ of the tension in $\text{N}$
Principle
The force and its components form a right triangle whose hypotenuse is the force. The horizontal component is the leg adjacent to the angle $\alpha$.
Free-body diagram
The force $\overline{T}$ of the rope as one solid arrow from the attachment point obliquely up at an angle of $40^\circ$, and the components $T_x$ and $T_y$ with dashed lines from the same point.
Formula
$T_x = T\cos\alpha$
Conversions
The force is in newtons and the angle in degrees, so no conversions are needed; the calculator is in DEG mode.
Substitution
$T_x$$= 250\,\text{N}\cdot\cos 40^\circ$$= 191.5\,\text{N}$Answer
$190\,\text{N}$
Or $1.9\cdot 10^{2}\,\text{N}$, two significant figures, because the angle $40^\circ$ is given with two figures and the force $250\,\text{N}$ is not more precise than that.
Reasonableness
The component is smaller than the force $250\,\text{N}$, as it should be. The angle is less than $45^\circ$, so the horizontal component is larger than the vertical component $250\,\text{N}\cdot\sin 40^\circ \approx 161\,\text{N}$.
- Exercise 6-6
A box slides to the right on a horizontal floor and slows down because of friction from the floor. Air drag is ignored.
Which forces are drawn in the free-body diagram of the box?
Answer
Weight, normal force and friction, because each force has a cause
Show the solution
Find
which forces are drawn in the free-body diagram when a box slides on a horizontal floor and slows down because of friction.
Principle
The free-body diagram includes only the forces that act on the chosen object, and every force has a cause: the Earth, a surface, a string or a pusher. Velocity and acceleration are not forces. According to Newton's second law, an object can move without a force in the direction of motion.
Model answer
Three forces act on the box. The Earth pulls the box down (weight), the floor supports it upward (normal force) and the floor resists the sliding to the left (friction). The weight and the normal force are equally long, because the box does not move in the vertical direction, and friction is the only horizontal force, so it slows the box down. There is no motion force pointing to the right, because there is no object that would cause it. Velocity is not drawn as a force. If needed, it is drawn separately beside the free-body diagram. The normal force must not be left out, even though it cancels the weight.
- Exercise 6-7
Two people push a cabinet on a horizontal floor in the same direction, one with a force of $120\,\text{N}$ and the other with $85\,\text{N}$. The friction force from the floor on the cabinet is $150\,\text{N}$ in the opposite direction. The weight and the normal force cancel each other.
How large is the net force acting on the cabinet?
Answer
$55\,\text{N}$
Show the solution
Given
pushing forces $F_1 = 120\,\text{N}$ and $F_2 = 85\,\text{N}$ in the same direction, friction force $F_\mu = 150\,\text{N}$ in the opposite direction; the weight and the normal force cancel each other
Find
the magnitude of the net force $\Sigma F$ in $\text{N}$
Principle
The net force is the vector sum of the forces. The forces lie on the same line, so the sum is calculated from the horizontal components, with $x$ pointing in the direction of the pushing. In the vertical direction the net force is zero.
Free-body diagram
The cabinet as a rectangle, $x$ in the direction of the pushing and $y$ up. Two pushing forces $\overline{F}_1$ and $\overline{F}_2$ forward, the friction force $\overline{F}_\mu$ backward, and the weight $\overline{G}$ down and the normal force $\overline{N}$ up with equal lengths. The friction force is longer than either pushing force but shorter than their sum.
Formula
$\Sigma F_x = F_1 + F_2 - F_\mu$
Conversions
All forces are in newtons, so no conversions are needed.
Substitution
$\Sigma F_x$$= 120\,\text{N} + 85\,\text{N} - 150\,\text{N}$$= 55\,\text{N}$Answer
$55\,\text{N}$
In the direction of the pushing, two significant figures like the given forces.
Reasonableness
The pushing forces total $205\,\text{N}$, which is greater than the friction $150\,\text{N}$, so the net force points in the direction of the pushing and the cabinet accelerates. If the pushing forces totalled less than $150\,\text{N}$, the cabinet would slow down or stay at rest. The net force is not the sum of the magnitudes of the forces, $355\,\text{N}$.
- Exercise 6-8
A box is on a horizontal floor. It is pulled with a string obliquely upward, but the box stays at rest on the floor.
How does the normal force $N$ from the floor compare with the weight $G$ of the box?
Answer
$N$ is smaller than the weight, because the string supports part of the weight
Show the solution
Find
how the normal force $N$ from the floor compares with the weight $G$ of the box when the box is pulled with a string obliquely upward and does not move.
Principle
The box is at rest, so according to Newton's second law the sum of the forces is zero also in the vertical direction. There are three forces in the vertical direction: the weight down, and the normal force and the vertical component of the tension of the string up.
Model answer
The equation in the vertical direction is $N + T\sin\alpha - G = 0$, so $N = G - T\sin\alpha$. The vertical component of the string supports part of the weight, and the floor needs to support only the rest, so the normal force is smaller than the weight. The normal force is therefore not always equal to the weight: it is exactly as large as the balance in the vertical direction requires. If the string pulled horizontally, the normal force would be equal to the weight, and if the string pulled the box obliquely downward, the normal force would be greater than the weight. The normal force would be zero only when the vertical component of the string were equal to the whole weight, and then the box would be about to lift off the floor.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 6-9Core
A sled of mass $15\,\text{kg}$ is pulled on horizontal snow by a rope that makes an angle of $30^\circ$ with the horizontal. The tension in the rope is $60\,\text{N}$, and the sled moves at constant velocity. Draw the free-body diagram of the sled in your notebook before you open the solution.
How large is the friction force from the snow?
Answer
$52\,\text{N}$
Show the solution
Given
$m = 15\,\text{kg}$, $T = 60\,\text{N}$, $\alpha = 30^\circ$, the sled moves at constant velocity
Find
the friction force $F_\mu$ from the snow in $\text{N}$
Principle
In uniform motion the acceleration is zero, so the sum of the forces is zero also in the direction of motion: friction is equal to the horizontal component of the force of the rope.
Free-body diagram
The sled as a rectangle, $x$ in the direction of motion and $y$ up. The weight $\overline{G}$ down, the normal force $\overline{N}$ from the snow up and shorter than the weight, the tension $\overline{T}$ of the rope obliquely up and forward, and the friction force $\overline{F}_\mu$ from the snow backward. The component $T_x$ is dashed.
Formula
$x$: $T\cos\alpha - F_\mu = 0$, so $F_\mu = T\cos\alpha$
Conversions
The force is in newtons, so no conversions are needed. The mass is not needed, because friction comes from the horizontal equation.
Substitution
$F_\mu$$= 60\,\text{N}\cdot\cos 30^\circ$$= 51.96\,\text{N}$Answer
$52\,\text{N}$
Two significant figures like the force $60\,\text{N}$.
Reasonableness
Friction is a little smaller than the force of the rope, $60\,\text{N}$, because only part of the force of the rope pulls forward. From the vertical direction the normal force is
$N$$= mg - T\sin\alpha$$\approx 147\,\text{N} - 30\,\text{N}$$= 117\,\text{N}$smaller than the weight.
- Exercise 6-10Core
A person of mass $60\,\text{kg}$ stands in an elevator. The elevator starts moving upward with an acceleration of $1.2\,\text{m/s}^2$.
How large is the normal force from the floor on the person during the acceleration?
Answer
$660\,\text{N}$
Show the solution
Given
$m = 60\,\text{kg}$, acceleration upward $a = 1.2\,\text{m/s}^2$; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ of the floor in $\text{N}$
Principle
The person accelerates upward with the elevator, so the net force points up and by Newton's second law is $ma$.
Free-body diagram
The person as a rectangle, $y$-axis up. The weight $\overline{G}$ down and the normal force $\overline{N}$ up, the normal force longer. The acceleration is drawn separately, not as a force.
Formula
$N - mg = ma$, so $N = m(g + a)$
Conversions
The mass is in kilograms and the accelerations in $\text{m/s}^2$, so no conversions are needed.
Substitution
$N$$= 60\,\text{kg}\cdot(9.81\,\text{m/s}^2 + 1.2\,\text{m/s}^2)$$= 60\,\text{kg}\cdot 11.01\,\text{m/s}^2$$= 660.6\,\text{N}$Answer
$660\,\text{N}$
Or $6.6\cdot 10^{2}\,\text{N}$, two significant figures like the mass $60\,\text{kg}$ and the acceleration $1.2\,\text{m/s}^2$.
Reasonableness
The weight is about $590\,\text{N}$, and the normal force is about $70\,\text{N}$ larger, which is
$ma$$= 60\,\text{kg}\cdot 1.2\,\text{m/s}^2$$= 72\,\text{N}$When the elevator starts upward you feel heavier, because the floor presses on your feet more than usual.
- Exercise 6-11
A box of mass $40\,\text{kg}$ is pushed on a horizontal floor from a handle with a force of $120\,\text{N}$ that points obliquely downward at an angle of $30^\circ$. The friction force from the floor on the box is $80\,\text{N}$.
How large is the acceleration of the box?
Answer
$0.60\,\text{m/s}^2$
Show the solution
Given
$m = 40\,\text{kg}$, $F = 120\,\text{N}$ at an angle $\alpha = 30^\circ$ below the horizontal, $F_\mu = 80\,\text{N}$
Find
the acceleration $a$ of the box in $\text{m/s}^2$
Principle
The box accelerates horizontally. By Newton's second law the sum of the horizontal force components is $ma$. The vertical equation is not needed, because friction is given.
Free-body diagram
The box as a rectangle, $x$ in the direction of motion and $y$ up. The weight $\overline{G}$ down, the normal force $\overline{N}$ from the floor up and longer than the weight, the force $\overline{F}$ of the handle obliquely down and forward, and the friction force $\overline{F}_\mu$ backward. The components $F_x$ and $F_y$ are dashed.
Formula
$x$: $F\cos\alpha - F_\mu = ma$, so $a = \dfrac{F\cos\alpha - F_\mu}{m}$
Conversions
The forces are in newtons and the mass in kilograms, so the result comes directly in $\text{m/s}^2$.
Substitution
$a$$= \dfrac{120\,\text{N}\cdot\cos 30^\circ - 80\,\text{N}}{40\,\text{kg}}$$= \dfrac{103.9\,\text{N} - 80\,\text{N}}{40\,\text{kg}}$$= 0.598\,\text{m/s}^2$Answer
$0.60\,\text{m/s}^2$
Two significant figures like the data.
Reasonableness
The horizontal component $104\,\text{N}$ is only a little larger than friction, $80\,\text{N}$, so the net force $24\,\text{N}$ is small and a box of $40\,\text{kg}$ accelerates slowly. If the whole force $120\,\text{N}$ acted forward, the acceleration would be $1.0\,\text{m/s}^2$.
- Exercise 6-12
Two children pull a sled of $50\,\text{kg}$ on ice with ropes that are horizontal and perpendicular to each other. One pulls with a force of $30\,\text{N}$ and the other with $40\,\text{N}$. Friction is ignored.
How large is the acceleration of the sled?
Answer
$1.0\,\text{m/s}^2$
Show the solution
Given
$m = 50\,\text{kg}$, $F_1 = 30\,\text{N}$ and $F_2 = 40\,\text{N}$ horizontal, perpendicular to each other; no friction
Find
the magnitude $a$ of the acceleration of the sled in $\text{m/s}^2$
Principle
The net force is the vector sum of the forces, and by Newton's second law it determines the acceleration. Vertically the weight and the normal force from the ice cancel each other.
Free-body diagram
The sled as a rectangle seen from above, $x$ in the direction of the force $\overline{F}_1$ and $y$ in the direction of the force $\overline{F}_2$. Seen from the side, the weight $\overline{G}$ down and the normal force $\overline{N}$ up, equally long.
Formula
$\Sigma F_x = F_1$ and $\Sigma F_y = F_2$, so $\Sigma F = \sqrt{F_1^2 + F_2^2}$. From Newton's second law $\Sigma F = ma$, so $a = \dfrac{\Sigma F}{m}$
Conversions
The forces are in newtons and the mass in kilograms, so no conversions are needed.
Substitution
$\Sigma F$$= \sqrt{(30\,\text{N})^2 + (40\,\text{N})^2}$$= 50\,\text{N}$$a$$= \dfrac{50\,\text{N}}{50\,\text{kg}}$$= 1.0\,\text{m/s}^2$Answer
$1.0\,\text{m/s}^2$
Two significant figures like the forces $30\,\text{N}$ and $40\,\text{N}$.
Reasonableness
The net force $50\,\text{N}$ is larger than either force but smaller than their sum $70\,\text{N}$, because the forces point in different directions. The acceleration points between the forces, closer to the larger force.
- Exercise 6-13
Force sensors are used to study a ring that stays at rest supported by three ropes. Rope A pulls the ring horizontally to the left and rope C vertically down. Rope B pulls the ring obliquely up and to the right at an angle $\alpha$ with the horizontal. The masses of the ropes and the mass and weight of the ring are ignored. The table has sample data from three measurements in which sensors A, B and C show the forces $F_\text{A}$, $F_\text{B}$ and $F_\text{C}$.
a) Explain why the sum of the forces acting on the ring is zero, and write the equations of motion in the $x$ and $y$ directions. Draw the free-body diagram of the ring as a rectangle.
b) Calculate the angle $\alpha$ from the forces $F_\text{A}$ and $F_\text{C}$ of each measurement and the mean of the three angles.
c) Check the result by calculating the force $F_\text{B}$ from the components of the first measurement and comparing it with the reading of the sensor.
The answer box takes the mean of the angles in part b).
Forces of three ropes on the ring (sample data) $F_\text{A}\,(\text{N})$ $F_\text{C}\,(\text{N})$ $F_\text{B}\,(\text{N})$ 3.0 2.5 3.9 4.5 3.8 5.9 6.0 5.1 7.9 Answer
$\alpha_\text{avg} = 40.1\,{}^\circ \approx 40\,{}^\circ$
Show the solution
Given
the ring is at rest; measurements $(F_\text{A}, F_\text{C}, F_\text{B})$: $(3.0$; $2.5$; $3.9)\,\text{N}$, $(4.5$; $3.8$; $5.9)\,\text{N}$ and $(6.0$; $5.1$; $7.9)\,\text{N}$
Find
the mean $\alpha_\text{avg}$ of the angle $\alpha$ in $^\circ$
Principle
The ring is at rest, so according to Newton's second law the sum of the forces acting on it is zero in both the horizontal and the vertical direction. Rope B is oblique, so its force is divided into components.
Free-body diagram
The ring as a rectangle, $x$ to the right and $y$ up. The tension $\overline{F}_\text{A}$ of rope A to the left and the tension $\overline{F}_\text{C}$ of rope C down. The force $\overline{F}_\text{B}$ of rope B starts obliquely up and to the right at an angle $\alpha$, and its components $F_\text{B}\cos\alpha$ and $F_\text{B}\sin\alpha$ are drawn with dashed lines. The components are as long as $F_\text{A}$ and $F_\text{C}$. The weight of the ring is not drawn, because it is ignored in the task.
Reading the graph
The forces $F_\text{A}$ and $F_\text{C}$ of each measurement are read from the table. The reading of sensor B is also taken for comparison.
Formula
$x$: $F_\text{B}\cos\alpha - F_\text{A} = 0$ and $y$: $F_\text{B}\sin\alpha - F_\text{C} = 0$, so
$\alpha$$= \arctan\dfrac{F_\text{C}}{F_\text{A}}$Conversions
The forces are in newtons and the angle is calculated in degrees, so no conversions are needed; the calculator is in DEG mode.
Substitution
The angles are
$$\begin{aligned} \alpha_1 &= \arctan\dfrac{2.5}{3.0} = 39.8\,{}^\circ \\ \alpha_2 &= \arctan\dfrac{3.8}{4.5} = 40.2\,{}^\circ \\ \alpha_3 &= \arctan\dfrac{5.1}{6.0} = 40.4\,{}^\circ \end{aligned}$$
The sum is $120.4\,{}^\circ$, so the mean is
$\alpha_\text{avg}$$= 120.4\,{}^\circ : 3$$= 40.1\,{}^\circ$Check with the first measurement:
$F_\text{B}$$= \sqrt{(3.0\,\text{N})^2 + (2.5\,\text{N})^2}$$= 3.9\,\text{N}$Answer
a) The free-body diagram and the equations are above. b) $\alpha_\text{avg} = 40.1\,{}^\circ \approx 40\,{}^\circ$, to two significant figures like the readings of the forces. c) The calculated $F_\text{B} = 3.9\,\text{N}$ is the same as the reading of the sensor.
Reasonableness
The three angles differ from each other by at most $0.6\,{}^\circ$, so they describe the same angle and the small variation is random variation of the measurement. The forces grow in the same proportion, and the ratio $F_\text{C} : F_\text{A}$ stays at about $0.84$. The angle is less than $45\,{}^\circ$, so $F_\text{A} > F_\text{C}$ as in the table. The force $F_\text{B}$ is greater than either component but smaller than their sum, $5.5\,\text{N}$.
- Exercise 6-14
A box of mass $12\,\text{kg}$ is kicked into motion on a horizontal floor. The box slides to the right and slows down. The friction force from the floor on the box is $30\,\text{N}$ to the left. Air drag is ignored. A student solves the acceleration as follows:
“The box moves to the right, so the force from the kick, $40\,\text{N}$ to the right, acts on it. The net force is $40\,\text{N} - 30\,\text{N} = 10\,\text{N}$ to the right, so $a = 10\,\text{N} : 12\,\text{kg} \approx 0.83\,\text{m/s}^2$ to the right and the box accelerates.”
a) What is the student's error? Draw the correct free-body diagram of the box.
b) Calculate the magnitude of the acceleration of the box correctly.
The answer box takes the magnitude of the acceleration in part b).
Answer
$a = 2.5\,\text{m/s}^2$
Show the solution
Given
$m = 12\,\text{kg}$, friction force $F_\mu = 30\,\text{N}$ to the left, the box slides to the right and slows down; there is no air drag
Find
a) the student's error; b) the magnitude of the acceleration $a$ in $\text{m/s}^2$
Principle
The free-body diagram includes only forces that act on the object and have a cause. The student drew a “force from the kick” on the box, although after the kick no object pushes the box to the right, and concluded the direction of the acceleration from the direction of the motion, although the box slows down. The force of the kick acts on the box only while the foot touches it. When the box has left the foot, only friction affects its horizontal motion.
Free-body diagram
The box as a rectangle, $x$ to the right and $y$ up. The weight $\overline{G}$ down and the normal force $\overline{N}$ from the floor up with equal lengths, and the friction force $\overline{F}_\mu$ to the left. No force is drawn to the right. The velocity to the right and the acceleration to the left are drawn separately beside the diagram.
Formula
$x$: $-F_\mu = m a_x$, so $a_x = -\dfrac{F_\mu}{m}$ and the magnitude of the acceleration is $a = \dfrac{F_\mu}{m}$
Conversions
The force is in newtons and the mass in kilograms, so the result comes directly in $\text{m/s}^2$.
Substitution
$a$$= \dfrac{30\,\text{N}}{12\,\text{kg}}$$= 2.5\,\text{m/s}^2$Answer
a) An extra force that has no cause was included in the free-body diagram. b) $a = 2.5\,\text{m/s}^2$ to the left, that is, against the motion, two significant figures like the given values. The box slows down.
Reasonableness
The net force $30\,\text{N}$ points to the left, so the acceleration points to the left and the box slows down, as the task says. The student's result, according to which the box accelerates, contradicts this. Friction is only about a quarter of the weight of the box, $118\,\text{N}$.
- Exercise 6-15
A student studies experimentally how two oblique forces combine. A ring is on a horizontal table, so its weight and the normal force from the table cancel each other. She attaches three force sensors to the ring. Sensors 1 and 2 pull the ring with forces $F_1 = 6.0\,\text{N}$ and $F_2 = 8.0\,\text{N}$, and the angle between their directions is $\varphi$. Sensor 3 pulls the ring in the direction opposite to the sum of the forces $\overline{F}_1$ and $\overline{F}_2$, and the ring stays at rest. The table has sample data from three measurements.
a) Explain why the reading $F_3$ of sensor 3 is equal to the magnitude of the sum of the forces $\overline{F}_1$ and $\overline{F}_2$.
b) Calculate the magnitude of the sum of the forces $\overline{F}_1$ and $\overline{F}_2$ when $\varphi = 60\,{}^\circ$. Choose the $x$ axis in the direction of the force $\overline{F}_1$.
c) Compare the calculated value with the measured one and tell what the small difference may be due to.
The answer box takes the magnitude of the sum in part b).
Reading of sensor 3 at different angles (sample data) $\varphi\,({}^\circ)$ $F_3\,(\text{N})$ 30 13.4 60 12.3 90 10.1 Answer
$\Sigma F \approx 12\,\text{N}$
Show the solution
Given
$F_1 = 6.0\,\text{N}$, $F_2 = 8.0\,\text{N}$, the angle between their directions $\varphi = 60\,{}^\circ$; from the table $F_3 = 12.3\,\text{N}$ when $\varphi = 60\,{}^\circ$
Find
the magnitude of the sum of the forces $\overline{F}_1$ and $\overline{F}_2$, $\Sigma F$, in $\text{N}$ when $\varphi = 60\,{}^\circ$
Principle
a) The ring is at rest, so according to Newton's second law the sum of the forces $\overline{F}_1$, $\overline{F}_2$ and $\overline{F}_3$ is zero. Therefore the force of sensor 3 is equal in magnitude and opposite in direction to the sum of the forces $\overline{F}_1$ and $\overline{F}_2$, and its reading tells the magnitude of the sum. b) The sum of oblique forces is calculated by components.
Free-body diagram
The ring as a rectangle, $x$ in the direction of the force $\overline{F}_1$. The force $\overline{F}_1$ starts straight in the direction of the $x$ axis and the force $\overline{F}_2$ at an angle of $60\,{}^\circ$ upward. The force $\overline{F}_3$ starts in the direction opposite to their sum. The components $F_{2x}$ and $F_{2y}$ are drawn with dashed lines.
Reading the graph
The row $\varphi = 60\,{}^\circ$ is read from the table: $F_3 = 12.3\,\text{N}$.
Formula
$\Sigma F_x = F_1 + F_2\cos\varphi$ and $\Sigma F_y = F_2\sin\varphi$, so $\Sigma F = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2}$
Conversions
The forces are in newtons and the angle in degrees, so no conversions are needed.
Substitution
First the components:
$$\begin{aligned} \Sigma F_x &= 6.0\,\text{N} + 8.0\,\text{N}\cdot\cos 60\,{}^\circ \\ &= 10.0\,\text{N} \\ \Sigma F_y &= 8.0\,\text{N}\cdot\sin 60\,{}^\circ \\ &= 6.93\,\text{N} \end{aligned}$$
Then the magnitude:
$\Sigma F$$= \sqrt{(10.0\,\text{N})^2 + (6.93\,\text{N})^2}$$= 12.2\,\text{N}$Answer
a) The ring is at rest, so the force $\overline{F}_3$ cancels the sum of the forces $\overline{F}_1$ and $\overline{F}_2$. b) $\Sigma F \approx 12\,\text{N}$, two significant figures like the given forces. c) The measured $12.3\,\text{N}$ differs from the calculated $12.2\,\text{N}$ by about $0.1\,\text{N}$, that is, about one per cent; the difference is of the order of the measurement uncertainty. It may be due to inaccuracy in reading the angle, zeroing of the sensors, or the ring not being completely at rest.
Reasonableness
The sum $12.2\,\text{N}$ is smaller than the sum of the magnitudes of the forces, $14.0\,\text{N}$, and greater than the greater force $8.0\,\text{N}$, as it should be at the angle $60\,{}^\circ$. In the table the reading decreases as the angle grows ($13.4\,\text{N}$, $12.3\,\text{N}$, $10.1\,\text{N}$), and at the angle $90\,{}^\circ$ the sum is $\sqrt{6.0^2 + 8.0^2}\,\text{N} = 10\,\text{N}$ as in the measurement.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 6-16
A person of mass $70\,\text{kg}$ stands in an elevator that moves upward but slows down as it stops. The deceleration is $1.5\,\text{m/s}^2$. Draw the free-body diagram of the person in your notebook, and mark the velocity and the acceleration separately beside it, before you open the solution.
How large is the normal force from the floor?
Answer
$580\,\text{N}$
Show the solution
Given
$m = 70\,\text{kg}$, the elevator moves upward and slows down, magnitude of the acceleration $1.5\,\text{m/s}^2$; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ of the floor in $\text{N}$
Principle
An elevator moving upward is slowing down, so its acceleration points down even though the velocity points up. By Newton's second law the net force points in the direction of the acceleration, that is down.
Free-body diagram
The person as a rectangle, $y$-axis up. The weight $\overline{G}$ down and the normal force $\overline{N}$ up, the normal force shorter. The velocity $\overline{v}$ up (blue) and the acceleration $\overline{a}$ down (orange) are drawn beside the diagram as thin arrows, not as forces. The component of the acceleration is $a_y = -1.5\,\text{m/s}^2$.
Formula
$N - mg = m a_y$, so $N = m(g + a_y)$
Conversions
The mass is in kilograms and the accelerations in $\text{m/s}^2$, so no conversions are needed.
Substitution
$N$$= 70\,\text{kg}\cdot(9.81\,\text{m/s}^2 - 1.5\,\text{m/s}^2)$$= 70\,\text{kg}\cdot 8.31\,\text{m/s}^2$$= 581.7\,\text{N}$Answer
$580\,\text{N}$
Or $5.8\cdot 10^{2}\,\text{N}$, two significant figures like the mass $70\,\text{kg}$ and the acceleration $1.5\,\text{m/s}^2$.
Reasonableness
The weight is about $690\,\text{N}$, and the normal force is smaller, because the net force points down. When the elevator stops at the top you feel lighter for a moment. If you got a normal force larger than the weight, the direction of the acceleration would have been chosen by the velocity, which is a common mistake.
- Exercise 6-17
A curling stone of mass $19\,\text{kg}$ leaves the thrower's hand at a velocity of $2.0\,\text{m/s}$ and slides straight until it stops after $20\,\text{s}$. Assume that the stone slows down uniformly and that air drag is negligibly small.
How large is the friction force from the ice on the stone during the slide?
Answer
$1.9\,\text{N}$
Show the solution
Given
$m = 19\,\text{kg}$, initial velocity $v_0 = 2.0\,\text{m/s}$, final velocity $v = 0$, time $\Delta t = 20\,\text{s}$; the deceleration is uniform
Find
the friction force $F_\mu$ in $\text{N}$
Principle
The speed of the stone changes uniformly, so the acceleration is $a = \dfrac{\Delta v}{\Delta t}$ (chapter Acceleration and uniformly accelerated motion). By Newton's second law the net force in the direction of motion is $ma$, and the only horizontal force is friction.
Free-body diagram
The stone as a rectangle, $x$ in the direction of motion and $y$ up. The weight $\overline{G}$ down, the normal force $\overline{N}$ of the ice up and the friction force $\overline{F}_\mu$ backward. There is no forward force, because the thrower is no longer touching the stone.
Formula
$a_x = \dfrac{v - v_0}{\Delta t}$ and $x$: $-F_\mu = m a_x$, so $F_\mu = \dfrac{m(v_0 - v)}{\Delta t}$
Conversions
The mass is in kilograms, the velocities in $\text{m/s}$ and the time in seconds, so no conversions are needed.
Substitution
$F_\mu$$= \dfrac{19\,\text{kg}\cdot(2.0\,\text{m/s} - 0)}{20\,\text{s}}$$= 19\,\text{kg}\cdot 0.10\,\text{m/s}^2$$= 1.9\,\text{N}$Answer
$1.9\,\text{N}$
Two significant figures like the mass, the velocity and the time.
Reasonableness
The weight of the stone is about $190\,\text{N}$, and friction is only about one percent of it, which suits smooth ice. The component of the acceleration is negative because the stone slows down, and friction points backward as in the diagram.
- Exercise 6-18
A helium balloon whose mass with its rubber and gas is $4.0\,\text{g}$ has risen to the ceiling of a room and is pressed against it at rest. The buoyancy on the balloon from the air is $0.10\,\text{N}$. The string is not attached. Draw the free-body diagram of the balloon and calculate the force with which the ceiling acts on the balloon.
Answer
$0.061\,\text{N} \approx 0.06\,\text{N}$
Show the solution
Given
$m = 4.0\,\text{g}$, buoyancy $F_\text{b} = 0.10\,\text{N}$, the balloon is at rest against the ceiling; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ of the ceiling in $\text{N}$
Principle
The balloon is at rest, so by Newton's second law the sum of the forces is zero. The ceiling touches the balloon from above, so its normal force is perpendicular to the ceiling and pushes the balloon away from the ceiling, that is down.
Free-body diagram
The balloon as a rectangle, $y$-axis up. Buoyancy $\overline{F}_\text{b}$ up, the weight $\overline{G}$ down and the normal force $\overline{N}$ of the ceiling down. The buoyancy arrow is as long as the weight and normal force arrows together. There is no air drag, because the balloon is not moving.
Formula
$F_\text{b} - mg - N = 0$, so $N = F_\text{b} - mg$
Conversions
$m$$= 4.0\,\text{g}$$= 4.0 \cdot 0.001\,\text{kg}$$= 0.0040\,\text{kg}$Substitution
$N$$= 0.10\,\text{N} - 0.0040\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 0.10\,\text{N} - 0.03924\,\text{N}$$= 0.06076\,\text{N}$Answer
$0.061\,\text{N} \approx 0.06\,\text{N}$
The normal force is a difference, and in a difference the number of decimals decides: the buoyancy $0.10\,\text{N}$ is known to a hundredth of a newton.
Reasonableness
The buoyancy is larger than the weight, $0.039\,\text{N}$, so without the ceiling the balloon would rise. The ceiling stops it by pushing down exactly by the excess: $0.039\,\text{N} + 0.061\,\text{N} = 0.10\,\text{N}$. Here the normal force points down, not up.
- Exercise 6-19
A box of mass $25\,\text{kg}$ is pulled on a horizontal floor with a rope that makes an angle of $30\,{}^\circ$ with the horizontal. The tension in the rope is $120\,\text{N}$ all the time. The box starts from rest, and a motion sensor measures its velocity. The table has sample data from the measurement. The mass of the rope and air drag are ignored.
a) Draw the free-body diagram of the box and write the equations of motion in the $x$ and $y$ directions.
b) Determine the acceleration of the box from the velocities in the table.
c) Calculate the friction force from the floor on the box.
d) Calculate the normal force from the floor and explain why it is smaller than the weight.
The answer box takes the friction force in part c).
Velocity of the box while the rope pulls (sample data) $t\,(\text{s})$ $v\,(\text{m/s})$ 0.0 0.00 1.0 0.78 2.0 1.63 3.0 2.38 4.0 3.22 5.0 4.01 Answer
$F_\mu \approx 84\,\text{N}$
Show the solution
Given
$m = 25\,\text{kg}$, $T = 120\,\text{N}$, $\alpha = 30\,{}^\circ$, the box starts from rest; from the table $v = 0$ at $t = 0$ and $v = 4.01\,\text{m/s}$ at $t = 5.0\,\text{s}$; $g = 9.81\,\text{m/s}^2$
Find
the friction force $F_\mu$ in $\text{N}$ and the normal force $N$
Principle
The box accelerates in the horizontal direction and does not move in the vertical direction. The acceleration is obtained from the change in velocity (chapter Acceleration and uniformly accelerated motion), and Newton's second law is written by components: in the horizontal direction the net force is $ma$, in the vertical direction zero.
Free-body diagram
The box as a rectangle, $x$ in the direction of motion and $y$ up. The weight $\overline{G}$ down and the normal force $\overline{N}$ from the floor up and shorter than the weight, the tension $\overline{T}$ of the rope obliquely up and forward, and the friction force $\overline{F}_\mu$ backward. The components $T_x$ and $T_y$ with dashed lines. The acceleration forward is drawn separately, not as a force.
Reading the graph
The velocity grows by about the same amount every second: the changes of consecutive seconds are $0.78$, $0.85$, $0.75$, $0.84$ and $0.79\,\text{m/s}$. The acceleration is therefore approximately constant, and it is obtained from the change over the whole interval,
$\Delta v$$= 4.01\,\text{m/s} - 0$$= 4.01\,\text{m/s}$$\Delta t = 5.0\,\text{s}$.
Formula
$a = \dfrac{\Delta v}{\Delta t}$, $x$: $T\cos\alpha - F_\mu = ma$ and $y$: $N + T\sin\alpha - mg = 0$, so $F_\mu = T\cos\alpha - ma$ and $N = mg - T\sin\alpha$
Conversions
All quantities are in SI units, and $\text{N} / \text{kg} = \text{m/s}^2$.
Substitution
The acceleration is
$a$$= \dfrac{4.01\,\text{m/s}}{5.0\,\text{s}}$$= 0.802\,\text{m/s}^2$The friction force:
$$\begin{aligned} F_\mu &= 120\,\text{N}\cdot\cos 30\,{}^\circ \\ &\quad - 25\,\text{kg}\cdot 0.802\,\text{m/s}^2 \\ &= 103.9\,\text{N} - 20.05\,\text{N} \\ &= 83.9\,\text{N} \end{aligned}$$
The normal force:
$$\begin{aligned} N &= 25\,\text{kg}\cdot 9.81\,\text{m/s}^2 \\ &\quad - 120\,\text{N}\cdot\sin 30\,{}^\circ \\ &= 245.25\,\text{N} - 60\,\text{N} \\ &= 185\,\text{N} \end{aligned}$$
Answer
a) The free-body diagram and the equations are above. b) $a \approx 0.80\,\text{m/s}^2$. c) $F_\mu \approx 84\,\text{N}$, two significant figures like the mass and the angle. d) $N \approx 185\,\text{N} \approx 1.9\cdot 10^2\,\text{N}$, two significant figures like the mass and the angle. The normal force is smaller than the weight, $245\,\text{N}$, because the vertical component of the rope, $T\sin\alpha = 60\,\text{N}$, supports part of the weight: the floor needs to support only the rest, $N + T\sin\alpha = mg$.
Reasonableness
The horizontal component of the rope, $104\,\text{N}$, is greater than the friction $84\,\text{N}$, so the net force $20\,\text{N}$ points forward and the box accelerates, as the table shows. The net force $ma = 20\,\text{N}$ is about a fifth of the horizontal component. Check in the vertical direction:
$185\,\text{N} + 60\,\text{N}$$= 245\,\text{N}$$= mg$ - Exercise 6-20
A crane lifts a load of mass $80\,\text{kg}$ from rest straight up and stops it at the top. The lift takes $8.0\,\text{s}$ and has three phases: the load accelerates between the moments $0.0\,\text{s}$ and $2.0\,\text{s}$, moves at constant velocity between $2.0\,\text{s}$ and $6.0\,\text{s}$, and decelerates to a stop between $6.0\,\text{s}$ and $8.0\,\text{s}$. The acceleration of each phase is approximately constant. A force sensor measures the force of the rope on the load, and the table has sample data of the readings. The mass of the rope and air drag are ignored.
a) Draw the free-body diagram of the load in the first phase and write the equation of motion.
b) Calculate the acceleration of the load in each phase using the mean of the readings of the phase.
c) Explain why the force of the rope in the third phase is smaller than the weight although the load moves upward.
d) How high is the load lifted?
The answer box takes the lifting height in part d).
Force of the rope on the load (sample data) $t\,(\text{s})$ $T\,(\text{N})$ 0.5 881 1.0 878 1.5 883 3.0 786 4.0 783 5.0 785 6.5 691 7.0 688 7.5 690 Answer
$h \approx 14.4\,\text{m}$
Show the solution
Given
$m = 80\,\text{kg}$, the load starts from rest and stops; phases $0$–$2.0\,\text{s}$, $2.0$–$6.0\,\text{s}$ and $6.0$–$8.0\,\text{s}$; the force of the rope from the table; $g = 9.81\,\text{m/s}^2$
Find
the lifting height $h$ in $\text{m}$ and the acceleration in each phase
Principle
The weight and the tension of the rope act on the load. According to Newton's second law the acceleration is obtained from the difference of the forces, and the direction of the acceleration is the direction of the net force: up when the force of the rope is greater than the weight, and down when it is smaller than the weight. When the acceleration is known, the height is obtained from the distances of the phases of the motion (chapter Acceleration and uniformly accelerated motion).
Free-body diagram
In the first phase the load is a rectangle, $y$ up. The weight $\overline{G}$ down and the tension $\overline{T}$ of the rope up, the tension longer than the weight. The velocity up and the acceleration up are drawn separately beside the diagram, not as forces. In the second phase the arrows are equally long, and in the third phase the tension is shorter than the weight.
Reading the graph
The readings of the table are divided into phases according to the moments $2.0\,\text{s}$ and $6.0\,\text{s}$. The sum of the readings of phase 1 is $881\,\text{N} + 878\,\text{N} + 883\,\text{N} = 2642\,\text{N}$, so the mean is $880.7\,\text{N}$. In phase 2 the sum is $2354\,\text{N}$ and the mean $784.7\,\text{N}$, and in phase 3 the sum is $2069\,\text{N}$ and the mean $689.7\,\text{N}$.
Formula
$T - mg = m a_y$, so $a_y = \dfrac{T - mg}{m}$. The greatest velocity is $v_\text{max} = a_1 \Delta t_1$, and the distances are $h_1 = \tfrac12 a_1 \Delta t_1^2$, $h_2 = v_\text{max}\,\Delta t_2$ and $h_3 = \tfrac12 v_\text{max}\,\Delta t_3$, because in the third phase the velocity decreases uniformly to zero. The height is $h = h_1 + h_2 + h_3$
Conversions
The forces are in newtons, the mass in kilograms and the times in seconds, so no conversions are needed. The weight is
$mg$$= 80\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 784.8\,\text{N}$Substitution
The accelerations:
$$\begin{aligned} a_1 &= \frac{880.7 - 784.8}{80}\,\frac{\text{m}}{\text{s}^2} \\ &= 1.20\,\text{m/s}^2 \\ a_2 &= \frac{784.7 - 784.8}{80}\,\frac{\text{m}}{\text{s}^2} \\ &\approx 0 \\ a_3 &= \frac{689.7 - 784.8}{80}\,\frac{\text{m}}{\text{s}^2} \\ &= -1.19\,\text{m/s}^2 \end{aligned}$$
The greatest velocity is
$v_\text{max}$$= 1.198\,\text{m/s}^2\cdot 2.0\,\text{s}$$= 2.397\,\text{m/s}$The distances:
$$\begin{aligned} h_1 &= \tfrac12\cdot 1.198\,\tfrac{\text{m}}{\text{s}^2}\cdot (2.0\,\text{s})^2 \\ &= 2.40\,\text{m} \\ h_2 &= 2.397\,\tfrac{\text{m}}{\text{s}}\cdot 4.0\,\text{s} \\ &= 9.59\,\text{m} \\ h_3 &= \tfrac12\cdot 2.397\,\tfrac{\text{m}}{\text{s}}\cdot 2.0\,\text{s} \\ &= 2.40\,\text{m} \end{aligned}$$
The height is
$h$$= 2.40\,\text{m} + 9.59\,\text{m} + 2.40\,\text{m}$$= 14.38\,\text{m}$Answer
a) The free-body diagram and the equation are above. b) $a_1 \approx 1.2\,\text{m/s}^2$ up, $a_2 \approx 0$ and $a_3 \approx -1.2\,\text{m/s}^2$, that is, the acceleration is down. c) In the third phase the net force points down, so the force of the rope is smaller than the weight although the velocity points up: the load slows down. The direction of the acceleration is the direction of the net force, not the direction of the motion. d) $h \approx 14.4\,\text{m}$, which to two significant figures is $14\,\text{m}$ like the mass $80\,\text{kg}$; the more precise value $14.4\,\text{m}$ is accepted in the answer box.
Reasonableness
In the uniform phase the force of the rope, $785\,\text{N}$, is equal to the weight, $785\,\text{N}$, as it should be. The changes in velocity cancel each other: in phase 3 the velocity decreases by $1.19\,\text{m/s}^2\cdot 2.0\,\text{s} = 2.38\,\text{m/s}$, that is, almost from $2.40\,\text{m/s}$ to zero, so the lift ends at rest as in the task. The height is also the product of the greatest velocity and a time: the acceleration and deceleration phases together cover the same distance as uniform velocity in $2.0\,\text{s}$, so
$h$$= v_\text{max}\cdot 6.0\,\text{s}$$= 2.397\,\text{m/s}\cdot 6.0\,\text{s}$$= 14.4\,\text{m}$