Weight, friction and the inclined plane
Why does a moving box stay in place on a slanted loading ramp, but start to slide when the ramp is made a little steeper?
Prior knowledge
- Drawing a free-body diagram and the equations of motion $\Sigma F_x = m a_x$ and $\Sigma F_y = m a_y$chapter Free-body diagrams and the combined effect of forces
- The components $F\cos\alpha$ and $F\sin\alpha$ of an oblique forcechapter Free-body diagrams and the combined effect of forces
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 3.
Think about it
A mover lifts a box onto the loading ramp of a van. The box may be empty or packed full. It stays in place even though the ramp is slanted. When the lower end of the ramp is moved closer to the van, the ramp gets steeper. At some point the box suddenly starts to slide down.
What holds the box in place on a gentle ramp, and why is it no longer enough on a steep ramp?
Optional Recall: the normal force on a slope
Predict
Make your predictions before investigating 2 questions
Prediction 1: friction on a tilting plank
Prediction 2: a heavy and a light box
Investigate
A box is on a plank whose slope you can change. Left: a snapshot and a graph of the friction force. Right: the free-body diagram of the box, to scale.
Simulation
How to use
Adjust the angle $\alpha$ of the plank with the slider. Choose the Mass and the Surfaces. The free-body diagram and the readings update at once. If the box slides, press Start to see the sliding, and Reset to return the box to the start. The graph marks the friction force at every angle you have tried. The box Show the components of the weight shows the components as dashed lines. Speed slows down or speeds up the motion without changing the readings.
Assumptions of the model
The values are model values calculated by the simulation. The box is at rest at first. It stays in place if the static friction needed is at most $\mu_0 N$; otherwise it slides, and kinetic friction $\mu N$ acts on it. The coefficients of friction are model values: wood on wood $\mu_0 = 0.50$ and $\mu = 0.30$, plastic on wood $\mu_0 = 0.30$ and $\mu = 0.20$, rubber on wood $\mu_0 = 0.80$ and $\mu = 0.60$. Air drag is ignored. The blue arrow of the velocity is in the snapshot, not in the free-body diagram, because velocity is not a force.
How the calculation works
The components of the weight are $G_x = mg\sin\alpha$ down the slope and $G_y = mg\cos\alpha$ toward the surface, and $g = 9.81\,\text{m/s}^2$. In the direction perpendicular to the surface the box does not accelerate, so $N = mg\cos\alpha$. If $G_x \le \mu_0 N$, the box stays in place, the static friction is $F_\mu = G_x$ and $a = 0$. Otherwise $F_\mu = \mu N$ and along the slope $G_x - F_\mu = ma$, so $a = g(\sin\alpha - \mu\cos\alpha)$.
Experiment 1: tilting the plank
Test the first prediction. The starting state selects the wood-on-wood surfaces and the mass $2.0\,\text{kg}$. Change only the angle $\alpha$. Record a measurement at least at the angles $0^\circ$, $10^\circ$ and $20^\circ$. Then increase the angle one degree at a time until the box starts to slide, and also record the last angle at which it stays in place and the first angle at which it slides.
Initial state: surfaces: wood on wood, mass $m$ = $2\,\text{kg}$, angle $\alpha$ = $0^\circ$. Kept constant: surfaces, mass $m$. Varied: angle $\alpha$.
| angle $\alpha$ | $G_x$ $(\text{N})$ | $N$ $(\text{N})$ | $F_\mu$ $(\text{N})$ | $a$ $(\text{m/s}^2)$ |
|---|
Find in the table the limit at which the box starts to slide. Use it to determine the coefficient of static friction $\mu_0$ between the box and the plank.
Hint 1: a guiding question
Just before sliding the static friction is at its greatest, that is $F_\mu = \mu_0 N$. Which forces are then in balance along the slope?
Hint 2: what is the connection
Along the slope $mg\sin\alpha_0 = \mu_0\,mg\cos\alpha_0$, and the mass cancels.
Hint 3: an intermediate step
$\mu_0 = \tan\alpha_0$, where $\alpha_0$ is halfway between the last angle at which the box stays in place and the first angle at which it slides.
Show the solution
Given
from the table the box stays in place at $26^\circ$ and slides at $27^\circ$, so the limiting angle is $\alpha_0 \approx 26.5^\circ$; $m = 2.0\,\text{kg}$
Find
the coefficient of static friction $\mu_0$ (no unit)
Principle
Just before sliding the box is still at rest, so the sum of the forces is zero, and the static friction is at its greatest, $F_\mu = \mu_0 N$. Newton's second law along the slope and perpendicular to the surface gives the coefficient.
Free-body diagram
The box as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the surface, outward. The weight $\overline{G}$ straight down, its components $G_x = mg\sin\alpha_0$ and $G_y = mg\cos\alpha_0$ dashed, the normal force $\overline{N}$ perpendicular to the surface, outward, and the static friction $\overline{F}_\mu$ up the slope.
Formula
$y$: $N - mg\cos\alpha_0 = 0$ and $x$: $mg\sin\alpha_0 - \mu_0 N = 0$, so
Conversions
The angle is in degrees, and the tangent is calculated with the calculator in degree mode; no other conversions are needed.
Substitution
$\mu_0 = \tan 26.5^\circ = 0.4986$
Answer
$0.50$
Two significant figures, because the limiting angle is known only to the accuracy of a degree: $\tan 26^\circ = 0.49$ and $\tan 27^\circ = 0.51$.
Reasonableness
In the table, just before sliding the friction is $8.6\,\text{N}$ and the normal force $17.6\,\text{N}$, and their ratio is $0.49$, the same result. The coefficient of friction of wood on wood is typically $0.3$–$0.6$, so the value is plausible.
Experiment 2: the mass changes
Test the second prediction. The starting state selects the wood-on-wood surfaces and the angle $35^\circ$, at which the box slides. Change only the mass and record a measurement for all three masses. You can watch the sliding by pressing Start.
Initial state: surfaces: wood on wood, angle $\alpha$ = $35^\circ$, mass $m$ = $2\,\text{kg}$. Kept constant: surfaces, angle $\alpha$. Varied: mass $m$.
| mass $m$ | $G_x$ $(\text{N})$ | $F_\mu$ $(\text{N})$ | $a$ $(\text{m/s}^2)$ |
|---|
Conclusion
Using your measurements, explain why the box stays in place on a gentle plank but starts to slide on a steeper one, and why the mass does not affect the acceleration.
Model answer
At rest the static friction acting on the box equals the component $G_x = mg\sin\alpha$ that pulls it down the slope, so it grows with the angle: in the model values $3.4\,\text{N}$ at $10^\circ$ and $8.6\,\text{N}$ at $26^\circ$. At the same time the normal force $N = mg\cos\alpha$ decreases, and with it the upper limit $\mu_0 N$ of static friction. When $G_x$ exceeds the upper limit, at about $27^\circ$, the box starts to slide, and the friction drops to the kinetic friction $\mu N$, in the model values $5.2\,\text{N}$. Both $G_x$ and the friction are proportional to the mass, so the mass cancels, and at $35^\circ$ the acceleration is $3.22\,\text{m/s}^2$ for all masses, which shows in the table of experiment 2. That the limiting angle does not depend on the mass follows from the condition $\tan\alpha_0 = \mu_0$, which contains no mass; this was not measured.
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
Students pull a wooden block on a horizontal table with a spring balance, horizontally and at a constant velocity. They add weights on top of the block and record the reading of the balance.
| $m$ (kg) | $0.50$ | $1.00$ | $1.50$ | $2.00$ |
|---|---|---|---|---|
| $F$ (N) | $1.7$ | $3.5$ | $5.1$ | $6.9$ |
Determine the coefficient of kinetic friction $\mu$ between the block and the table as the mean of the measurements.
Hint 1: a guiding question
The block moves at a constant velocity, so the force of the balance equals the kinetic friction. What is the normal force on a horizontal table?
Hint 2: what is the connection
for each row.
Hint 3: an intermediate step
For example from the first row
Calculate all four and their mean.
Show the solution
Given
the masses in the table $0.50$–$2.00\,\text{kg}$ and the balance readings $1.7$–$6.9\,\text{N}$; the block moves horizontally at a constant velocity; $g = 9.81\,\text{m/s}^2$
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
The block moves at a constant velocity, so the sum of the forces is zero: in the horizontal direction the force of the balance equals the kinetic friction and in the vertical direction the normal force equals the weight. The kinetic friction is $F_\mu = \mu N$.
Formula
$F - \mu N = 0$ and $N - mg = 0$, so $\mu = \dfrac{F}{mg}$
Conversions
The masses are in kilograms and the forces in newtons, so no conversions are needed.
Substitution
similarly $\mu_2 = 0.357$, $\mu_3 = 0.347$ and $\mu_4 = 0.352$; the mean
Answer
$0.35$
Two significant figures like the balance readings, for example $1.7\,\text{N}$.
Reasonableness
All four ratios are close to each other, so the friction really is proportional to the normal force. When the mass becomes four times as great, the reading grows from $1.7\,\text{N}$ to $6.9\,\text{N}$, that is about four times as great. The coefficient of friction of wood on wood is typically $0.3$–$0.6$.
Theory summary
Key idea
Weight
The weight $\overline{G}$ of an object is the force with which the Earth pulls the object. It always points straight down, also on a slope, and in a free-body diagram it is drawn from the center of the object. The magnitude of the weight is the product of the mass and the acceleration of free fall, $mg$. Mass describes the inertia of an object, that is how hard it is to change its state of motion, and it is the same everywhere. Weight, on the other hand, depends on the place: on the surface of the Moon $g = 1.62\,\text{m/s}^2$, so the weight of a $60\,\text{kg}$ person there is about $97\,\text{N}$, whereas on Earth it is about $590\,\text{N}$.
Static friction and kinetic friction
When two surfaces touch each other, they resist sliding relative to each other. This force parallel to the surface is friction $\overline{F}_\mu$. The surfaces are not perfectly smooth: tiny bumps catch on each other, and there are attractive forces between the particles of the surfaces. The harder the surfaces are pressed against each other, that is the greater the normal force $N$ is, the greater the friction can be.
When a box on a floor is pulled only weakly, it does not move. Then static friction acts on it, equal to the pull and in the opposite direction, so the net force is zero. Static friction is therefore not constant but adapts to the situation. It does have an upper limit, though. When the pull exceeds it, the box starts to slide, and the friction changes into kinetic friction. Kinetic friction is usually smaller than the greatest static friction, and at ordinary speeds it is approximately constant.
In the graph the normal force of the floor is $100\,\text{N}$. Static friction grows with the pull up to $50\,\text{N}$, where the box starts to move. A sliding box is acted on by a kinetic friction of $35\,\text{N}$, even if the pull is increased. This is why a heavy cabinet is hardest to get moving, and once it is moving it goes more easily.
The inclined plane
On an inclined plane, that is a slope, the weight still points straight down, but the surface is slanted. The normal force is perpendicular to the surface, and the friction is parallel to the surface. For calculation the weight is split into two components: the component along the slope tends to take the object down the slope, and the component perpendicular to the surface presses the object against the surface. The components are not new forces. They replace the weight in the calculation, and in a free-body diagram they are drawn as dashed lines.
The steeper the slope, the greater the component along the slope and the smaller the perpendicular component. When only the weight and the normal force act perpendicular to the surface and the object does not accelerate in that direction, the normal force decreases together with the perpendicular component, and with it the upper limit of static friction. This is why a box starts to slide on a steep slope: the component pulling it down the slope grows greater than the greatest static friction that the ramp can give.
Friction does not always oppose motion
Friction opposes the sliding of surfaces relative to each other, not necessarily the motion of the object. When a pickup truck starts to move with an acceleration, a box on its bed accelerates along with it. When air drag is ignored, the only horizontal force on the box is the friction of the bed, so it must point forward, in the direction of motion: the bed tends to slide forward from under the box, and static friction prevents this. Likewise, when we walk, the foot pushes the ground backward, and the static friction of the ground pushes the foot forward. On mirror-smooth ice, where there is very little friction, getting started is therefore difficult.
Besides friction, other contact forces can act on an object. Buoyancy is the force with which a liquid or a gas pushes upward an object immersed in it, and the drag of the medium is the force with which air or water opposes the motion of an object. Air drag is treated in more detail in the chapter Air drag and falling in air. In this chapter air drag and buoyancy are ignored.
The friction force and the inclined plane
The magnitude of the weight is
$$G = mg$$
where $G$ is the weight (N), $m$ the mass (kg) and $g$ the acceleration of free fall, $9.81\,\text{m/s}^2$ on the surface of the Earth.
Kinetic friction is proportional to the normal force:
$$F_\mu = \mu N$$
where $F_\mu$ is the magnitude of the kinetic friction (N), $\mu$ the coefficient of kinetic friction (no unit) and $N$ the magnitude of the normal force (N). Static friction is at most the product of the normal force and the coefficient of static friction $\mu_0$:
$$F_{\mu 0} \le \mu_0 N$$
where $F_{\mu 0}$ is the magnitude of the static friction (N). The coefficients of friction depend on the materials and the condition of the surfaces, and usually $\mu < \mu_0$. The formulas are empirical models that describe dry surfaces well. In them the friction does not depend on the area of contact or on the sliding speed.
On an inclined plane whose angle of inclination is $\alpha$, the $x$-axis is chosen along the slope and the $y$-axis perpendicular to the surface. The components of the weight are
$$\begin{aligned} G_x &= G\sin\alpha \\ G_y &= G\cos\alpha \end{aligned}$$
where $G_x$ is the component along the slope (N) and $G_y$ the component perpendicular to the surface (N). The angle between the weight and the component $G_y$ is the same $\alpha$ as the angle of inclination of the slope, because each is perpendicular to its counterpart: the weight to the horizontal and $G_y$ to the slope.
If only the weight and the normal force act perpendicular to the surface and the object does not accelerate in that direction, the normal force equals $G_y$: $N = mg\cos\alpha$, which is less than the weight. (A pulling or pushing force directed toward or away from the surface would change the normal force.) The static friction acting on an object at rest equals $G_x$. The object stays in place as long as $mg\sin\alpha \le \mu_0\,mg\cos\alpha$, that is $\tan\alpha \le \mu_0$. The mass cancels, so according to this model a heavy and a light box start to slide on the same surface at the same angle.
Determining the coefficient of friction
In the diagrams and in the simulation, the static friction acting on an object at rest is denoted $\overline{F}_\mu$; its magnitude is $F_{\mu 0} \le \mu_0 N$, and while sliding $F_\mu = \mu N$.
The coefficient of static friction is obtained from an experiment in which a board is tilted slowly until the object starts to slide. At this limiting angle $\alpha_0$ the static friction is at its greatest, so $\mu_0 = \tan\alpha_0$. The coefficient of kinetic friction is obtained for example by pulling the object on a horizontal table with a spring balance, horizontally and at a constant velocity. Then the pull equals the kinetic friction and the normal force equals the weight, so $\mu = F : mg$. When the measurement is repeated with different masses, the coefficient of friction is the slope of the straight line through the origin fitted to the $(N, F_\mu)$ graph.
Common misconception
Many people think that the friction force is always $\mu_0 N$ or $\mu N$ and always against the motion. Static friction, however, is only as large as the situation requires: if a box is pushed with a force of $20\,\text{N}$ and it does not move, the static friction is $20\,\text{N}$, even if its upper limit is greater. Friction opposes the sliding of surfaces relative to each other, so it can also point in the direction of motion, as on the bed of an accelerating truck. Another common mistake is to take the normal force to be always equal to the weight. On a slope $N = mg\cos\alpha$ is less than the weight, when only the weight and the normal force act perpendicular to the surface and the object does not accelerate in that direction, and the weight still points straight down and not perpendicular to the slope.
Summary
- The magnitude of the weight is $G = mg$, and the weight always points straight down. Mass is the same everywhere, but weight depends on the acceleration of free fall.
- Friction is parallel to the surface and opposes the sliding of surfaces relative to each other, not always the motion of the object.
- Static friction adapts to the situation: $F_{\mu 0} \le \mu_0 N$. A sliding object is acted on by kinetic friction $F_\mu = \mu N$, and usually $\mu < \mu_0$.
- On a slope the weight is split into the components $G_x = mg\sin\alpha$ and $G_y = mg\cos\alpha$, and the normal force is $N = mg\cos\alpha$, when only the weight and the normal force act perpendicular to the surface and the object does not accelerate in that direction. The components are drawn as dashed lines, and they are not new forces.
- The coefficient of static friction is obtained from the limiting angle at which the object starts to slide: $\mu_0 = \tan\alpha_0$.
Optional Return to your predictions
Compare your predictions with the measurements. What did you learn about the magnitude of static friction? Also return to the box at the start: from the condition $\tan\alpha_0 = \mu_0$, deduce why an empty box and a box packed full start to slide at the same angle of the ramp.
Would you change your prediction?
Apply
Worked example
Problem
A moving box of mass $18\,\text{kg}$ slides down a loading ramp. The ramp is at an angle of $25^\circ$ to the horizontal, and the coefficient of kinetic friction between the box and the ramp is $0.30$. How large is the acceleration of the box, and how large is the friction force? Air drag is ignored.
Step 1/8
Given and asked. $m = 18\,\text{kg}$, $\alpha = 25^\circ$, $\mu = 0.30$ and $g = 9.81\,\text{m/s}^2$. The acceleration $a$ and the friction force $F_\mu$ are asked.
Step 2/8
Principle. The box is chosen as the object. The Earth (weight) and the ramp (normal force and kinetic friction) act on it. The box slides down, so the kinetic friction points up the slope. Newton's second law is written along the slope, where the box accelerates, and perpendicular to the surface, where it does not accelerate: $a_y = 0$.
Step 3/8
Free-body diagram. The box is drawn as a rectangle along the slope, and the forces start from their points of application. The axes are chosen so that $x$ points down the slope, that is in the direction of the acceleration, and $y$ perpendicular to the surface, outward. The weight is split into the components $G_x = mg\sin\alpha$ and $G_y = mg\cos\alpha$, drawn with a dashed line.
Step 4/8
The formula solved. The equations of motion are read from the diagram and the asked quantities are solved:
$$\begin{aligned} &x\text{:}\ \ mg\sin\alpha - \mu N = ma \\ &y\text{:}\ \ N - mg\cos\alpha = 0 \\ &F_\mu = \mu mg\cos\alpha \\ &a = g(\sin\alpha - \mu\cos\alpha) \end{aligned}$$
Step 5/8
Unit conversions. None are needed: all quantities are in SI units, and the coefficient of friction has no unit.
Step 6/8
Substitution. The numerical values are substituted with their units:
$$\begin{aligned} &F_\mu = \mu mg\cos\alpha \\ &\phantom{F_\mu} = 0.30\cdot 18\,\text{kg} \\ &\quad\ \cdot 9.81\,\text{m/s}^2\cdot\cos 25^\circ \\ &\phantom{F_\mu} \approx 48.0\,\text{N} \\[4pt] &a = g(\sin\alpha - \mu\cos\alpha) \\ &\phantom{a} = 9.81\,\text{m/s}^2 \\ &\quad\ \cdot(\sin 25^\circ - 0.30\cdot\cos 25^\circ) \\ &\phantom{a} \approx 1.48\,\text{m/s}^2 \end{aligned}$$
Step 7/8
Result. The acceleration of the box is $a \approx 1.5\,\text{m/s}^2$ down the slope and the friction force is $F_\mu \approx 48\,\text{N}$, to two significant figures like the data $\mu = 0.30$ and $m = 18\,\text{kg}$.
Step 8/8
Reasonableness. Without friction the acceleration would be $g\sin 25^\circ \approx 4.1\,\text{m/s}^2$, so friction reduces it, as it should. The friction force $48\,\text{N}$ is less than the component along the slope $G_x \approx 75\,\text{N}$, so the net force points down the slope and the box accelerates down. The mass does not affect the acceleration, because both $G_x$ and $F_\mu$ are proportional to the mass.
Find the error
Choose the correct free-body diagram
Apply to a new situation
A moving walkway at an airport carries an $8.0\,\text{kg}$ suitcase upward at a constant velocity. The belt is at an angle of $15^\circ$ to the horizontal, and the suitcase does not slip on the belt.
How large is the friction force of the belt on the suitcase? Answer in newtons.
Hint 1: a guiding question
Draw the free-body diagram of the suitcase. The suitcase moves upward at a constant velocity, so the net force is zero. Which force can cancel the component of the weight along the slope?
Hint 2: what is the connection
Choose $x$ up along the belt. Along the slope $F_\mu - mg\sin\alpha = 0$.
Hint 3: an intermediate step
Show the solution
Given
$m = 8.0\,\text{kg}$, the angle of the belt $\alpha = 15^\circ$, the suitcase moves at a constant velocity and does not slip on the belt; $g = 9.81\,\text{m/s}^2$
Find
the friction force of the belt $F_\mu$ in $\text{N}$
Principle
The suitcase moves at a constant velocity, so by Newton's second law the sum of the forces is zero. The suitcase does not slip relative to the belt, so static friction acts on it, and it is exactly as large as needed.
Free-body diagram
The suitcase as a rectangle along the belt, $x$ up along the belt and $y$ perpendicular to the belt, outward. The weight $\overline{G}$ straight down, its components $mg\sin\alpha$ down the belt and $mg\cos\alpha$ toward the belt dashed, the normal force $\overline{N}$ perpendicular to the belt, outward, and the static friction $\overline{F}_\mu$ up along the belt. The suitcase tends to slide down relative to the belt, so the friction points up, in the direction of motion.
Formula
$x$: $F_\mu - mg\sin\alpha = 0$, so $F_\mu = mg\sin\alpha$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
Answer
$20\,\text{N}$
Up along the belt, two significant figures like the mass $8.0\,\text{kg}$.
Reasonableness
The weight of the suitcase is about $80\,\text{N}$, and on a gentle belt the component along the slope is only about a quarter of it, so $20\,\text{N}$ is reasonable. The friction points in the direction of motion: it opposes the sliding of the suitcase relative to the belt, not the motion of the suitcase.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A board is tilted slowly, and an object on it starts to slide at the angle $\alpha_0$. Why is the coefficient of static friction $\tan\alpha_0$?
Answer
Just before sliding the static friction is at its greatest, $\mu_0 N$, and it cancels the component of the weight along the slope: $mg\sin\alpha_0 = \mu_0\,mg\cos\alpha_0$. The mass cancels, so
$\mu_0$$= \dfrac{\sin\alpha_0}{\cos\alpha_0}$$= \tan\alpha_0$ - A walker steps forward, and the foot pushes the ground backward. In which direction does the friction force of the ground act on the foot, and why?
Answer
Forward, in the direction of motion. The foot tends to slide backward relative to the ground, and static friction opposes this sliding. Friction opposes the sliding of the surfaces relative to each other, not the motion of the walker: it is friction that carries the walker forward.
- The mass of a person is $60\,\text{kg}$. How large are his or her mass and weight on the surface of the Moon, when $g_\text{Moon} = 1.62\,\text{m/s}^2$?
Answer
The mass is the same as on Earth, $60\,\text{kg}$. The weight is
$G$$= mg_\text{Moon}$$= 60\,\text{kg}\cdot 1.62\,\text{m/s}^2$$\approx 97\,\text{N}$whereas on Earth it is about $590\,\text{N}$.
1. Concept
2. Calculation
A book is given a push on a tilted desktop, and it slides down at a constant velocity when the angle of the desktop is $17^\circ$.
How large is the coefficient of kinetic friction between the book and the desktop?
Hint 1: a guiding question
Draw the free-body diagram of the book. In uniform motion the net force is zero, also along the slope.
Hint 2: what is the connection
Along the slope $mg\sin\alpha - \mu N = 0$ and in the perpendicular direction $N = mg\cos\alpha$.
Hint 3: an intermediate step
The mass cancels: $\mu = \tan\alpha$.
Show the solution
Given
the angle of the desktop $\alpha = 17^\circ$, the book slides down at a constant velocity
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
In uniform motion the acceleration is zero, so by Newton's second law the sum of the forces is zero in both directions. The book slides, so kinetic friction $F_\mu = \mu N$ acts on it.
Free-body diagram
The book as a rectangle along the desktop, $x$ down along the desktop and $y$ perpendicular to the desktop, outward. The weight $\overline{G}$ straight down and its components $mg\sin\alpha$ and $mg\cos\alpha$ dashed, the normal force $\overline{N}$ perpendicular to the desktop, outward, and the kinetic friction $\overline{F}_\mu$ up along the desktop, because the book slides down.
Formula
$y$: $N - mg\cos\alpha = 0$ and $x$: $mg\sin\alpha - \mu N = 0$, so
Conversions
The angle is in degrees; no other conversions are needed, and the mass cancels.
Substitution
$\mu = \tan 17^\circ = 0.3057$
Answer
$0.31$
Two significant figures like the angle $17^\circ$.
Reasonableness
The coefficient is positive and less than $1$, as is typical between a book and wood. A smaller angle would mean a smaller coefficient of friction: on ice a book would slide at a constant velocity already at a very gentle angle.
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 $20\,\text{kg}$ is pulled by a rope up a $20^\circ$ slope at a constant velocity. The rope is parallel to the slope, and the coefficient of kinetic friction between the box and the slope is $0.25$. Draw the free-body diagram of the box on paper: the box as a rectangle and every force from its point of application. Then write the equations of motion along the slope and perpendicular to it and calculate the tension of the rope.
Model diagram
The Earth, the slope and the rope act on the box. The weight $\overline{G}$ points straight down from the center. The slope acts from the lower surface with two forces: the normal force $\overline{N}$ points perpendicular to the surface, outward, and the kinetic friction $\overline{F}_\mu$ down the slope, because the box slides up. The tension $\overline{T}$ of the rope points up the slope. The box moves at a constant velocity, so the net force is zero. When $x$ points up the slope and $y$ out of the surface: $y$: $N - mg\cos 20^\circ = 0$, so $N = 20\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot\cos 20^\circ \approx 184\,\text{N}$ and $F_\mu = \mu N \approx 46\,\text{N}$. $x$: $T - mg\sin 20^\circ - \mu N = 0$, so $T = mg\sin 20^\circ + \mu mg\cos 20^\circ \approx 67\,\text{N} + 46\,\text{N} \approx 113\,\text{N}$. The rope has to overcome both the component of the weight along the slope and the friction.
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.
- Weight
- The gravitational force the Earth exerts on an object.$$G = mg$$
- Inclined plane
- The situation becomes easier when the axes are chosen along the plane and the forces are split into components.
- Friction
- A force between surfaces in contact that opposes their relative motion or the start of it.
- Static friction
- Prevents surfaces from sliding over each other and adjusts to the size needed, up to its maximum.
- Kinetic friction
- Opposes the motion of surfaces sliding over each other (sliding friction).$$F_{\mu} = \mu N$$
- Coefficient of friction
- A dimensionless number that depends on the pair of materials and describes the friction between the surfaces.
- Buoyancy
- The upward force a liquid or gas exerts on an object.
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 8-1Core
The combined mass of an astronaut and the spacesuit is $160\,\text{kg}$. On the surface of the Moon the acceleration of free fall is $1.62\,\text{m/s}^2$.
How large is the combined weight of the astronaut and the suit on the surface of the Moon?
Answer
$259\,\text{N}$
Show the solution
Given
$m = 160\,\text{kg}$, the acceleration of free fall on the Moon $g_\text{Moon} = 1.62\,\text{m/s}^2$
Find
the weight on the surface of the Moon $G$ in $\text{N}$
Principle
Weight is the force with which a celestial body pulls an object, and its magnitude is the product of the mass and the acceleration of free fall. The mass is the same on the Moon as on Earth, but the acceleration of free fall is smaller on the Moon.
Free-body diagram
The astronaut as a rectangle on the surface of the Moon: the weight $\overline{G}$ from the center down and the normal force $\overline{N}$ from the lower surface up, equally long. What is asked is the magnitude of the weight $mg_\text{Moon}$.
Formula
$G = m g_\text{Moon}$
Conversions
The mass is in kilograms and the acceleration in $\text{m/s}^2$, so no conversions are needed.
Substitution
$G$$= 160\,\text{kg}\cdot 1.62\,\text{m/s}^2$$= 259.2\,\text{N}$Answer
$259\,\text{N}$
Three significant figures like the acceleration of free fall $1.62\,\text{m/s}^2$.
Reasonableness
On Earth the weight would be $160\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 1570\,\text{N}$. The acceleration of free fall on the Moon is about a sixth of the value on Earth, and $1570\,\text{N} : 6 \approx 260\,\text{N}$, the same result.
- Exercise 8-2Core
A sled of mass $30\,\text{kg}$ slides on horizontal snow. The coefficient of kinetic friction between the sled and the snow is $0.12$. Draw the free-body diagram of the sled.
How large is the friction force of the snow on the sled?
Answer
$35\,\text{N}$
Show the solution
Given
$m = 30\,\text{kg}$, $\mu = 0.12$, the sled slides on horizontal snow; $g = 9.81\,\text{m/s}^2$
Find
the kinetic friction $F_\mu$ in $\text{N}$
Principle
The sled slides, so kinetic friction $F_\mu = \mu N$ acts on it. In the vertical direction the sled does not accelerate, so by Newton's second law the normal force equals the weight.
Free-body diagram
The sled as a rectangle, $x$ in the direction of motion and $y$ up. The weight $\overline{G}$ from the center down, the normal force $\overline{N}$ of the snow from the lower surface up and equally long as the weight, the kinetic friction $\overline{F}_\mu$ from the rear corner of the lower surface backward. There is no force pushing forward.
Formula
$y$: $N - mg = 0$, so $N = mg$ and $F_\mu = \mu N = \mu mg$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$F_\mu$$= 0.12\cdot 30\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 35.316\,\text{N}$Answer
$35\,\text{N}$
Two significant figures like the coefficient of friction $0.12$ and the mass $30\,\text{kg}$.
Reasonableness
The weight of the sled is about $290\,\text{N}$, and the friction is about an eighth of it, as the small coefficient of friction $0.12$ requires. The friction slows the sled down:
$a$$= F_\mu : m$$\approx 1.2\,\text{m/s}^2$ - Exercise 8-3
A box of mass $40\,\text{kg}$ is pushed on a horizontal floor with a horizontal force of $120\,\text{N}$, and it slides at a constant velocity.
How large is the coefficient of kinetic friction between the box and the floor?
Answer
$0.31$
Show the solution
Given
$m = 40\,\text{kg}$, the horizontal pushing force $F = 120\,\text{N}$, the box slides at a constant velocity; $g = 9.81\,\text{m/s}^2$
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
In uniform motion the sum of the forces is zero, so the kinetic friction equals the push and the normal force equals the weight. The kinetic friction is $F_\mu = \mu N$.
Free-body diagram
The box as a rectangle, $x$ in the direction of motion and $y$ up. The pushing force $\overline{F}$ forward, the kinetic friction $\overline{F}_\mu$ backward and equally long, the weight $\overline{G}$ down and the normal force $\overline{N}$ up, equally long.
Formula
$x$: $F - \mu N = 0$ and $y$: $N - mg = 0$, so $\mu = \dfrac{F}{mg}$
Conversions
The mass is in kilograms and the force in newtons, so no conversions are needed.
Substitution
$\mu$$= \dfrac{120\,\text{N}}{40\,\text{kg}\cdot 9.81\,\text{m/s}^2}$$= \dfrac{120\,\text{N}}{392.4\,\text{N}}$$= 0.3058$Answer
$0.31$
Two significant figures like the mass $40\,\text{kg}$.
Reasonableness
The coefficient of friction has no unit, because it is the ratio of two forces. The push $120\,\text{N}$ is a little under a third of the weight of the box $392\,\text{N}$, so the coefficient is a little over $0.3$, a common value for wood against a floor.
- Exercise 8-4
A book of mass $0.80\,\text{kg}$ is at rest on a desktop that is tilted at an angle of $20^\circ$ to the horizontal. Draw the free-body diagram of the book.
How large is the normal force of the desktop on the book?
Answer
$7.4\,\text{N}$
Show the solution
Given
$m = 0.80\,\text{kg}$, the angle of the desktop $\alpha = 20^\circ$, the book is at rest; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ in $\text{N}$
Principle
The book does not move in the direction perpendicular to the desktop, so by Newton's second law the sum of the forces in that direction is zero. The normal force cancels the perpendicular component of the weight.
Free-body diagram
The book as a rectangle along the desktop, $x$ down along the desktop and $y$ perpendicular to the desktop, outward. The weight $\overline{G}$ straight down and its components $mg\sin\alpha$ and $mg\cos\alpha$ dashed, the normal force $\overline{N}$ perpendicular to the desktop, outward, and the static friction $\overline{F}_\mu$ up along the desktop.
Formula
$y$: $N - mg\cos\alpha = 0$, so $N = mg\cos\alpha$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$N$$= 0.80\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot\cos 20^\circ$$= 7.848\,\text{N}\cdot 0.9397$$= 7.375\,\text{N}$Answer
$7.4\,\text{N}$
Two significant figures like the mass $0.80\,\text{kg}$.
Reasonableness
The weight of the book is $7.8\,\text{N}$, and the normal force is a little smaller, because the desktop is slanted. On a horizontal desktop the normal force would be the whole weight, on a vertical one zero.
- Exercise 8-5
A box of mass $12\,\text{kg}$ is on a slope whose angle of inclination is $30^\circ$.
How large is the component of the weight of the box along the slope?
Answer
$59\,\text{N}$
Show the solution
Given
$m = 12\,\text{kg}$, the angle of inclination of the slope $\alpha = 30^\circ$; $g = 9.81\,\text{m/s}^2$
Find
the component of the weight along the slope $G_x$ in $\text{N}$
Free-body diagram
The box as a rectangle along the slope, $x$ down the slope. The weight $\overline{G}$ straight down and its components $G_x$ and $G_y$ dashed; the other forces do not affect the asked component.
Principle
The weight points straight down. When the $x$-axis is chosen along the slope, the component of the weight in its direction is obtained from a right triangle in which the angle between the weight and the perpendicular component is the angle of inclination $\alpha$ of the slope.
Formula
$G_x = G\sin\alpha = mg\sin\alpha$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$G_x$$= 12\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot\sin 30^\circ$$= 117.72\,\text{N}\cdot 0.5$$= 58.86\,\text{N}$Answer
$59\,\text{N}$
Two significant figures like the mass $12\,\text{kg}$.
Reasonableness
At $30^\circ$ the sine is exactly $0.5$, so the component along the slope is half of the weight $118\,\text{N}$. The component is always smaller than the weight itself.
- Exercise 8-6
A box is at rest on a horizontal floor. It is pushed horizontally with a force of $20\,\text{N}$, but it does not move.
The greatest possible value of the static friction between the box and the floor is $60\,\text{N}$. Which statement about the static friction $F_{\mu 0}$ exerted by the floor on the box is correct?
Answer
$F_{\mu 0} = 20\,\text{N}$, because at rest the static friction is equal to the push.
Show the solution
Find
the magnitude of the friction force exerted by the floor on the box while the box stays in place
Principle
The box is at rest, so by Newton's second law the sum of the horizontal forces is zero. Static friction adapts to the situation: it is equal in magnitude to the push and opposite in direction, as long as the push does not exceed the maximum value of static friction.
Model answer
The magnitude of the friction force is $20\,\text{N}$, because the box is at rest and the friction cancels the push. The maximum value $60\,\text{N}$ is only an upper limit, and the push $20\,\text{N}$ stays below it. The claim that the friction is always equal to the maximum value is wrong, because static friction is only as large as the situation requires. The claim that there is no friction at all is wrong, because it is exactly the friction that keeps the box from moving. The claim that the friction is the difference between the maximum value and the push is wrong, because at rest the friction and the push are equal.
- Exercise 8-7
A $50\,\text{kg}$ box is at rest on a horizontal floor. The coefficient of static friction between the box and the floor is $0.45$.
How large a horizontal force must at least be applied to the box for it to start moving?
Answer
$2.2\cdot 10^2\,\text{N}$
Show the solution
Given
$m = 50\,\text{kg}$, $\mu_0 = 0.45$, the box is at rest on a horizontal floor; $g = 9.81\,\text{m/s}^2$
Find
the smallest horizontal force $F$ in $\text{N}$ with which the box starts to move
Principle
The box starts to move when the push exceeds the maximum value of static friction $\mu_0 N$. In the limiting case the push is equal to this maximum value. In the vertical direction the box does not accelerate, so $N = mg$.
Free-body diagram
The box as a rectangle, $x$ in the direction of the push and $y$ up. The pushing force $\overline{F}$ forward, the static friction $\overline{F}_\mu$ backward and, in the limiting case, equally long, the weight $\overline{G}$ down and the normal force $\overline{N}$ up, equally long.
Formula
$y$: $N - mg = 0$ and $x$: $F - \mu_0 N = 0$, so $F = \mu_0 mg$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$F$$= 0.45\cdot 50\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 220.7\,\text{N}$Answer
$2.2\cdot 10^2\,\text{N}$
Or about $220\,\text{N}$, two significant figures like the coefficient of friction $0.45$ and the mass $50\,\text{kg}$. With a smaller force the box stays in place.
Reasonableness
The weight of the box is $490\,\text{N}$, and the required force is $0.45$ times that, as the coefficient of friction requires. After the box has started to move, the kinetic friction is usually smaller than $\mu_0 N$, so a smaller force is enough to keep it moving.
- Exercise 8-8
A body is at rest on an inclined plane at an angle of $30^\circ$. Only the weight $\overline{G}$, the normal force $\overline{N}$ and friction act on the body.
Which statement about the magnitude $N$ of the normal force and the magnitude $G$ of the weight is correct?
Answer
$N < G$, because the normal force cancels only the perpendicular component of the weight.
Show the solution
Find
the ratio of the magnitude of the normal force $N$ to the magnitude of the weight $G$ for a body at rest on an inclined plane
Principle
Choose the $y$-axis perpendicular to the surface. In that direction the body does not accelerate, so the sum of the forces is zero. Perpendicular to the surface only the normal force and the component of the weight $G_y = mg\cos\alpha$ act, and friction is parallel to the surface. The weight points straight down, and it is divided into components along the slope and perpendicular to it.
Model answer
The normal force is smaller than the weight: $N = mg\cos\alpha < G$. The normal force cancels only the component of the weight perpendicular to the surface, and the component along the slope is cancelled by friction. At the angle $30^\circ$ we get $N = G\cos 30^\circ \approx 0.87\,G$. The claim that the normal force and the weight are always opposite and equal holds only on a horizontal surface. The claim $N > G$ is wrong, because friction does not press the body against the surface. The claim $N = 0$ is wrong, because the component of the weight $G_y$ presses the body against the surface.
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 8-9Core
A sled starts from rest and slides down a hill with an angle of inclination of $12^\circ$, and reaches a velocity of $3.0\,\text{m/s}$ in $4.0\,\text{s}$. The acceleration is constant.
How large is the coefficient of kinetic friction between the sled and the snow? Air drag is ignored.
Answer
$0.13$
Show the solution
Given
$\alpha = 12^\circ$, the sled starts from rest, $v = 3.0\,\text{m/s}$ in the time $t = 4.0\,\text{s}$, the acceleration is constant, air drag is ignored; $g = 9.81\,\text{m/s}^2$
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
The acceleration is obtained from the change of velocity, and Newton's second law along the slope gives the friction. The sled slides, so kinetic friction $F_\mu = \mu N$ acts on it. In the perpendicular direction the sled does not accelerate.
Free-body diagram
The sled as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the surface, outward. The weight $\overline{G}$ straight down and its components dashed, the normal force $\overline{N}$ perpendicular to the surface, outward, and the kinetic friction $\overline{F}_\mu$ up the slope. The acceleration $\overline{a}$ is drawn separately as an orange arrow down the slope.
Formula
$a = \dfrac{v - 0}{t}$, $y$: $N - mg\cos\alpha = 0$ and $x$: $mg\sin\alpha - \mu N = ma$, so
$\mu$$= \dfrac{g\sin\alpha - a}{g\cos\alpha}$Conversions
All quantities are in SI units, so no conversions are needed, and the mass cancels.
Substitution
$a$$= \dfrac{3.0\,\text{m/s}}{4.0\,\text{s}}$$= 0.75\,\text{m/s}^2$$\mu$$= \dfrac{9.81\,\text{m/s}^2\cdot\sin 12^\circ - 0.75\,\text{m/s}^2}{9.81\,\text{m/s}^2\cdot\cos 12^\circ}$$= \dfrac{2.040 - 0.75}{9.596}$$= 0.134$Answer
$0.13$
Two significant figures like the velocity $3.0\,\text{m/s}$ and the time $4.0\,\text{s}$.
Reasonableness
Without friction the acceleration would be $g\sin 12^\circ \approx 2.0\,\text{m/s}^2$, so friction reduces it, as it should. The coefficient $0.13$ is smaller than $\tan 12^\circ = 0.21$, at which the sled would slide at a constant velocity, and it fits between a plastic sled and snow ($0.05$–$0.3$).
- Exercise 8-10Core
A warehouse worker pulls a $40\,\text{kg}$ pallet along a concrete floor with a horizontal rope. The tension of the rope is $150\,\text{N}$, and the coefficient of kinetic friction between the pallet and the floor is $0.30$.
How large is the acceleration of the pallet?
Answer
$0.81\,\text{m/s}^2$
Show the solution
Given
$m = 40\,\text{kg}$, the horizontal tension of the rope $T = 150\,\text{N}$, $\mu = 0.30$; $g = 9.81\,\text{m/s}^2$
Find
the acceleration $a$ in $\text{m/s}^2$
Principle
The pallet slides, so kinetic friction $\mu N$ acts on it. In the vertical direction it does not accelerate, so $N = mg$. In the horizontal direction, by Newton's second law, the difference between the tension and the friction accelerates the pallet.
Free-body diagram
The pallet as a rectangle, $x$ in the direction of the pull and $y$ up. The tension of the rope $\overline{T}$ forward, the kinetic friction $\overline{F}_\mu$ from the rear corner of the lower surface backward and shorter, the weight $\overline{G}$ down and the normal force $\overline{N}$ up, equally long.
Formula
$y$: $N - mg = 0$ and $x$: $T - \mu N = ma$, so $a = \dfrac{T - \mu mg}{m}$
Conversions
All quantities are in SI units, so no conversions are needed.
Substitution
$a$$= \dfrac{150\,\text{N} - 0.30\cdot 40\,\text{kg}\cdot 9.81\,\text{m/s}^2}{40\,\text{kg}}$$= \dfrac{150\,\text{N} - 117.72\,\text{N}}{40\,\text{kg}}$$= 0.807\,\text{m/s}^2$Answer
$0.81\,\text{m/s}^2$
Two significant figures like the coefficient of friction $0.30$ and the mass $40\,\text{kg}$.
Reasonableness
Without friction the acceleration would be $150\,\text{N} : 40\,\text{kg} \approx 3.8\,\text{m/s}^2$. The friction $118\,\text{N}$ cancels most of the force of the rope, so the acceleration stays under $1\,\text{m/s}^2$, which is reasonable.
- Exercise 8-11
Below is example data: the kinetic friction of a wooden block measured at different normal forces. The graph shows the measurement points. Determine from the graph the coefficient of kinetic friction between the block and the table.
Answer
$0.21$
Show the solution
Given
the five measurement points $(N, F_\mu)$ of the graph of the example data: $(5\,\text{N}$; $1.1\,\text{N})$, $(10\,\text{N}$; $2.1\,\text{N})$, $(15\,\text{N}$; $3.3\,\text{N})$, $(20\,\text{N}$; $4.2\,\text{N})$ and $(25\,\text{N}$; $5.4\,\text{N})$
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
Kinetic friction is proportional to the normal force, $F_\mu = \mu N$, so the $(N, F_\mu)$ graph is a straight line through the origin whose slope is $\mu$. The slope is read from a line fitted to the points, not from two neighbouring measurement points, because measurement points are usually not exactly on the same line because of measurement uncertainty.
Reading the graph
A line is drawn through the origin so that it passes as close as possible to all the points. A point far away is read from the line, for example at the normal force $N = 25\,\text{N}$ the value of the line is about $F_\mu = 5.35\,\text{N}$.
Formula
$\mu = \dfrac{\Delta F_\mu}{\Delta N}$, when $\Delta F_\mu$ and $\Delta N$ are read from the fitted line
Conversions
Both axes are in newtons, so the units cancel and no conversions are needed.
Substitution
$\mu$$= \dfrac{5.35\,\text{N} - 0\,\text{N}}{25\,\text{N} - 0\,\text{N}}$$= 0.214$Answer
$0.21$
Two significant figures, because the friction forces have been measured to two figures and the line is drawn by hand.
Reasonableness
The individual points give the ratios $0.21$–$0.22$, for example $4.2\,\text{N} : 20\,\text{N} = 0.21$, so the slope of the fitted line falls between them. The coefficient has no unit, as the ratio of two forces should not.
- Exercise 8-12
A moving box of mass $10\,\text{kg}$ is placed at rest on a plank that is at an angle of $35^\circ$ to the horizontal. The coefficient of static friction between the box and the plank is $0.60$.
Calculate the greatest possible value of static friction and compare it with the component of the weight along the slope. By how many newtons does the component exceed the greatest static friction? (If the number were negative, the box would stay in place.)
Answer
$56.3\,\text{N} - 48.2\,\text{N} = 8.1\,\text{N}$
Show the solution
Given
$m = 10\,\text{kg}$, $\alpha = 35^\circ$, $\mu_0 = 0.60$, the box is placed at rest; $g = 9.81\,\text{m/s}^2$
Find
the difference $G_x - F_\text{max}$ in $\text{N}$, where $F_\text{max}$ is the greatest value of static friction; the sign of the difference tells whether the box stays in place
Principle
Static friction is at most $\mu_0 N$. The box stays in place only if this upper limit is enough to cancel the component of the weight along the slope. In the perpendicular direction the box does not accelerate, so $N = mg\cos\alpha$.
Free-body diagram
The box as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the surface, outward. The weight $\overline{G}$ straight down and its components $mg\sin\alpha$ and $mg\cos\alpha$ dashed, the normal force $\overline{N}$ perpendicular to the surface, outward, and the friction $\overline{F}_\mu$ up the slope.
Formula
$y$: $N - mg\cos\alpha = 0$, so the greatest static friction is $F_\text{max} = \mu_0 mg\cos\alpha$, and it is compared with the component $G_x = mg\sin\alpha$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$F_\text{max}$$= 0.60\cdot 10\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot\cos 35^\circ$$= 48.2\,\text{N}$$G_x$$= 10\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot\sin 35^\circ$$= 56.3\,\text{N}$so $G_x - F_\text{max} = 8.1\,\text{N}$
Answer
the component exceeds the greatest static friction by $56.3\,\text{N} - 48.2\,\text{N} = 8.1\,\text{N}$, two significant figures like the coefficient of friction $0.60$. The greatest static friction $48\,\text{N}$ is smaller than the component $56\,\text{N}$, so the box does not stay in place but starts to slide. The difference is positive, so the friction is not enough.
Reasonableness
The same conclusion follows from the angle: $\tan 35^\circ = 0.70$ is greater than $\mu_0 = 0.60$, so the plank is steeper than the limiting angle $\tan^{-1} 0.60 \approx 31^\circ$.
- Exercise 8-13
Students pulled a wooden block on a horizontal table with a spring balance, horizontally and at a constant velocity. Weights were added on top of the block, and for each mass the pull shown by the balance was read. Below is example data.
Draw the $(N, F_\mu)$ graph and fit a straight line through the origin to it. Determine the coefficient of kinetic friction between the block and the table from the graph. The normal force is calculated from the mass.
Combined mass $m$ of the block and the weights and the pull $F$ of the balance (example data) $m\,(\text{kg})$ $F\,(\text{N})$ 0.20 0.85 0.40 1.62 0.60 2.50 0.80 3.31 1.00 4.02 Answer
$0.42$
Show the solution
Given
the five measurements $(m, F)$ of the example data: $(0.20\,\text{kg}$; $0.85\,\text{N})$, $(0.40\,\text{kg}$; $1.62\,\text{N})$, $(0.60\,\text{kg}$; $2.50\,\text{N})$, $(0.80\,\text{kg}$; $3.31\,\text{N})$ and $(1.00\,\text{kg}$; $4.02\,\text{N})$; the block is pulled horizontally at a constant velocity; $g = 9.81\,\text{m/s}^2$
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
At a constant velocity the sum of the forces is zero, so the pull is equal to the kinetic friction, $F = F_\mu$. On a horizontal table the normal force is equal to the weight, $N = mg$. Because $F_\mu = \mu N$, the $(N, F_\mu)$ graph is a straight line through the origin whose slope is $\mu$.
Reading the graph
First calculate the normal forces: $N = mg$ gives the values $1.96\,\text{N}$, $3.92\,\text{N}$, $5.89\,\text{N}$, $7.85\,\text{N}$ and $9.81\,\text{N}$. Draw the points $(N, F_\mu)$ and fit to them a straight line through the origin that passes as close as possible to all the points. Read a distant point from the line: at the normal force $N = 9.81\,\text{N}$ the value of the line is about $F_\mu = 4.1\,\text{N}$.
Formula
$\mu = \dfrac{\Delta F_\mu}{\Delta N}$, when $\Delta F_\mu$ and $\Delta N$ are read from the fitted line
Conversions
Both axes are in newtons, so the units cancel. The masses are in kilograms, so no conversions are needed.
Substitution
$\mu$$= \dfrac{4.1\,\text{N} - 0\,\text{N}}{9.81\,\text{N} - 0\,\text{N}}$$= 0.418$Answer
$0.42$
Two significant figures, because the pulls were measured to three figures but the line is drawn by hand.
Reasonableness
The individual points give ratios $F : N$ between $0.41$ and $0.43$, for example $4.02\,\text{N} : 9.81\,\text{N} = 0.41$, so the slope of the fitted line falls between them. The coefficient has no unit, and it is of the usual order of magnitude for wood on a table.
- Exercise 8-14
A force sensor was used to pull a $0.75\,\text{kg}$ wooden block on a horizontal table. The pull was horizontal all the time. The pull was increased slowly until the block started to move, and after that the block was pulled horizontally at a constant velocity. The block moved at a constant velocity from the moment $t = 3.5\,\text{s}$ onward. Below is example data.
Draw the $(t, F)$ graph and read from the table the greatest reading as the greatest value of static friction and the pulling force during the uniform motion. Determine the coefficient of static friction $\mu_0$ and the coefficient of kinetic friction $\mu$.
Write the coefficient of kinetic friction $\mu$ in the answer box.
Pulling force $F$ measured by the sensor as a function of time $t$ (example data) $t\,(\text{s})$ $F\,(\text{N})$ 0.0 0.0 0.5 0.7 1.0 1.4 1.5 2.1 2.0 2.8 2.5 3.3 3.0 2.3 3.5 2.2 4.0 2.1 4.5 2.2 Answer
$\mu = 0.29$
Show the solution
Given
$m = 0.75\,\text{kg}$, the pulling force $F$ of the example data as a function of time $t$; the greatest reading $3.3\,\text{N}$ at the moment $t = 2.5\,\text{s}$, the readings during uniform motion $2.2\,\text{N}$, $2.1\,\text{N}$ and $2.2\,\text{N}$ at the moments $3.5\,\text{s}$–$4.5\,\text{s}$; $g = 9.81\,\text{m/s}^2$
Find
the coefficient of kinetic friction $\mu$ (no unit) and the coefficient of static friction $\mu_0$
Principle
Before the block starts to move, the pull increases and the static friction increases with it. When the pull exceeds the maximum value of static friction $\mu_0 N$, the block starts to move, and the peak of the graph is this maximum value. At a constant velocity the sum of the forces is zero, so the pull is equal to the kinetic friction $\mu N$. On a horizontal table $N = mg$.
Reading the graph
The pull increases linearly up to the moment $t = 2.5\,\text{s}$, where the peak of the graph is $3.3\,\text{N}$. After the peak the pull drops and stays almost constant, about $2.2\,\text{N}$, while the block moves at a constant velocity. The sum of the constant-pull readings is $2.2\,\text{N} + 2.1\,\text{N} + 2.2\,\text{N} = 6.5\,\text{N}$, so their mean is $6.5\,\text{N} : 3 = 2.17\,\text{N}$.
Formula
$N = mg$, $\mu_0 = \dfrac{F_{\text{peak}}}{N}$ and $\mu = \dfrac{F_{\text{steady}}}{N}$
Conversions
The mass is in kilograms and the forces in newtons, so no conversions are needed.
Substitution
$N$$= 0.75\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 7.36\,\text{N}$$\mu_0$$= \dfrac{3.3\,\text{N}}{7.36\,\text{N}}$$= 0.449$$\mu$$= \dfrac{2.17\,\text{N}}{7.36\,\text{N}}$$= 0.295$Answer
the coefficient of kinetic friction is $\mu = 0.29$ and the coefficient of static friction $\mu_0 = 0.45$, two significant figures like the mass $0.75\,\text{kg}$ and the force readings.
Reasonableness
The coefficient of kinetic friction is smaller than the coefficient of static friction, $0.29 < 0.45$, as is usual. This is why the pull drops after the block has started to move: keeping it moving takes a smaller force than getting it moving. The coefficients are of the usual order of magnitude for wood on a table.
- Exercise 8-15
The coefficient of static friction was determined by an experiment: a wooden board was tilted slowly until a wooden block on it started to slide, and the angle was read from an angle gauge. The experiment was repeated five times. Below is example data.
Justify why the coefficient of static friction is $\mu_0 = \tan\alpha_0$, where $\alpha_0$ is the angle at which the block starts to slide. Calculate the mean of the angles and use it to determine the coefficient of static friction.
Write the coefficient of static friction $\mu_0$ in the answer box.
The angle $\alpha_0$ at which the block starts to slide (example data) Measurement $\alpha_0\,({}^\circ)$ 1 24 2 26 3 25 4 27 5 25 Answer
$0.47$
Show the solution
Given
the five angles of the example data: $24^\circ$, $26^\circ$, $25^\circ$, $27^\circ$ and $25^\circ$
Find
the coefficient of static friction $\mu_0$ (no unit)
Principle
As the board tilts, the component of the weight along the slope $G_x = mg\sin\alpha$ increases, and the static friction increases with it. The block starts to slide when the component exceeds the maximum value of static friction $\mu_0 N$. In the perpendicular direction the block does not accelerate, so $N = mg\cos\alpha$. At the limiting angle $\alpha_0$, $mg\sin\alpha_0 = \mu_0\,mg\cos\alpha_0$. The mass cancels, and dividing both sides gives $\mu_0 = \tan\alpha_0$.
Formula
$\mu_0 = \tan\alpha_0$, where $\alpha_0$ is the mean of the measurements
Conversions
The angles are in degrees, so no conversions are needed. The calculator must be in degree mode.
Substitution
The sum of the angles is $24^\circ + 26^\circ + 25^\circ + 27^\circ + 25^\circ = 127^\circ$, so the mean is
$\alpha_{0,\text{mean}}$$= 127^\circ : 5$$= 25.4^\circ$$\mu_0 = \tan 25.4^\circ = 0.4748$
Answer
$0.47$
Two significant figures, as the precision of the mean of the angles gives.
Reasonableness
The individual angles give $\tan 24^\circ = 0.45$ and $\tan 27^\circ = 0.51$, so the mean $0.47$ lies between them. The mass of the block is not needed in the experiment, because it cancels. The result is of the usual order of magnitude for the coefficient of static friction between wood and wood.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 8-16
On the bed of a pickup truck there is a $50\,\text{kg}$ box that is not fastened. The truck starts to move on a horizontal road with an acceleration of $3.5\,\text{m/s}^2$.
What is the smallest coefficient of static friction between the bed and the box that keeps the box from sliding on the bed?
Answer
$0.36$
Show the solution
Given
$m = 50\,\text{kg}$, the acceleration of the truck and the box $a = 3.5\,\text{m/s}^2$ forward, the road is horizontal; $g = 9.81\,\text{m/s}^2$
Find
the smallest value of the coefficient of static friction $\mu_0$ (no unit)
Principle
If the box does not slide, it accelerates along with the truck. The only horizontal force on the box is the static friction of the bed, so by Newton's second law it must be $ma$ and point forward, in the direction of motion. Static friction can be at most $\mu_0 N$.
Free-body diagram
The box as a rectangle, $x$ forward and $y$ up. The weight $\overline{G}$ from the center down, the normal force $\overline{N}$ of the bed from the lower surface up and equally long, the static friction $\overline{F}_\mu$ from the front corner of the lower surface forward. The acceleration is drawn separately as a thin orange arrow, not as a force.
Formula
$y$: $N - mg = 0$ and $x$: $F_\mu = ma$. The condition $F_\mu \le \mu_0 N$ gives $ma \le \mu_0 mg$, so $\mu_0 \ge \dfrac{a}{g}$
Conversions
The accelerations are in $\text{m/s}^2$, so no conversions are needed, and the mass cancels.
Substitution
$\mu_0$$= \dfrac{3.5\,\text{m/s}^2}{9.81\,\text{m/s}^2}$$= 0.357$Answer
$0.36$
Two significant figures like the acceleration $3.5\,\text{m/s}^2$.
Reasonableness
The friction points in the direction of motion, because the bed tends to slide forward from under the box. The mass cancels, so the result holds for a light and a heavy box. On wet plywood the coefficient can be below $0.36$, and then the box slides backward relative to the bed: this is why cargo is fastened.
- Exercise 8-17
A puck is pushed up a $20^\circ$ slope and leaves the hand at a velocity of $3.0\,\text{m/s}$. The coefficient of kinetic friction between the puck and the slope is $0.25$.
How far up the slope does the puck slide before it stops? Air drag is ignored.
Answer
$0.80\,\text{m}$
Show the solution
Given
the angle of the slope $\alpha = 20^\circ$, the initial velocity $v_0 = 3.0\,\text{m/s}$ up the slope, $\mu = 0.25$, the final velocity $v = 0$; $g = 9.81\,\text{m/s}^2$
Find
the sliding distance $s$ in $\text{m}$
Principle
When sliding up, both the component of the weight along the slope and the kinetic friction point down the slope, so the puck decelerates uniformly. The acceleration is obtained from Newton's second law, and the distance from the equation of uniformly accelerated motion $v^2 = v_0^2 - 2as$, where $a$ is the magnitude of the deceleration.
Free-body diagram
The puck as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the surface, outward. The weight $\overline{G}$ straight down and its components dashed, the normal force $\overline{N}$ perpendicular to the surface, outward, and the kinetic friction $\overline{F}_\mu$ down the slope, because the puck slides up. The velocity is up the slope, but it is not a force.
Formula
$y$: $N = mg\cos\alpha$ and $x$: $mg\sin\alpha + \mu mg\cos\alpha = ma$, so $a = g(\sin\alpha + \mu\cos\alpha)$. From the equation of motion $v^2 = v_0^2 - 2as$ and the condition $v = 0$ we get $s = \dfrac{v_0^2}{2a}$, where $a = g(\sin\alpha + \mu\cos\alpha)$
Conversions
All quantities are in SI units, so no conversions are needed.
Substitution
$a$$= 9.81\,\text{m/s}^2\cdot(\sin 20^\circ + 0.25\cdot\cos 20^\circ)$$= 5.660\,\text{m/s}^2$$s$$= \dfrac{(3.0\,\text{m/s})^2}{2\cdot 5.660\,\text{m/s}^2}$$= 0.795\,\text{m}$Answer
$0.80\,\text{m}$
Two significant figures like the velocity $3.0\,\text{m/s}$ and the coefficient of friction $0.25$.
Reasonableness
Without friction the deceleration would be $g\sin 20^\circ \approx 3.4\,\text{m/s}^2$ and the distance about $1.3\,\text{m}$; friction shortens the distance. After stopping, the puck stays on the slope only if the coefficient of static friction is at least $\tan 20^\circ \approx 0.36$; otherwise it starts to slide back down.
- Exercise 8-18
On Earth a box slides down a $30^\circ$ slope with an acceleration of $2.0\,\text{m/s}^2$.
How large would the acceleration of the box be on the same slope and the same surface on the Moon, where the acceleration of free fall is $1.62\,\text{m/s}^2$? The Moon has no air, and air drag is ignored on Earth too.
Answer
$0.33\,\text{m/s}^2$
Show the solution
Given
on Earth $\alpha = 30^\circ$ and $a_\text{Earth} = 2.0\,\text{m/s}^2$ down the slope, on the Moon the same slope and surface, $g = 9.81\,\text{m/s}^2$ and $g_\text{Moon} = 1.62\,\text{m/s}^2$; there is no air drag
Find
the acceleration on the Moon $a_\text{Moon}$ in $\text{m/s}^2$
Principle
The box slides, so kinetic friction $\mu N$ acts on it up the slope. The coefficient of friction is not given, but it is a property of the surface and the same on Earth and on the Moon. The acceleration on Earth gives $\mu$, and on the Moon the same law gives the new acceleration. In the perpendicular direction the box does not accelerate.
Free-body diagram
The box as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the surface, outward. The weight $\overline{G}$ straight down and its components dashed, the normal force $\overline{N}$ perpendicular to the surface, outward, and the kinetic friction $\overline{F}_\mu$ up the slope.
Formula
$y$: $N - mg\cos\alpha = 0$ and $x$: $mg\sin\alpha - \mu N = ma$, so
$a$$= g(\sin\alpha - \mu\cos\alpha)$$\mu$$= \tan\alpha - \dfrac{a_\text{Earth}}{g\cos\alpha}$Because the expression in the parentheses is the same in both places,
$a_\text{Moon}$$= a_\text{Earth}\cdot\dfrac{g_\text{Moon}}{g}$Conversions
All quantities are in SI units, so no conversions are needed, and the mass cancels.
Substitution
$\mu$$= \tan 30^\circ - \dfrac{2.0\,\text{m/s}^2}{9.81\,\text{m/s}^2\cdot\cos 30^\circ}$$= 0.577 - 0.235$$= 0.342$and
$a_\text{Moon}$$= 1.62\,\text{m/s}^2\cdot(\sin 30^\circ - 0.342\cdot\cos 30^\circ)$$= 0.330\,\text{m/s}^2$Directly from the ratio:
$2.0\,\text{m/s}^2\cdot\dfrac{1.62}{9.81}$$= 0.330\,\text{m/s}^2$Answer
$0.33\,\text{m/s}^2$
Two significant figures like the acceleration on Earth $2.0\,\text{m/s}^2$.
Reasonableness
On the Moon both the weight and the friction are about a sixth of the values on Earth, so the acceleration is also about a sixth. The box slides on the Moon too, because the condition $\tan\alpha > \mu$ does not depend on the acceleration of free fall.
- Exercise 8-19
The braking distance of a passenger car is calculated with a model in which the wheels lock and kinetic friction acts between the tires and the road. Below is example data on the coefficients of kinetic friction between the tires and the road.
The car brakes from a speed of $80\,\text{km/h}$ to a stop. Air drag is ignored, and the acceleration is constant.
a) Determine the braking distance on dry horizontal asphalt.
b) Determine the braking distance on wet horizontal asphalt.
c) The wet road slopes downhill at an angle of $6.0^\circ$. Determine the braking distance downhill and justify why the mass of the car is not needed.
Write the braking distance of part c) in the answer box.
Coefficient of kinetic friction $\mu$ between the tires and the road (example data) Road surface $\mu$ Dry asphalt 0.75 Wet asphalt 0.45 Answer
$73\,\text{m}$
Show the solution
Given
the initial velocity $v_0 = 80\,\text{km/h}$, the final velocity $v = 0$, constant acceleration, the wheels lock; the coefficients of kinetic friction of the example data: dry asphalt $\mu = 0.75$, wet asphalt $\mu = 0.45$; in part c) a wet downhill slope $\alpha = 6.0^\circ$; $g = 9.81\,\text{m/s}^2$
Find
the braking distance $s$ in $\text{m}$, in parts a) dry horizontal, b) wet horizontal and c) wet downhill
Principle
The locked wheels slide on the road, so kinetic friction $\mu N$ acts on the car against the motion. By Newton's second law the friction gives the deceleration, and the distance is obtained from the equation of uniformly accelerated motion $v^2 = v_0^2 - 2as$, where $a$ is the magnitude of the deceleration. In the perpendicular direction the car does not accelerate, so $N = mg\cos\alpha$.
Free-body diagram
The car of part c) as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the road, outward. The weight $\overline{G}$ straight down and its components dashed, the normal force $\overline{N}$ perpendicular to the road, outward, and the kinetic friction $\overline{F}_\mu$ up the slope. The velocity is down the slope, but the deceleration is up the slope.
Formula
on a horizontal road $N = mg$ and $\mu mg = ma$, so $a = \mu g$. Downhill $x$: $\mu mg\cos\alpha - mg\sin\alpha = ma$, so $a = g(\mu\cos\alpha - \sin\alpha)$. The distance is $s = \dfrac{v_0^2}{2a}$
Conversions
$v_0$$= 80\,\text{km/h}$$= 80 : 3.6\,\text{m/s}$$= 22.22\,\text{m/s}$and $v_0^2 = 493.8\,\text{m}^2/\text{s}^2$.
Substitution
Part a):
$a_a$$= 0.75\cdot 9.81\,\text{m/s}^2$$= 7.358\,\text{m/s}^2$$s_a$$= \dfrac{493.8\,\text{m}^2/\text{s}^2}{2\cdot 7.358\,\text{m/s}^2}$$= 33.6\,\text{m}$Parts b) and c):
$$\begin{aligned} a_b &= 0.45 \cdot 9.81\,\text{m/s}^2 = 4.415\,\text{m/s}^2 \\ s_b &= \frac{493.8\,\text{m}^2/\text{s}^2}{2 \cdot 4.415\,\text{m/s}^2} = 55.9\,\text{m} \\[4pt] a_c &= 9.81\,\text{m/s}^2 \\ &\quad \cdot (0.45\cos 6.0^\circ - \sin 6.0^\circ) \\ &= 3.365\,\text{m/s}^2 \\ s_c &= \frac{493.8\,\text{m}^2/\text{s}^2}{2 \cdot 3.365\,\text{m/s}^2} = 73.4\,\text{m} \end{aligned}$$
Answer
a) $34\,\text{m}$, b) $56\,\text{m}$ and c) $73\,\text{m}$ (calculated $73.4\,\text{m}$). The result of part c) is rounded to two significant figures like the coefficient of friction $0.45$ and the angle $6.0^\circ$. The mass of the car is not needed, because it cancels: both the friction and the component of the weight are proportional to the mass, so the deceleration does not depend on it.
Reasonableness
On the wet road the distance is $55.9 : 33.6 = 1.66$ times that on the dry road, and the same ratio follows from the coefficients of friction $0.75 : 0.45 = 1.67$. Downhill the distance is longer than on the horizontal wet road, $73\,\text{m} > 56\,\text{m}$, because the component of the weight down the slope reduces the deceleration. The deceleration $3.365\,\text{m/s}^2$ is still positive, because $\mu = 0.45 > \tan 6.0^\circ = 0.11$.
- Exercise 8-20
A box of mass $20\,\text{kg}$ starts from rest and slides down a slope of $30^\circ$. The coefficient of kinetic friction between the box and the slope is $0.35$, and air drag is ignored.
A student solved the problem like this: “The normal force is equal to the weight, $N = mg$. The equation of motion along the slope is $mg\sin\alpha - \mu mg = ma$, so $a = g(\sin\alpha - \mu) \approx 1.5\,\text{m/s}^2$.”
a) Explain the student’s error and draw the free-body diagram of the box.
b) Calculate the correct acceleration of the box.
c) How long a distance does the box slide during the first $3.0\,\text{s}$?
Write the distance of part c) in the answer box.
Answer
$8.7\,\text{m}$
Show the solution
Given
$m = 20\,\text{kg}$, $\alpha = 30^\circ$, $\mu = 0.35$, the box starts from rest, $t = 3.0\,\text{s}$; in the student's solution $N = mg$ and the result $1.5\,\text{m/s}^2$; $g = 9.81\,\text{m/s}^2$
Find
the student's error, the correct acceleration $a$ and the distance $s$ in the time $3.0\,\text{s}$ in $\text{m}$
Principle
The student used the whole weight as the normal force, $N = mg$. On a slope only the weight and the normal force act perpendicular to the surface, and the box does not accelerate in that direction, so $N = mg\cos\alpha$, which is smaller than the weight. The student's friction $\mu mg$ is therefore too large and the acceleration too small. The correct acceleration is obtained from Newton's second law along the slope, and the distance from the equation of uniformly accelerated motion with the initial velocity zero.
Free-body diagram
The box as a rectangle along the slope, $x$ down the slope and $y$ perpendicular to the surface, outward. The weight $\overline{G}$ straight down and its components dashed, the normal force $\overline{N}$ perpendicular to the surface, outward, and shorter than the weight, the kinetic friction $\overline{F}_\mu$ up the slope. The acceleration is drawn separately down the slope.
Formula
$y$: $N - mg\cos\alpha = 0$ and $x$: $mg\sin\alpha - \mu N = ma$, so $a = g(\sin\alpha - \mu\cos\alpha)$. The distance is $s = \tfrac{1}{2}at^2$, because the initial velocity is zero
Conversions
All quantities are in SI units, so no conversions are needed, and the mass cancels.
Substitution
$a$$= 9.81\,\text{m/s}^2\cdot(\sin 30^\circ - 0.35\cdot\cos 30^\circ)$$= 1.931\,\text{m/s}^2$and
$$\begin{aligned} s &= \tfrac{1}{2} \cdot 1.931\,\text{m/s}^2 \cdot (3.0\,\text{s})^2 \\ &= 8.69\,\text{m} \end{aligned}$$
Answer
the student's error is $N = mg$ on a slope, the correct normal force is $N = mg\cos\alpha \approx 170\,\text{N}$ and the correct acceleration $a \approx 1.9\,\text{m/s}^2$. The box slides a distance of $8.7\,\text{m}$ in the time $3.0\,\text{s}$, two significant figures like the coefficient of friction $0.35$ and the time $3.0\,\text{s}$.
Reasonableness
The student's acceleration $1.5\,\text{m/s}^2$ is too small, because the friction $\mu mg = 69\,\text{N}$ was calculated too large; the correct friction is $\mu mg\cos\alpha = 59\,\text{N}$. Without friction the acceleration would be $g\sin 30^\circ \approx 4.9\,\text{m/s}^2$, so friction reduces it. The box accelerates down while sliding, because $\tan 30^\circ = 0.58 > \mu = 0.35$; according to the problem it starts to move, so $\mu_0 < \tan 30^\circ$.
Photographs
- Michał Frąckowiak - Snow Sledding. Michał Frąckowiak. source, CC BY-SA 2.0
