PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: friction, circular motion and gravitation (PHYS 101)

This sheet is not a summary of the chapter: you already have the lecture notes. It answers one question, what makes students lose marks on friction, circular motion and gravitation in PHYS 101 at McGill University, and what exact line to write instead. Everything here is algebra, proportions and right-angle trigonometry, with no derivative and no integral anywhere, because that is the course PHYS 101 actually is.

Two reversals carry the whole chapter, and almost every trap below is one of them in a new costume. Static friction is an inequality, not a formula: it takes whatever value keeps the object still, and $\mu_{s}N$ is only the largest value it may take. The centripetal force is not a force: it is the name of the resultant, along the radius, of the real forces already drawn. Get those two straight and the banked curve taken too slowly, the bucket of water, the centrifuge, the spinning space habitat and the orbiting satellite all become the same three lines of work.

Updated on October 25, 2026

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The thread of the chapter

The centripetal force is not one more arrow to add to your diagram, it is the NAME of the resultant of the arrows already on it. Static friction works the same way: it has no value of its own, it takes whatever value stops the sliding, and $\mu_{s}N$ is only the ceiling it may not pass.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

Before this chapter, you need

• Newton's laws and free-body diagrams

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Static friction ceiling
Coefficient of kinetic friction
Angle of repose
Centripetal acceleration
Banked curve and design speed
Conical pendulum
Vertical circular motion
Newton's law of universal gravitation
Surface gravity of a planet
Kepler's third law
Apparent weight
Artificial gravity

The essentials

On your formula sheet: this chapter's lines, and when each one holds

• $F_{fr,k}=\mu_{k}F_{N}$ : Only while the surfaces slide on each other. $F_{N}$ comes from the free-body diagram: it is not $mg$ on a slope or under a slanted pull.
• $F_{fr,s}\le\mu_{s}F_{N}$ : An inequality: static friction takes whatever value keeps the object at rest, up to $\mu_{s}F_{N}$ . The equality holds only at the point of slipping.
• $v=\omega r$ : $\omega$ in rad/s, never in rpm: multiply revolutions per minute by $\frac{2\pi}{60}$ .
• $a=\frac{v^{2}}{r}$ : Directed toward the centre. It is not a force: on the diagram, the real forces (tension, normal, friction, gravity) supply $\frac{mv^{2}}{r}$ between them.
• $\tan\theta=\frac{v^{2}}{rg}$ : Only at the design speed of a frictionless bank. Slower, friction acts up the slope; faster, down the slope.
• $F=\frac{Gm_{1}m_{2}}{r^{2}}$ : $r$ is measured between CENTRES: the radius of the planet plus the altitude, never the altitude alone.
• $G=6.67\times10^{-11}$ : In N m²/kg². At a planet's surface $GM=gR^{2}$ , which avoids looking up the planet's mass.
• $\frac{T^{2}}{r^{3}}=\frac{4\pi^{2}}{GM}$ : A circular orbit around a central mass $M$ , the mass at the CENTRE, not the satellite's. Two satellites of the same body share the same $\frac{T^{2}}{r^{3}}$ .

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam. This sheet writes $f_{k}$ , $f_{s}$ and $N$ where the formula sheet writes $F_{fr,k}$ , $F_{fr,s}$ and $F_{N}$ : same forces.

Static friction is an inequality, kinetic friction is a value

• At rest: $f_{s}$ takes whatever value balances the other forces along the surface, and it obeys $f_{s} \le \mu_{s}N$ .
• On the point of slipping, and only there: $f_{s} = \mu_{s}N$ . The words that announce it are on the point of, about to, minimum, maximum.
• Sliding: $f_{k} = \mu_{k}N$ , directed against the motion, no longer adjustable, and $\mu_{k} < \mu_{s}$ for the same pair of surfaces.
• The test that opens every friction problem: compare the force demanded along the surface with $\mu_{s}N$ , then choose which of the two lines above to write.
• $N$ is not always $mg$ . On an incline $N = mg\cos\theta$ ; on a banked curve at its design speed $N = \frac{mg}{\cos\theta}$ ; in a lift $N = m(g \pm a)$ .

Friction rises along the line

f = F

as long as the block holds, reaches

110.25

N, then DROPS to

78.4

N the moment it slides: the block leaps forward with no extra push.

A single number, $\mu_{s}N$ , plays two different parts: it is a CEILING when the object is at rest and a VALUE only at the instant it lets go. Half the marks lost on friction come from treating the ceiling as a value.

Friction on the diagram: the two standard cases

• Friction lies ALONG the surface and the normal force perpendicular to it; both come from the same contact, and on a free-body diagram both start at the centre of the body.
• It opposes the sliding, or, for a body at rest, the sliding that would happen without it: on an incline at rest, $f_{s}$ points UP the slope.
• Pulled at constant speed on a level floor: $T = f_{k} = \mu_{k}mg$ . At rest on an incline: $f_{s} = mg\sin\theta$ , as long as that stays under $\mu_{s}mg\cos\theta$ .

Left,

10

kg pulled at constant speed: the pull matches

f_{k} = 29.4

N. Right,

4.0

kg at rest at

25^{\circ}

f_{s} = mg\sin 25^{\circ} = 16.6

N points up the slope.

A block sliding UP an incline has friction pointing DOWN the slope, adding to $mg\sin\theta$ : always ask which way the surface is rubbing.

Centripetal is a direction, never a new force

• $a_{c} = \frac{v^{2}}{r} = \frac{4\pi^{2}r}{T^{2}}$ , always pointing at the centre of the circle.
• The method, unchanged from chapter 4: draw the real forces, point one axis at the centre, write $\sum F_{\text{radius}} = \frac{mv^{2}}{r}$ .
• The other axis carries NO acceleration in a horizontal circle: on it, $\sum F = 0$ as in any statics problem.
• Uniform means constant SPEED. The velocity changes direction at every instant, and that is the whole reason there is an acceleration.
• Nothing on a correct free-body diagram is labelled $F_{c}$ , and nothing points away from the centre.

Uniform circular motion seen from above: the velocity is tangent and keeps the same length, the acceleration always points at the centre, and the two are perpendicular at every point.

Use $\frac{4\pi^{2}r}{T^{2}}$ whenever the data are a period or a rotation rate, and $\frac{v^{2}}{r}$ only when a speed is given. The two forms hold different things fixed, and picking the wrong one turns a doubling into a halving.

Banked curve and conical pendulum: one diagram

• Axes horizontal and vertical, NOT along the road: the acceleration is horizontal and points at the centre of the turn.
• Banked curve without friction: $N\cos\theta = mg$ vertically and $N\sin\theta = \frac{mv^{2}}{r}$ horizontally, so $\tan\theta = \frac{v^{2}}{rg}$ , with no mass left in it.
• Conical pendulum: the same two equations with $T$ in place of $N$ , $T\cos\theta = mg$ and $T\sin\theta = \frac{mv^{2}}{r}$ , $\theta$ measured from the vertical.
• Below the design speed the bank pushes inward too hard and static friction points UP the slope, $f = m\left(g\sin\theta - \frac{v^{2}}{r}\cos\theta\right)$ ; above it, DOWN. The formula sheet's $\tan\theta = \frac{v^{2}}{rg}$ is the frictionless line only.

Left, a car on a

12^{\circ}

bank, to scale:

N\cos\theta

holds the weight,

N\sin\theta

turns the car. Right, the conical pendulum, where the string plays the part of the road.

On a frictionless banked curve $N = \frac{mg}{\cos\theta}$ is LARGER than $mg$ , the reverse of an incline. Writing $mg\cos\theta$ there is the classic slip.

Vertical circles: the crest, the dip and the loop

• Point one axis at the centre of the circle and write $\Sigma F_{\text{toward centre}} = \frac{mv^{2}}{r}$ with the real forces only.
• Over a crest the centre is below: $mg - N = \frac{mv^{2}}{r}$ , and $N$ drops to zero at $v = \sqrt{gr}$ . In a dip the centre is above: $N - mg = \frac{mv^{2}}{r}$ , and the seat presses harder.
• At the top of a loop the track is above the car, so $N$ and $mg$ BOTH point at the centre: $N + mg = \frac{mv^{2}}{r}$ , and the slowest safe speed is again $\sqrt{gr}$ .

Forces to scale, with

\frac{v^{2}}{r} = \frac{g}{2}

on the crest and in the dip and

\frac{3g}{2}

at the top of the loop:

a

always points at the centre, and

N

completes it.

The apparent weight is $N$ in all three cases: a pilot pulling out of a dive feels heavy, a car cresting a hill feels light.

Gravitation: three formulas and one distance

• $F = G\frac{m_{1}m_{2}}{r^{2}}$ with $G = 6.67\times 10^{-11}$ N m $^2$ /kg $^2$ , and $r$ measured CENTRE to centre.
• At the surface of a body of mass $M$ and radius $R$ : $g = \frac{GM}{R^{2}}$ . Mass is in kilograms everywhere, weight is a force in newtons.
• Circular orbit: $v = \sqrt{\frac{GM}{r}}$ and $T^{2} = \frac{4\pi^{2}}{GM}r^{3}$ . The orbiting mass cancels out of both.
• Inverse SQUARE: distance $\times 2$ divides the force by $4$ , distance $\times 3$ divides it by $9$ . To halve $g$ you go out by $\sqrt{2}$ .
• Apparent weight is the NORMAL force: $N = m\left(g \pm \frac{v^{2}}{r}\right)$ , plus at the bottom of a dip, minus at the top of a crest.

Not to scale: the Earth pulls the satellite and the satellite pulls the Earth with the same

G\frac{Mm}{r^{2}}

, a third-law pair on two bodies;

r

runs centre to centre.

Orbital speed DROPS as you climb, since $v \propto 1/\sqrt{r}$ : the geostationary satellite moves at $3.1$ km/s against $7.7$ km/s for the space station. Any result of yours that still contains the satellite mass has an algebra error in it.

Artificial gravity: the floor supplies the circle

• On a ring of radius $R$ spinning at $\omega$ , the floor is the only force on anything resting on it, and it points at the axis: $N = m\omega^{2}R$ . No gravity is involved.
• Earth-like load on the rim: $\omega^{2}R = g$ , so $\omega = \sqrt{g/R}$ . For $R = 224$ m, $\omega = 0.209$ rad/s, one turn in $30$ s.
• On a deck at distance $r$ from the axis, turning with the same $\omega$ : $N = m\omega^{2}r = mg\,\frac{r}{R}$ . A deck at $0.60\,R$ gives $0.60\,g$ , and on the axis there is no load at all.
• At fixed $\omega$ use $a = \omega^{2}r$ , which grows with $r$ . The form $\frac{v^{2}}{r}$ falls with $r$ only at fixed speed, which a rigid habitat does not have.

Forces to scale: the floor pushes each crate toward the axis, with

N = mg

on the main deck and

N = 0.60\,mg

on the deck at

0.60\,R

;

N

and

a

always point at the axis.

What feels like weight on the rim is the floor pushing you INWARD. Nothing presses you outward: without the floor you would carry on in a straight line, along a tangent.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which friction force to write, for a crate of $25$ kg on level ground

Take $\mu_{s} = 0.45$ and $\mu_{k} = 0.32$ , so $N = 245$ N, the ceiling is $\mu_{s}N = 110.25$ N and the sliding value is $\mu_{k}N = 78.4$ N. Read the line that matches what the crate is DOING, never the line that matches the word friction.

What the crate does	What decides the value	The friction force
at rest, pushed below the ceiling	Newton's first law along the surface	$f_{s} = F$
Example: Pushed with $F = 60$ N: $f_{s} = 60$ N and $a = 0$ , because $60 < 110.25$ .
at rest, on the point of slipping	the ceiling is exactly reached	$f_{s} = \mu_{s}N = 110.25$ N
Example: The largest push the crate takes without moving is $0.45 \times 245 = 110.25$ N.
sliding	friction has locked at its kinetic value	$f_{k} = \mu_{k}N = 78.4$ N
Example: Pushed with $F = 130$ N: $a = \frac{130 - 78.4}{25} = 2.06$ m/s $^2$ .
at rest, and you write $f = \mu_{s}N$ anyway	nothing, the equality is not available yet	no value rule that does not exist
Example: Writing $110.25$ N under a push of $60$ N leaves a net force of $-50.25$ N on a crate that is standing perfectly still. Same form, other result: Under $F = 60$ N the friction is $60$ N; under $F = 110$ N it is $110$ N. Same ceiling of $110.25$ N, two different friction forces, so the ceiling cannot be the answer. What to do: Run the comparison first: is the demanded force below $\mu_{s}N$ or above it? Then write $f_{s} = F$ or $f_{k} = \mu_{k}N$ .

The same three lines govern a block on an incline, with $N = mg\cos\theta$ in place of $mg$ , and a coin on a turntable, where the demanded force is $\frac{mv^{2}}{r}$ instead of a push.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Static friction set equal to its ceiling on a block that is merely at rest

2 to 4 points, and every later number of the problem inherits the wrong friction

What not to write

The crate does not move, so $f_{s} = \mu_{s}N = 0.45 \times 245 = 110.25$ N.

What to write

The crate does not move, so the forces along the floor cancel: $f_{s} = F = 60$ N. The ceiling $\mu_{s}N = 110.25$ N is not reached.

Why: $f_{s} \le \mu_{s}N$ is an INEQUALITY. The equality holds at one single instant, the one where the block lets go, and the question says so in words: on the point of, about to slide, minimum force.

2. An invented centripetal arrow added to a diagram that was already complete

the whole question, because the diagram now balances and predicts a straight line

What not to write

The car is turning, so the forces on it are the weight, the normal force, friction and the centripetal force $F_{c} = \frac{mv^{2}}{r}$ .

What to write

The forces on it are the weight, the normal force and friction. Their resultant along the radius IS the centripetal force, and it equals $\frac{mv^{2}}{r}$ .

Seen from behind, centre to the left. Correct: three real forces, friction already supplying

\frac{mv^{2}}{r}

. Wrong: the black Fc is that same force drawn twice.

Why: A force is something another object exerts. Circular motion supplies no new object, so it supplies no new force. Writing $F_{c}$ as a fourth arrow counts friction twice, and the sum along the radius comes out zero when it should be $\frac{mv^{2}}{r}$ .

3. The incline formula for the normal force carried over to a banked curve

3 points, and the error is visible without a calculator

What not to write

On the banked curve, $N = mg\cos 12^{\circ} = 1200 \times 9.8 \times 0.978 = 11\,500$ N.

What to write

At the design speed no friction is needed and the car has no vertical acceleration, so $N\cos 12^{\circ} = mg$ and $N = \frac{mg}{\cos 12^{\circ}} = 12\,023$ N, LARGER than the weight.

Why: On an incline the block accelerates ALONG the surface, so the axes follow the surface and the weight is split. On a banked curve the acceleration is HORIZONTAL, so the axes stay horizontal and vertical and the NORMAL force is split. The road holds the car up and turns it at the same time, so it must push harder than the weight.

4. A larger radius read as a smaller acceleration

2 points, and a conclusion that is exactly the reverse of what happens

What not to write

Since $a_{c} = \frac{v^{2}}{r}$ , the coin further from the axis of the turntable has the smaller acceleration and is the safer one.

What to write

The turntable rate is fixed, so use $a_{c} = \frac{4\pi^{2}r}{T^{2}}$ : doubling $r$ from $0.12$ m to $0.24$ m DOUBLES $a_{c}$ , from $2.66$ to $5.33$ m/s $^2$ .

Why: $\frac{v^{2}}{r}$ falls with $r$ only at constant SPEED. On a turntable the speed is not constant between the two positions, the period is: the outer coin travels twice as far in the same time. Ask what the problem holds fixed before choosing the form.

5. An altitude fed into the law of gravitation in place of a distance from the centre

the whole question, for an answer out by a factor of about

290

What not to write

At $400$ km above the ground, $g = \frac{GM}{(4\times 10^{5})^{2}} = 2.5\times 10^{3}$ m/s $^2$ .

What to write

$r = R_{E} + h = 6.37\times 10^{6} + 4.00\times 10^{5} = 6.77\times 10^{6}$ m, so $g = \frac{3.982\times 10^{14}}{(6.77\times 10^{6})^{2}} = 8.69$ m/s $^2$ .

Why: In $F = G\frac{m_{1}m_{2}}{r^{2}}$ , $r$ is the separation of the two CENTRES. Add the radius of the planet to the altitude every time, and run the order-of-magnitude check: no gravitational field near a planet is thousands of times stronger than at its surface.

6. Weightlessness in orbit blamed on an absence of gravity

2 points on a conceptual question that turns up on almost every final

What not to write

Astronauts float inside the station because there is no gravity $400$ km above the ground.

What to write

There $g$ is still $8.69$ m/s $^2$ , $89$ per cent of its value at the ground, and it is what holds the station in orbit. They float because the station and everything in it fall together, so nothing presses on the floor: $N = 0$ .

Why: What your body feels is never $g$ , it is the NORMAL force. It is larger than $mg$ at the bottom of a dive, smaller at the top of a crest, and zero in free fall, while $g$ barely changes in any of those situations.

7. A square root taken where Kepler's third law asks for a cube root

3 points, although the order of magnitude catches it at once

What not to write

From $T^{2} = \frac{4\pi^{2}}{GM}r^{3}$ , a $24$ h period gives $r = \sqrt{\frac{GM\,T^{2}}{4\pi^{2}}} = 2.7\times 10^{11}$ m.

What to write

$r = \left(\frac{GM\,T^{2}}{4\pi^{2}}\right)^{1/3} = 4.22\times 10^{7}$ m, which is an altitude of about $35\,900$ km.

Why: The law is cubic in $r$ and quadratic in $T$ , so isolating $r$ needs a cube root. The check costs one second: a satellite of the Earth cannot orbit at $2.7\times 10^{11}$ m, which is about $700$ times the distance to the Moon.

8. An outward force invented to keep the water in the bucket

2 points here, and the same mistake ruins the centrifuge and the orbit questions

What not to write

At the top of the swing the centrifugal force is larger than the weight, which is what keeps the water in the bucket.

What to write

At the top, tension and weight BOTH point down, toward the centre: $T + mg = \frac{mv^{2}}{L}$ . The water stays in as long as $v \ge \sqrt{gL} = 3.28$ m/s with $L = 1.10$ m.

At the top both arrows point at the centre and ADD, so the rope can go slack; at the bottom they OPPOSE, so the rope carries more than the weight. Same rope, same bucket, two different equations.

Why: No object exerts an outward force on the water. The bucket bottom is ABOVE the water at the top of the circle and can only push it downward. That push plus the weight is exactly what the circle demands, and at $v_{\min}$ the push is zero and gravity does all of it.

9. Friction on a banked curve pointed toward the centre when the car is too slow

3 points, because the sign of

f

carries into

N

and into

\mu_{s}

What not to write

The car takes the icy ramp at $9.0$ m/s, below its design speed, so friction points down the slope and helps it turn.

What to write

Below the design speed the bank alone pushes inward too hard, so friction points UP the slope: $f = m\left(g\sin\theta - \frac{v^{2}}{r}\cos\theta\right) = 1.33\times 10^{3}$ N, which needs $\mu_{s} \ge 0.113$ .

Same ramp, forces to scale except friction, drawn three times longer: below the design speed

f

points up the slope, at it

f = 0

, above it down the slope;

a

always points at the centre.

Why: Friction opposes the sliding that would happen without it. Slower than the design speed, the car would slide DOWN the bank, toward the inside; faster, UP and out. Check the two limits: at $v = 0$ it is a car parked on an incline, $f = mg\sin\theta$ up the slope, and at the design speed $f = 0$ .

Which method to choose

Which friction expression to write

Decide on what the problem says about MOTION along the surface, never on the word friction.

If the object is at rest and stays at rest → $f_{s}$ is whatever balances the other forces along the surface, then check $f_{s} \le \mu_{s}N$

Example: pushed with $60$ N on a level floor: $f_{s} = 60$ N, ceiling $110.25$ N, so it holds

the single most profitable line of the chapter, and the one most often skipped
If the words on the point of, about to move, minimum or maximum appear → $f_{s} = \mu_{s}N$ , the one situation where the equality holds

Example: tilting a plank until the block lets go: $\tan\theta_{s} = \mu_{s} = 0.60$ , so $\theta_{s} = 31.0^{\circ}$

this is also how $\mu_{s}$ is measured in the laboratory, and the angle does not depend on the mass
If the object is already sliding → $f_{k} = \mu_{k}N$ , directed against the motion

Example: $0.32 \times 245 = 78.4$ N, so a push of $130$ N gives $a = 2.06$ m/s $^2$

friction stops adjusting here, which is why the block leaps forward at the moment it lets go
If the object turns without sliding on the surface that carries it → static friction again, and it IS the centripetal force: $\frac{mv^{2}}{r} \le \mu_{s}N$

Example: coin on a turntable, $a_{c} = 2.66$ m/s $^2$ , so $\mu_{s} \ge 0.272$

the mass cancels, so a heavy car and a light car leave the road at the same speed

On an incline replace $N = mg$ by $N = mg\cos\theta$ before anything else. Forgetting the cosine multiplies both the ceiling and the kinetic value by the same wrong factor, so the error survives to the end of the problem undetected.

Which circular-motion equation to write

The equation is always $\sum F = \frac{mv^{2}}{r}$ along the radius. What changes is which forces sit on that radius, and that is decided by WHERE the centre is.

If flat curve, centre horizontal, friction the only horizontal force → $\frac{mv^{2}}{r} \le \mu_{s}mg$ , so $v_{\max} = \sqrt{\mu_{s}gr}$

Example: $r = 60$ m on dry asphalt, $\mu_{s} = 0.70$ : $v_{\max} = 20.3$ m/s, about $73$ km/h

no mass in the answer, so the loaded truck and the small car go off at the same speed
If banked curve or conical pendulum, one tilted force and no friction → $X\cos\theta = mg$ and $X\sin\theta = \frac{mv^{2}}{r}$ , divide to get $\tan\theta = \frac{v^{2}}{rg}$

Example: $\theta = 12^{\circ}$ and $r = 85$ m give $v = 13.3$ m/s, about $47.9$ km/h

$X$ is the normal force for the car and the tension for the pendulum; in both, $X > mg$
If top of a vertical circle, the centre is BELOW → $T + mg = \frac{mv^{2}}{r}$ , and $T = 0$ gives $v_{\min} = \sqrt{gr}$

Example: $L = 1.10$ m: $v_{\min} = 3.28$ m/s, below which the rope goes slack

the same condition, $v = \sqrt{gr}$ , is apparent weightlessness at the crest of a hill
If bottom of a vertical circle, of a dip or of a dive, the centre is ABOVE → $N - mg = \frac{mv^{2}}{r}$ , so $N = m\left(g + \frac{v^{2}}{r}\right)$

Example: $180$ m/s on a $650$ m arc: $\frac{N}{mg} = 1 + \frac{v^{2}}{rg} = 6.09$

this is where ropes break and pilots grey out, never at the top
If orbit, gravity the only force → $G\frac{Mm}{r^{2}} = \frac{mv^{2}}{r}$ , so $v = \sqrt{\frac{GM}{r}}$ and $T^{2} = \frac{4\pi^{2}}{GM}r^{3}$

Example: $r = 6.77\times 10^{6}$ m gives $v = 7.67$ km/s and $T = 92.4$ min

the orbiting mass cancels in the first line, so it must be absent from every answer

Draw the centre of the circle on your sketch before writing anything. Four of these five branches are told apart by that single dot, and the two vertical-circle cases differ only by which side of the object it sits on.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a friction problem

When to use it: Any problem naming a coefficient of friction, on level ground or on an incline.

1 Draw the free-body diagram and name the surface: friction acts along it, the normal force across it.
2 Compute $N$ FIRST and write the line: $N = mg$ on level ground, $N = mg\cos\theta$ on an incline. Both coefficients multiply it.
3 Compute the ceiling $\mu_{s}N$ and compare it with the force demanded along the surface. Write the comparison as a sentence, it is worth a mark on its own.
4 Choose the branch: at rest, $f_{s}$ equals the demanded force; sliding, $f_{k} = \mu_{k}N$ against the motion.
5 Apply $\sum F = ma$ along the surface and check that the sign of $a$ matches the direction you expected.

Concluding sentence

Since $F = 60$ N is below the ceiling $\mu_{s}N = 110.25$ N, the crate stays at rest and the friction force on it is $f_{s} = 60$ N.

The trap: The step everybody drops is the third. Without the written comparison, the marker cannot tell whether you chose the right branch or guessed it, and a correct final number in a problem where the block slides earns partial credit at best.

Marking: 1 point for the diagram and $N$ , 1 point for the ceiling and the comparison, 2 points for the correct branch, 1 point for Newton's second law and the sign.

Writing up a circular motion problem

When to use it: Anything moving on a circle: a curve, a pendulum tracing a cone, a bucket, a centrifuge, a satellite.

1 Draw ONLY the real forces. If you cannot name the object exerting an arrow, that arrow does not exist.
2 Mark the centre of the circle on the sketch and draw the axis from the object toward it.
3 Write $\sum F_{\text{radius}} = \frac{mv^{2}}{r}$ , counting each force positive toward the centre and negative away from it.
4 Write the second equation on the other axis, where the acceleration is ZERO in a horizontal circle.
5 Solve, then check that the mass has cancelled everywhere it should, and that any normal force or tension you found is positive.

Concluding sentence

Taking the direction of the centre as positive, the resultant of the real forces along the radius is $T + mg = \frac{mv^{2}}{L}$ , which gives $T = 25.8$ N.

The trap: The sacrificed step is the second. Without the centre drawn, the top and the bottom of a vertical circle get the same equation, and the two differ only by the sign of $mg$ : writing $T - mg$ at the top turns a tension of $25.8$ N into one of $84.6$ N.

Marking: 1 point for a diagram with no invented force, 1 point for the axis and the centre, 2 points for the radial equation with its signs, 1 point for the second axis.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Did the mass survive?

Go back over your final expression and look for $m$ . It must be absent from a slipping angle, a maximum speed on a curve, a banked design speed, a minimum speed at the top of a circle, an orbital speed and an orbital period. If $m$ is still there, you divided somewhere you should have cancelled.

$v_{\max} = \sqrt{\mu_{s}gr} = \sqrt{0.70 \times 9.8 \times 60} = 20.3$ m/s, the same for a $1300$ kg car and a $30\,000$ kg truck.
Is the normal force on the right side of the weight?

On an incline $N$ is SMALLER than $mg$ , by the factor $\cos\theta$ . On a banked curve at its design speed, at the bottom of a dip and in a lift accelerating upward, $N$ is LARGER than $mg$ . Getting the wrong side means you split the wrong vector, and it is visible before any arithmetic.

Banked at $12^{\circ}$ with $m = 1200$ kg: $N = 12\,023$ N against a weight of $11\,760$ N, so $N$ is larger, as it must be.
Is the friction you wrote under its ceiling?

After finding $f_{s}$ , compute $\mu_{s}N$ and compare. A static friction above the ceiling means the object is in fact sliding and the whole problem must be redone with $\mu_{k}$ . A static friction exactly equal to the ceiling means you assumed the limiting case without being told to.

On a plank at $25^{\circ}$ with $\mu_{s} = 0.60$ : $mg\sin 25^{\circ} = 16.6$ N against a ceiling of $21.3$ N, so the block holds.
Does the order of magnitude make sense?

Surface gravity runs from about $1$ to $25$ m/s $^2$ in the Solar System. A low orbit takes about $90$ minutes and a geostationary orbit exactly $24$ h. A car on a curve leaves the road at tens of km/h, not hundreds. One glance at these ranges catches a cube root taken as a square root, or an altitude used as a distance from the centre.

On Mars, $g = \frac{GM}{R^{2}} = 3.73$ m/s $^2$ , about $0.38$ of the Earth's value, which is the figure quoted everywhere.

The typical problem, taken apart

The curve at 73 km/h, and what banking actually buys you

A flat curve of radius $r = 60$ m is covered in dry asphalt, $\mu_{s} = 0.70$ . A car of mass $1300$ kg takes it.

Find the largest speed at which the car holds the curve, and say whether that speed depends on the mass of the car.

Then compare with the same curve banked at $15^{\circ}$ , first on dry asphalt and then in freezing rain, where $\mu_{s}$ falls to $0.10$ .

Seen from above, the centre of the turn is level with the car and the only horizontal force available is friction: the whole problem is

f_{s} = \frac{mv^{2}}{r}

with

f_{s}

under its ceiling.

Step 1

Forces on the car: the weight $mg$ down, the normal force $N$ up, friction $f_{s}$ horizontal. Nothing else.

Why

Naming the three real forces and stopping there is the step being marked. The car is turning, so a fourth arrow is tempting, and it would be the same friction counted twice.

Step 2

Vertical axis, no acceleration: $N = mg = 1300 \times 9.8 = 12\,740$ N.

Why

The curve is FLAT, so the car goes neither up nor down and the vertical equation is a statics equation. This is also the line that feeds the friction ceiling, so it comes first.

Step 3

Radial axis, pointing at the centre: $f_{s} = \frac{mv^{2}}{r}$ , with the constraint $f_{s} \le \mu_{s}N = 0.70 \times 12\,740 = 8918$ N.

Why

Here is the whole chapter in one line. Friction is not given a value, it is told what it must supply, and separately told what it may not exceed. The inequality is what produces a maximum speed at all.

Step 4

Combine: $\frac{mv^{2}}{r} \le \mu_{s}mg$ , the mass cancels, $v_{\max} = \sqrt{\mu_{s}gr} = \sqrt{0.70 \times 9.8 \times 60} = \sqrt{411.6} = 20.3$ m/s, that is $73.0$ km/h.

Why

The mass cancels because it sits on both sides, in the demand $\frac{mv^{2}}{r}$ and in the supply $\mu_{s}mg$ . That is why a heavier vehicle is no safer on a curve, which is the answer to the second question and is worth stating in words.

Step 5

Banked at $15^{\circ}$ with no help from friction: $\tan 15^{\circ} = \frac{v^{2}}{rg}$ , so $v = \sqrt{60 \times 9.8 \times 0.2679} = 12.6$ m/s, about $45.2$ km/h.

Why

A different mechanism, same three lines: split $N$ instead of splitting nothing, and divide the two equations. Note that the axes are horizontal and vertical again, NOT along the road surface.

Step 6

In freezing rain, $\mu_{s} = 0.10$ : flat, $v_{\max} = \sqrt{0.10 \times 9.8 \times 60} = 7.7$ m/s, about $27.6$ km/h. Banked, the design speed $45.2$ km/h needs no friction at all.

Why

This is why curves are banked. On dry asphalt friction alone already gives $73$ km/h, more than the bank provides; the bank earns its keep the day friction disappears, where it turns $28$ km/h into $45$ km/h with nothing gripping.

The conclusion, written out

On dry asphalt the car holds the flat curve up to $20.3$ m/s, that is $73.0$ km/h, whatever its mass; banking the same curve at $15^{\circ}$ gives a design speed of $12.6$ m/s, that is $45.2$ km/h, which still works when the road ices over and the flat curve is down to $27.6$ km/h.

The classic mistake on this problem: The classic loss is to add the two effects by writing $v_{\max} = \sqrt{\mu_{s}gr} + \sqrt{rg\tan\theta}$ . Speeds do not add like that. A banked curve WITH friction is a single free-body diagram in which $N$ and $f_{s}$ both have horizontal components, and it is solved with the same two axes, never by adding two separate answers.

Learn by heart

• $f_{s} \le \mu_{s}N$ is an INEQUALITY; $f_{k} = \mu_{k}N$ is an equality. The equality for $f_{s}$ holds only on the point of slipping.
• A block lets go of an incline at $\tan\theta_{s} = \mu_{s}$ , an angle that does not contain the mass.
• $a_{c} = \frac{v^{2}}{r} = \frac{4\pi^{2}r}{T^{2}}$ : the first form for a given speed, the second for a given period or rotation rate.
• Centripetal force is the RESULTANT along the radius, never a fourth arrow. Nothing on a correct diagram points outward.
• Flat curve $v_{\max} = \sqrt{\mu_{s}gr}$ ; frictionless banked curve $\tan\theta = \frac{v^{2}}{rg}$ with $N = \frac{mg}{\cos\theta}$ , LARGER than $mg$ .
• Vertical circle: $T + mg = \frac{mv^{2}}{r}$ at the top, $T - mg = \frac{mv^{2}}{r}$ at the bottom, and $v_{\min} = \sqrt{gr}$ when the rope goes slack.
• $F = G\frac{m_{1}m_{2}}{r^{2}}$ with $r$ from CENTRE to centre; $g = \frac{GM}{R^{2}}$ at the surface; $v = \sqrt{\frac{GM}{r}}$ and $T^{2} = \frac{4\pi^{2}}{GM}r^{3}$ in orbit.
• Apparent weight $N = m\left(g \pm \frac{v^{2}}{r}\right)$ : weightlessness means $N = 0$ , never $g = 0$ .
• Banked curve below its design speed: friction points UP the slope, $f = m\left(g\sin\theta - \frac{v^{2}}{r}\cos\theta\right)$ ; above it, DOWN.
• Spinning habitat: $N = m\omega^{2}r$ , $\omega = \sqrt{g/R}$ for $1$ g on the rim, and the load on a deck is proportional to its distance from the axis.

Frequently asked questions

Is the centripetal force a real force in PHYS 101?

No, and saying so is worth marks. The real forces are the ones other objects exert: gravity, the normal force, tension, friction. Circular motion adds none of them. The phrase centripetal force is the name given to the resultant of those real forces, measured along the direction of the centre, and its size is the mass times the speed squared divided by the radius. Draw the real forces, point an axis at the centre, and set their sum equal to that quantity. If you add a separate arrow for it, you have counted one force twice.

When can I write that friction equals mu times the normal force?

Always for kinetic friction, once the object is actually sliding. For static friction, only at the single instant when the object is on the point of slipping, which the problem announces with words such as on the point of, about to move, minimum force or maximum speed. In every other case the object is at rest and static friction simply balances whatever is pushing along the surface, so you find it from the condition that the forces cancel, not from a formula.

Why are astronauts weightless if gravity is still there?

Because what your body feels is the floor pushing on you, not gravity itself. Four hundred kilometres up, the strength of gravity is still about eighty nine per cent of its value on the ground, and that is exactly what keeps the station on its circular path. The station, the astronaut and the sandwich are all falling toward the Earth with the same acceleration, so the floor never has to push on anyone. The normal force is zero, and that sensation is what we call weightlessness.

Do I need calculus for the circular motion chapter of PHYS 101?

No. PHYS 101 is the algebra based mechanics course at McGill, and every result in this chapter comes from Newton's second law, right angle trigonometry and ordinary algebra. Most solutions you will find online are written for a calculus based course and open by differentiating a position vector, which is not the method you are being examined on. If a solution starts with a derivative, it is not a PHYS 101 solution even when its final number happens to be correct.

Does a heavier car go off a curve before a lighter one?

No, and the reason is worth understanding. The heavier car needs a larger friction force to turn, because the force required grows with the mass. But it also presses harder on the road, so the largest friction available grows with the mass in exactly the same proportion. The two effects cancel and the mass disappears from the answer, leaving a maximum speed that depends only on the coefficient of friction, the radius of the curve and the strength of gravity.

How does artificial gravity work on a rotating space station?

The station spins, and the floor of its outer rim has to push everything resting on it toward the axis to keep it moving on a circle. That push is exactly what a bathroom scale reads, so it feels like weight. For a ring a couple of hundred metres across, one turn every thirty seconds or so gives the same push as gravity on Earth. A deck built closer to the axis turns at the same rate on a smaller circle, so its floor pushes less, in proportion to the distance: a deck at sixty per cent of the radius gives sixty per cent of the weight.

Practise it

Corrected exercises: Friction, circular motion and gravitation, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Previous sheet Newton's laws and free-body diagrams Next sheet Work, energy and power

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-friction-circular-gravitation. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: friction, circular motion and gravitation (PHYS 101)

The essentials

On your formula sheet: this chapter's lines, and when each one holds

Static friction is an inequality, kinetic friction is a value

Friction on the diagram: the two standard cases

Centripetal is a direction, never a new force

Banked curve and conical pendulum: one diagram

Vertical circles: the crest, the dip and the loop

Gravitation: three formulas and one distance

Artificial gravity: the floor supplies the circle

The rules in table form

Which friction force to write, for a crate of 252525 kg on level ground

The mistakes that cost marks

1. Static friction set equal to its ceiling on a block that is merely at rest

2. An invented centripetal arrow added to a diagram that was already complete

3. The incline formula for the normal force carried over to a banked curve

4. A larger radius read as a smaller acceleration

5. An altitude fed into the law of gravitation in place of a distance from the centre

6. Weightlessness in orbit blamed on an absence of gravity

7. A square root taken where Kepler's third law asks for a cube root

8. An outward force invented to keep the water in the bucket

9. Friction on a banked curve pointed toward the centre when the car is too slow

Which method to choose

Which friction expression to write

Which circular-motion equation to write

How the answer is expected to be written

Writing up a friction problem

Writing up a circular motion problem

Check before you hand in

The typical problem, taken apart

The curve at 73 km/h, and what banking actually buys you

Learn by heart

Frequently asked questions

Is the centripetal force a real force in PHYS 101?

When can I write that friction equals mu times the normal force?

Why are astronauts weightless if gravity is still there?

Do I need calculus for the circular motion chapter of PHYS 101?

Does a heavier car go off a curve before a lighter one?

How does artificial gravity work on a rotating space station?

Corrected exercises: Friction, circular motion and gravitation, PHYS 101 at McGill

See also

Looking for a PHYS 101 tutor in Montreal?

Which friction force to write, for a crate of $25$ kg on level ground