Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: dynamics, Newton's laws and friction (203-SN1-RE)

Q: Does centrifugal force exist?

Not in the CEGEP mechanics course, which works in an inertial frame. An object in a curve is not in equilibrium: it accelerates toward the centre, and the only forces to put on the diagram are real ones, such as friction, the normal force or tension. The sensation of being pushed outward comes from the passenger's inertia, not from a force acting on them.

Q: When is the normal force equal to the weight?

Only on a horizontal surface, when no other force has a vertical component and the object does not accelerate vertically. On a slope it equals the weight times the cosine of the angle. If you pull or push at an angle, or if the object is in an accelerating elevator, it changes again. The normal force is always calculated, never assumed.

Q: How do I know whether an object will slide or stay at rest?

You compare two numbers. First the net driving force without friction, that is, the sum of the forces along the possible slipping. Then the ceiling of static friction, equal to the coefficient of static friction times the normal force. If the driving force exceeds the ceiling, the object slides and you start again with the kinetic coefficient. Otherwise it stays at rest and the friction equals exactly the driving force.

Q: What is the minimum speed at the top of a vertical loop?

It is the square root of the product of the gravitational acceleration and the radius. You get it by writing that the contact force becomes zero: gravity then supplies the centripetal acceleration all by itself. Careful, the condition is not a zero speed but a zero contact force, which is very different.

Q: Why is the tension not equal to the weight of the hanging mass?

Because that equality holds only at equilibrium. As soon as the mass goes down while accelerating, the rope pulls less hard than the weight, otherwise the mass would not go down. The tension is then the mass times the difference between the gravitational acceleration and the acceleration of the system, and it always stays between zero and the weight.

Q: How do I handle two blocks joined by a rope and a pulley?

You write one free-body diagram per block, never a single one for both. You orient the positive axis of each block in the assumed direction of motion, you project, then you add the rope constraint: same magnitude of acceleration and same tension if the pulley is ideal. By adding the two equations, the tension disappears and the acceleration comes out on its own.

In 203-SN1-RE (formerly 203-NYA), dynamics adds almost nothing to what you saw in secondary school, in terms of formulas: $\Sigma\vec{F}=m\vec{a}$ and friction, that is all. What changes is the difficulty of the setups: two coupled bodies, a banked curve, a vertical circle, a terminal speed. In each of them, the error never comes from the calculation, it comes from the free-body diagram.

This sheet gathers what really costs marks on this chapter: invented forces, the assumed normal force, friction written at its maximum for no reason, and the confusion between a force and a net force. Each trap is written as it appears on a student paper. It is written for CEGEP students in the Science program (Sciences de la nature).

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A force enters an equation only if it appears on the free-body diagram of ONE isolated body. Centripetal force is never an extra force: it is the name the net force takes when the path is curved.

This chapter is part of 203-SN1-RE, Mechanics

The essentials

The method, always the same, in this order

• ISOLATE one body, and only one. The free-body diagram is not a drawing of the scene: it is the list of forces acting ON that body.
• The forces that exist in this course: the weight $m\vec{g}$ , the normal force $\vec{N}$ , friction $\vec{f}$ , tension $\vec{T}$ , an applied force, a fluid drag force. Nothing else.
• CHOOSE the axes, preferably with one axis along the acceleration: along the slope on an incline; toward the centre on a circle.
• PROJECT, then write $\Sigma F_{x}=ma_{x}$ and $\Sigma F_{y}=ma_{y}$ . Two equations per body, as many systems as bodies.
• A centrifugal force, a force of motion, an inertial force: none appears in an inertial frame, so never in this course.

The weight (red) is the only force at a distance in this course; the normal force (blue) and the tension (green) exist only through contact. Each diagram keeps only the forces acting ON the block.

For two bodies joined by an inextensible rope, the constraint is written before the equations: same magnitude of acceleration, same magnitude of tension if the pulley is ideal. It is what makes the system solvable.

Friction changes its rule depending on whether you slip

• Static: $f_{s}\le\mu_{s}N$ . As long as nothing slips, $f_{s}$ equals exactly what is needed to balance, NOT $\mu_{s}N$ .
• $\mu_{s}N$ is a CEILING, reached only at the instant of breakaway, that is, when the statement says “just before slipping” or “at minimum”.
• Kinetic: $f_{k}=\mu_{k}N$ , constant, opposite to the relative SLIPPING, with always $\mu_{k}<\mu_{s}$ .
• The normal force $N$ equals $mg$ only on a horizontal surface, with no other vertical force and no vertical acceleration. Everywhere else it comes out of the $y$ equation.
• On an inclined plane alone: $N=mg\cos\theta$ , and the object stays at rest as long as $\tan\theta\le\mu_{s}$ , independently of its mass.

The weight is resolved into two arrows shorter than itself:

N

balances only

mg\cos\theta

, so

N<mg

. As long as nothing slips,

f

exactly balances

mg\sin\theta

, not

\mu_{s}N

The test for staying at rest is always written in the same order: compute the net driving force excluding friction, compute $\mu_{s}N$ , compare. Writing the acceleration before this comparison means answering a question that has not been asked yet.

Two bodies joined by a rope: Atwood and the table

• Atwood machine, two masses hanging on either side of an ideal pulley: same $T$ on both sides, same magnitude of acceleration, and $a=\dfrac{(m_{2}-m_{1})g}{m_{1}+m_{2}}$ .
• Block on a table joined to a hanging mass: on the table, $T-f=m_{1}a$ with $N=m_{1}g$ ; for the hanging mass, $m_{2}g-T=m_{2}a$ .
• As soon as the system accelerates, $T$ lies strictly between what holds back and what pulls: $T=m_{2}g$ is true only at equilibrium.

To scale: on each block, the tension (green) is smaller than the weight of the mass going down and larger than what holds it back, the weight of

m_{1}

or the friction

f

Apparent weight, in an elevator

• The scale measures $N$ , not the weight. Axis upward: $N-mg=ma$ , so $N=m(g+a)$ .
• Only the DIRECTION of the acceleration matters: upward, $N>mg$ ; downward, $N<mg$ ; zero, $N=mg$ , at rest as well as at constant speed.
• In free fall, $a=-g$ and $N=0$ : the weight has not disappeared, the contact has.

Same weight in all three cases: only the normal force changes, longer than

mg

when

\vec{a}

points up, shorter when

\vec{a}

points down, equal at constant speed.

Circular motion, where the chapter tips over

• Curved path and constant speed: $a=\dfrac{v^{2}}{r}$ directed TOWARD THE CENTRE, so $\Sigma F_{\text{toward the centre}}=\dfrac{mv^{2}}{r}$ .
• This right-hand side is not a force: it is $ma$ . It is written to the right of the equals sign, never in the diagram.
• Banked curve without friction: $\tan\theta=\dfrac{v^{2}}{rg}$ , obtained by projecting $N$ onto the horizontal and the vertical.
• Vertical circle, at the highest point: the weight AND the tension (or the normal force) both point toward the centre, so $T+mg=\dfrac{mv^{2}}{r}$ . The minimum speed corresponds to $T=0$ , that is $v_{min}=\sqrt{gr}$ .
• At the lowest point of the same circle: $T-mg=\dfrac{mv^{2}}{r}$ , so $T$ is greatest there. The two points are not handled with the same equation.

Banked curve without friction:

N

leans toward the centre, its vertical part carries

mg

and its horizontal part is the net force

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

(black dotted). No centrifugal force.

Choosing the positive axis TOWARD THE CENTRE in circular motion removes half the sign errors: the acceleration is always positive, and each force takes its sign according to whether it points toward the centre or not.

The hump and the conical pendulum: everything reads toward the centre

• Top of a hump of radius $r$ : the centre is BELOW the car, so $mg-N=\dfrac{mv^{2}}{r}$ and $N<mg$ . The car leaves the ground when $N=0$ , that is $v=\sqrt{gr}$ .
• Conical pendulum: $T\cos\theta=mg$ vertically, $T\sin\theta=\dfrac{mv^{2}}{R}$ toward the axis, with $R=L\sin\theta$ .
• In both cases, the sum of the forces directed toward the centre equals $\dfrac{mv^{2}}{r}$ : no extra force enters the diagram.

At the top of the hump,

N<mg

and the net force points down toward the centre; in the conical pendulum,

T

carries

mg

and its horizontal part is the net force toward the axis.

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. Adding a centrifugal force to the free-body diagram of an object in a curve

the whole question, 5 to 8 marks

What not to write

“The car is in equilibrium in the curve: the centripetal force toward the centre balances the centrifugal force toward the outside.”

What to write

“The car is not in equilibrium, it accelerates toward the centre. Friction is the only horizontal force: $f=\dfrac{mv^{2}}{r}$ .”

A single horizontal force, friction, and it points toward the centre: it is the net force

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

. Nothing balances it, the car is not in equilibrium.

Why: Writing the two forces equal and opposite amounts to writing $\Sigma F=0$ , so $a=0$ , so a straight path. The calculation then proves that the car does not turn.

2. Setting the normal force equal to the weight when a force is applied at an angle

3 marks, and an acceleration that is always underestimated

What not to write

“We pull the $20$ kg crate with $80$ N at $30°$ above the horizontal: $N=mg=196$ N and $f=\mu_{k}\times 196$ .”

What to write

“The $y$ equation gives $N+80\sin 30°-196=0$ , so $N=156$ N, and $f=\mu_{k}\times 156$ . The pull lightens the crate.”

The pull has a vertical component of

80\sin 30°=40

N that lightens the crate:

N=196-40=156

N, blue arrow shorter than the weight. Arrows drawn to scale.

Why: The normal force is an unknown, never a given: it is read from the equation perpendicular to the motion. Any force with a vertical component changes it, one way or the other.

3. Writing friction at its maximum on a body that does not move

3 marks, and a nonzero net force on a body at rest

What not to write

“The $10$ kg crate is pushed with $20$ N and does not move, $\mu_{s}=0.5$ : so $f=0.5\times 98=49$ N.”

What to write

“It does not move, so $\Sigma F=0$ and $f=20$ N. The $49$ N is the CEILING that friction could reach, not its value here.”

Why: A friction of $49$ N against a push of $20$ N would give a net force of $29$ N backward: the crate would move backward all by itself under the effect of friction.

4. Adding the two forces of an action-reaction pair on the same body

the whole reasoning question, often 4 marks

What not to write

“The horse pulls the cart and the cart pulls the horse with the same force: they cancel, so nothing can move.”

What to write

“These two forces act on DIFFERENT bodies, so they never appear in the same diagram. On the horse, what matters is the friction of the ground forward against the pull of the cart backward.”

The two green 700 N arrows form the action-reaction pair: one acts on the cart, the other on the horse. On the horse alone, 900 N forward against 700 N: it moves.

Why: The third law relates two bodies, the second law describes a single body. Mixing up the two levels is the most persistent conceptual error of the chapter, and it comes back on every exam in a different form.

5. Taking the tension equal to the weight of the hanging mass in a system that accelerates

3 marks, and the rest of the system thrown off

What not to write

“The $4.0$ kg mass hangs at the end of the rope, so $T=4.0\times 9.8=39.2$ N.”

What to write

“The mass goes down while accelerating, so $m_{2}g-T=m_{2}a$ and $T=m_{2}(g-a)<m_{2}g$ . With $a=1.60$ m/s $^{2}$ , $T=32.8$ N.”

Why: $T=mg$ is true only at equilibrium. As soon as there is acceleration, the tension lies strictly between the two weights of a two-mass system: it is, incidentally, the best plausibility check on the result.

6. Looking for the minimum speed at the top of a vertical circle by setting the speed to zero

the whole question, 5 marks

What not to write

“At the top, the ball is about to fall, so its minimum speed is zero.”

What to write

“The limiting condition is $N=0$ , not $v=0$ : gravity alone must supply the centripetal acceleration, so $mg=\dfrac{mv^{2}}{r}$ and $v_{min}=\sqrt{gr}$ .”

At the top, both arrows point down: they ADD toward the centre, so

T

can vanish without the ball falling. At the bottom, they oppose each other, and that is where the rope is most tense.

Why: With $v=0$ at the top, there is no circular path at all: the object leaves the track and falls. The physical condition concerns the contact, hence the contact force.

7. Mixing up the frictionless banked-curve formula and the case with friction

the whole question, 6 marks

What not to write

“Curve banked at $15°$ with $\mu_{s}=0.60$ and $r=80$ m: $\tan 15°=\dfrac{v^{2}}{rg}$ gives $v=14.5$ m/s, which is the maximum speed.”

What to write

“ $\tan\theta=\dfrac{v^{2}}{rg}$ gives the IDEAL speed, the one where friction is zero. The maximum speed is obtained by putting $f=\mu_{s}N$ back into both projections, and here it is $28.5$ m/s.”

At the maximum speed, friction points DOWN the slope and adds its horizontal part to that of

N

: the net force exceeds that of the frictionless curve.

Why: The frictionless formula is a special case, not a general formula. A statement that supplies $\mu_{s}$ is telling you exactly that it must not be used: an unused given is almost always the sign of a wrong method.

8. Looking for the terminal speed by setting the speed to zero instead of the acceleration

the whole question, 4 marks

What not to write

“At terminal speed the motion stops, so $v=0$ and $mg=bv$ gives $v=0$ .”

What to write

“At terminal speed the ACCELERATION vanishes, not the speed: $mg-bv_{T}=0$ , so $v_{T}=\dfrac{mg}{b}$ . The motion continues, at constant speed.”

The weight does not change,

bv

grows with the speed: at

v_{T}

the two arrows are equal, the net force is zero and the speed stops growing, without vanishing.

Why: “Terminal” describes the value toward which the speed tends, not a stop. It is the same confusion as between $v=0$ and $a=0$ at the top of a throw, transposed to air resistance.

9. Using the total mass of the system in the equation of a single body

3 marks

What not to write

“Three blocks of $2$ , $3$ and $5$ kg pushed by $30$ N: the force that the first exerts on the second is $30$ N.”

What to write

“The common acceleration is $a=\dfrac{30}{10}=3.0$ m/s $^{2}$ . We then isolate blocks $2$ and $3$ : the force of the first on the second is $(3+5)\times 3.0=24$ N.”

We isolate blocks 2 and 3 (dashed frame): the only horizontal force pushing them is that of block 1,

(3+5)\times 3.0=24

N, not the 30 N applied to block 1.

Why: The total mass is used to find the common acceleration, never to calculate an internal force. A contact force is always read by isolating what lies AHEAD of it in the direction of motion.

Which method to choose

Which friction to write

The exact wording of the statement about the relative motion between the surfaces

If the statement says “slides”, “moves over”, “is pulled at constant speed” → $f=\mu_{k}N$ , opposite to the slipping

Example: crate pulled over a floor: $f=\mu_{k}N$ even if $a=0$
If the statement says “stays at rest”, “does not move”, “without slipping” → $f$ is an unknown determined by $\Sigma F=ma$ , never $\mu_{s}N$

Example: crate pushed with $20$ N that does not move: $f=20$ N
If the statement says “just before slipping”, “minimum force”, “maximum angle”, “maximum speed” → $f=\mu_{s}N$ , the only case where the equality is legitimate

Example: breakaway angle: $\tan\theta_{max}=\mu_{s}$
If the statement asks WHETHER the object moves → compare the net driving force with $\mu_{s}N$ before any other equation

Example: $19.6$ N against a ceiling of $10.2$ N: it slides
If a wheel rolls without slipping → the friction is STATIC at the contact point, and it dissipates nothing

that is what lets a car accelerate without its tires heating up

In every case, $N$ is calculated first, and never by assuming $N=mg$ : it comes out of the equation perpendicular to the motion, once all the forces are projected.

Which body to isolate, and how many equations to write

The number of bodies and the nature of the question

If a single body, a single acceleration → one diagram, two projections

Example: block on a rough inclined plane
If several joined bodies, and the question is about the common acceleration → treat the whole as a single body of total mass, the internal forces disappearing

Example: three blocks pushed: $a=\dfrac{F}{m_{1}+m_{2}+m_{3}}$
If the question is about a TENSION or an internal contact force → isolate one of the bodies, necessarily: internal forces exist only in an individual diagram

Example: $T=m_{2}(g-a)$ for the hanging mass
If the bodies are joined by a rope passing over a pulley → one diagram per body, plus the rope constraint: same $|a|$ , same $T$

Example: inclined plane plus hanging mass, a system of two equations
If one block rests on another block → two diagrams, and the normal force between them is NOT the weight of the top block as soon as there is vertical acceleration

the force that sets the top block in motion is friction, the only horizontal force acting on it

Practical rule: as many unknowns as equations. Two bodies give four projections; if you write three, an equation is missing, and it is almost always the rope constraint.

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.

Solving a system of two bodies joined by a rope

When to use it: Two masses are joined by a rope over an ideal pulley, and the acceleration, the tension, or both are asked for.

1 Choose an assumed DIRECTION OF MOTION and state it, for example “we assume that $m_{2}$ goes down”. All the projections follow from it, and a negative result will simply signal that the real direction is the other one.
2 Draw one free-body diagram per body, putting in it only the forces acting on that body.
3 For each body, orient the positive axis in the assumed direction of motion, then project: $\Sigma F=ma$ along this axis, and $\Sigma F=0$ perpendicular to it to get $N$ .
4 Write the rope constraint: $|a_{1}|=|a_{2}|=a$ and $T_{1}=T_{2}=T$ if the pulley is ideal.
5 Add the two equations of motion to eliminate $T$ and get $a$ , then substitute $a$ into one of them to get $T$ .
6 Check that $T$ lies between the values it would take if each body were alone, and conclude with units.

Concluding sentence

“With the positive axis in the assumed direction of motion and the constraint $|a_{1}|=|a_{2}|$ , the sum of the two equations gives $a=1.60$ m/s $^{2}$ ; substituting into the equation of the hanging mass, $T=32.8$ N, a value comfortably between $0$ and $m_{2}g=39.2$ N.”

The trap: Step 4. One student in three writes three equations for four unknowns, forgets the rope constraint, and declares the system unsolvable.

Marking: Typically 2 marks for the two diagrams, 3 for the projections, 1 for the constraint, 2 for the solution, 1 for the conclusion and units.

Deciding whether a body starts moving

When to use it: The statement asks “does the block move”, “does the system stay at rest”, or supplies both $\mu_{s}$ and $\mu_{k}$ .

1 Calculate the net driving force excluding friction, that is, the sum of the forces along the possible motion, friction left out.
2 Calculate $N$ from the perpendicular equation, then the ceiling $\mu_{s}N$ .
3 Compare the two numbers, and write the comparison explicitly.
4 If the driving force is less than the ceiling, conclude that it stays at rest and give $f$ equal to the driving force. Otherwise, start again with $f=\mu_{k}N$ and calculate the acceleration.

Concluding sentence

“The net driving force is $19.6$ N and the static ceiling $\mu_{s}N=10.2$ N; since $19.6>10.2$ , the system starts moving and we continue with kinetic friction.”

The trap: Using $\mu_{k}$ for the test. The test is ALWAYS done with $\mu_{s}$ , and $\mu_{k}$ is used only afterwards, once the motion is established.

Check before you hand in

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

The tension bracketed by the two weights

In a two-mass system, the tension is always strictly between the physically possible extreme values. A tension greater than the weight of the hanging mass is certainly wrong.

A $T=45$ N for a hanging mass of $4.0$ kg, whose weight is $39.2$ N: the rope would be pulling the mass upward while it goes down.
The limiting case $\theta=0$ on an inclined plane

Replace the angle by zero in the final result. The normal force must give back $mg$ , the driving component must vanish, and the acceleration must become that of the horizontal plane.

An expression that gives $N=mg\sin\theta$ gives back $N=0$ on the flat: sine and cosine have been swapped.
The sign of the acceleration against the assumed direction

A negative $a$ is not a mistake: it says that the motion is in the other direction. But the friction must then be redone, because it opposes the REAL motion, so it changes direction too.

An $a=-0.8$ m/s $^{2}$ obtained with friction oriented according to the initial assumption: the result is not simply to be sign-flipped, the equation has to be redone.
The order of magnitude of the acceleration

A partial free-fall acceleration never exceeds $9.8$ m/s $^{2}$ in a system where gravity is the only driving force. Any larger value signals an error in mass or sign.

An $a=14$ m/s $^{2}$ on an Atwood machine: impossible, the maximum value there is $g$ , reached only if one of the masses is zero.
The unused given

Reread the statement and tick off each given in the calculation. A supplied $\mu_{s}$ that is never used almost always signals that an ideal case was treated instead of the real one.

A banked curve solved by $\tan\theta=v^{2}/rg$ when the statement gave $\mu_{s}=0.60$ : the formula was that of the frictionless case.

The typical problem, taken apart

An inclined plane joined by a pulley to a hanging mass

A block $m_{1}=4.0$ kg rests on a plane inclined at $30°$ . A rope leaving the block passes over an ideal pulley at the top of the plane and supports a block $m_{2}=4.0$ kg hanging in the air.

The coefficients of friction between $m_{1}$ and the plane are $\mu_{s}=0.30$ and $\mu_{k}=0.20$ . Determine whether the system starts moving, then the acceleration and the tension. Take $g=9.8$ m/s $^{2}$ .

Two opposing drivers in the same system: the weight of

m_{2}

pulls the block UP the slope,

m_{1}g\sin 30°

pulls it down. Friction chooses its direction only afterwards.

Step 1

Assumed direction: $m_{2}$ goes down, so $m_{1}$ goes up along the slope. Positive axis chosen in this direction for both bodies.

Why

Stating the direction before any calculation makes every sign justifiable. Without this sentence, the marker cannot know whether a negative sign is an error or information.

Step 2

Test for staying at rest. Net driving force excluding friction: $m_{2}g-m_{1}g\sin 30°=39.2-19.6=19.6$ N. Normal force: $N=m_{1}g\cos 30°=33.9$ N, so $\mu_{s}N=0.30\times 33.9=10.2$ N.

Why

This comparison is the only question that can be answered before knowing $a$ . Skipping it risks calculating an acceleration for a system that does not move.

Step 3

Since $19.6\ \text{N}>10.2\ \text{N}$ , the system starts moving, and we switch to kinetic friction: $f_{k}=\mu_{k}N=0.20\times 33.9=6.79$ N, directed DOWN the slope since $m_{1}$ goes up.

Why

The direction of friction is decided after the direction of motion, never before. It is the only force in the diagram whose direction depends on the answer to the previous question.

Step 4

Hanging block: $m_{2}g-T=m_{2}a$ , that is $39.2-T=4.0\,a$ . Block on the slope: $T-m_{1}g\sin 30°-f_{k}=m_{1}a$ , that is $T-19.6-6.79=4.0\,a$ .

Why

One equation per body, each written in its own frame but with the same $a$ and the same $T$ : it is the constraint of the inextensible rope and the ideal pulley that allows this identification.

Step 5

Sum of the two equations: $39.2-19.6-6.79=8.0\,a$ , so $a=\dfrac{12.8}{8.0}=1.60$ m/s $^{2}$ .

Why

The addition eliminates $T$ with no algebraic manipulation: that is the whole point of having oriented both axes in the assumed direction of motion.

Step 6

Substitute into the first equation: $T=39.2-4.0\times 1.60=32.8$ N.

Why

We substitute into the simplest equation, the one containing neither friction nor angle: fewer terms, fewer chances to slip.

Step 7

Check with the other equation: $T=19.6+6.79+4.0\times 1.60=32.8$ N, and $0<32.8<39.2$ N.

Why

Recomputing $T$ with the equation we did not use checks the whole system at once. Bracketing by the weight of $m_{2}$ is the plausibility check that takes a second.

The conclusion, written out

“Since the net driving force of $19.6$ N exceeds the static ceiling of $10.2$ N, the system starts moving: block $m_{2}$ goes down with an acceleration of $1.60$ m/s $^{2}$ and the rope is under a tension of $32.8$ N.”

The classic mistake on this problem: Going straight to the two Newton equations with $f=\mu_{s}N$ or with no friction at all, without having done the test for staying at rest. With $m_{2}=3.0$ kg instead of $4.0$ kg, the same method would have produced an acceleration for a system that is perfectly at rest.

Learn by heart

• Free-body diagram: ONE body, isolated, and only the forces acting ON it.
• $N=mg$ only on a horizontal plane, with no extra vertical force and no vertical acceleration.
• Static friction: $f_{s}\le\mu_{s}N$ , equality only at the threshold. Kinetic: $f_{k}=\mu_{k}N$ , always.
• Inclined plane: $mg\sin\theta$ along the slope, $mg\cos\theta$ perpendicular, breakaway at $\tan\theta=\mu_{s}$ .
• Circular motion: $\Sigma F_{\text{toward the centre}}=\dfrac{mv^{2}}{r}$ . This term is $ma$ , never a force in the diagram.
• Vertical circle: at the top $T+mg=\dfrac{mv^{2}}{r}$ and $v_{min}=\sqrt{gr}$ ; at the bottom $T-mg=\dfrac{mv^{2}}{r}$ .
• Ideal banked curve: $\tan\theta=\dfrac{v^{2}}{rg}$ . If $\mu_{s}$ is given, this formula is not the one asked for.
• Terminal speed: $a=0$ , not $v=0$ . So $mg=bv_{T}$ or $mg=bv_{T}^{2}$ depending on the model.

Frequently asked questions

Does centrifugal force exist?

Not in the CEGEP mechanics course, which works in an inertial frame. An object in a curve is not in equilibrium: it accelerates toward the centre, and the only forces to put on the diagram are real ones, such as friction, the normal force or tension. The sensation of being pushed outward comes from the passenger's inertia, not from a force acting on them.

When is the normal force equal to the weight?

Only on a horizontal surface, when no other force has a vertical component and the object does not accelerate vertically. On a slope it equals the weight times the cosine of the angle. If you pull or push at an angle, or if the object is in an accelerating elevator, it changes again. The normal force is always calculated, never assumed.

How do I know whether an object will slide or stay at rest?

You compare two numbers. First the net driving force without friction, that is, the sum of the forces along the possible slipping. Then the ceiling of static friction, equal to the coefficient of static friction times the normal force. If the driving force exceeds the ceiling, the object slides and you start again with the kinetic coefficient. Otherwise it stays at rest and the friction equals exactly the driving force.

What is the minimum speed at the top of a vertical loop?

It is the square root of the product of the gravitational acceleration and the radius. You get it by writing that the contact force becomes zero: gravity then supplies the centripetal acceleration all by itself. Careful, the condition is not a zero speed but a zero contact force, which is very different.

Why is the tension not equal to the weight of the hanging mass?

Because that equality holds only at equilibrium. As soon as the mass goes down while accelerating, the rope pulls less hard than the weight, otherwise the mass would not go down. The tension is then the mass times the difference between the gravitational acceleration and the acceleration of the system, and it always stays between zero and the weight.

How do I handle two blocks joined by a rope and a pulley?

You write one free-body diagram per block, never a single one for both. You orient the positive axis of each block in the assumed direction of motion, you project, then you add the rope constraint: same magnitude of acceleration and same tension if the pulley is ideal. By adding the two equations, the tension disappears and the acceleration comes out on its own.

Practise it

Corrected exercises: Dynamics, Newton's laws and friction, 203-SN1-RE

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

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys1-dynamics-newtons-laws. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: dynamics, Newton's laws and friction (203-SN1-RE)

The essentials

The method, always the same, in this order

Friction changes its rule depending on whether you slip

Two bodies joined by a rope: Atwood and the table

Apparent weight, in an elevator

Circular motion, where the chapter tips over

The hump and the conical pendulum: everything reads toward the centre

The mistakes that cost marks

1. Adding a centrifugal force to the free-body diagram of an object in a curve

2. Setting the normal force equal to the weight when a force is applied at an angle

3. Writing friction at its maximum on a body that does not move

4. Adding the two forces of an action-reaction pair on the same body

5. Taking the tension equal to the weight of the hanging mass in a system that accelerates

6. Looking for the minimum speed at the top of a vertical circle by setting the speed to zero

7. Mixing up the frictionless banked-curve formula and the case with friction

8. Looking for the terminal speed by setting the speed to zero instead of the acceleration

9. Using the total mass of the system in the equation of a single body

Which method to choose

Which friction to write

Which body to isolate, and how many equations to write

How the answer is expected to be written

Solving a system of two bodies joined by a rope

Deciding whether a body starts moving

Check before you hand in

The typical problem, taken apart

An inclined plane joined by a pulley to a hanging mass

Learn by heart

Frequently asked questions

Does centrifugal force exist?

When is the normal force equal to the weight?

How do I know whether an object will slide or stay at rest?

What is the minimum speed at the top of a vertical loop?

Why is the tension not equal to the weight of the hanging mass?

How do I handle two blocks joined by a rope and a pulley?

Corrected exercises: Dynamics, Newton's laws and friction, 203-SN1-RE

See also

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