PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: Newton's laws and free-body diagrams (PHYS 101)

Q: What forces go on a free-body diagram in PHYS 101?

Only forces with an identifiable source. One arrow for every object that touches the body, a normal force from a surface, a tension from a cord, a push from a hand, plus one arrow for gravity, which acts at a distance. Nothing else. There is no force of motion pointing along the velocity, and the product of mass and acceleration is not a force either: it is the result of the arrows you have drawn, so it belongs on the other side of the equation.

Q: Why is the normal force not always equal to the weight?

Because the normal force is not a property of the object, it is whatever the equation perpendicular to the surface requires. On a level floor with nothing else pulling vertically, that equation happens to give the weight. Tilt the surface and it gives the weight times the cosine of the angle. Pull on a rope at an angle and part of the load is carried by the rope. Accelerate the floor upward, as in a lift, and the normal force grows. Same object, same weight, different normal force every time.

Q: Why do action and reaction forces not cancel each other?

Because they act on two different bodies, and forces are only added inside the equation of one body. When two skaters push each other apart, the force on the first appears in the first skater's equation and the force on the second appears in the second skater's equation. They are never on the same line, so they never cancel. Equal forces on unequal masses even give unequal accelerations, which is why the lighter skater flies off faster.

Q: How do you choose axes for a block on a frictionless incline?

Put one axis along the slope and the other perpendicular to it. The block cannot leave the surface, so its acceleration is entirely along the slope and has one component instead of two. The normal force is already on the perpendicular axis, and only the weight has to be split, into a part along the slope carrying the sine of the angle and a part perpendicular carrying the cosine. Horizontal and vertical axes work too, but they double the work.

Q: Is the tension the same on both sides of an ideal pulley?

Yes. An ideal pulley is massless and its axle is frictionless, so it needs nothing to spin and it only changes the direction of the pull. That is what lets a single symbol for the tension appear in the equation of the block on the table and in the equation of the hanging block, which is exactly what makes the system solvable. A real pulley with mass would give different tensions on its two sides, and that belongs to the rotation chapter.

This sheet is not a summary of the course, you already have that. It answers one question: what makes a student lose marks on Newton's laws in PHYS 101, and what exact gesture prevents each loss.

Everything here works with algebra and right-angle trigonometry only. No derivative and no integral appear, because PHYS 101 is the mechanics course taken without calculus at McGill University, and a solution that begins by differentiating is not a PHYS 101 solution even when its final number is right.

The thread of the chapter

The free-body diagram is drawn BEFORE the first equation, never after it. Drawn after, it is built to match the equation already written, and a missing force stays missing. Two rules keep it honest: one arrow per contact or per field and nothing else, and one body isolated at a time.

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

Before this chapter, you need

• Units and vectors
• Motion in a straight line

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

Newton's first law and inertia
Newton's second law
Newton's third law
Action-reaction pair
Free-body diagram
Mass and weight
Normal force
Tension in an ideal cord
Ideal pulley
Frictionless inclined plane
Apparent weight
Static equilibrium

The essentials

The three laws, in the form you actually write

• First law: with $\Sigma \vec{F} = \vec{0}$ the velocity does not change. At rest and at a constant $12$ m/s are the SAME mechanical state, and no equation ever contains the speed.
• Second law: $\Sigma \vec{F} = m\vec{a}$ , used as two scalar equations, $\Sigma F_{x} = ma_{x}$ and $\Sigma F_{y} = ma_{y}$ . A force perpendicular to an axis contributes zero to that axis.
• Third law: if A pushes B with $\vec{F}$ , then B pushes A with $-\vec{F}$ . Same magnitude, opposite direction, TWO DIFFERENT BODIES.
• Weight: $W = mg$ with $g = 9.80$ m/s $^2$ , always, whatever the motion. A $12$ kg body weighs $117.6$ N in a lift, on a slope and in free fall.

Everything else in the chapter is these three lines plus a choice of axes.

Building the diagram: four gestures, in this order

• 1. Isolate ONE body and redraw it alone, as a box or a dot. The floor, the cord and the hand leave the drawing and come back as arrows.
• 2. One arrow per CONTACT: a surface gives a normal force, a cord gives a tension, a hand gives a push. One arrow per FIELD: gravity gives the weight.
• 3. Name the source of every arrow. An arrow whose source you cannot name does not exist, and that single test kills the invented force of motion.
• 4. Draw the acceleration BESIDE the diagram, clearly marked as an acceleration, never among the forces. $m\vec{a}$ is the result, not a force.

On the left the situation, on the right the same crate ISOLATED: the floor has become

\vec{N}

, the hand has become

\vec{F}

, the Earth has become

\vec{W}

, and nothing else appears.

The diagram comes first. Written after the equation, it is drawn to look like the equation, and it can no longer catch anything.

Where each force gets its value

• Weight: read it, $W = mg$ . It is the only force of the chapter with a formula of its own.
• Normal force: it has NO formula. It comes out of the equation perpendicular to the surface, and it changes as soon as anything else has a component along that perpendicular.
• Tension: it also has no formula. In a light cord over an ideal pulley it has the same value everywhere, so one symbol $T$ appears in the equations of both bodies.
• Applied force: read it from the statement, then split it on your axes if it is oblique.

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.

The normal force on the SAME 12 kg body, case by case

The weight is $117.6$ N in all six lines. The normal force is not.

Situation	Equation perpendicular to the surface	$N$
Level ground, nothing else vertical	$N - mg = 0$	$117.6$ N
Example: $N = 12 \times 9.80 = 117.6$ N
Cord pulling at $30^{\circ}$ above, $T = 85$ N	$N + T\sin 30^{\circ} - mg = 0$	$75.1$ N
Example: $N = 117.6 - 42.5 = 75.1$ N
Cord pushing at $30^{\circ}$ below, $T = 85$ N	$N - T\sin 30^{\circ} - mg = 0$	$160.1$ N
Example: $N = 117.6 + 42.5 = 160.1$ N
Frictionless incline at $28^{\circ}$	$N - mg\cos 28^{\circ} = 0$	$103.8$ N
Example: $N = 117.6 \times 0.8829 = 103.8$ N
Lift accelerating upward at $1.30$ m/s $^2$	$N - mg = ma$	$133.2$ N
Example: $N = 12 \times (9.80 + 1.30) = 133.2$ N
Lift in free fall	$N - mg = -mg$	$0$ N
Example: $N = 12 \times (9.80 - 9.80) = 0$ N

Six lines, one body, one weight, six different normal forces. This is why $N = mg$ is not a law but the answer to one particular case.

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. A body on a tilted surface, or with something else pulling vertically

2 marks, and every later line built on it

What not to write

$N = mg = 117.6$ N, written straight from the mass

What to write

$N = mg\cos 28^{\circ} = 103.8$ N, read off the equation perpendicular to the surface

The same

12

kg body in three situations: the normal arrow is shorter on the slope and longer in the accelerating lift, while the weight is

117.6

N in all three.

Why: The normal force is not a property of the body, it is whatever the perpendicular equation requires. On a slope, part of the weight is aimed ALONG the surface and the surface does nothing about it, so it has less to hold. The error is invisible on a gentle ramp, $738$ N instead of $764$ N at $15^{\circ}$ , which is exactly what makes the habit dangerous.

2. An accelerating body, and a fourth arrow along the motion

the whole question, because every equation inherits the extra term

What not to write

An arrow $\vec{F}_{\text{motion}}$ pointing forward, or $m\vec{a}$ drawn among the forces

What to write

Three arrows only, each with a named source, and the acceleration drawn BESIDE the body

Why: A force needs an object that exerts it. Nothing is in front of the crate, so nothing pushes it forward. Drawing $m\vec{a}$ leads to writing $F - mg - ma = 0$ , which counts the acceleration twice: once on the left as an invented force, once on the right where it belongs.

3. Two bodies pushing on each other

3 marks, and the reasoning is unusable for the rest of the problem

What not to write

The two forces are equal and opposite, so they cancel and the system does not move

What to write

The two forces act on two different bodies, so they never appear in the same equation

Why: Forces are added in the equation of ONE body. The $96$ N on skater B gives $a_{B} = 2.00$ m/s $^2$ and the $96$ N on skater A gives $a_{A} = 1.33$ m/s $^2$ : same force, different masses, different accelerations. Drawing both skaters on one picture is what produces the false cancellation; two separate diagrams make it impossible.

4. Naming the partner of a force

2 marks, every term

What not to write

The partner of the weight of the book is the normal force from the table

What to write

The partner of the weight is the pull of the book ON THE EARTH; the normal force is a different force that happens to balance it

Why: Read the force as a sentence and swap the two nouns: the Earth pulls the book becomes the book pulls the Earth. A partner acts on the OTHER body and never appears in the equation of this one. Test it: put the book in an accelerating lift and the normal force changes while the weight does not, so they cannot be a pair.

5. A hanging mass on a cord while the system accelerates

4 marks, the answer to the whole problem

What not to write

$T = m_{2}g = 29.4$ N, the tension equals the hanging weight

What to write

$T = m_{2}(g - a) = 16.8$ N, strictly less than $m_{2}g$ as long as the block falls

Why: If $T$ equalled $m_{2}g$ the hanging block would have zero net force and would not move at all, while a tension pulled the other block along a frictionless table. The tension equals the hanging weight only in equilibrium. Quick check on every pulley problem: $T$ must sit strictly between the two extreme values the two bodies would give alone.

6. Splitting the weight on axes tilted along an incline

2 marks, and an acceleration larger than $g$ on a steep slope, which is absurd

What not to write

$W_{\parallel} = mg\cos\theta$ and $W_{\perp} = mg\sin\theta$

What to write

$W_{\parallel} = mg\sin\theta$ and $W_{\perp} = mg\cos\theta$

Why: Do not memorise it, test it on a flat surface: at $\theta = 0$ nothing pulls the block along the ground, and $\sin 0 = 0$ , so the along-slope component is the one carrying the sine. Second test, on the answer itself: $a = g\sin\theta$ must always be smaller than $g$ .

7. Adding two cords that pull at an angle to each other

2 marks, and a traction force overestimated by 10 percent

What not to write

$T_{1} + T_{2} = 31.4 + 31.4 = 62.7$ N along the axis

What to write

$2T\cos 25^{\circ} = 56.8$ N along the axis, the perpendicular parts cancelling

Why: Forces add head to tail, and magnitudes add only when the forces are parallel. Each cord spends $T\sin 25^{\circ} = 13.3$ N pulling sideways, cancelled by its symmetric neighbour, and contributes only $T\cos 25^{\circ} = 28.4$ N along the axis. Open the angle to $40^{\circ}$ with the same mass and the pull drops to $48.0$ N.

Which method to choose

Which axes, decided on the SHAPE of the statement

Look at the surface the body is on, not at the topic of the chapter.

The three shapes the first three branches decide between: a flat surface, an incline, and a point held by cords. Nothing about the topic is involved, only the geometry.

If The body slides on a level surface, or moves straight up and down → Axes horizontal and vertical. Only an oblique applied force gets split.
If The body stays on an incline at $\theta$ → Axis $x$ along the slope, axis $y$ perpendicular to it. Then $\vec{N}$ and the acceleration are already on an axis, and only $\vec{W}$ is split, into $mg\sin\theta$ and $mg\cos\theta$ .
If A point is held at rest by two or more cords → Axes horizontal and vertical, or along the axis of symmetry. Two equations, both equal to zero.
If Two bodies are joined by a cord over a pulley → One set of axes PER BODY, each with its positive direction chosen along that body's own motion, so that a single $a$ appears in both equations.

Choosing the axes along the motion is what makes the perpendicular equation read $\Sigma F = 0$ , which is where the normal force comes from.

Which equation to write once the diagram is drawn

Count the bodies and look at the acceleration.

If The body is at rest or moves at constant velocity → $\Sigma F_{x} = 0$ and $\Sigma F_{y} = 0$ . Two equations, so two unknowns can be found.
If One body accelerates in a known direction → $\Sigma F = ma$ along that direction, $\Sigma F = 0$ perpendicular to it.
If Two bodies are joined by an inextensible cord → One equation per body, the SAME $a$ in both because the cord does not stretch, the SAME $T$ in both because the pulley is ideal. Add the equations to eliminate $T$ .
If The question asks for a force of contact between two bodies → Find the acceleration first, on the system as a whole, then write the equation of the SMALLER body alone: that is the equation in which the contact force appears.

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.

The five lines a marker expects on a dynamics question

When to use it: Every problem where a force or an acceleration is asked for.

1 Name the body you are isolating: write it, one line, for instance the block on the incline.
2 Draw the free-body diagram, one arrow per contact and per field, each arrow named.
3 State the axes and the positive direction, for instance $x$ down the slope, $y$ perpendicular to it.
4 Write the two scalar equations before substituting any number.
5 Substitute, compute, and check the sign against the direction you chose.

Concluding sentence

“Isolating the 12 kg block, with $x$ down the slope: $\Sigma F_{x} = mg\sin 28^{\circ} = ma$ , hence $a = g\sin 28^{\circ} = 4.60$ m/s $^2$ down the slope.”

The trap: Substituting the numbers into the first line loses the structure: a marker who cannot see $\Sigma F_{x} = ma$ written symbolically cannot give the method marks, even when the final number is right.

Marking: Typically 1 mark for the diagram, 1 for the axes, 2 for the two equations, 1 for the numerical answer with its unit and direction.

Naming an action and reaction pair in one sentence

When to use it: Whenever the question says identify the reaction force, or asks why the forces do not cancel.

1 Write the force as a sentence with two nouns: the table pushes the book upward with $8.0$ N.
2 Swap the two nouns and keep the magnitude: the book pushes the table downward with $8.0$ N.
3 Say which body each force acts on, and conclude that they are never in the same equation.

Concluding sentence

“The partner of the normal force of the table on the book is the force of the book on the table, $8.0$ N downward; it acts on the table, so it never appears in the equation of the book.”

The trap: Answering the weight of the book: that force acts on the SAME body, so it is not a partner. Its own partner is the pull of the book on the Earth.

Marking: 2 marks, one for the magnitude and direction, one for naming the body acted on.

Check before you hand in

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

Every arrow has a source

Point at each arrow and name the object that exerts it. If you cannot, erase it.

On a crate pushed on a frictionless floor there are exactly 3 arrows, not 4.
Only one body on the drawing

If two bodies are on the same sheet, you will add forces that belong to different equations.

Two skaters pushing each other need 2 separate diagrams, each with its own $96$ N.
The normal force came from an equation

Look back one line: $N$ must appear as the result of the perpendicular equation, never copied from $mg$ .

At $28^{\circ}$ you should see $103.8$ N, not $117.6$ N.
The acceleration is smaller than $g$ on a slope

A frictionless incline gives $a = g\sin\theta$ , so it is always below $9.80$ m/s $^2$ .

$4.60$ m/s $^2$ at $28^{\circ}$ is plausible; $11$ m/s $^2$ means the sine and cosine were swapped.
The tension sits between the two extremes

With a hanging block, $T$ must be strictly less than its weight while it falls, and strictly more than the resisting weight on the other side.

$24.5 < 32.7 < 39.2$ in the worked example; a $T$ outside that bracket is an error.

The typical problem, taken apart

A block on a smooth incline pulled by a hanging mass

A $5.0$ kg block sits on a frictionless incline at $30^{\circ}$ . A light cord runs from it, parallel to the slope, over an ideal pulley at the top, and down to a $4.0$ kg block hanging in the air.

Find the acceleration of the system, the tension in the cord, and the normal force on the block on the incline. Take $g = 9.80$ m/s $^2$ .

Two bodies, one cord: the same tension

T

at both ends and the same

a

in magnitude, because the pulley is ideal and the cord does not stretch.

Step 1

Isolate the $5.0$ kg block. Axes: $x$ UP the slope, $y$ perpendicular. Forces: $\vec{W}_{1}$ , $\vec{N}$ , $\vec{T}$ .

Why

Choosing $x$ up the slope, in the direction the block will actually move, means every positive answer later confirms the guess instead of forcing a re-reading.

Step 2

Split the weight: $W_{1\parallel} = m_{1}g\sin 30^{\circ} = 24.5$ N down the slope, $W_{1\perp} = m_{1}g\cos 30^{\circ} = 42.4$ N into the slope.

Why

The weight is the only oblique force once the axes are tilted. That is the whole reason for tilting them.

Step 3

Two equations for block 1: $T - m_{1}g\sin 30^{\circ} = m_{1}a$ and $N - m_{1}g\cos 30^{\circ} = 0$ .

Why

The perpendicular equation is where the normal force comes from, and it is worth writing even when the question does not ask for $N$ : it is one mark and it is free.

Step 4

Isolate the $4.0$ kg block, positive downward: $m_{2}g - T = m_{2}a$ , that is $39.2 - T = 4.0a$ .

Why

Same $T$ because the pulley is ideal, same $a$ because the cord is inextensible. Choosing down as positive HERE and up the slope as positive THERE is consistent, since those two directions correspond through the cord.

Step 5

Add the two motion equations: $m_{2}g - m_{1}g\sin 30^{\circ} = (m_{1} + m_{2})a$ , so $a = \dfrac{39.2 - 24.5}{9.0} = 1.63$ m/s $^2$ .

Why

Adding eliminates $T$ without any substitution. This is why both equations are written symbolically before a single number goes in.

Step 6

Back-substitute: $T = m_{2}(g - a) = 4.0 \times 8.17 = 32.7$ N, and $N = m_{1}g\cos 30^{\circ} = 42.4$ N.

Why

Use one equation to solve and the OTHER to check: $T - 24.5 = 8.17$ N and $m_{1}a = 5.0 \times 1.63 = 8.17$ N, they agree, so no sign was dropped.

The conclusion, written out

“The system accelerates at $1.63$ m/s $^2$ , the hanging block going down and the block on the incline going up the slope; the tension is $32.7$ N and the normal force on the incline is $42.4$ N.”

The classic mistake on this problem: The classic error here is to write $T = m_{2}g = 39.2$ N. The bracket check refutes it at once: the tension has to be smaller than $39.2$ N, otherwise the hanging block could not accelerate downward, and larger than $24.5$ N, otherwise the block on the incline could not be pulled up. The correct $32.7$ N sits inside.

Learn by heart

• $g = 9.80$ m/s $^2$ , and $W = mg$ whatever the motion. A $12$ kg body weighs $117.6$ N even in free fall.
• The normal force has no formula: it is the result of the equation perpendicular to the surface.
• On an incline at $\theta$ : $W_{\parallel} = mg\sin\theta$ , $W_{\perp} = mg\cos\theta$ , and $a = g\sin\theta$ when there is no friction.
• In a lift, the scale reads $N = m(g + a)$ with $a$ counted positive upward. Free fall gives $N = 0$ .
• An action and reaction pair: same magnitude, opposite direction, two different bodies, never in the same equation.
• Ideal pulley: the tension is the same on both sides. Inextensible cord: the same $a$ in magnitude for both bodies.
• While a hanging block falls, its cord tension is strictly less than its weight.

Frequently asked questions

What forces go on a free-body diagram in PHYS 101?

Only forces with an identifiable source. One arrow for every object that touches the body, a normal force from a surface, a tension from a cord, a push from a hand, plus one arrow for gravity, which acts at a distance. Nothing else. There is no force of motion pointing along the velocity, and the product of mass and acceleration is not a force either: it is the result of the arrows you have drawn, so it belongs on the other side of the equation.

Why is the normal force not always equal to the weight?

Because the normal force is not a property of the object, it is whatever the equation perpendicular to the surface requires. On a level floor with nothing else pulling vertically, that equation happens to give the weight. Tilt the surface and it gives the weight times the cosine of the angle. Pull on a rope at an angle and part of the load is carried by the rope. Accelerate the floor upward, as in a lift, and the normal force grows. Same object, same weight, different normal force every time.

Why do action and reaction forces not cancel each other?

Because they act on two different bodies, and forces are only added inside the equation of one body. When two skaters push each other apart, the force on the first appears in the first skater's equation and the force on the second appears in the second skater's equation. They are never on the same line, so they never cancel. Equal forces on unequal masses even give unequal accelerations, which is why the lighter skater flies off faster.

How do you choose axes for a block on a frictionless incline?

Put one axis along the slope and the other perpendicular to it. The block cannot leave the surface, so its acceleration is entirely along the slope and has one component instead of two. The normal force is already on the perpendicular axis, and only the weight has to be split, into a part along the slope carrying the sine of the angle and a part perpendicular carrying the cosine. Horizontal and vertical axes work too, but they double the work.

Is the tension the same on both sides of an ideal pulley?

Yes. An ideal pulley is massless and its axle is frictionless, so it needs nothing to spin and it only changes the direction of the pull. That is what lets a single symbol for the tension appear in the equation of the block on the table and in the equation of the hanging block, which is exactly what makes the system solvable. A real pulley with mass would give different tensions on its two sides, and that belongs to the rotation chapter.

Practise it

Corrected exercises: Newton's laws and free-body diagrams, PHYS 101

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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Revision sheet: Newton's laws and free-body diagrams (PHYS 101)

The essentials

The three laws, in the form you actually write

Building the diagram: four gestures, in this order

Where each force gets its value

The rules in table form

The normal force on the SAME 12 kg body, case by case

The mistakes that cost marks

1. A body on a tilted surface, or with something else pulling vertically

2. An accelerating body, and a fourth arrow along the motion

3. Two bodies pushing on each other

4. Naming the partner of a force

5. A hanging mass on a cord while the system accelerates

6. Splitting the weight on axes tilted along an incline

7. Adding two cords that pull at an angle to each other

Which method to choose

Which axes, decided on the SHAPE of the statement

Which equation to write once the diagram is drawn

How the answer is expected to be written

The five lines a marker expects on a dynamics question

Naming an action and reaction pair in one sentence

Check before you hand in

The typical problem, taken apart

A block on a smooth incline pulled by a hanging mass

Learn by heart

Frequently asked questions

What forces go on a free-body diagram in PHYS 101?

Why is the normal force not always equal to the weight?

Why do action and reaction forces not cancel each other?

How do you choose axes for a block on a frictionless incline?

Is the tension the same on both sides of an ideal pulley?

Corrected exercises: Newton's laws and free-body diagrams, PHYS 101

See also

Looking for a physics tutor in Montreal?