Chap. 3 – Vectors and 2-d motion (show slides as you move on)
Vectors are represented by arrows. The length of the arrow corresponds to the magnitude of the vector,
on a certain scale. The direction is represented with an angle in respect to the
positive x-axis. The convention for angles is that:
o angles measured CCW from the + x-axis are positive
o angles measured CW from the + x-axis are negative
In our tests, students will be asked to use the two methods shown below for vector
operations, as they complement each the other. The graphical method gives approximate
answers, but it allows you to check for “ballpark numbers”. Whereas the components’
method will give precise answers (but the arctan function can give ambiguous values
that only the drawing or graphical method can help to sort out).
Graphical method for vector operations (+, –, or multiply by a constant)
Instructor: just make drawings and give the explanations orally as you go.
1.
2.
3.
4.
5.
6.
7.
Draw a Cartesian plane (x, y axes, with origin at 0,0).
Draw the first vector in the operation with its tail starting at the origin of the Cartesian plane.
Obey the vector’s scaled magnitude (use ruler) and direction (use protractor).
Connect the tail of the second vector to the arrowhead of the first vector, maintaining its
magnitude and direction as well: this is called the “head-to-tail” technique. Note that the direction
(angle) of this second vector must be measured in respect to a new Cartesian plane drawn at the
arrowhead of the first vector.
And so on, if there are more vectors to add, keep connecting successive heads to tails.
The result of the operation is called the resultant vector 𝑅⃗ .
To find 𝑅⃗ , you must connect the origin of the very first vector (0,0), to the arrowhead of the last
vector drawn. 𝑅⃗ points toward the last vector drawn.
The magnitude of R is calculated on the same scale as the other vectors used. The direction of 𝑅⃗ is
measured in respect to the + x-axis on the original Cartesian plane drawn at the beginning.
Important:
o
o
o
o
negative vectors are vectors flipped by 180 (same magnitude, but opposite direction).
Vectors should only be added or subtracted if they have the same units.
Equal vectors must have the same magnitude AND direction (together, both conditions).
To multiply a vector by a constant number, simply multiply its length by the constant.
The magnitude will change, but not the direction.
Examples of drawings in the slides (also more examples must be drawn on the blackboard).
End of the chapter problems: 2, 7
(move faster in this part, focus more on the components’ method below)
Method of the components for vector operations (+, –, or multiply by a constant)
From the figures:
𝐴 = 𝐴𝑥 + 𝐴𝑦
(vector form)
2
2
𝐴 = √𝐴𝑥 + 𝐴𝑦 (magnitude only)
𝐴𝑥 = 𝐴 𝑐𝑜𝑠 𝜃
𝐴𝑦 = 𝐴 𝑠𝑖𝑛 𝜃
Only if  is measured
from + x-axis.
𝐴𝑥 𝑎𝑛𝑑 𝐴𝑦 are 1-d vectors that can be + or –
to indicate their direction. The cos or sin functions
will tell whether + or –, but only if measured from
+ x-axis. Remember that 1-d vectors use +/– to indicate
direction, whereas 2-d vectors will need angles for the
final description of the direction.
Handout with exercises about vector components.
⃗ , for example when 𝑅⃗ = 𝐴 + 𝐵
⃗:
Often we want to find the magnitude and direction of ⃗𝑹
Then:
⃗𝑥
𝑅⃗𝑥 = 𝐴𝑥 + 𝐵
⃗𝑦
𝑅⃗𝑦 = 𝐴𝑦 + 𝐵
𝑅 = √𝑅𝑥2 + 𝑅𝑦2
⃗𝑦
𝑅
𝜃 = arctan 𝑅⃗
𝑥
(magnitude)
(direction) Note: 𝑅⃗𝑥,𝑦 are 1-d vectors!
Very important:
The arctan function only returns angles within +90 >  > –90, i.e., only 1st and 4th quadrants. What to do
when the resultant 𝑅⃗ is in the 2nd or 3rd quadrants? Add 180 to the value given by the calculator.
Students must verify where 𝑅⃗ is going to be (what quadrant, and therefore what angle to expect) by
making a drawing (graphical method), or simply plotting the components 𝑅⃗𝑥 and 𝑅⃗𝑦 . Angles can be + or –.
In practice, when solving problems with the components’ method, instruct the students to always build
this helpful table:
X-component
Y-component
𝐴
𝐴𝑥 =
𝐴𝑦 =
⃗
𝐵
⃗𝑥 =
𝐵
⃗𝑦 =
𝐵
𝐶
𝐶𝑥 =
𝐶𝑦 =
…
⃗𝑅𝑥 = 𝐴𝑥 + 𝐵
⃗ 𝑥 + 𝐶𝑥
𝑅⃗𝑥 =
…
⃗𝑅𝑦 = 𝐴𝑦 + 𝐵
⃗ 𝑦 + 𝐶𝑦
𝑅⃗𝑦 =
etc
𝑅⃗
End of the chapter problems: 13, 19, 20, 17
Students must practice more problems at home asap, as they find 2-d vectors challenging, in general.
3.4 Motion in 2-d, projectile motion
The most important fact about 2-d motion is to
remember that x and y motions are independent from each other.
So, students should be alerted to the fact that they are
essentially solving two problems (vertical and horizontal)
at the same time.
We have a simple demo in class that allows a simultaneous drop
of a metal sphere vertically, while another sphere falls with an
initial horizontal velocity. They both land on table at the same time, but ask the students what they think
before the demo.
Allow students to have a new index card with the following equations (remind students that 1-d
components can be + or –, pay attention to the signs!):
Where
Horizontal
𝑎𝑥 = 0
Vertical
𝑎𝑦 = −9.8 m/s2
𝑣𝑥 = 𝑣0𝑥
𝑣𝑦 = 𝑣0𝑦 − 9.8 ∆𝑡
∆𝑥 = 𝑣0𝑥 ∆𝑡
∆𝑦 = 𝑣0𝑦 ∆𝑡 − 4.9 ∆𝑡 2
𝑣0𝑥 = 𝑣0 cos 𝜃
2
𝑣𝑦2 = 𝑣0𝑦
− 19.6 ∆𝑦
𝑣0𝑦 = 𝑣0 sin 𝜃
𝑣 = √𝑣𝑥2 + 𝑣𝑦2
⃗𝑦
𝑣
Emphasize that
𝑣0 ≠ 𝑣0𝑥 ≠ 𝑣0𝑦
𝜃 = arctan 𝑣⃗
𝑥
End of the chapter problems: 22, 23, 28, 29 (ST), 32 (ST), 58, 65 (ST)
In the slides:
o quick mathematical demonstration why the trajectory is a parabola
o demonstration of the Range Equation (two possible angles with the same range: solve problem 69).
Explain to students that they should not copy the Range Equation in their notecard, it will not be
necessary, but it is important to see how and in what case it can be derived. Emphasis on the fact that
the Range Equation can only be used when y = 0 for the complete trajectory.
In general, we skip Section 3.5 Relative Velocity in PHY101 because of time constraints.
Test 2.
Chap. 4 – Laws of Newton (show slides as you move on)
We will study now WHY things move.
Note that in the slides from now on, there are no end-of-chapter problems to display. Students should
read the problems in their own textbooks.
Characteristics of the force vectors we will be using this semester:
⃗ 𝑵 𝒐𝒓 𝒏
⃗ ): needs a surface. Points at a direction perpendicular to the surface, and
o Normal Force (𝑭
outward. Draw example of book on the table, and hand pressing against a vertical wall. There is no
formula for 𝐹𝑁 , its value will be found through comparison and balance with other forces.
⃗ ): needs a string/cord/rope. Tension pulls, it does not push. Along a string, we will draw
o Tension (𝑻
the tension force on both extremities of the string (vectors of opposite direction), pulling in. Along the
string, therefore, the sum of all forces will be zero (TT=0): we are thus assuming our
string/cord/rope will not expand during the experiment. If there are multiple strings in a problem, do
⃗ , its value will be found through comparison
not assume the same value of T. There is no formula for 𝑇
and balance with other forces.
⃗ 𝒈 ):
o Force of gravity (𝑭
𝐹𝑔 = 𝐺
𝑀𝑚
= 𝑚𝑔
𝑑2
where G is the Universal Gravitation Constant 6.67  10-11 Nm2/kg2.
𝐹𝑔 is always present, even in space at the ISS for example g = 8.7 m/s2.
𝐹𝑔 always points vertically down, towards the center of the planet.
𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑦𝑜𝑢𝑟 𝑠𝑒𝑛𝑠𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝐹𝑁
"real" 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑦𝑜𝑢 = 𝐹𝑔
⃗ , 𝒄𝒂𝒏 𝒃𝒆 ⃗𝒇𝒔 𝒐𝒓 ⃗𝒇𝒌 ): needs two solid surfaces in contact. It always points against
o Force of Friction (𝒇
the motion or, in cases where there is no motion yet, it points against the tendency of motion. To
figure out what is the tendency of motion, ask yourself: if there were no friction, where would this
body go? That’s the direction of the tendency of motion. Friction will be against that direction.
𝑓𝑠,𝑘 = 𝜇𝑠,𝑘 ∙ 𝐹𝑁
where  is a coefficient. This is what you need to know now, more about it later on.
⃗ ): we will use this general symbol for any applied contact force, such as a push, etc.
o Applied force (𝑭
⃗ 𝑫 ): essentially, we will not use this in PHY101, as it is not
o Drag force or force of air resistance (𝑭
very important in most of our problems because most objects being analyzed are small, dense, and
moving with small speeds.
o Net Force (𝐹𝑛𝑒𝑡 ): this is not a real force, you will NOT draw this force. It represents the mathematical
vector combination of all forces acting on the body: it is a resultant force. Always calculate one for
each direction x and y. Then later, if necessary, you may combine the directions.
𝐹𝑛𝑒𝑡𝑋 = ∑ 𝐹𝑋
𝐹𝑛𝑒𝑡𝑌 = ∑ 𝐹𝑌
There are many other types of forces, but we will concentrate on the forces that affect macroscopic
motion in mechanics.
Newton’s 1st Law of motion:
𝐼𝑓 𝐹𝑛𝑒𝑡𝑋 = 0 𝑡ℎ𝑒𝑛 𝑣𝑥 = 0 𝑜𝑟 𝑣𝑋 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 , 𝑖. 𝑒., 𝑎𝑋 = 0, 𝑎𝑛𝑑 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎.
𝐼𝑓 𝐹𝑛𝑒𝑡𝑌 = 0 𝑡ℎ𝑒𝑛 𝑣𝑌 = 0 𝑜𝑟 𝑣𝑌 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 , 𝑖. 𝑒., 𝑎𝑌 = 0, 𝑎𝑛𝑑 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎.
Important:
the vector velocity 𝑣 is only constant when moving along a straight line. Later on we will see what
happens when moving in a circular path (then the 2nd Law is necessary).
Newton’s 2nd Law of motion:
𝐼𝑓 𝐹𝑛𝑒𝑡𝑋 ≠ 0 𝑡ℎ𝑒𝑛 𝑎𝑥 ≠ 0 , 𝑎𝑛𝑑 𝐹𝑛𝑒𝑡𝑋 = 𝑚 𝑎𝑥 .
𝐼𝑓 𝐹𝑛𝑒𝑡𝑌 ≠ 0 𝑡ℎ𝑒𝑛 𝑎𝑌 ≠ 0 , 𝑎𝑛𝑑 𝐹𝑛𝑒𝑡𝑌 = 𝑚 𝑎𝑌 .
UNITS:
1 𝑛𝑒𝑤𝑡𝑜𝑛 = 1 𝑁 = 1 𝑘𝑔
𝑚
𝑠2
1 N = 0.225 lb
Notice that pound (lb) is a unit of force (like newtons), and not mass (kg, slug).
Show the Force Diagram slides. These are exercises in drawing force diagrams and writing the equations
for Fnet for each case.
End of chapter problems: 3, 6, 10, 5, 7, 12, 15
Newton’s 3rd Law of motion:
Forces occur in pairs. Let’s call the pair action-reaction, or 𝐹1→2 the force that body 1 exerts on body 2,
and 𝐹2→1 the force that body 2 exerts on body 1. When bodies exchange forces, they will experience the
same magnitude of F, but with opposite direction, acting on the different bodies.
𝐹1→2 = − 𝐹2→1 (vector form)
|𝐹 |1→2 = |𝐹 |2→1 (in magnitude)
Steps to solve problems:
1. read carefully
2. draw a free-body diagram (explain to students that this is not optional, they must draw it) for each
body involved in the problem
3. choose the appropriate law to apply in x, y directions
4. If two or more bodies analyzed in the problem are in contact with each other, then the 3rd Law needs
to be used.
5. write an equation for 𝐹𝑛𝑒𝑡𝑋 and another equation for 𝐹𝑛𝑒𝑡𝑌 .
End of chapter problems: 17, 20, 19 (ST), 21, 25, 26 (ST), 28, 29, 36
It is important to have a lot of homework problems at this stage.
4.6 Force of Friction between two solid surfaces
Static friction force (when 𝐹 ≠ 0 𝑏𝑢𝑡 𝑣 = 0)
𝑓𝑠 = 𝜇𝑠 ∙ 𝐹𝑁
(magnitude)
The direction of 𝑓𝑠 always points against the tendency of motion. Imagine there was no friction, then
where would be object move due to the applied force 𝐹 ? Well, that’s the direction of the tendency of
motion, even though v = 0.
𝑓𝑠 is a self-adjusting force, that will increase as your applied force 𝐹 increases – but up to a certain point!
There is a maximum value for 𝑓𝑠 , and beyond this value, motion will start. The values for the coefficient 𝜇𝑠
found in tables are for this maximum force case: they are “peak” values.
Slides: show graph of 𝑓𝑠 transitioning to 𝑓𝑘 .
Kinetic friction force (when 𝑣 ≠ 0)
𝑓𝑘 = 𝜇𝑘 ∙ 𝐹𝑁
(magnitude)
This time, motion will be obvious. 𝑓𝑘 points in the opposite direction of motion.
The coefficients  are dimensionless. Their values are in general less than 1, and never negative.
Friction is a complex combination between roughness of surface and chemical or electromagnetic
interactions in the molecular and atomic levels.
In general, 𝜇𝑠 > 𝜇𝑘 because when the surfaces are kept in motion, the distance between them will reach
some average that is greater than the average distance when they are at rest. Greater separation means
that only the peaks of the surfaces are coming into contact. There will be a reduced attraction between
the molecules of the two objects. This results in a lower component of force antiparallel to the direction
of motion, i.e., less friction.
Reference https://www.physicsforums.com/threads/why-is-the-kinetic-friction-always-smaller-than-the-static-friction.140426/
End of chapter problems: 40, 45 (ST), 44, 62, 54, 57, 66
Review problems, to be solved by students alone, preferably in class: 59, 60, 79
Test 3.