PHYS 207.013

Chapter 5

Newton laws

Instructor: Dr. Bianco

TAs: Joey Betz; Lily Padlow

University of Delaware - Spring 2021

https://slides.com/federicabianco/phys20713_5

forces

So far we have describe how things happen, with forces we can explain why things happen

H&R CH5 Newton Laws

WHAT WE OBSERVE?

We experience that it is harder to lift or move heavier obects

H&R CH5 Newton Laws

principle

H&R CH4 motion in 2D and 3D

the force required to move an object is proportional to its mass

principle

H&R CH4 motion in 2D and 3D

the force required to move an object is proportional to its mass

Forces are related to masses

\vec{F} \propto \vec{m}

WHAT WE OBSERVE?

Things that are still stay still until we act on them.

H&R CH5 Newton Laws

WHAT WE OBSERVE?

Things that are still stay still until we act on them.

H&R CH5 Newton Laws

WHAT WE OBSERVE?

but also things that move continue moving unless we act on them

it is just rare to be in the position to observe a body on which nothing acts

H&R CH5 Newton Laws

principle

H&R CH4 motion in 2D and 3D

if no forces act on a body its velocity does not change - either in magnitude or direction

principle

H&R CH4 motion in 2D and 3D

if no forces act on a body its velocity does not change - either in magnitude or direction

Forces are related to acceleration

\vec{F} \propto \vec{a}

historical perspective

H&R CH5 Newton Laws

Newton's idea

The Principia states Newton's laws of motion, forming the foundation of classical mechanics; Newton's law of universal gravitation; and a derivation of Kepler's laws of planetary motion which Kepler first obtained empirically.

H&R CH5 Newton Laws

Newton's idea

and put together these observations about how the world works with calculs

\int v(t) dt = ...

\frac{dv(t)}{dt} = ...

H&R CH5 Newton Laws

Newton law I

A body that is at rest or moving at a uniform rate in a straight line will remain in that state until some force is applied to it.

\vec{F}_\mathrm{net} = 0

if there is not acceleration in the motion

H&R CH5 Newton Laws

Newton law II

\vec{F} = m\vec{a}

H&R CH5 Newton Laws

Newton law II

H&R CH4 motion in 2D and 3D

The vector acceleration of a body is proportional to the vector force applied to the body

\vec{F} = m\vec{a}

Newton law II

H&R CH4 motion in 2D and 3D

The vector acceleration of a body is proportional to the vector force applied to the body

\vec{F} = m\vec{a}

direction + magnitude

Newton law II

H&R CH4 motion in 2D and 3D

The acceleration of a body is proportional to the magnitude of the force applied to the body in the direction of that force

\vec{F} = m\vec{a}

Newton law II

H&R CH4 motion in 2D and 3D

The acceleration of a body is inversely proportional to its mass

\vec{F} = m\vec{a}

a = \frac{F}{m}

definition:

1~\mathrm{Newton} (1N) ~\mathrm{is~the~force~required}\\ \mathrm{to~accelerate}~ 1 kg ~\mathrm{by} 1 \frac{m}{s^2}

1 N = 1 kg\frac{m}{s^2}

H&R CH5 Newton Laws

S.I. unit of force

from Ch1 - 7 fundamental quantities

Length - meter (m)

Time - second (s)

Mass - kilogram (Kg)

Luminous intensity - candela (cd)

Temperature - kelvin (K)

Amount of substance - mole (mole)

Electric current - ampere (A)

}

velocity

acceleration

}

force

H&R CH1 measuring things - including length

\left[\frac{m}{s}\right] ; \left[\frac{km}{h}\right] \\ \left[\frac{m}{s^2}\right] ; \left[ \frac{km}{h^2}\right]

\left[N=\frac{kg~m}{s^2}\right]

density

\left[\rho=\frac{kg}{m^3}\right]

DIMENSION - UNIT

1 N = 1 kg\frac{m}{s^2}

given the same 1N force, what will the acceleration be if the mass is 2kg?

question

H&R CH5 Newton Laws

question

1 N = 1 kg\frac{m}{s^2}

given the same 1N force, what will the acceleration be if the mass is 2kg?

given the same 1N force, what must the mass be if the force accelerates it by 2 ?

\frac{m}{s^2}

H&R CH5 Newton Laws

If the earth exert a force pulling us toward its center, and

why are we not sinkging toward the center of the earth?

\vec{F} \propto \vec{a}

Newton II:

\vec{F} = m\vec{a}

\vec{F} = m\vec{g}

H&R CH5 Newton Laws

If the earth exert a force pulling us toward its center, and

why are we not sinkging toward the center of the earth?

\vec{F} \propto \vec{a}

Newton II:

\vec{F} = m\vec{a}

fact:

\vec{a} = 0

\vec{F} = m\vec{g}

H&R CH5 Newton Laws

If the earth exert a force pulling us toward its center, and

why are we not sinkging toward the center of the earth?

\vec{F} \propto \vec{a}

Newton II:

\vec{F} = m\vec{a}

fact:

\vec{a} = 0

Newton I:

\mathrm{if~}\vec{a} = 0 ~=>~ \vec{F}_\mathrm{net} = 0

\vec{F} = m\vec{g}

H&R CH5 Newton Laws

If the earth exert a force pulling us toward its center, and

why are we not sinkging toward the center of the earth?

\vec{F} \propto \vec{a}

Newton II:

\vec{F} = m\vec{a}

fact:

\vec{a} = 0

Newton I:

\mathrm{if~}\vec{a} = 0 ~=>~ \vec{F}_\mathrm{net} = 0

\vec{F} = m\vec{g}

\vec{F}_\mathrm{net} = \sum_i \vec{F_i}

because forces are vectors:

H&R CH5 Newton Laws

If the earth exert a force pulling us toward its center, and

why are we not sinkging toward the center of the earth?

\vec{F} \propto \vec{a}

Newton II:

\vec{F} = m\vec{a}

fact:

\vec{a} = 0

Newton I:

\mathrm{if~}\vec{a} = 0 ~=>~ \vec{F}_\mathrm{net} = 0

\vec{F}_\mathrm{net} = \sum_i \vec{F_i}

There must be another force that compensated fo the pull of the Earth: the Normal force

\vec{F} = m\vec{g}

because forces are vectors:

H&R CH5 Newton Laws

Newton law III

when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction.

\vec{F}_{1->2} = -\vec{F}_{2->1}

H&R CH5 Newton Laws

simple problem

a hockey puck (m=100 g, no friction) is hit from a static position in the center of the field to a velocity of (v = 100 km/h ~ 25 m/s). With which force was it hit?

H&R CH5 Newton Laws

v_f = 25 \frac{m}{s}

\vec{F} = m\vec{a} = 0.1*25 kg \frac{m}{s^2} = 2.5N

m = 0.1 kg\\

v_f = v_0 + at\\ \mathrm{assume~the~puck~accelerates~to}\\ ~v_f~\mathrm{in~1sec}

v_0 = 0\\

for a proper solution of this problem https://www.youtube.com/watch?v=WyTS7NIw0TU

••6 In a two-dimensional tug-ofwar,

Alex, Betty, and Charles pull

horizontally on an automobile tire at

the angles shown in the overhead

view of Fig. 5-30. The tire remains

stationary in spite of the three pulls.

Alex pulls with force FA of magnitude

220 N, and Charles pulls with

force FC of magnitude 170 N. Note

that the direction of FC is not given.

What is the magnitude of Betty’s

force FB

Normal force, tension

what is the hardest part of a snatch?

Sukanya Srisurat

H&R CH5 Newton Laws

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_T = -m\vec{g}

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

tension: force along a rope

H&R CH5 Newton Laws

this was supposed to be a demo...

vectors and scalars

F1 \cos(\theta) + F2 \cos(\theta) = -mg

F1 = F2

F1 = \frac{mg}{2\cos(\theta)}

}

F = mg

F = -mg

F = mg

F = -\frac{mg}{2}

F = \frac{mg}{2 \cos {\theta}}

F = -\frac{mg}{2 \cos {\theta}}

F = mg

\frac{mg}{2}

H&R CH5 Newton Laws

\theta

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

FBD: Free Body Diagram

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

FBD: Free Body Diagram

\vec{F}_g = -m\vec{g}

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

FBD: Free Body Diagram

\theta

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

FBD: Free Body Diagram

\theta

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

\theta

FBD: Free Body Diagram

F_N = - mg \cos{\theta}

\vec{F}_{N} = -m\vec{g} \cos{\theta}

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

\theta

FBD: Free Body Diagram

F_N = - mg \cos{\theta}

F_{\mathrm{net}~y'} = - mg \cos{\theta} + mg \cos{\theta} = 0

\vec{F}_{N} = -m\vec{g} \cos{\theta}

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

FBD: Free Body Diagram

\theta

\vec{F\mathrm{net}} = -m\vec{g} \sin{\theta}

F_{\mathrm{net}, x'} = - mg \sin{\theta} \hat{i} = 0

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

FBD: Free Body Diagram

\theta

\vec{F\mathrm{net}} = -m\vec{g} \sin{\theta}

F_{\mathrm{net}, x'} = - mg \sin{\theta} \hat{i} = 0

v_{x'}(t) = v_{0,x'} +a t

Normal force, tension

\vec{F}_g = m\vec{g}

\vec{F}_N = -m\vec{g}

normal force: contact force that is perpendicular to the surface

H&R CH5 Newton Laws

\vec{F}_g = m\vec{g}

KEY POINTS:

Forces are vectors and all vector rules apply

Newton I: A body's velocity does not change unless a force is applied: v=const =>

Newton III: For every action, there is an equal and opposite reaction.

\vec{F}_{1->2} = - \vec{F}_{2->1}

\vec{F} = m\vec{a}

\vec{F}_{net} = \sum{\vec{F_i}}

Forces and Newton Laws

Newton II: A force is proportional to the acceleration and the mass of a body

H&R CH5 Newton Laws

\sum\vec{F}_i = 0

KEY POINTS:

Forces are vectors and all vector rules apply

Newton I: A body's velocity does not change unless a force is applied: v=const =>

Newton III: For every action, there is an equal and opposite reaction.

\vec{F}_{1->2} = - \vec{F}_{2->1}

\vec{F} = m\vec{a}

\vec{F}_{net} = \sum{\vec{F_i}}

Forces and Newton Laws

Newton II: A force is proportional to the acceleration and the mass of a body

H&R CH5 Newton Laws

\sum\vec{F}_i = 0

how to solve problems with force

4) identify if there system is initially at rest or in motion and if you need to choose the appropriate Chap2 equation of motion now that you have the acceleration from the forces.

1) make your Free Body Diagram identifying all forces acting on the body

3) identify if the system is in equilibrium or the forces are unbalanced and cause acceleration

\vec{F} = m\vec{a}

Forces and Newton Laws

2) identify all angles and where to use cosine and sine and the component of all forces

H&R CH5 Newton Laws

\sum_i \vec{F}_i = 0

example: tension on an inclined plane

H&R CH5 Newton Laws

UNIFORM CIRCULAR MOTION

\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}

H&R CH4 motion in 2D and 3D

\theta

Uniform cirular motion

1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed

acceleration:

|a| = \frac{|v_c|^2}{r}

UNIFORM CIRCULAR MOTION

\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}

H&R CH4 motion in 2D and 3D

\theta

Uniform cirular motion

1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed

acceleration:

|a| = \frac{|v_c|^2}{r}

2. By Newton's II law: F = ma if there is an acceleration there is a force!

Force:

F = m|a| = m\frac{|v_c|^2}{r}

UNIFORM CIRCULAR MOTION

\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}

H&R CH4 motion in 2D and 3D

\theta

Uniform cirular motion

1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed

acceleration:

|a| = \frac{|v_c|^2}{r}

2. By Newton's II law: F = ma if there is an acceleration there is a force!

Force:

F = m|a| = m\frac{|v_c|^2}{r}

F \propto \frac{1}{r}

UNIFORM CIRCULAR MOTION

\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}

H&R CH4 motion in 2D and 3D

\theta

Uniform cirular motion

1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed

acceleration:

|a| = \frac{|v_c|^2}{r}

2. By Newton's II law: F = ma if there is an acceleration there is a force!

Force:

F = m|a| = m\frac{|v_c|^2}{r}

F \propto \frac{1}{r}

UNIFORM CIRCULAR MOTION

F = m|a| = m\frac{|v_c|^2}{r}

F \propto \frac{1}{r}

exam instructions

EXAM PREP

exam instructions

What to have handy: remember that these exams are open-book, so also have whatever helps you solve problems handy! Whether its a comfort blanket, your favorite pen, or a cheat sheet with more equations

I recommend you have the equations of 1D motion handy https://slides.com/federicabianco/phys20713_2#/6 - all ND problems we are doing can be solved as a sets of 1D problem along each axis
I recommend you have handy the value of sines, cosines, tangents for some common angles https://slides.com/federicabianco/phys20713_3#/3/4 because it is faster to look them up than to plug them into the calculator and the cheat sheet tells you more clearly if your result is in radians or degrees of angle
A few other equations were studied and used in solving problems so far:
- the Range equation https://slides.com/federicabianco/phys20713_4#/9/2,
- the equations that govern circular motion https://slides.com/federicabianco/phys20713_4/edit#/11/6 and
- Newton’s laws of motion https://slides.com/federicabianco/phys20713_5/live?context=editing#/13/1

EXAM PREP

exam instructions

Write down all the information neatly at the beginning of the problem: what is the text of the problem telling you? What else do you know. For example

if it is a projectile motion problem you know other things about other forces/accelerations at play that the problem wont typically tell you: that gravity is acting on your projectile.
If it is circular motion you know that the velocity and acceleration have a certain relation with the radius to keep the object in the circular orbit and that the velocity and acceleration have to point in a specific direction at every instant
If it is a force problem you know that if the system is static all forces are balanced (SumF=0)

EXAM PREP

exam instructions

Draw the system: Take your time with this: it is going to save time when you work on solving the problem!

Identify the most convenient axis system to solve your problem: where does the y axis point (e.g. up or down with gravity?), what vectors are the x and y axis aligned with?
Identify with respect to what reference point the angles are given to you: if it is with respect to the axis or to a preferred direction in the problem (e.g. the forward direction, or the East-West directions). If you get this wrong the whole problem will be different and wrong.
Draw all the vectors that are described in the problem and those that are not but are implicitly known to be there (e.g. acceleration due to gravity). I recommend you use different colors if you can here, so that you do not get confused (you can raw each vector in a different color and its components with the same color along the axes for example)

Remember dimensional analysis: lots of people are losing points in the quiz cause they have the answer in the wrong units, and therefore the wrong answer!

All trigonometric functions on your calculator will give you/expect as input angles in radians unless you specify degrees. Remember 1 rad = 180/3.1415 degrees
If you input in your problem g=9.8 then you are using meters/second! That is the right unit to use - but remember: all other values will have to be input as meters, seconds, etc, and the result will be in meters, seconds, etc, not in feet or miles!!
When you figure out what equation to use, do a quick check if the dimensions are coming up to be what you expect when you solve it for the relevant quantity

EXAM PREP

exam instructions

EXAM PREP

Strategy:

Look at what the problem tells you (your data) and the equations that we studied (your tools) and

figure out which equations can give you the quantity you are asked to find. You need as many equations as unknowns.

It may take more than one step: the equations that contain the quantity you are after may also contain other unknown quantities: find an equation that will give the other unknowns and solve the equations one after the other.

Example: you may need to solve an equation of motion to find the terminal velocity, then plug that number in (or that function, that may have unknown quantities) into another equation to find an unknown (like time, or acceleration)

so long as the first equation is solved in terms of the relevant unknowns you can substitute and obtain an equation that has one fewer unknown

phys207 Newton laws

PHYS 207.013

Chapter 5

forces

WHAT WE OBSERVE?

principle

principle

WHAT WE OBSERVE?

WHAT WE OBSERVE?

WHAT WE OBSERVE?

principle

principle

historical perspective

Newton's idea

Newton's idea

Newton law I

Newton law II

Newton law II

Newton law II

Newton law II

Newton law II

definition:

question

question

Newton law III

simple problem

Normal force, tension

Normal force, tension

Normal force, tension

this was supposed to be a demo...

vectors and scalars

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

Normal force, tension

UNIFORM CIRCULAR MOTION

UNIFORM CIRCULAR MOTION

UNIFORM CIRCULAR MOTION

UNIFORM CIRCULAR MOTION

UNIFORM CIRCULAR MOTION