Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

Lecture Capture:

http://echo360.uml.edu/danylov2013/physics1fall.html

Lecture 7

Chapter 4

Dynamics: Newton’s Laws of Motion

09.25.2013 Physics I

Solving Problems using Newton’s laws

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Chapter 4. Sections 4.7-4.8 Types of forces Free-body diagrams Solving Problem using Newton’s laws

Outline

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Know the forces : find the motion

Know the motion : find the forces

Applying Newton’s Laws

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Solving Problems with Newton’s Laws: Free-Body Diagrams 1. Draw a sketch.

2. For one object, draw a free-body diagram, showing all the forces acting on the object. Make the magnitudes and directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one.

3. Resolve vectors into components.

4. Apply Newton’s second law to each component.

5. Solve.

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Possible forces acting on a “system”

Gravity Fg=mg Normal Force FN Tension Ft

Applied Force F Internal Forces (F12, F21) cancel each other (N. 3rd Law)

The block is moving at constant speed, so

it must have no net force on it. The forces

on it are N (up) and mg (down), so N = mg,

just like the block at rest on a table.

ConcepTest 1 Going Up I A block of mass m rests on the floor of

an elevator that is moving upward at constant speed. What is the relationship between the force due to gravity and the normal force on the block?

1) N > mg

2) N = mg

3) N < mg (but not zero)

4) N = 0

5) depends on the size of the elevator

m

v

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Two blocks problem

x

y F

m1

There are two objects, draw a separate diagram for each one.

∑ = xx amF 1

m2

Given: F, m1, m2

Find: a, F12 , F21 (contact forces between m1 and m2)

1221 FF

−=

F12 F21

1221 FF =

m1 amFF 112 =−

∑ = xx amF 2 amF 221 =

Add them amamFFF 212112 +=+−

Internal forces cancel each other!!!!!!!!!!!!! Cancel due to Newton’s Third Law

m2

21 mmFa+

=

(2) (1)

Put it into (2) amF 221 =21

2

mmFm

+= 21F=

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Two blocks problem

x

y F

Treat, m=m1+m2, as the system (one big block)

maF =

21 mmFa+

=

Given: F, m1, m2

Find: a (common acceleration)

Apply N. 2nd law to m x component of

N. 2nd law

amm )( 21 +=

m=m1+m2

We can now forget about the internal forces

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

ay = 0 (no motion in y direction) yN mamgF =−

x

y

Fg=mg

m

FN

Y equation gives a normal force

mgFN =

Two blocks problem: y-equation Box is a described

object

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Similar Problems Two boxes are held together by a cable and pulled by a string on a frictionless table.

Ft

Ft

x

y F

m1 m2

F12 F21

y

Ft1

Ft2

mg

Ft1

mg

012 =−− tt FmgF

Hanging buckets.

01 =− mgFt

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

mg

Fn

Choice of axes (friendly advice)

Typically, equations simplify if one axis chosen to be along observed accelerated motion

Replace forces with their components along chosen axes θ

θ

Block slides down an incline

yn mamgF =−+ θcos ay = 0 (no y-motion)

θcosmgFn =

Given: m, θ, no friction

Find: a, normal force

In case 1, the force F is pushing down (in

addition to mg), so the normal force

needs to be larger. In case 2, the force F

is pulling up, against gravity, so the

normal force is lessened.

ConcepTest 2 Normal Force

Case 1

Case 2

Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle θ. In which case is the normal force greater?

A) case 1

B) case 2

C) it’s the same for both

D) depends on the magnitude of the force F

E) depends on the ice surface

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Atwood machine Two boxes of mass m1 and m2 are held together by a cable that passes over a massless frictionless pulley.

m1

m2

a) a, acceleration of each box b) Ft, tension in cord

Find:

Pulleys just redirect the tension The force of tension has the same

magnitude throughout the cable The magnitude of the acceleration is the

same for both objects They move in opposite directions

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

For m1 amgmFt 11 =−

ammgmm )()( 2112 +=−

)(22 amgmFt −=−

gmmmma

)()(

21

12

+−

=

)(1 agmFt +=

Atwood machine

m1

Ft

m2g

m2

Ft

m1g

y

a

a

For m2

The magnitude of the acceleration is the same for both objects

aaa == 12

(2) (1)

(1)-(2)

Limit: if m2=0, then a=g m1 would be a freely falling object If m1=m2 , then a=0

Using (1)

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Another typical problem A 5-kg block (m2) sits on an inclined plane tilted at an angle of 30º. It is attached via a massless cord to a 2-kg block (m2) lying on a flat surface. All surfaces are frictionless.

– Draw a free body diagram for each block (include coordinate system). – What are the magnitude of the Normal Forces on each block? – Find: a) acceleration of the blocks; b) tension in the cord

m1

θ m1g

m2g

FN1

FN2 Ft

x y

𝒎𝟐𝒈𝐬𝐬𝐬𝜽

Eq. for m1

yN amgmF 111 =−amFt 1=

y x

0

yN amgmF 222 cos =− θamFgm t 22 sin =−θ

y x

0 Eq. for m2

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Another typical problem (cont.) gmFN 11 =

amFt 1=

θcos22 gmFN =

Add them: ammgm )(sin 212 +=θ

21

2 sinmm

gma+

=θ θsin

21

211 g

mmmmamFt +

==

(2)

(1)

m1

θ m1g

m2g

FN1

FN2 Ft

x y

𝒎𝟐𝒈𝐬𝐬𝐬𝜽

amFgm t 22 sin =−θ

F12 F21

The forces are equal and opposite by Newton’s Third Law!

ConcepTest 3 Bowling vs. Ping-Pong I

In outer space, a bowling ball and a Ping-Pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare?

A) the bowling ball exerts a greater force on the Ping-Pong ball

B) the Ping-Pong ball exerts a greater force on the bowling ball

C) the forces are equal D) the forces are zero because they

cancel out E) there are actually no forces at all

The forces are equal and opposite—

this is Newton’s Third Law!! But the

acceleration is F/m and so the smaller

mass has the bigger acceleration.

In outer space, gravitational forces exerted by a bowling ball and a Ping-Pong ball on each other are equal and opposite. How do their accelerations compare?

A) they do not accelerate because they are weightless

B) accelerations are equal, but not opposite

C) accelerations are opposite, but bigger for the bowling ball

D) accelerations are opposite, but bigger for the Ping-Pong ball

E) accelerations are equal and opposite

ConcepTest 4 Bowling vs. Ping-Pong II

F12 F21

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Summary of Chapter 4 Newton’s first law: If the net force on an object is zero, it will

remain either at rest or moving in a straight line at constant speed.

Newton’s second law:

Newton’s third law:

Weight is the gravitational force on an object.

Free-body diagrams are essential for problem-solving. Do one object at a time, make sure you have all the forces, pick a coordinate system and find the force components, and apply Newton’s second law along each axis.

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 7

Thank you See you on Monday