chapter 4 dynamics: newton’s laws of...

Chapter 4

Dynamics: Newton’s Laws of Motion

• Force

• Newton’s First Law of Motion

• Mass

• Newton’s Second Law of Motion

• Newton’s Third Law of Motion

• Weight – the Force of Gravity; and the Normal Force

• Applications Involving Friction, Inclines

AssignmentAssignment 55

Textbook (Giancoli, 6 th edition). Due on Oct. 28 th.

http://ilc2.phys.uregina.ca/~barbi/academic/phys109 /2009/phys109.html

1. On page 101 of Giancoli, problem 47.

2. On page 102 of Giancoli ,problem 52.

3. On page 102 of Giancoli, problem 60.

4. A block is pressed against a vertical wall by a force P, as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The magnitude of the force is different in the two cases, while the directional angle is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.250. The weight of the block is 39.0 N, and the directional angle for the force P is θ = 300 . Determine the magnitude of P when the block slides

(a) up the wall and(b) down the wall.

MidtermMidterm

Date: October 19, 1pm – 2:15pm. Location: EA-106.

Remind:

- There are four problems. One of them will be based on problems found in assignment 5. The midterm includes everything covered until today , includingfriction and inclines (chapters 1 to 4 ).

- You ARE allowed to use a calculator . An equation sheet will be provided (link athttp://ilc2.phys.uregina.ca/~barbi/academic/phys109 /2009/phys109.html )

- Solutions of the assignments can be found on the webpage listed above. You can also find a link to old midterms from this webpage.

- I will be exceptionally available for discussions, questions, etc between 14:30 and 15:00h tomorrow , and between 12:30 and 15:00h on Monday .

- Dr. Ouimet will offer a review session on October 19 th (Monday) between 4:00pm-6:00pm in CL-112 . Next Wednesday’s tutorial session has been cancelled

- Remember, your marked assignments are available for you . Feel free to come anytime to my office (LB-212) if you have failed to pick it up in classroom .

Problem 4-38 (textbook Suppose that you are standing on a train accelerating (horizontally) at 0.20 g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

Tension in a Flexible CordTension in a Flexible Cord

Problem 4.38 (textbook) :

A free-body diagram for you as you stand on the train is shown.

You do not accelerate vertically, and so

NF mg=mgr

NFr

frFr

The maximum static frictional force is , and that must be greater than or equal to the force needed to accelerate you.

The static coefficient of friction must be at least 0.20 for you not to slide.

s NFµ

fr N 0.20 0.20s s sF ma F ma mg ma a g g gµ µ µ≥ → ≥ → ≥ → ≥ = =

FrictionFriction and and InclinedInclined PlanesPlanes

Remember:

(4.5)

(4.6)


The problem with inclined plans are no different from the problems we have dealt with so far.

However, remember:

1- The normal force is ALWAYS perpendicular to the plane where the object lies on.

� So, on an inclined plane it is not going to be in the vertical direction.� So, on an inclined plane it is not going to be in the vertical direction.

(See next slide)


Three forces can ALWAYS be identified acting on an object moving on an inclined plane:

• Gravity (vertical);

• Friction (along the surface);

• Normal (perpendicular to the surface).friend 1

friend 2

So, it is convenient to define your coordinate system such that the x direction coincides with the inclined line , and y in the direction perpendicular to the plane .

You can surely have other applied forces:

� for instance, you (or you and a group of friends) can push or pull this object.

you

Problem 4-41 (textbook A 15.0-kg box is released on a 32º incline and accelerates down the incline at 0.30 m/s2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?

Friction and Inclined PlanesFriction and Inclined Planes

= 0.30 m/s2

32o

= 0.30 m/s2


A free-body diagram for the box is shown.

Write Newton’s 2nd law for each direction:

Notice that the sum in the y direction is 0, since

y

xθ

θ

mgr

NFr

frFr

fr

N

sin

cos 0

x x

y y

F mg F ma

F F mg ma

θ

θ

= − =

= − = =∑∑

Notice that the sum in the y direction is 0, since there is no motion (and hence no acceleration) in the y direction. Solve for the force of friction.

Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from the y direction force equation above.

mg

( ) ( )( )fr

2 o 2

fr

sin

sin 15.0 kg 9.80 m s sin 32 0.30 m s 73.40 N 73 N

x

x

mg F ma

F mg ma

θ

θ

− = →

= − = − = ≈

( )( )( )fr

fr N 2 o

73.40 Ncos 0.59

cos 15.0 kg 9.80 m s cos 32k k k

FF F mg

mgµ µ θ µ

θ= = → = = =

Problem 4-65 (textbook A bicyclist of mass 65 kg (including the bicycle) can coast down a 6.0º hill at a steady speed of 6.0 Km/h because of air resistance. How much force must be applied to climb the hill at the same speed and same air resistance?

Tension in a Flexible CordTension in a Flexible Cord


Consider a free-body diagram for the cyclist coasting downhill at a constant speed. Here we call Ffr the friction due to air resistance (and not sliding or static friction).

Since there is no acceleration, the net force in each direction must be zero . Write Newton’s 2nd law for the x direction.

This establishes the size of the air friction force at 6.0 km/h, and so can be used in the next part.

fr frsin 0 sinxF mg F F mgθ θ= − = → =∑ y

θ

NFr

frFr

θ6.0 km/h, and so can be used in the next part.

Now consider a free-body diagram for the cyclist climbing the hill. Fp is the force pushing the cyclist uphill. Again, write Newton’s 2nd law for the x direction, with a net force of 0.

xθ

mgr

θ

y

x

θθ

mgr

NFr

frFr

PFr

fr Psin 0 xF F m g Fθ= + − = →∑

( )( )( )P fr

2 o 2

sin 2 sin

2 65 kg 9.8 m s sin 6.0 1.3 10 N

F F mg mgθ θ= + =

= = ×

Chapter 5

Circular Motion; Gravitation

• Kinematics of Uniform Circular Motion

• Dynamics of Uniform Circular Motion

• Highway Curves, Banked and Unbanked

• Non-uniform Circular Motion

• Centrifugation

Will be covered after chapter 7

• Centrifugation

• (5.6) Newton’s Law of Universal Gravitation

• (5.7) Gravity Near the Earth’s Surface

Newton’s Law of Universal GravitationNewton’s Law of Universal Gravitation

If the force of gravity is being exerted on objects on Earth, what is the origin of that force?

Newton’s realization was that the forceNewton’s realization was that the forcemust come from the Earth .

He further realized that this force must be what keeps the Moon in its orbit .


More than that: the Earth exerts a downward force on you , and you exert an upward force on the Earth .

When there is such a difference in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant .

Therefore, the gravitational force must be proporti onal to both masses.


By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses .

Summary: The gravitational force is directly proportional to the masses and inversely proportional to the square of their distance.

Or, as stated in Newton’s law of universal gravitation :

“ Every particle in the universe attracts every other particle with a force that is

proportional to the products of their masses and in versely proportional to the

square of the distance between them. This forces ac ts along the line joining

the two particles.”


The magnitude of this force is given by:

Where:

(4.7)


We know the relation between force and acceleration: . But what is a in

?

If m2 is the object exerting a force on m1 , then the gravitational acceleration felt by object 1 can be identified as:

(4.8)⇒ ⇒

The opposite is also true: object 2 will feel an acceleration

due to the gravitational force applied by object 1.

You can then write for the force acting on m1 and for the force acting on m2 .

Gravity Near the Earth’s SurfaceGravity Near the Earth’s Surface

An object of mass m ON the surface of the Earth will feel a force given by:

Where,

mE = mass of the Earth ;

rE = 6.38 x 106 m = radius of the Earth.

g has been measured and it is known to be 9.80 m/s2.

(4.9)

g has been measured and it is known to be 9.80 m/s2.

In fact, the value of g can be considered constant at any position near the Earth’s surface (this is what we have been assuming without much discussion so far).

Note that knowing G, rE and g, we can calculate the mass of the Earth. From eq. 4.9:

More accurate calculations lead to mE = 5.974 x 1024 Kg


Problem 5-31 (textbook): A hypothetical planet has a radius 1.5 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?

Solution :

The acceleration due to gravity at any location on or above the surface of a planet is given by

2

planet Planetg G M r=

where r is the distance from the center of the planet to the location in question.

( )

22Planet Earth Earth

planet Earth22 2 2 2 2

EarthEarth

1 1 9.8 m s4.4 m s

1.5 1.5 1.51.5

M M Mg G G G g

r RR= = = = = =


Problem 5-34 (textbook): Calculate the effective value of g, the acceleration of gravity, at (a) 3200 m, and (b) 3200 km, above the Earth’s surface.

Solution :

The acceleration due to gravity at any location at or above the surface of a planet is given by

where r is the distance from the center of the planet to the location in question. For this problem,

(a)

2

p lan et P lanetg G M r=

2 4

P lanet E arth 5 .97 10 kgM M= = ×

(a)

(b)

6

Earth 3200 m 6.38 10 m 3200 mr R= + = × +

( ) ( )( )

24

11 2 2 2Earth

22 6

5.97 10 kg6.67 10 N m kg 9.77 m s

6.38 10 m 3200 m

Mg G

r−

×= = × =

× +�

6 6 6

Earth 3200 km 6.38 10 m 3.20 10 m 9.58 10 mr R= + = × + × = ×

( ) ( )( )

24

11 2 2 2Earth

22 6

5.97 10 kg6.67 10 N m kg 4.34 m s

9.58 10 m

Mg G

r−

×= = × =

×�

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