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Lecture 8 Applications of Newton’s Laws. (Chapter 6). Announcements. Assignment #4: due tomorrow night, 11:59pm. Midterm Exam #1 scores are up: Class Average 75.0%. Reading and Review. Static friction (  s = 0.4 ). T. m. Will It Budge?.

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Lecture 8 Applications of Newton’s Laws

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Lecture 8 applications of newton s laws

Lecture 8

Applications of Newton’s Laws

(Chapter 6)


Lecture 8 applications of newton s laws

Announcements

Assignment #4: due tomorrow night, 11:59pm

Midterm Exam #1 scores are up:

Class Average 75.0%


Lecture 8 applications of newton s laws

Reading and Review


Lecture 8 applications of newton s laws

Static friction(s= 0.4)

T

m

Will It Budge?

a) moves to the left, because the force of static friction is larger than the applied force

b) moves to the right, because the applied force is larger than the static friction force

c) the box does not move, because the static friction force is larger than the applied force

d) the box does not move, because the static friction force is exactly equal the applied force

e) The answer depends on the value for μk.

A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?


Lecture 8 applications of newton s laws

Static friction(s= 0.4)

T

m

Will It Budge?

a) moves to the left, because the force of static friction is larger than the applied force

b) moves to the right, because the applied force is larger than the static friction force

c) the box does not move, because the static friction force is larger than the applied force

d) the box does not move, because the static friction force is exactly equal the applied force

e) The answer depends on the value for μk.

A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

The static friction force has a maximum of sN= 40 N. The tension in the rope is only30 N. So the pulling force is not big enough to overcome friction.

Follow-up: What happens if the tension is 35 N? What about 45 N?


Tension

T

T

Fs

W

W

Tension


Lecture 8 applications of newton s laws

T

T

W

2.00 kg

W

Translational

equilibrium?

Tension in the rope?


Lecture 8 applications of newton s laws

m1 : x :

m2 : y :

y :

fk


Lecture 8 applications of newton s laws

m

a

F = 98 N

Over the Edge

a) case (1)

b) acceleration is zero

c) both cases are the same

d) depends on value of m

e) case (2)

In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes.

m

a

10 kg

Case (1)

Case (2)


Lecture 8 applications of newton s laws

m

T

a

W

F = 98 N

Over the Edge

a) case (1)

b) acceleration is zero

c) both cases are the same

d) depends on value of m

e) case (2)

In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes.

In case (2) the tension is 98 N due to the hand. In case (1) the tension is less than 98 N because the block is accelerating down. Only if the block were at rest would the tension be equal to 98 N.

m

a

10 kg

Case (1)

Case (2)

a>0 downward

implies T<W


Tension1

Ty

T

T

T

T

W

Tension

Force is always along a rope


Lecture 8 applications of newton s laws

Springs

Hooke’s law for springs states that the force increases with the amount the spring is stretched or compressed:

The constant k is called the spring constant.


Lecture 8 applications of newton s laws

Springs

Note: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite.


Lecture 8 applications of newton s laws

A mass M hangs on spring 1, stretching it length L1

Springs and Tension

Mass M hangs on spring 2, stretching it length L2

S1

Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?

S2

a) (L1 + L2) / 2

b) L1 or L2, whichever is smaller

c) L1 or L2, whichever is bigger

d) depends on which order the springs are attached

e) L1 + L2


Lecture 8 applications of newton s laws

A mass M hangs on spring 1, stretching it length L1

Fs=T

Springs and Tension

W

Mass M hangs on spring 2, stretching it length L2

S1

Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?

S2

a) (L1 + L2) / 2

b) L1 or L2, whichever is smaller

c) L1 or L2, whichever is bigger

d) depends on which order the springs are attached

e) L1 + L2


Lecture 8 applications of newton s laws

A mass M hangs on spring 1, stretching it length L1

Fs=T

Springs and Tension

W

Mass M hangs on spring 2, stretching it length L2

S1

Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?

S2

a) (L1 + L2) / 2

b) L1 or L2, whichever is smaller

c) L1 or L2, whichever is bigger

d) depends on which order the springs are attached

e) L1 + L2

Spring 1 supports the weight.

Spring 2 supports the weight.

Both feel the same force, and

stretch the same distance as before.


Lecture 8 applications of newton s laws

Instantaneous acceleration

Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing:

From Lecture 3


Lecture 8 applications of newton s laws

The magnitude of this centripetal force is given by:

If the speed is constant, the direction of the force and the acceleration is towards the center of the circle.

a

a

For circular motion problems, it is often convenient to choose coordinate axes with one pointing along the direction of this centripetal force

Circular Motion

An object moving in a circle must have a force acting on it; otherwise it would move in a straight line.


Lecture 8 applications of newton s laws

Circular Motion

This force may be provided by the tension in a string, the normal force, or friction, among others.


Examples of centripetal force

Examples of centripetal force

no friction is needed to hold the track!

when


Lecture 8 applications of newton s laws

Circular Motion

An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well:


Lecture 8 applications of newton s laws

A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?


Lecture 8 applications of newton s laws

To support mass M, the necessary tension is:

necessary centripetal force:

Only force on puck is tension in the string!

A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?


Circular motion and apparent weight

Circular motion and apparent weight

This normal force is the apparent, or perceived, weight


Key points re circular motion

Key points re: circular motion

1)If object moving in a circular path

Then

2) is NOT a separate force; it represents

the sum of the physical forces acting on m


Going in circles i

Going in Circles I

a) N remains equal to mg

b) N is smaller than mg

c) N is larger than mg

d) none of the above

You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?


Going in circles i1

Going in Circles I

a) N remains equal to mg

b) N is smaller than mg

c) N is larger than mg

d) none of the above

You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?

You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg.

Follow-up: Where is N larger than mg?


Vertical circular motion

Vertical circular motion

http://www.youtube.com/watch?v=BHu8LAWSKxU


Vertical circular motion1

So, as long as:

vertical (down)

at C:

Centripetal acceleration must be

C

at the top, then N>0 and pointing down.

(now apparent weight is in the opposite direction to true weight!)

B

horizontal

vertical (up)

A

Vertical circular motion

Condition for falling: N=0


Lecture 8 applications of newton s laws

The Centrifuge

Other common examples:

  • spin cycle on washing machine

  • salad spinner

  • artificial gravity on giant space station in show on the SciFi channel


Barrel of fun

c

e

b

a

d

Barrel of Fun

A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?


Barrel of fun1

c

e

b

a

d

Barrel of Fun

A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically.

Follow-up: What happens if the rotation of the ride slows down?


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