lecture 8 applications of newton’s laws

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

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.

Tm

Static friction (s= 0.4)

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 only 30 N.

So the pulling force is not big

enough to overcome friction.

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.

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

Tm

Static friction (s= 0.4)

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?

TensionT

Fs

W

W

T

2.00 kg

Tension in the rope?

Translationalequilibrium?

W

W

TT

y :m1 : x :

m2 : y :

fk

Over the Edge

m

10 kg a

m

a

F = 98 N

Case (1) Case (2)

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.

Over the Edge

m

10 kg a

m

a

F = 98 N

Case (1) Case (2)

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.

W

T

a>0 downwardimplies T<W

TensionForce is always along a rope

W

TTTy

TT

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.

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

Springs and TensionA mass M hangs on spring 1, stretching it length L1

Mass M hangs on spring 2, stretching it length L2

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

S1

S2

a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2




S1

S2

W

Fs=T





Spring 1 supports the weight.Spring 2 supports the weight.Both feel the same force, and stretch the same distance as before.

S1

S2

W

Fs=T


Instantaneous acceleration

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

From

Lectu

re

3 Fr

om Le

cture

3

Circular MotionAn object moving in a circle must have a force

acting on it; otherwise it would move in a straight line.If the speed is constant, the direction of the force and the acceleration is towards the center of the circle.

The magnitude of this centripetal force is given

by:

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

aa

Circular Motion

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

Examples of centripetal force

whenno friction is needed to hold the track!

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:

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?

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?

necessary centripetal force:

Only force on puck is tension in the string!

To support mass M, the necessary tension is:

Circular motion and apparent weight

This normal force is the apparent, or perceived, weight

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

2c

m v tF t r

cF

c cF f

Going in Circles IGoing 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 IGoing 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

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



Vertical circular motion

A

B

C vertical (down)

vertical (up)

horizontal

Centripetal acceleration must be

Condition for falling: N=0

at C:

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

So, as long as:

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

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 FunBarrel 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? a b c d e

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.

Barrel of FunBarrel 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?

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

a b c d e

lecture 8 applications of newton’s laws

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