Phys 207 Announcements z z z

z z

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Hwk4 submission deadline = 1st thing Monday morning Quizzes in DSC next week as usual Exam 1 on Mar. 11th, NEXT FRIDAY! ÍBring Student or Photo ID ÍBring a Calculator Today’s Agenda Finish up motion in a circle Friction ÍWhat is it? ÍHow do we characterize it? ÍModel of friction ÍStatic & Kinetic friction Some problems involving friction

1

Forces and Angles z

A box is suspended from the ceiling by two ropes making an angle θ with the horizontal. What is the tension in each rope?

θ

θ m

2

Page 1

Forces and Angles z

Draw a FBD:

T1

T2 j

T2sin θ

T1sin θ

θ T1cos θ

θ T2cos θ

i

mg z

Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0 Fx,NET = T1cos θ - T2cos θ = 0

T1 = T2

Fy,NET = T1sin θ + T2sin θ - mg = 0

T1 = T2 =

mg 2 sin θ 3

Problem: Motion in a Circle z

A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. ÍWhat is the tension T in the string at the top of the rock’s trajectory? v T

R

4

Page 2

Motion in a Circle... z z z

Draw a Free Body Diagram (pick y-direction to be down): We will use FNET = ma (surprise) y First find FNET in y direction: mg

FNET = mg +T

T

5

Motion in a Circle... FNET = mg +T v z

y

Acceleration in y direction: mg

ma = mv2 / R

T

F = ma

mg + T = mv2 / R R T = mv2 / R - mg

6

Page 3

Motion in a Circle... z

What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? Íi.e. find v such that T = 0.

v

mv2

/ R = mg + T

mg

v2 / R = g

T= 0

v = Rg z

Notice that this does not depend on m.

R

7

Lecture 11, Act 1 Motion in a Circle z

A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground? v mg N R

Rg (a) v = mRg (b) v = m

(c) v = Rg 8

Page 4

Lecture 11, Act 1 Solution z

mv2 / R = mg - N

z

For N = 0: v = Rg

v

mg N R

9

New Topic: Friction z

What does it do? ÍIt opposes relative motion!

z

How do we characterize this in terms we have learned? ÍFriction results in a force in the direction opposite to the direction of relative motion! j

N FAPPLIED ma fFRICTION

i

mg 10

Page 5

Surface Friction... z

Friction is caused by the “microscopic” interactions between the two surfaces:

11

Surface Friction... z

Force of friction acts to oppose relative motion: ÍParallel to surface. ÍPerpendicular to Normal force.

j

N F ma fF

i

mg 12

Page 6

Model for Sliding Friction z

The direction of the frictional force vector is perpendicular to the normal force vector N.

z

The magnitude of the frictional force vector |fF| is proportional to the magnitude of the normal force |N |. Í |fF| = µK | N | ( = µK | mg | in the previous example) ÍThe “heavier” something is, the greater the friction will be...makes sense!

z

The constant µK is called the “coefficient of kinetic friction.”

These relations are all useful APPROXIMATIONS to messy reality. 13

Model... z

Dynamics: i: j: so

F − µKN = ma N = mg

F − µKmg = ma

j

N F ma µK mg

i

mg 14

Page 7

Lecture 11, Act 2 Forces and Motion A box of mass m1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (µk = 0.51) on top of a second box having mass m2 = 3 kg, which in turn slides on a frictionless floor. ÍWhat is the acceleration of the second box ? (a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 3.0 m/s2

z

T

slides with friction (µk=0.51 )

m1

a=?

m2

slides without friction 15

Lecture 11, Act 2 Solution z

First draw FBD of the top box:

N1 T

f = µKN1 = µKm1g

m1

m1g

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Page 8

Lecture 11, Act 2 Solution z

Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.

z

As we just saw, this force is due to friction:

f2,1 = µKm1g

m1 f1,2

m2

17

Lecture 11, Act 2 Solution z

Now consider the FBD of box 2:

f1,2 = µkm1g

N2

m2

m1g m2g 18

Page 9

Lecture 11, Act 2 Solution z

Finally, solve F = ma in the horizontal direction: a=

µKm1g = m2a

1 .5 kg m1 × 0.51 × 9 .81 m s 2 µk g = 3 kg m2

a = 2.5 m/s2

f1,2 = µKm1g

m2

19

Inclined Plane with Friction: z

Draw free-body diagram:

ma

µKN

j

N θ mg θ i 20

Page 10

Inclined plane... z

Consider i and j components of

FNET = ma :

µKN ma j

i

mg sin θ − µKN = ma

j

N = mg cos θ

mg sin θ − µKmg cos θ = ma

N θ mg i

θ

a / g = sin θ − µKcos θ

mg cos θ

mg sin θ 21

Static Friction... z

So far we have considered friction acting when the two surfaces move relative to each other- I.e. when they slide.. ÍWe also know that it acts when they move together: the ‘static” case.

z

In these cases, the force provided by friction will depend on the OTHER forces on the parts of the system. N F

j i

fF

mg 22

Page 11

Static Friction… (with one surface stationary) z

Just like in the sliding case except a = 0. i: F − fF = 0 j: N = mg

z

While the block is static: fF = F

N

j

F

i

fF

mg 23

Static Friction… (with one surface stationary) z

The maximum possible force that the friction between two objects can provide is fMAX = µSN, where µs is the “coefficient of static friction.” ÍSo fF ≤ µS N. ÍAs one increases F, fF gets bigger until fF = µSN and the object starts to move.

N F

j i

fF

mg 24

Page 12

Static Friction... z

µS is discovered by increasing F until the block starts to slide: i: FMAX − µSN = 0 j: N = mg µS = FMAX / mg j

N FMAX

µSmg

i

mg 25

Lecture 11, Act 3 Forces and Motion z z

A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is µs = 0.4. A rope is attached to the box and pulled at an angle of θ = 30o above horizontal with tension T = 40 N. ÍDoes the box move?

(a) yes

(b) no (c) too close to call

T static friction (µs = 0.4 )

m

θ 26

Page 13

Lecture 11, Act 3 Solution z

Pick axes & draw FBD of box:

z

Apply FNET = ma

y

x

y: N + T sin θ - mg = maY = 0 N = mg - T sin θ

N

= 80 N

T

x: T cos θ - fFR = maX The box will move if T cos θ - fFR > 0

θ

m

fFR

mg 27

Lecture 11, Act 3 Solution

y

x

y: N = 80 N x: T cos θ - fFR = maX The box will move if T cos θ - fFR > 0

N

T cos θ = 34.6 N

T

fMAX = µsN = (.4)(80N) = 32 N

fMAX = µsN

So T cos θ > fMAX and the box does move

θ

m

mg 28

Page 14

Static Friction: z

We can also consider µS on an inclined plane.

θ z

In this case, the force provided by friction will depend on the angle θ of the plane.

29

Static Friction... z

The force provided by friction, fF , depends on θ.

fF ma = 0

(block is not moving)

mg sin θ − ff = 0

j

N θ

(Newton’s 2nd Law along x-axis)

mg θ i

30

Page 15

Static Friction... z

We can find µs by increasing the ramp angle until the block slides: mg sin θ − ff = 0 In this case: ff = µSN = µSmg cos θM

µSN

mg sin θM − µSmg cos θM = 0

j N θM mg θ

µS = tan θM

i 31

Additional comments on Friction: z

Since fF = µN , the force of friction does not depend on the area of the surfaces in contact. (This is a surprisingly good rule of thumb, but not an exact relation)

z

By definition, it must be true that system (think about it...).

µS ≥ µK

for any

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Page 16

Aside: z

Graph of Frictional force vs Applied force:

fF = µSN fF = µKN fF fF = FA FA 33

Problem: Box on Truck z

A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is µS. ÍWhat is the maximum acceleration a that the truck can have without the box slipping?

m

µS

a

34

Page 17

Problem: Box on Truck z

Draw Free Body Diagram for box: ÍConsider case where fF is max... (i.e. if the acceleration were any larger, the box would slip). N j i fF = µSN

mg 35

Problem: Box on Truck z

Use FNET = ma for both i and j components Íi µSN = maMAX Íj N = mg

aMAX = µS g

N j

aMAX i fF = µSN

mg 36

Page 18

Lecture 11, Act 4 Forces and Motion z

An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. What is the direction of the static frictional force? µS

a

Ff Ff

(a)

Ff (b)

(c) 37

Lecture 11, Act 4 Solution z

First consider the case where the inclined plane is not accelerating. N Ff

mg

z

All the forces add up to zero!

N mg Ff 38

Page 19

Lecture 11, Act 4 Solution z

If the inclined plane is accelerating, the normal force decreases and the frictional force increases, but the frictional force still points along the plane: N Ff a mg

z

N

All the forces add up to ma! ÍF = ma ÍThe answer is (a)

mg

ma

Ff 39

Problem: Putting on the brakes z

Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since µS > µK .

z

The driver of a car moving with speed vo slams on the brakes. The coefficient of static friction between the wheels and the road is µS . What is the stopping distance D?

ab

vo v=0

D 40

Page 20

Problem: Putting on the brakes z

Use FNET = ma for both i and j components Íi µSN = ma Íj N = mg N

a = µS g

j a fF = µSN

i mg

41

Problem: Putting on the brakes

z

As in the last example, find ab = µSg. Using the kinematic equation: v2 - v02 = 2a( x -x0 )

z

In our problem:

z

ab

0 - v02 = − 2ab( D )

vo v=0

D 42

Page 21

Problem: Putting on the brakes z z

0 - v02 = − 2ab( D )

In our problem: Solving for D: D=

Putting in ab = µSg

ab

D=

v 02 2 ab

v 02 2 µs g

vo v=0

D 43

Recap of today’s lecture z

z

Friction ÍWhat is it? ÍHow do we characterize it? ÍModel of friction. ÍStatic & Kinetic friction.

(Text: 5-8)

(Text: 5-8)

Some problems involving friction. ÍBox on truck. ÍBraking distance.

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