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Physics 111: Lecture 8, Pg 1

Static Friction…Static Friction…

We can also consider S on an inclined plane.

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


Static Friction...Static Friction...

mg

N

ma = 0 (block is not moving)

The force provided by friction, fF , depends on .

fF

mg sin ff

(Newton’s 2nd Law along x-axis)

ii

jj


Static Friction...Static Friction...

We can find s by increasing the ramp angle until the block slides:

M mg

N

SN

In this case:

mg sin MSmg cos M

Stan Mii

jj

mg sin ff

ffSN Smg cos M

Blocks


Problem: Box on TruckProblem: Box on Truck

A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is S.What is the maximum acceleration aa that the truck can

have without the box slipping?

m S

aa



Draw Free Body Diagram for box:

Consider case where fF is max...(i.e. if the acceleration were any larger, the box would slip).

N

fF = SN mg

ii

jj



Use FNET = ma for both ii and jj components

ii SN = maMAX

jj N = mg

aMAX = S g N

fF = SN mg

aMAX

ii

jj


Forces and MotionForces and Motion

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?

(a) (b) (c)

Ff

Ff Ff

S aa


SolutionSolution

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

mg

Ff

N All the forces add up to zero!

mg

NFf


mg

NFf

SolutionSolution

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

aa

All the forces add up to ma!F = maThe answer is (a) mg

Ff

Nma


Problem: Putting on the brakesProblem: Putting on the brakes

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

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 S . What is the stopping distance D?

ab

vo

v = 0

D

Wheel



N

fF = SN mg

Use FNET = ma for both ii and jj components

ii SN = ma jj N = mg

a = S g

a

ii

jj



As in the last example, find ab = Sg. Using the kinematic equation: v2

- v02 = 2a( x -x0 )

In our problem: 0 - v02 = 2ab( D )

ab

vo

v = 0

D



In our problem: 0 - v02 = 2ab( D )

Solving for D:

Putting in ab = Sg

ab

vo

v = 0

D

b

20

a2

vD =

Dv

gs

02

2


Atwood’s Machine:Atwood’s Machine:

Find the accelerations, a1 and a2, of the masses. What is the tension in the string T ?

Masses m1 and m2 are attached to an ideal massless string and hung as shown around an ideal massless pulley.

Fixed Pulley

m1

m2

j

a1

a2

T1T2


Atwood’s Machine...Atwood’s Machine... Draw free body diagrams for each object Applying Newton’s Second Law: ( jj -components)

T1 - m1g = m1a1

T2 - m2g = m2a2

But T1 = T2 = T since pulley is ideal

and a1 = -a2 = -a.since the masses are connected by the string

m2gm1g

Free Body Diagrams

T1 T2

ja1 a2


T - m1g = -m1 a (a)

T - m2g = m2 a (b) Two equations & two unknowns

we can solve for both unknowns (T and a).

subtract (b) - (a): g(m1 - m2 ) = a(m1+ m2 )

a =

add (b) + (a): 2T - g(m1 + m2 ) = -a(m1 - m2 ) =

T = 2gm1m2 / (m1 + m2 )

Atwood’s Machine...Atwood’s Machine...

21

221

mm

)mm(g

+

--

g)mm(

)mm(

21

21

+-


Atwood’s Machine...Atwood’s Machine...

m1

m2

j

a

a

TT

So we find:

am m

m mg

( )

( )1 2

1 2

g)mm(

mm2T

21

21

+=

Atwood’s Machine


Is the result reasonable?Is the result reasonable? Check limiting cases!Check limiting cases!

Special cases:

i.) m1 = m2 = m a = 0 and T = mg. OK!

ii.) m2 or m1 = 0 |a| = g and T= 0. OK!

Atwood’s machine can be used to determine g (by measuring the acceleration a for given masses).

a)mm(

)mm(g

12

12 +=

-

am m

m mg

( )

( )1 2

1 2

gmm

mm2T

21

21

)(


Attached bodies on two inclined planesAttached bodies on two inclined planes

All surfaces frictionless

m1

m2

smooth peg

1 2


How will the bodies move?How will the bodies move?

From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law:

Taking “x” components:

1) T1 - m1g sin 1 = m1 a1X

2) T2 - m2g sin 2 = m2 a2X

But T1 = T2 = T

and a1X = -a2X = a

(constraints)

m1

yx

T1

N

m1g

m2

m2g

T2

Nx y

1

2


Solving the equationsSolving the equations

Using the constraints, solve the equations.

T - m1gsin 1 = -m1 a (a)

T - m2gsin 2 = m2 a (b)

Subtracting (a) from (b) gives:

m1gsin 1 - m2gsin 2 = (m1+m2 )a

So:

am m

m mg

1 1 2 2

1 2

sin sin


Special Case 1:Special Case 1:

m1m2

1 2

m1 m2

If 1 = 0 and 2 = 0, a = 0.

Boring

a

m m

m mg

1 1 2 2

1 2

sin sin



If 1 = 90 and 2 = 90, am m

m mg

( )

( )1 2

1 2

m2

TT

m1

Atwood’s Machine

m1m2

1 2

a

m m

m mg

1 1 2 2

1 2

sin sin



If 1 = 0 and 2 = 90, am

m mg

2

1 2( )

m1

m2

Lab configuration

m1m2

1 2

a

m m

m mg

1 1 2 2

1 2

sin sin

Air-track


Two-body dynamicsTwo-body dynamics

In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless.

(a)(a) Case (1) (b)(b) Case (2) (c)(c) same

m

10kga

m

a

F = 98.1 N

Case (1) Case (2)


Lecture 8, Lecture 8, Act 2Act 2 SolutionSolution

m

10kga

Add (a) and (b):

98.1 N = (m + 10kg)a

kg10m

N198a

.

Note:kg10m

mN198T

.

(a)

(b)

T = ma (a)

(10kg)g -T = (10kg)a (b)

For case (1) draw FBD and write FNET = ma for each block:


SolutionSolution

The answer is (b) Case (2)

T = 98.1 N = mam

N198a

. For case (2)

m

10kga

m

a

F = 98.1 N

Case (1) Case (2)

kg10m

N198a

.

m

N198a

.


Problem: Two strings & Two Masses onProblem: Two strings & Two Masses onhorizontal frictionless floor:horizontal frictionless floor:

m2 m1T2 T1

Given T1, m1 and m2, what are a and T2?

T1 - T2 = m1a (a)

T2 = m2a (b)

Add (a) + (b):

T1 = (m1 + m2)a a

a

i i

21

1

mm

T

+=

Plugging solution into (b):

21

212 mm

mTT

+=


Two-body dynamicsTwo-body dynamics

Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings?

(a) T1 > T2 > T3 (b) T3 > T2 > T1 (c) T1 = T2 = T3

T3 T2 T13m 2m m

a


SolutionSolution

Draw free body diagrams!!

T33mT3 = 3ma

T3 T22mT2 - T3 = 2ma

T2 = 2ma +T3 > T3

T2 T1m

T1 - T2 = ma

T1 = ma + T2 > T2

T1 > T2 > T3


SolutionSolution

Alternative solution:

T3 T2 T13m 2m m

a

Consider T1 to be pulling all the boxes

T3 T2 T13m 2m m

a

T2 is pulling only the boxes of mass 3m and 2m

T3 T2 T13m 2m m

a

T3 is pulling only the box of mass 3m

T1 > T2 > T3


Problem: Rotating puck & weight.Problem: Rotating puck & weight.

A mass m1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m2.What is the tension (T) in the string?What is the speed (v) of the sliding mass?

m1

m2

v

R


Problem: Rotating puck & weight...Problem: Rotating puck & weight...

Draw FBD of hanging mass:Since R is constant, a = 0.

so T = m2g

m2

m2g

T

m1

m2

v

R

T


Problem: Rotating puck & weight...Problem: Rotating puck & weight...

Draw FBD of sliding mass:

m1

T = m2g

v gRm

m 2

1

m1g

N

m1

m2

v

R

T

Use F = T = m1a

where a = v2 / R

m2g = m1v2 / R

T = m2g

Puck

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