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Pulley:

F

A

B

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Identify: When B moves, friction acts on the surface separating A and B, pointing to the left. We should draw a free-body diagrams for each block. Constant speed=equilibrium in x.!

Setup: DRAW free-body diagram for A and B. Starts with A (reason: it is a simpler system with regard to forces)!

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Setup for A: I can find out the friction force (which contributes to forces in B).

NBA Normal force that B exerts on A

fBA Friction to left as A moves relative to B to right

NBA

T fBA

WA

T Tension force to the right (note I didn’t label A or B, why?)

Normal force that B exerts on A WA

Execute: let’s do Newton II:!In the y direction:!

!

Fy = may"

!

NBA"W

a= 0

!

NBA

=Wa

=1.40N

!

fBA = NBAµ =1.40 " 0.3 = 0.42N

So friction is:!

!

T " fBA = 0# T = 0.42N

Now, in the x direction, since uniform motion, tension force is equal and opposite to friction:!

!

Fx

= max" = 0

B'

Setup for B: Draw a free-body diagram,

NAB Normal force that A exerts on B,

Equal and opposite to wA!

fAB Friction to right as B moves relative to A to left

Weight of B WB

fAB

WB

T

n

Normal force ground acts on B n

B

!

Fx

= max"

!

fAB = NABµ = NBAµ = 0.42N (Newton III, pointing to right)

fGB

fGB Friction from ground to B, pointing to the right as B moves to left

Now the real trick in this problem, calculating the friction between B and ground!!

!

fGB = NGBµ = (WA +WB )µ

!

= 0.42N +WBµ

!

= 0.42 + 4.2 " 0.3 =1.68N

!

F = T + fAB + FGB

!

F = 0.42 + 0.42 +1.68 = 2.52 N

F

WA

E'

Dynamics of Uniform Circular Motion

!

! a

c=

v2

! r

If you recall, one of the multiple choices that were given in the notes asks about the acceleration of the system. Let’s revisit the problem and think of it in dynamics:!

!

! F

net=

! F " = m

! a

c= m

v2

! r

This means that the centripetal acceleration is the result of NET FORCE

(gravity, thrust of the propeller, etc.). It is on the right side of the NEWTON II!

!

T =2"r

v

(period)

!

! F

net=4" 2!

r

T2

m

In terms of constant period of the circular motion T, the

net force >'

Example: Uniform rotations in the vertical plane

Example: Trinity runs with uniform speed along the interior of a cylindrical pipe in Machine City. She weighs 50 kg and moves at 12 m/s. The radius of the pipe is 2.0 m. !What is the normal force Trinity exerts on !the wall of the pipe at A (bottom) and B

(top)?!

A

B

Identify: uniform circular motion, same circular acceleration pointing to center. Forces involved gravity, normal force from pipe wall to Trinity.!

Setup+execute: DRAW free-body diagram for A and B. !

Starts with A: No horizontal force. Take up as positive!

!

Fx

= max" = 0

!

Fy = may" = mv2

r

A

N

W

!

! a y

L'

!

N "mg = mv2

r

!

N = m(g +v2

r) = 50 " (9.81+

122

2) = 4090.90N

Case at Point B: !

!

Fx

= max" = 0

!

Fy = may" = mv2

r B

N

W

!

! a

!

"N "mg = "mv2

r(negative since Up is positive direction)

!

"N = "mv2

r+ mg

!

"N = m(g "v2

r) = 50 # (9.81"12

2/2) = "3109.50N

!

N = 3109.50N

This indicates that the normal force is acting downward with smaller magnitude than the magnitude of upward normal force at the bottom of the pipe! The fact N is positive because I assumed it to be a positive quantity which separates from its Downward direction indicated by “-”.! M'

“Rough” Flat Curves vs. Banked Curves

Two common types of problem in circular motion are (1) to solve for the maximum/minimum speed of moving body on a flat curve with friction (to keep it from slipping off path) and (2) to solve for the angle/speed of a banked curve (with or w/o friction) to keep a moving body within a circular path. !

Flat Curve: In INDI-500 a car is rounding a flat unbanked curve with a radius of 200 meters. If the coefficient of static friction (remember small feet on tires) is 0.3, what is the maximum speed Vmax that the car can safely round the curve without skidding?!

Solution: DRAW free-body diagram for the car. ! N

W

!

! a

c

fs

!

Fy = "W + N# = 0

!

Fx = fs = max" = mvmax

2

r

To solve for Vmax, we need friction (static), To find friction we need Normal force, which can be found by y component. !

N'

!

"W + N = 0Y component

!

N = mg

X component

!

fs = mvmax

2

r

!

µs" N = m

vmax

2

r

!

µs "mg = mvmax

2

r

!

vmax

= r "µs " g

!

vmax

= 200 " 0.3" 9.81

!

vmax

= 24.26 m/s

Note: (1) mass does not matter -! Formula 1 car or Monster truck make no

difference to maximum allowable speed! (2) What happens if it is watery surface (rain, snow) --! coef of friction

decreases, which means maximum allowable speed also lowers! O'

Banked Curve: The surface of a banked circular track makes an angle of ! degrees with ground. Now, suppose a

car is rounding this track with a radius of r meters (considered at the center line of the car). Let’s assume the

normal force is N (perpendicular to track surface, inclined to flat ground), What is the maximum speed Vmax that the car can safely round the curve without skidding?!

Solution: DRAW free-body diagram for the car. !

r

!

N

W

!

! a

c

fs

!

!

P'

N

W

!

! a

c

fs

!

!

First, lets consider the simple case that there is no friction (icy day). What can I conclude or find? !

IF THERE IS NO FRICTION, if I know the weight W (mass x g), I can find Normal force given the angle since the vertical component forces of W and Ncos(!) balances. Knowing N I can further find friction (assuming coefficient of friction is known). I can further find Vmax knowing fs.!Now, lets go back to this problem with friction!

!

"W " f s sin# + N cos# = 0Y component

X component

!

vmax =rN(sin" + µcos")

m!

"mg" Nµsin# + N cos# = 0

!

m =N(cos" #µsin")

g

!

N sin" + f s cos" = mvmax

2

r <?'

!

vmax =rN(sin" + µcos")

m

!

vmax =rN(sin" + µcos")g

N(cos" #µsin")=

r(sin" + µcos")g

(cos" #µsin")!

m =N(cos" #µsin")

g

Mass again doesn’t matter, we got the following for mass,

Substitute into velocity expression,

Again, only the inclined angles, radius, g and friction

coefficient matter to the velocity (max without skidding)

<<'

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CDE(F(G'Identify: Pulley problem. The weight of the object causes the middle pulley to move to the right through tension forces on the angled at 20o.!

foot

m

Fixed horizontal bar

fixed

fixed moveable

20o

20o

Setup: Draw a free-body diagram for the foot inside the strap.!

foot 20o

T

T

!

! a

x

<B'

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!

Fy" = T sin(20) #T sin(20) = 0

!

T = mg

Execute:!

In the y-direction!

In the x-direction, positive to right!

!

Fx" = T cos(20) + T cos(20)

!

165 = 2T cos(20)

!

T =165

2cos20

!

= 87.79N

foot 20o

T

T

!

m = T /g = 87.79 /9.81= 8.95 kg

Mass of the weight: !

!

T = mg

<E'

Supplementary: To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by x = (9.0!103 m/s2) t2

– (8.0"104 m/s3) t3. The object leaves the end of the barrel at t=0.025 s. a) How long must the gun barrel be? b) What will be the speed of the objects as they leave the end of the barrel? c) What net force must be exerted on a 1.50 kg object at i) t=0 and ii) t=0.025 s?!

R'

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