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•  Recitation homework due Thursday/Friday •  Exam 1 scores are on D2L •  There may be a curve for the final score but there are no

individual curves for each exam. •  Questions on the answers and solutions can be addressed to

professors or TA’s.

Friction

Exam scores0

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10 20 30 40 50 60 70 80 90 100

Stud

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Average: 71 Median: 72

•  Questions on your score should be addressed to Professor Kinney.

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About the exam •  I believe the exam was a fair representation of what was

covered in lecture, CAPA, and recitation. It was also similar to the practice exams on D2L.

•  I expect all of you to be able to solve every problem on the exam.

•  You should go over the solutions to make sure you understand the problem and where you might have gone wrong so that you can avoid the same mistakes in the future.

•  At least two questions from each midterm will appear on the final

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Learning assistant info session •  There will be an info session on the LA program on

Monday, February 23 from 5 to 6:30pm in UMC 235. •  Refreshments will be served. •  Great for people interested in teaching and learning

about education. Nearly all participants find it to be very rewarding. Feel free to chat with your LA’s during recitation about their experiences.

•  Applications are accepted between 2/23 and 3/9. •  More info available by contacting Steve Pollock at

steven.pollock@colorado.edu or by visiting the web site: laprogram.colorado.edu

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Force of kinetic friction is It depends on which depends on the materials and on the normal force n which is the force pushing the materials together.

is the coefficient of kinetic friction

Friction Experiments have found two main types of sliding friction

Static friction is the force exerted when the two objects are not in motion relative to each other (no slipping)

Kinetic friction is the force exerted when the two objects are in motion relative to each other (slipping)

.nf kk µ=kµ m

v m

nnf kk µ=

Fg =mg

kµ

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Formula for static friction is similar: .

Static friction nf ss µ≤

The static friction will oppose forces which try to slide one object across another. However, once the maximum possible static friction force is exceeded the object will “break loose” and start moving.

)( nsµ

m extF

m

nnFf ss µ≤= ext

Fg =mg

extF

If the block is not moving then:

is the coefficient of static friction sµ

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Static and Kinetic Friction

Applied force

Forc

e of

fric

tion

kinetic friction

nf ss µ=

nf kk µ=

Start with object at rest feeling a normal force of n

Start applying a force perpendicular to n.

Up to a force of µsn static friction prevents movement

After movement starts, frictional force is reduced to µkn.

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Clicker question 1 Set frequency to BA

A stationary block rests on an inclined plane. The coefficient of static friction is µs = 0.2. The normal, gravity, and static friction forces are shown. Which set of relations applies to this situation? A. B. C. D. θθ cos and sin mgfmgn ==

θθ sin and cos mgfmgn ==

nfmgn sµθ == and tan

θ mgθcosmg

θsinmg

θ θ−90

x

y

Fnet,x =mgsinθ − f = 0 so f =mgsinθ

Fnet,y = n−mgcosθ = 0 so n =mgcosθ

n

mg

f

θnfmgn sµ== and

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Propulsion To walk (or run), we push against the earth with leg muscles and by the 3rd law, the earth pushes against us.

Because of this, it is impossible to walk on a frictionless surface.

A car works the same way with tires instead of feet.

Is it static friction or kinetic friction that propels us when walking or driving?

Static. The two surfaces are not sliding.

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A thief forgot to bring a safe cracking kit or dolly so is trying to drag a 200 kg safe up a ramp at an angle of θ=20°. The safe is currently not moving while he pulls on the rope with a force of 400 N. What is the magnitude and direction of the force of friction on the safe?

Clicker question 2 Set frequency to BA

Free clicker question. Full credit for any answer!

A. Friction force is down the ramp B. Friction force is up the ramp C. Friction force is zero D. Need to know the coefficient of friction E. Need to know the mass of the thief

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Decompose forces into x & y components

With our coordinate choice, only need to worry about gravity

The 200 kg safe is not moving while the thief pulls on the rope with a force of 400 N. What is the magnitude and direction of the force of friction on the safe (use g=10 m/s2)?

°20Fg =mg

f

nT

Always need a free body diagram

Pick a coordinate system

Fnet,x =mgsinθ + f −T

Fg,x =mgsinθFg,y =mgcosθOnly now do we

apply Newton’s laws and solve. Fnet,x =maxFnet,y =may

0sin =−+ Tfmg θ

0=xmaθsinmgTf −=

and

N 28420sinm/s 10kg 200N 400 2 −=°⋅⋅−=f

so that

It is in the opposite direction!

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A.  |a2| > |a1| B.  |a2| < |a1| C.  |a2| = |a1| in the same direction D.  |a2| = |a1| but directions are opposite

A mass slides down a rough inclined plane with some non-zero acceleration of a1. The same mass is shoved up the same incline with a large, brief initial push. As the mass moves up the incline, its acceleration is a2. How do a1 and a2 compare?

1v 2v

Clicker question 3 Set frequency to BA

The force of gravity is the same: down the slope θsinmg

The force of friction has the same magnitude in both cases but opposite direction (opposite the velocity)

In 2, both forces are in the same direction so net force is greater so acceleration is greater.

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Cornering problem A car is taking a corner with radius of 100 m at a speed of 56 mph (25 m/s). On dry pavement the rubber/asphalt coefficient of static friction is 0.8. Will the car hold the line or start skidding?

May need two free body diagrams for a single object due to 3D nature.

nf

mg

Need 3D coordinate system for these types of problems.

r

t

r

z

r

z

There are no forces in the t direction so only need a rz free body diagram.

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Cornering problem A car is taking a corner with radius of 100 m at a speed of 56 mph (25 m/s). On dry pavement the rubber/asphalt coefficient of static friction is 0.8. Will the car hold the line or start skidding?

nf

mg rz We have the free body diagram so now

we can apply Newton’s 2nd law in r and z.

mgn = so

Fnet,r = f

Fnet,z = n−mgSum the forces in z:

The acceleration in r is arad

0=−mgn

radmaf =rmvf

2=

Sum the forces in r :

Acceleration in z is 0 so

but remember rva2

rad = so so

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Cornering problem A car is taking a corner with radius of 100 m at a speed of 56 mph (25 m/s). On dry pavement the rubber/asphalt coefficient of static friction is 0.8. Will the car hold the line or start skidding?

mgn = rmvf

2=So far we have found and

We also know that for static friction: nf ss µ≤

The maximum possible friction force is µsn. Is this sufficient to keep the car going around the curve?

22max, m/s 84.7m/s 8.98.0 ⋅=⋅⋅=== mmmgµnµf sss (max friction force)

( ) 22

2m/s 25.6m 100

m/s 25 ⋅=⋅== mmrmvf (actual friction force needed)

nf sµ≤ So and therefore the car will hold the corner.