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2.001 - MECHANICS AND MATERIALS I

Lecture #

9/18/

Prof. Carol Livermore

TOPIC: FRICTION

EXAMPLE: Box on floor

μs =Coefficient of Static Friction

FBD

Equation of equilibrium

Fy = 0

N − W = 0

N = W

Fx = 0

T − F = 0

T = F

At impending motion only:

F = μsN

For well lubricated, μs ≈ 0 .05.

For very clean surfaces μs ≈ 0. 4 − 1.

After it starts to move:

F = μkN

μk = Coefficient of kinetic friction.

μk < μs

EXAMPLE: Block on an inclined plane

Q: At what angle (α) does the block slide down the plane?

FBD:

So:

L 2

a = tan α 2

The resultant of the normal force and frictional force act directly ”below” the

center of mass.

EXAMPLE

Q: For what range of W 0 is the block in equilibrium?

FBD

Case 1: Impending motion is down the plane.

Fx = 0

T 1 + F 1 − W sin α = 0

Fy = 0

N 1 − W cos α = 0

Case 2: Impending motion is up the plane.

Fx = 0

T 2 + F 2 − W sin α = 0

Fy = 0

N 2 − cos α = 0

What about T?

FBD of Cable

Look at differential element

The block will be stable against downward motion when:

W 0 ≤ W sin α + μsW cos α

So it is stable when:

W (sin α − μs cos α) ≤ W 0 ≤ W^ (sin^ α^ +^ μs cos^ α)

What about pulley with friction? (^) Look at a differential element.

Recall a rope around a rod.

Fx = 0

T (θ) cos

dθ

2

−T (θ + dθ) cos

dθ

2

−dF = 0

Fy = 0

( ) ( ) dθ dθ dN − T (θ) sin −T (θ + dθ) sin −dF = 0 2 2

dθ dθ sin ≈ 2 2

dθ cos ≈ 1 2

dT = T (θ + dθ) − T (θ) ⇒ T (θ + dθ) = T (θ) + dT = T + dT

So:

T (θ) − T (θ + dθ) − dF = 0

dT = −dF

T dθ dθ dN − + (T + dT ) = 0 2 2

T + dT → 0

dN − T dθ = 0

With impending motion:

dF = μsdN

dT = −μsdN

dT dN = − μs

Substitute:

dT − − T dθ = 0 μs

Thus:

dT = −μsdθ T

Integrate:

∫ (^) T 2 ∫ (^) φ dT = −μsdθ T 1 T 0 [ ]T 2

ln T = −μsφ T 1

ln

T 2

= −μsφ T 1

T 2 = exp(−μsφ) T 1

T 2 = T 1 exp(−μsφ)

This is known as the capstan effect.

EXAMPLE: Boat on a dock

φ = 3(2π) ≈ 20

μs = 0. 4

T 2 = T 1 exp(−8)

T 2 ≈ T 1