chap85.doc

6
Belt Friction Let's look at a flat belt passing over a drum Let's take a look at a differential element Motion is assumed to be impending. dN dF  s  µ = The normal force is a differential force because it acts on a differential element of area. 0 2 cos ) ( 2 cos 0 =       + +       = Σ θ  µ θ d dT T dN d T  F  s  x 0 2 sin ) ( 2 sin 0 = +       +       = Σ dN d dT T d T  F  y θ θ For small angles cos(d θ ) = 1 sin(d θ ) = d θ 0 = + + Σ dT T dN T  F  s  x µ  dN dT  s  µ = T 1 P P' β θ d θ P 1 P 2 O dN d θ µ s dN T+dT T d θ/2 d θ/2 x y r 

Transcript of chap85.doc

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Belt Friction

Let's look at a flat belt passing over a drum

Let's take a look at a differential element

Motion is assumed to be impending.dN dF 

 s µ =∴

The normal force is a differential force because it acts on a differential element of area.

02

cos)(2

cos

0

=   

  ++− 

  

  −

θ  µ 

θ  d dT T dN 

d T 

 F 

 s

 x

02

sin)(2

sin

0

=+   

  +− 

  

  −

dN d 

dT T d 

 F  y

θ θ 

For small angles

cos(d θ ) = 1sin(d θ ) = d θ 

0=++−−⇒Σ dT T dN T  F   s x µ 

  dN dT  s

 µ =

T1

P P'

β

θ

d θ

P 1 P 2

O

dN

d θ

µs

dN

T+dTT

d θ/2 d θ/2x

y

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02

)(2

=+   

  +− 

  

  −⇒Σ dN 

d dT T 

d T  F  y

θ θ 

  0222

=+   

  − 

  

  − 

  

  − dN 

d dT 

d T 

d T 

θ θ θ 

02 =   

   θ d 

dT  (2 small numbers squared = 0)

Thus: θ d T dN  =

Substituting:

θ  µ 

θ  µ 

d T 

dT 

d T dT 

 s

 s

=

=

Integrating:

θ  µ β 

d T 

dT 

Os

T  ∫ ∫  =2

1

 BT T S 

 µ =− 12 lnln

Thus:

 BT 

T  s µ =  

 

  

 

1

2ln

OR 

 B seT 

T  µ =1

2

These formulas apply to:

1). Flat belts passing over fixed drums.

2). Ropes wrapped around a post or capstan.3). band brakes- the drum is about to rotate while the band remains fixed.

4). Problems involving belt drives. Both the pulley and belt rotate; determine whether the

 belt will slip/move to the pulley.

- Should only be used if the belt, rope, or brake is about to slip.

- In the equations 2T   will always be larger than 1

T   . So 2T   represents the tension which pulls

and 1T   is the resisting tension.- β must be in radius. B may be larger than π 2 . If a rope is wrapped around a post n times,

nπ β  2= .

- If the belt, rope is actually slipping, k  µ  should be used.

-If no slipping occurs and is not about to slip, none of these formulas should be used.

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V-belts

Proceeding as before:

= )2/(sin

1

2 α 

β  µ  s

eT 

α

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Examples

1). Given: A hawser thrown from a ship to a pier is wrapped 2 full turns around a capstan. Thetension in the hawser is 7500 N; by exerting a force of 150 N on its free end, a dock worker can

 just keep the hawser from slipping.

Find: a). Determine the coefficient of friction between the hawser and capstan.

 b). Determine the tension in the hawser that could be resisted by 150 N force if thehawser were wrapped 3 full turns around the capstan.

a) Slipping is impending

311.0

turn

rad2turns2

150

7500ln

ln1

2

=

   

  =

=

S  BT 

 µ 

π  µ 

 µ 

 b)

kN7.52

150

ln

2

)2)(3)(311(.2

1

2

=

=

=

eT 

 BT 

π  

 µ 

150 N

7500 N

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2). Given: A flat belt is used to transmit the 30 ft lb torque developed by an electric motor. The

drum in contact with the belt has a diameter of 8 in and .30.0=S 

 µ 

Find: Determine the minimum allowable value of the tension in each part of the belt if the belt is

not to slip.

lbs6.158

lbs6.68

90311.2

90

90

in12

ft1

2

in8)(30

)(

=

=

+=

+=

=−

   

  

   

  

 −=

−=

Α

ΒΑ

ΒΑ

ΒΑ

T T 

T T 

T T 

T T 

r T T  M 

 B

 B B

 B A

3). Given: Example problem #2, using a V-belt with 36=α 

60o

40 o

A

B

60 o

TA

40 o

TB

20 o

30 o

70 o

β

 AT T  =

2

 BT T  =

1

160=β 

( )

ΒΑ

Β

Α

=

=

=

T T 

eT 

eT 

T  s

311.2

9/83.0

1

2

π 

β  µ 

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Find: Determine minimum allowable tension

ΒΑ

Β

Α

=

=

=

T T 

eT 

eT 

T  s

045.15

18sin/)9/8(3.0

)2/sin(/

1

2

π 

α β  µ 

90

90

12

4)(30

)(

+==−

  

 

 

 

−=

−=

ΒΑ

ΒΑ

ΒΑ

T T 

T T 

T T 

r T T  M 

 B A