Belt Friction

10/10/2010

ROPES AND FLAT BELTS Weight W is held in static equilibrium by a rope that passes over a peg. If peg is frictionless: P = W to maintain static equilibrium

peg

Chapter 7(Cont’n) Belt Friction

ROPES AND FLAT BELTS

D•

C

•

Weight W is held in static equilibrium by a rope that passes over a peg.

50N

If peg is frictionless: P = W

peg B•

A•

If the contact surface between the peg and the rope has friction : P  W to maintain equilibrium. There is a possibility that P < W and still maintain equilibrium

Determine the weight W to maintain equilibrium if pulleys are frictionless: W = 400N

W

7.64

Analysis of Belt Friction When motion is impending: Thin, flat belt

T2  T1es

s

The 30N weight is attached to a rope that runs over a fixed cylinder. The coefficient of static friction between the rope and the cylinder is 0.3. Determine the range of the force P for which the system will be at rest.

T2  T1es

s

Where: T1= tension in the slack side

CW motion: T2= tension in the tight side T2 > T 1 S= coefficient of static friction  = angle of wrap = radians ( radians = 180 deg.)

CCW motion:

P  30e0.3

P  77N

30  Pe0.3

P  11.7N

System is at rest if: 11.7N  P  77N

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10/10/2010

7.59 The rope is wrapped 1 and ½ times around a circular peg. What is ?

  1.5( 2)

How many turns of rope around the capstan are needed for the 60N force to resist the 9000-N pull of a docked ship? The coefficient of friction between the capstan and the rope is 0.2.

  3 or:

  n( 2) n = no. of turns 2 radians = 1 revolution = 360 deg.

Other Applications of Belt Friction:

Capstan – a device for fastening a ship to the dock.

7.59 s=0.2

Capstan – a device for fastening a ship to the dock.

T2  T1es

9000  60e0.2

Belt Drives – the friction between the belt and the pulleys enables power to be transmitted between rotating shafts.

  25.05 radians

Solving:

  n( 2)

Band Brakes – use friction between a band (belt) and a cylindrical drum to reduce the speed of a rotating machinery.

25.05  n( 2) n  3.98 or 4 turns 4 turns are required to hold the ship.

s=0.15

Problem A

The rope running over two fixed cylinders carries the 4-kg mass at one end. Determine the force P that must be applied to the other end to initiate motion. The coefficient of friction between the rope and the cylinders is 0.15. s=0.15 A

•

•

T T

s=0.15

4(9.81)

B

•

s=0.15

B

•

T

Cylinder A:

Cylinder B:

40o

T2  T1es T  4(9.81)e0.15A

T2  T1es

P  Te 0.15B

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10/10/2010

Determination of : 1.

2.

 •

Determination of A and B

•

3. 

•

•

 90o 

•

90o

40o

40o

•

T T

     180o  o   180 

     90o  o   180    2

  

40o





     90o    o   180 



A 



40o

4(9.81)



 

    A  90o  40o  o   180 

         o   180 

•

•



   B  40o  o   180 

130 180o

B 

40 180o



s=0.15

7.62 A

•

T T

•

s=0.15

4(9.81)

Cylinder A:

T2  T1e

B

Cylinder B:  s

T2  T1es

T  4(9.81)e T  4(9.81)e

0.15A

 130  0.15   180 

T  55.15N

P  Te

FIN.

0.15B

P  55.15e

 40  0.15   180 

P  61.24N

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