Belt friction lecture_new

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ApplicationsEngineering Mechanics: Belt Friction

Band brakes

Torque Transmission

Capstan(Use less smaller force to hold loads)

Pulley(Raising or lowering loads)

Theory• Relate T1 and T2 when belt is about to slide to right.

Engineering Mechanics: Belt Friction

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2 1T / T e s

Condition for impending slip for belt about to slide toward right (T2 > T1 )

T2 = Tension in the belt that pullsT1 =Tension in the belt that resist

Engineering Mechanics: Belt Friction Theory

Angle of contact shall be expressed in radianThe formula can be applied for

impending slip condition for problems involving:1. Flat belt passing over a fixed

cylindrical drum 2. Ropes/Belt wrapped around a

post or capstan or pulleys3. Belt drives, both pulley and

belt rotate 2 1T / T e k

If the belt or rope is actually slipping (relative slip between belt/rope and drum/pulley)

Problem-1A cord is wrapped twice around acapstan A and three times aroundsecond capstan B. Finally cord goesover a half barrel section and supportsa mass M of 500 kg. What is tension Trequired to maintain this load? Takecoefficient of friction 0.1 for allsurfaces of contact.


Maximum value of tension can befound out by upward impendingmotion of mass ‘M’ for whichMg<T.

Minimum value of tension can befound out by the impendingdownward motion of ‘M’ forwhich Mg>T .

Solution:

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Problem-1

For 181.15 <T<132811.04 N, the load will be held in place .


For impending downward motion of M ,

Mg > T2>T1> T

M=500 Kg, βA=4π, βB=6π, βpulley=0.5 πSOLUTION:

0.20s 0.15k

A cable passes around three 2-in.-radius pulleys and supports two blocks as shown. Two of the pulleys ( D and E) are locked to prevent rotation, while the third pulley is rotated slowly at a constant speed. Knowing that the coefficients of friction between the cable and the pulleys are

determine the largest weight which can be raised if pulley C is rotated clockwise .

Problem-2Engineering Mechanics: Belt Friction

Solution:Impending slip at C

Slips at D and E

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Note: 1sin 30 rad,2 6rr

2 ,2 6 3C D E

2 ,s CAW T e

1 2 1, 16 lbk D k ET T e T e

5 23 30.15 0.20

16 lb 16 lb 11.09 lbk E D s CAW e e e e

C is rotated clockwise slowly


Problem-2

Belt Drives

Driver Pulley

T2 >T1

Driven pulley

M1


Connected to electric motor (say). An external torque (M) is applied

Connected to a machine tool (say). Torque transferred to the pulley by the machine tool is M1

Torque TransmittedM1= (T2-T1)*0.2

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Problem-3A flat belt connects pulley A, which

drives the a machine tool, to pulley B,which is attached to shaft of an electricmotor. The coefficients of friction are s= 0.25 and k = 0.20 between bothpulleys and the belt. Knowing that themaximum allowable tension in the belt is600 N, determine the largest torquewhich can be exerted by the belt onpulley A.


Solution:Since slipping depends upon angle ofcontact and coefficient of friction ,pulley‘B’ will slip first.(Since it is having samecoefficient of friction but smaller angle ofcontact than pulley ‘A’).

For the maximum tension in one side ofbelt of pulley ‘B’, resistance tension inother side should be found out.

From moment about center of pulley ‘A’,maximum allowable torque can be foundout.

For no slip at ‘A’ , µs < (µs) max conditionshould be satisfied at pulley ‘A’.

Problem-3

.

Pulley-B


SOLUTION:

0 25

240 4 3

120 2 3

so

Ao

B

.

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Problem-3Here couple MA is applied to the pulley by the

machine tool to which it is attached and is equal

and opposite to the torque exerted by the belt.

Pulley-A


Problem-4

A differential band brake is used to controlthe speed of a drum which rotates at aconstant speed. Knowing that the coefficientof kinetic friction between the belt and thedrum is 0.30 and that a couple of magnitude125 lb-ft is applied to the drum, determinethe corresponding magnitude of the force Pthat is exerted on end D of the lever whenthe drum is rotating (a) clockwise, (b)counter clockwise.

Solution

• Relative slip between belt and drum

• Moment equilibrium of drum about center of drum will give tension generated in both side of the drum.

• Again moment equilibrium of lever about point B will give the force ‘P’


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Problem-4SOLUTION:


Problem-4

Lever
