+1 vote
in Farm Machinery by (38.9k points)
A multiple disc clutch, steel on bronze is to transmit 6 HP at 750 rpm. The inner radius of contact is 4cm and outer radius of contact is 7 cm. Coefficient of friction is 0.1 and the average allowable pressure is 3.5 kgf/cm^2. Find the actual axial force required.

(a) 2335.09 N

(b) 2590.90 N

(c) 1999.76 N

(d) 2143.65 N

I had been asked this question in semester exam.

Asked question is from Farm Tractor in chapter Farm Tractor of Farm Machinery

1 Answer

+2 votes
by (243k points)
selected by
 
Best answer
Right option is (b) 2590.90 N

The explanation is: P = \(\frac{2πNT}{60} \)

T = \(\frac{6*746*60}{6.28*750}\)=57 Nm

T = ημWReff

ηW = 2*5181.81

3.5 * 9.8*10^4 = \(\frac{W}{π[49-16]*10^{-4}} \)

W = 3554.16

N = 2.91

Actual axial force

T = ημWactReff

Wact = \(\frac{57*2}{4*0.1*0.1}\) = 2590.90 N

Related questions

We welcome you to Carrieradda QnA with open heart. Our small community of enthusiastic learners are very helpful and supportive. Here on this platform you can ask questions and receive answers from other members of the community. We also monitor posted questions and answers periodically to maintain the quality and integrity of the platform. Hope you will join our beautiful community
...