Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
7.5k views
in Physics by (58.2k points)

A pulley used to transmit power by means of ropes, has a diameter of 3.6m and has 15 groove of 45º angle. The angle of contact is 170º and the coefficient of friction between the ropes and the groove side is 0.28. The maximum possible tension in the ropes is 960N and the mass of the rope is 1.5Kg per m length. What is the speed of the pulley in rpm and the power transmitted if the condition of maximum power prevails?

1 Answer

+1 vote
by (66.1k points)
selected by
 
Best answer

Given data:

Dia. Of pulley(D) = 3.6m 

Number of groove(or ropes) = 15 

Angle of groove(2a) = 45°, α = 22.50º 

Angle of contact(θ) = 170° = 1700 X (Π/180°) = 2.97 rad 

Coefficient of friction(µ) = 0.28 

Max. Tension(Tmax) = 960N 

Mass of rope(m) = 1.5Kg per m length 

For maximum power: Tc = 1/3Tm 

= 1/3 X 960 = 320N ...(i)

Tm = T1 + TC

960 = T1 + 320 

T1 = 640N ...(ii)

Now Tc = (1/3)Tm = mV2

V = (Tm/3m)1/2

= [960/(3 X 1.5)]1/2

= 14.6m/sec ...(iiii)

Since V = πDN/60 = 14.6, 

N = 77.45R.P.M.

Now, Ratio of Tension in V-Belt:: T1/T2 = eµ.θ.cosec α 

640/T2 = e(0.28).(2.97).cosec22.5 

T2 = 73.08N ...(iv) 

Maximum power transmitted(P) = (T1–T2).v/1000 Kw 

P = [(640 – 73.08) X 14.6]/1000 KW 

P = 8.277KW 

Total maximum power transmitted = Power of one rope X No. of rope 

P = 8.277 X 15 = 124.16KW

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...