0 votes
in Physics by (58.6k points)

A belt drives a pulley of 200mm diameter such that the ratio of tensions in the tight side and slack side is 1.2. If the maximum tension in the belt is not to exceed 240KN. Find the safe power transmitted by the pulley at a speed of 60rpm.

1 Answer

+1 vote
by (66.3k points)
selected by
Best answer

Given that, 

D1 = Diameter of the driver = 200mm = 0.2m

T1/T2 = 1.2

Since between T1 and T2, T1 is always greater than T2,

Hence T1 = 240KN

N1 = Speed of the driver in R.P.M. = 60PRM

P = ?

We know that

T1/T2 = 1.2

T2 = T1/1.2 =240/1.2 = 200KN

 V = Velocity of the belt in m/sec

= πDN/60 m/sec, D is in meter and N is in RPM

= (3.14 X 0.2 X 60)/60 = 0.628 m/sec

P = (T1 - T2) X V

P = (240 - 200) X 0.628

P = 25.13KW

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.