Question
Download Solution PDFIf T and m represent the maximum tension and mass per unit length of a belt, then the maximum permissible speed of the belt is given by
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Power transmitted by a belt:
P = (T1 – T2)v
For maximum power:
T1 = T/3
where T1 = Tension in the tight side, T = Maximum tension to which the belt is subjected.
The velocity (maximum) of the belt for maximum power:
\(v = \sqrt {\frac{T}{{3m}}} \)
where m is mass of belt per unit length.
m = Area × length × density = b.t.l.ρ
Important Points
Centrifugal Tension (Tc):
Since the belt continuously runs over the pulleys, therefore some centrifugal force is caused, whose effect is to increase the tension on both the tight as well as the slack sides.
The tension caused by this centrifugal force is called centrifugal tension.
Condition for Maximum power transmitted by the belt is T = 3TC
Last updated on Jul 2, 2025
-> ESE Mains 2025 exam date has been released. As per the schedule, UPSC IES Mains exam 2025 will be conducted on August 10.
-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.