+1 vote
in Machine Dynamics by (34.3k points)
The effort of a Porter governor is equal to

(a) c(m – M)g

(b) c(m + M)g

(c) C/(m + M)g

(d) g

The question was posed to me during a job interview.

I would like to ask this question from Hartnell Governor topic in chapter Governors of Machine Dynamics

1 Answer

0 votes
by (38.6k points)
Correct answer is (b) c(m + M)g

Easy explanation: The effort of a Porter governor is equal to c(m + M)g.

The power of a Porter governor is equal to 4c^2/1 + 2c (m + M)gh.

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