+1 vote
in Machine Dynamics by (34.3k points)
A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the governor effort.

(a) 44.7 N

(b) 57.4 N

(c) 88.4 N

(d) 53.8 N

This question was addressed to me in class test.

The origin of the question is Effort and Power of a Porter Governor in portion Governors of Machine Dynamics

1 Answer

0 votes
by (38.6k points)
Right answer is (b) 57.4 N

Easiest explanation: If  c is the percentage change in speed

cN1 = N2 – N1 = 31.4

c = 31.4/161 = 0.195

P = c(mg +Mg +F)

= 0.195 (5 × 9.81 + 25 × 9.81 + 10) = 57.4 N.

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