Figure Q2 shows a Porter governor for which the speed range can be varied by means of the auxiliary spring, S. The spring force is transmitted to the sleeve by the arm AB which is pivoted at A. The two balls each of mass 0.4 kg are supported by links C1, C, C; and C4, each 90 mm in length. The sleeve carries a mass of 1.2 kg.

Figure Q2 shows a Porter governor for which the speed range can be varied by means of the auxiliary spring, S. The spring force is transmitted to the sleeve by the arm AB which is pivoted at A. The two balls each of mass 0.4 kg are supported by links C1, C, C; and C4, each 90 mm in length. The sleeve carries a mass of 1.2 kg.

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter7: Dry Friction

Section: Chapter Questions

Problem 7.5P: The contact surface between the 36-lb block and 20-lb homogenous cylinder is frictionless. Can the...

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Question

Chapter 04: Governors
If the force exerted by the spring S is P N,
then
force on sleeve = P N
Therefore total load on sleeve =P+ 0-9 x 9-81 = P+ 8-83 N
Example C04.01
The lower link C, is in equilibrium under the centrifugal force on the
ball, F, the weight of the ball, 3-532 N, half the load on the sleeve,
P+ 4:415 N, the tension in the upper link C, and the horizontal reaction
on the sleeve.
Question
Figure Q4.1(a) shows a Porter governor for which the speed range can be varied by
means of the auxiliary spring S. The spring force is transmitted to the sleeve by the arm
AB which is pivoted at A. The two balls each of mass 0.36 kg are supported by links Ci,
C2. C; and C4, each 90 mm in length. The sleeve carries a mass of 0.9 kg.
Taking moments about I, Figure Q4.1(b),
Fx XZ = 3-532 x IZ + (P+ 4-415) x IY
30mm
i.e.
0.36(Ф)? г.h - (3.6 х 9.81) (г-0.03) + (15.Р+ 4.415) х 2(г - 0.03)
90mm/C,
90mm
0-36( 27)
Xr xh= 3-532(r – 0-03) + (}P+ 4-415) x 2(r – 0-03)
E X,
i.e.
0-00395Nrh = (r – 0-03)(12-36 + 0•4P)
(1)
When N= 200 rev/min,
r= 0-075 m
30mm
C2
30mm
and
A h= V(0-092 – 0-045°) = 0-078 m
C4
3:532N
Therefore, from equation (1), P,= 20-45 N
When N= 220 rev/min,
h= 0-078 – 0-005 = 0-073 m
7.
and
1 = v(0-092 – 0-073²) + 0-03
V+4:415N
(ь)
k60mm 90mm
= 0-0826 m
: P= 23-9 N
Р, — Р,
sleeve movement
(a)
.. spring stiffnes:
Figure Q4.1
23-9 - 20-45
= 863 N/m
{x 0-01
The sleeve begins to rise when the balls revolve at 200 rev/min in a circle of 75 mm
radius. The speed of the governor is not to exceed 220 rev/min when the sleeve has risen
10 mm from its original position.
(b)
20-45
Vertical reaction at A at 200 rev/min ==
Determine
+ 4-415 + 3-532
(a) The necessary stiffness of the spring S, and
(b) The tension in the link C1 when the sleeve begins to rise.
= 21-04 N
0-09
: tengion in C, - 21-04 sec o = 21-04 x
0-078
Solution
= 13-9 N
(a)

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