Three belts move on two pulleys without slipping in the speed reduction systemshown in the figure. At the instant shown, the velocity of point A on the input bandis 5.5 ft/s to the right, decreasing at a rate of 9.9 ft/s2.3 to the lightsT4 in.↓8 in.BC(a) Determine for that instant the velocity of point C, in ft/s, on the exit band.b) Determine for that instant the acceleration of point C, in ft/s2, in theoutput band.(c) Determine for that instant the acceleration of point B, in ft/s2, at theoutput pulley.

Given Data: The velocity of point A VA=5.5 fts The acceleration of point A aA=9.9 fts2 The inner…

Answered: Three belts move on two pulleys without…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Three belts move on two pulleys without slipping in the speed reduction system
shown in the figure. At the instant shown, the velocity of point A on the input band
is 5.5 ft/s to the right, decreasing at a rate of 9.9 ft/s2.
3 to the lights
T
4 in.
↓
8 in.
B
C
(a) Determine for that instant the velocity of point C, in ft/s, on the exit band.
b) Determine for that instant the acceleration of point C, in ft/s2, in the
output band.
(c) Determine for that instant the acceleration of point B, in ft/s2, at the
output pulley.

Expert Solution

Step 1

Given Data:

The velocity of point A $V_{A} = 5.5 \frac{ft}{s}$

The acceleration of point A $a_{A} = 9.9 \frac{ft}{s^{2}}$

The inner radius of the left pulley $r_{1 L} = 2 i n = 2 i n \times \frac{1 f t}{12 i n} = 0.1667 ft$

The outer radius of the left pulley $r_{2 L} = 4 i n = 4 i n \times \frac{1 f t}{12 i n} = 0.333 ft$

The inner radius of the left pulley $r_{3 R} = 2 i n = 2 i n \times \frac{1 f t}{12 i n} = 0.1667 ft$

The outer radius of the left pulley $r_{4 R} = 4 i n = 4 i n \times \frac{1 f t}{12 i n} = 0.333 ft$

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