Question.01A pulley of 150 mm effective diameter running at 1500 rev/min drives a follower of 750 mmdiameter, the two shafts being parallel, 1 m apart, and the free parts of the belt consideredstraight. The belt has a mass of 0.5 kg/m and the maximum tension is to be 800 N.If u= 0.35, estimate the maximum tension difference allowing for inertia of the belts.If the belt has a cross-sectional area of 350 mm? and E for the material is 320 MN/m?, estimatea) the speed of the driven pulley at the maximum condition, andb) the power transmitted to it.375 mm75mm1m Chapter 03: Belt DrivesSolutionReferring to Figure Q1Example C03.010-375 – 0-075cos- 0-3Question= 72° 33' = 1-266 radA pulley of 150 mm effective diameter running at 1500 rev/min drives a follower of 750 mmdiameter, the two shafts being parallel, 1 m apart, and the free parts of the belt consideredstraight. The belt has a mass of 0.4 kg/m and the maximum tension is to be 720 N... angle of lap on smaller pulley,0 = 2-532 radIf u= 0.4, estimate the maximum tension difference allowing for inertia of the belts.x 0-075 = 11-78 m/s60V= 1500 xIf the belt has a cross-sectional area of 320 mm? and E for the material is 300 MN/m?, estimatethe speed of the driven pulley at the maximum condition and the power transmitted to it.: Te= my?from equation (9(3.1)= 0-4 x 11-782 = 55-6 N一。公-”from equation (9 (3.2)720 - 55-6i.e.= *x26S2 = 2-754375 mmT, – 55-6-75mmT= 296-8 N:. T-T= 720– 296-8 = 423-2 Nfrom which=1-(T, - T,)from equation (9 **(3.10)1m423-2320 x 10- x 300 × 10°e 11:728 m/sFigure Q111-72860= 298-6 rov/min27. speed of follower=0-375Power transmitted = 423-2 x298-6 ×10-375 ) = 4960 W

ameter, the two shafts being parallel, 1 m apart, and the free parts of the belt considered raight. The belt has a mass of 0.5 kg/m and the maximum tension is to be 800 N. µ= 0.35, estimate the maximum tension difference allowing for inertia of the belts. the belt has a cross-sectional area of 350 mm² and E for the material is 320 MN/m², estimate a) the speed of the driven pulley at the maximum condition, and b) the power transmitted to it. 375 mm -75 mm 1m

ameter, the two shafts being parallel, 1 m apart, and the free parts of the belt considered raight. The belt has a mass of 0.5 kg/m and the maximum tension is to be 800 N. µ= 0.35, estimate the maximum tension difference allowing for inertia of the belts. the belt has a cross-sectional area of 350 mm² and E for the material is 320 MN/m², estimate a) the speed of the driven pulley at the maximum condition, and b) the power transmitted to it. 375 mm -75 mm 1m

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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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

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Chapter 03: Belt Drives
Solution
Referring to Figure Q1
Example C03.01
0-375 – 0-075
cos
- 0-3
Question
= 72° 33' = 1-266 rad
A pulley of 150 mm effective diameter running at 1500 rev/min drives a follower of 750 mm
diameter, the two shafts being parallel, 1 m apart, and the free parts of the belt considered
straight. The belt has a mass of 0.4 kg/m and the maximum tension is to be 720 N.
.. angle of lap on smaller pulley,
0 = 2-532 rad
If u= 0.4, estimate the maximum tension difference allowing for inertia of the belts.
x 0-075 = 11-78 m/s
60
V= 1500 x
If the belt has a cross-sectional area of 320 mm? and E for the material is 300 MN/m?, estimate
the speed of the driven pulley at the maximum condition and the power transmitted to it.
: Te= my?
from equation (9
(3.1)
= 0-4 x 11-782 = 55-6 N
一。
公-”
from equation (9 (3.2)
720 - 55-6
i.e.
= *x26S2 = 2-754
375 mm
T, – 55-6
-75mm
T= 296-8 N
:. T-T= 720– 296-8 = 423-2 N
from which
=1-
(T, - T,)
from equation (9 **
(3.10)
1m
423-2
320 x 10- x 300 × 10°
e 11:728 m/s
Figure Q1
11-728
60
= 298-6 rov/min
27
. speed of follower=
0-375
Power transmitted = 423-2 x
298-6 ×
10-375 ) = 4960 W

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