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
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Chapter7: Dry Friction
Section: Chapter Questions
Problem 7.70P: Solve Sample Problem 7.16 if the contact pressure under the polisher varies parabolically from p0 at...
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![Question.01
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 0.5 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?, estimate
a) the speed of the driven pulley at the maximum condition, and
b) the power transmitted to it.
375 mm
75mm
1m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F953dbc5e-185b-4b91-9ca0-40c4eb259a44%2F6ec86e00-6a96-491a-9fa5-454fddf6c860%2Fc8mxlws_processed.png&w=3840&q=75)
Transcribed Image Text:Question.01
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 0.5 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?, estimate
a) the speed of the driven pulley at the maximum condition, and
b) the power transmitted to it.
375 mm
75mm
1m
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F953dbc5e-185b-4b91-9ca0-40c4eb259a44%2F6ec86e00-6a96-491a-9fa5-454fddf6c860%2Fcmjnrklq_processed.png&w=3840&q=75)
Transcribed Image Text: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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