Lesson 26.

Q1:- The coefficient of friction between belt and pulley is 0.39; angle of lap is 150°. The belt runs over pulley of diameter 500 mm at 150 rpm. Calculate power transmitted if maximum tension in the belt is 3000 N.

Sol:-

D = 500mm = 0.5m

N = 150rpm

µ= 0.39

L 26 eq.1

We know that velocity of Belt

L 26 eq.2

Power = (T1 – T2) V= (3000- 1367.3)3.925=6408.3 W = 6.40 KW

Q2:- The belt in 90mm wide and 10mm thick having density of1000kg/m3. The coefficient of friction between belt and pulley is 0.3. The maximum stress in the belt is 5mPa, Calculate maximum power the belt can transmit, if angle of lap is 120°.

Sol:- We know that maximum tension in Belt

T = t.b.t. = 5 X 106 X 0.09 X .01

=4500N

Mass of belt per meter length

L 26 eq.3

L 26 eq.4

L 26 eq.5

L 26 eq.6

L 26 eq.7

Last modified: Wednesday, 26 March 2014, 11:35 AM