Lecture 25 ENGR-1100 Introduction to Engineering Analysis

Today’s Lecture Outline Belt Friction Flat belts V-belts

Belt Friction

Flat Belt  F r = 0 - T sin -(T +  T) sin =0 PP  P= 2T sin +  T sin (a)

(T +  T) cos - T cos -  F = 0  F  = 0  T cos =  F (b)

Assuming slip is impending  F =  s  P  Tcos(  /2)=  s  sin(  /2)+  s  sin(  /2) Combining equation (a) and (b) yields:  P= 2T sin +  T sin (a)  T cos =  F (b) cos (  /2) =  s T + Dividing by 

cos  1 1 dT/d  =  s T +O(  T) dT/T = =  s d  Rearranging dT/d  =  s T TT  dT d  ln ( T 2 / T 1 ) =  s (  2 -  1 ) =  s  Integrating Where:  =  2 -  1 Or T 2 = T 1 e  s  (c) T1 T2 

V-Belt T 2 = T 1 e (  s ) enh  Where (  s ) enh =[  s /sin(  /2)] is the enhanced coefficient of friction.

Example 9-85 AA rope attached to a 500 lb block passes over a frictionless pulley and is wrapped for one full turn around a fixed post as shown in Fig. P9-85. If the coefficient of friction between the rope and the post is 0.25, determine: T(a) The minimum force P that must be used to keep the block from falling. ((b) The minimum force P that must be used to begin to raise the block.

Solution A free-body diagram for the block:   F y = T - W = 0 T 2 = T 1 e  =T 1 e 0.25(2  ) = T 1 (a) During lowering of the block: T= 500 lb From equation (c): P min = T 1 = =  lb T 2 = 500 lb

(b) During raising of the block: P min = T 2 = T 1 = (500)= lb  2410 lb T 1 = 500 lb

Class Assignment: Exercise set P9-86 please submit to TA at the end of the lecture AA rope attached to a 220-kg block passes over a fixed drum as shown in Fig. P9-86. If the coefficient of friction between the rope and the drum is 0.30, determine (a) The minimum force P that must be used to keep the block from falling. (b) The minimum force P that must be used to begin to raise the block. a)P min = 1347 N b) P min = 3460 N

Solution   F y = T - mg = 0 T = mg =220 (9.807) = 2158 N A free-body diagram for the block: T 2 = T 1 e  =T 1 e 0.30(  /2) = T 1 From equation (c): (a) During lowering of the block:T 2 = 2158 N P min = T 1 = =  1347 N

(b) During raising of the block: T 1 = 2158 N P min = T 2 = T 1 = (2158)  3460 N

The band brake of Fig. P-90 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is 0.35 and the weight of the handle is 15 N. If a force of 200 N is applied to the end of the handle, determine the maximum torque for which no motion occurs if the torque is applied (a) Clockwise (b) Counterclockwise Class assignment: Exercise set P9-90 please submit to TA at the end of the lecture T max = N  m T max = 686 N  m

Solution  = (2  ) = rad  = 0.35 (3.665) = T 2 = T 1 e  =T 1 e = 3.607T 1 From equation (c): (a) From a free-body diagramof the handle  M A = 200(675) - 15(225) - T (75) = 0 T = N T

T 1 = =486.6 N T max = (T 2 - T 1 )R = ( ) (0.150) = N  m (a) From a free-body diagramof the drum when the rotation of the drum is clockwise: T 2 =T T1 T2 T

T 1 = N T 2 = T 1 = (1755.0) = N T max = (T 2 - T 1 )R = ( ) (0.150) = 686 N  m (b) From a free-body diagramof the drum when the rotation of the drum is counterclockwise: T2 T1 T T 1 =T

Class assignment: Exercise set P9-89 please submit to TA at the end of the lecture Solution: a min =7.53 in.

Class assignment: Exercise set 9-96 A 100-N crate is sitting on a 200-N crate as shown in the figure. The rope that join the crates passes around a frictionless pulley and a fixed drum. The coefficient of friction is 0.3 at all surfaces. Determine the minimum force P that must be applied to the 200-N crate to start motion. P min =159 N