# Chapter3-Friction¶

## Ex1-pg102¶

In [1]:
##CHAPTER 3 ILLUSRTATION 1 PAGE NO 102
##TITLE:FRICTION
##FIRURE 3.16(a),3.16(b)
import math
##===========================================================================================
##INPUT DATA
P1=180.##                        PULL APPLIED TO THE BODY IN NEWTONS
theta=30.##                      ANGLE AT WHICH P IS ACTING IN DEGREES
P2=220.##                        PUSH APPLIED TO THE BODY IN NEWTONS
##Rn=                           NORMAL REACTION
##F=                            FORCE OF FRICTION IN NEWTONS
##U=                            COEFFICIENT OF FRICTION
##W=                            WEIGHT OF THE BODY IN NEWTON
##==========================================================================================
##CALCULATION
F1=P1*math.cos(theta/57.3)##              RESOLVING FORCES HORIZONTALLY FROM 3.16(a)
F2=P2*math.cos(theta/57.3)##              RESOLVING FORCES HORIZONTALLY FROM 3.16(b)
##                               RESOLVING FORCES VERTICALLY  Rn1=W-P1*sind(theta) from 3.16(a)
##                               RESOLVING FORCES VERTICALLY  Rn2=W+P1*sind(theta) from 3.16(b)
##                               USING THE RELATION F1=U*Rn1    &     F2=U*Rn2  AND SOLVING FOR W BY DIVIDING THESE TWO EQUATIONS
X=F1/F2##                        THIS IS THE VALUE OF   Rn1/Rn2
Y1=P1*math.sin(theta/57.3)
Y2=P2*math.sin(theta/57.3)
W=(Y2*X+Y1)/(1-X)##                BY SOLVING ABOVE 3 EQUATIONS
U=F1/(W-P1*math.sin(theta/57.3))##          COEFFICIENT OF FRICTION
##=============================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s '%('WEIGHT OF THE BODY =',W,'N''THE COEFFICIENT OF FRICTION =',U,'')

WEIGHT OF THE BODY = 989.9 NTHE COEFFICIENT OF FRICTION = 0.2


## Ex2-pg103¶

In [2]:
##CHAPTER 3 ILLUSRTATION 2 PAGE NO 103
##TITLE:FRICTION
##FIRURE 3.17
import math
##===========================================================================================
##INPUT DATA
THETA=45##                ANGLE OF INCLINATION IN DEGREES
g=9.81##                   ACCELERATION DUE TO GRAVITY IN N/mm**2
U=.1##                     COEFFICIENT FRICTION
##Rn=NORMAL REACTION
##M=MASS IN NEWTONS
##f=ACCELERATION OF THE BODY
u=0.##                      INITIAL VELOCITY
V=10.##                     FINAL VELOCITY IN m/s**2
##===========================================================================================
##CALCULATION
##CONSIDER THE EQUILIBRIUM OF FORCES PERPENDICULAR TO THE PLANE
##Rn=Mgcos(THETA)
##CONSIDER THE EQUILIBRIUM OF FORCES ALONG THE PLANE
##Mgsin(THETA)-U*Rn=M*f.............BY SOLVING THESE 2 EQUATIONS
f=g*math.sin(THETA/57.3)-U*g*math.cos(THETA/57.3)
s=(V**2-u**2)/(2*f)##                  DISTANCE ALONG THE PLANE IN metres
##==============================================================================================
##OUTPUT
print'%s %.1f %s'%('DISTANCE ALONG THE INCLINED PLANE=',s,' m')

DISTANCE ALONG THE INCLINED PLANE= 8.0  m


## Ex3-pg104¶

In [3]:
##CHAPTER 3 ILLUSRTATION 3 PAGE NO 104
##TITLE:FRICTION
##FIRURE 3.18
import math
##===========================================================================================
##INPUT DATA
W=500.##                  WEGHT IN NEWTONS
THETA=30.##               ANGLE OF INCLINATION IN DEGRESS
U=0.2##                  COEFFICIENT FRICTION
S=15.##                   DISTANCE IN metres
##============================================================================================
Rn=W*math.cos(THETA/57.3)##       NORMAL REACTION IN NEWTONS
P=W*math.sin(THETA/57.3)+U*Rn##   PUSHING FORCE ALONG THE DIRECTION OF MOTION
w=P*S
##============================================================================================
##OUTPUT
print'%s %.1f %s'%('WORK DONE BY THE FORCE=',w,' N-m')

WORK DONE BY THE FORCE= 5048.8  N-m


## Ex4-pg104¶

In [4]:
##CHAPTER 3 ILLUSRTATION 4 PAGE NO 104
##TITLE:FRICTION
##FIRURE 3.19(a)  &  3.19(b)
import math
##===========================================================================================
##INPUT DATA
P1=2000.##           FORCE ACTING UPWARDS WHEN ANGLE=15 degrees IN NEWTONS
P2=2300.##           FORCE ACTING UPWARDS WHEN ANGLE=20 degrees IN NEWTONS
THETA1=15.##         ANGLE OF INCLINATION IN 3.19(a)
THETA2=20.##         ANGLE OF INCLINATION IN 3.19(b)
##F1=               FORCE OF FRICTION IN 3.19(a)
##Rn1=              NORMAL REACTION IN 3.19(a)
##F2=               FORCE OF FRICTION IN 3.19(b)
##Rn2=              NORMAL REACTION IN 3.19(b)
##U=                 COEFFICIENT OF FRICTION
##===========================================================================================
##CALCULATION
##P1=F1+Rn1             RESOLVING THE FORCES ALONG THE PLANE
##Rn1=W*cosd(THETA1)....NORMAL REACTION IN 3.19(a)
##F1=U*Rn1
##BY SOLVING ABOVE EQUATIONS P1=W(U*cosd(THETA1)+sind(THETA1))---------------------1
##P2=F2+Rn2             RESOLVING THE FORCES PERPENDICULAR TO THE PLANE
##Rn2=W*cosd(THETA2)....NORMAL REACTION IN 3.19(b)
##F2=U*Rn2
##BY SOLVING ABOVE EQUATIONS P2=W(U*cosd(THETA2)+sind(THETA2))----------------------2
##BY SOLVING EQUATIONS 1 AND 2
X=P2/P1
U=(math.sin(THETA2/57.3)-(X*math.sin(THETA1/57.3)))/((X*math.cos(THETA1/57.3)-math.cos(THETA2/57.3)))##        COEFFICIENT OF FRICTION
W=P1/(U*math.cos(THETA1/57.3)+math.sin(THETA1/57.3))
##=============================================================================================
##OUTPUT
##print'%s %.1f %s'%('%f',X)
print'%s %.1f %s  %.1f %s '%('COEFFICIENT OF FRICTION=',U,'' 'WEIGHT OF THE BODY=',W,' N')

COEFFICIENT OF FRICTION= 0.3 WEIGHT OF THE BODY=  3927.0  N


## Ex5-pg105¶

In [5]:
##CHAPTER 3 ILLUSRTATION 5 PAGE NO 105
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=5.##                     DIAMETER OF SCREW JACK IN cm
p=1.25##                  PITCH IN cm
l=50.##                    LENGTH IN cm
U=.1##                    COEFFICIENT OF FRICTION
PI=3.147
##=============================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d)/57.3))
PY=math.atan(U/57.3)
P=W*math.tan((ALPHA+PY)*57.)
P1=P*d/(2.*l)
##=============================================================================================
##OUTPUT
print'%s %.1f %s '%('THE AMOUNT OF EFFORT NEED TO APPLY =',P1,' N')

THE AMOUNT OF EFFORT NEED TO APPLY = 180.4  N


## Ex6-pg106¶

In [6]:
##CHAPTER 3 ILLUSRTATION 6 PAGE NO 106
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=50.##                 DIAMETER OF SCREW IN mm
p=12.5##               PITCH IN mm
U=0.13##               COEFFICIENT OF FRICTION
PI=3.147
##===========================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d))/57.3)
PY=math.atan(U/57.3)
P=W*math.tan((ALPHA+PY)/57.3)##         FORCE REQUIRED TO RAISE THE LOAD IN N
T1=P*d/2.##                   TORQUE REQUIRED IN Nm
P1=W*math.tan((PY-ALPHA)/57.3)##        FORCE REQUIRED TO LOWER THE SCREW IN N
T2=P1*d/2.##                  TORQUE IN N
X=T1/T2##                     RATIOS REQUIRED
n=math.tan((ALPHA/(ALPHA+PY))/57.3)##    EFFICIENCY
##============================================================================================
print'%s %.1f %s'%('RATIO OF THE TORQUE REQUIRED TO RAISE THE LOAD,TO THE TORQUE REQUIRED TO LOWER THE LOAD =',X,'')

RATIO OF THE TORQUE REQUIRED TO RAISE THE LOAD,TO THE TORQUE REQUIRED TO LOWER THE LOAD = 4.1


## Ex7-pg107¶

In [7]:
##CHAPTER 3 ILLUSRTATION 7 PAGE NO 107
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=39.##                 DIAMETER OF THREAD IN mm
p=13.##                 PITCH IN mm
U=0.1##                COEFFICIENT OF FRICTION
PI=3.147
##===========================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d))/57.3)
PY=math.atan(U/57.3)
P=W*math.tan((ALPHA+PY)*57.3)##         FORCE IN N
T1=P*d/2.##                   TORQUE REQUIRED IN Nm
T=2.*T1##                      TORQUE REQUIRED ON THE COUPLING ROD IN Nm
K=2.*p##                      DISTANCE TRAVELLED FOR ONE REVOLUTION
N=20.8/K##                   NO OF REVOLUTIONS REQUIRED
w=2.*PI*N*T/100.##                 WORKDONE BY TORQUE
w1=w*(7500.-2500.)/2500.##      WORKDONE TO INCREASE THE LOAD FROM 2500N TO 7500N
n=math.tan(ALPHA/57.3)/math.tan((ALPHA+PY)/57.3)##    EFFICIENCY
##============================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s %.1f %s '%('workdone against a steady load of 2500N=',w,' N' 'workdone if the load is increased from 2500N to 7500N=',w1,' N' 'efficiency=',n,'')

workdone against a steady load of 2500N= 1025.5  Nworkdone if the load is increased from 2500N to 7500N= 2050.9  Nefficiency= 0.5


## Ex8-pg107¶

In [8]:
##CHAPTER 3 ILLUSRTATION 8 PAGE NO 107
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
W=50000.##                        WEIGHT OF THE SLUICE GATE IN NEWTON
P=40000.##                        POWER IN WATTS
N=580.##                          MAX MOTOR RUNNING SPEEED IN rpm
d=12.5##                         DIAMETER OF THE SCREW IN cm
p=2.5##                          PITCH IN cm
PI=3.147
U1=.08##                          COEFFICIENT OF FRICTION for SCREW
U2=.1##                           C.O.F BETWEEN GATES AND SCREW
Np=2000000.##                     NORMAL PRESSURE IN NEWTON
Fl=.15##                       FRICTION LOSS
n=1.-Fl##                       EFFICIENCY
ng=80.##                        NO OF TEETH ON GEAR
##===========================================================================================
##CALCULATION
TV=W+U2*Np##                     TOTAL VERTICAL HEAD IN NEWTON
ALPHA=math.atan((p/(PI*d))/57.3)##
PY=math.atan(U1/57.3)##
P1=TV*math.tan((ALPHA+PY)*57.3)##            FORCE IN N
T=P1*d/2./100.##                    TORQUE IN N-m
Ng=60000.*n*P*10**-3./(2.*PI*T)##              SPEED OF GEAR IN rpm
np=Ng*ng/N##                       NO OF TEETH ON PINION
##=========================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s '%('NO OF TEETH ON PINION =',np,' say ',np+1,'')

NO OF TEETH ON PINION = 19.8  say  20.8


## Ex9-pg108¶

In [9]:
##CHAPTER 3 ILLUSRTATION 9 PAGE NO 108
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=5.##                         MEAN DIAMETER OF SCREW IN cm
p=1.25##                      PITCH IN cm
dc=6.##                        MEAN DIAMETER OF COLLAR IN cm
U=.15##                       COEFFICIENT OF FRICTION OF SCREW
Uc=.18##                      COEFFICIENT OF FRICTION OF COLLAR
P1=100.##                      TANGENTIAL FORCE APPLIED IN NEWTON
PI=3.147
##============================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d))/57.3)##
PY=math.atan(U/57.3)##
T1=W*d/2*math.tan((ALPHA+PY)/100)*57.3##         TORQUE ON SCREW IN NEWTON
Tc=Uc*W*dc/2./100.##                      TORQUE ON COLLAR IN NEWTON
T=T1+Tc##                          TOTAL TORQUE
D=2.*T/P1/2.*100.##                       DIAMETER OF HAND WHEEL IN cm
##============================================================================================
##OUTPUT
print'%s %.1f %s'%('SUITABLE DIAMETER OF HAND WHEEL =',D,' cm')

SUITABLE DIAMETER OF HAND WHEEL = 111.4  cm


## Ex10-pg108¶

In [10]:
##CHAPTER 3 ILLUSRTATION 10 PAGE NO 108
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
PI=3.147
d=2.5##                    MEAN DIA OF BOLT IN cm
p=.6##                     PITCH IN cm
beeta=55/2.##               VEE ANGLE
dc=4.##                     DIA OF COLLAR IN cm
U=.1##                       COEFFICIENT OF FRICTION OF BOLT
Uc=.18##                      COEFFICIENT OF FRICTION OF COLLAR
W=6500.##                     LOAD ON BOLT IN NEWTONS
L=38.##                       LENGTH OF SPANNER
##=============================================================================================
##CALCULATION
##LET X=tan(py)/tan(beeta)
##y=tan(ALPHA)*X
PY=math.atan(U)*57.3
ALPHA=math.atan((p/(PI*d)))*57.3
X=math.tan(PY/57.3)/math.cos(beeta/57.3)
Y=math.tan(ALPHA/57.3)
T1=W*d/2.*10**-2*(X+Y)/(1.-(X*Y))##             TORQUE IN SCREW IN N-m
Tc=Uc*W*dc/2.*10**-2##                         TORQUE ON BEARING SERVICES IN N-m
T=T1+Tc##                                     TOTAL TORQUE
P1=T/L*100.##                                      FORCE REQUIRED BY @ THE END OF SPANNER
##=============================================================================================
##OUTPUT
print'%s %.1f %s'%('FORCE REQUIRED @ THE END OF SPANNER=',P1,' N')

FORCE REQUIRED @ THE END OF SPANNER= 102.3  N


## Ex11-pg109¶

In [11]:
##CHAPTER 3 ILLUSRTATION 11 PAGE NO 109
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d1=15.##                                 DIAMETER OF VERTICAL SHAFT IN cm
N=100.##                                 SPEED OF THE MOTOR rpm
U=.05##                                  COEFFICIENT OF FRICTION
PI=3.147
##==================================================================================
T=2./3.*U*W*d1/2.##                         FRICTIONAL TORQUE IN N-m
PL=2.*PI*N*T/100./60.##                         POWER LOST IN FRICTION IN WATTS
##==================================================================================
##OUTPUT
print'%s %.1f %s'%('POWER LOST IN FRICTION=',PL,' watts')

POWER LOST IN FRICTION= 524.5  watts


## Ex12-pg109¶

In [12]:
##CHAPTER 3 ILLUSRTATION 12 PAGE NO 109
##TITLE:FRICTION
import math
##===================================================================================
##INPUT DATA
PI=3.147
d2=.30##                             DIAMETER OF SHAFT IN m
N=75.##                              SPEED IN rpm
U=.05##                             COEFFICIENT OF FRICTION
p=300000.##                          PRESSURE AVAILABLE IN N/m**2
P=16200.##                           POWER LOST DUE TO FRICTION IN WATTS
##====================================================================================
##CaLCULATION
T=P*60./2./PI/N##                      TORQUE INDUCED IN THE SHFT IN N-m
##LET X=(r1**3-r2**3)/(r1**2-r2**2)
X=(3./2.*T/U/W)
r2=.15##                             SINCE d2=.30 m
c=r2**2.-(X*r2)
b= r2-X
a= 1.
r1=( -b+ math.sqrt (b**2. -4.*a*c ))/(2.* a);##     VALUE OF r1 IN m
d1=2*r1*100.##                               d1 IN cm
n=W/(PI*p*(r1**2.-r2**2.))
##================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s %.1f %s'%('EXTERNAL DIAMETER OF SHAFT =',d1,' cm''NO OF COLLARS REQUIRED =',n,'' '0 or ',n+1,'')

EXTERNAL DIAMETER OF SHAFT = 50.6  cmNO OF COLLARS REQUIRED = 5.1 0 or  6.1


## Ex13-pg111¶

In [13]:
##CHAPTER 3 ILLUSRTATION 13 PAGE NO 111
##TITLE:FRICTION
import math
##===================================================================================
##INPUT DATA
PI=3.147
ALPHA=120./2.##                               CONE ANGLE IN DEGREES
p=350000.##                                INTENSITY OF PRESSURE
U=.06
N=120.##                                    SPEED OF THE SHAFT IN rpm
##d1=3d2
##r1=3r2
##===================================================================================
##CALCULATION
##LET K=d1/d2
k=3.
Z=W/((k**2.-1.)*PI*p)
r1=3.*r2
T=2.*U*W*(r1**3.-r2**3.)/(3.*math.sin(60/57.3)*(r1**2.-r2**2.))##     total frictional torque in N
P=2.*PI*N*T/60000.##                            power absorbed in friction in kW
##================================================================================
print'%s %.1f %s %.1f %s %.1f %s'%('THE INTERNAL DIAMETER OF SHAFT =',r2*100,' cm' 'THE EXTERNAL DIAMETER OF SHAFT =',r1*100,' cm' 'POWER ABSORBED IN FRICTION =',P,' kW')

THE INTERNAL DIAMETER OF SHAFT = 4.8  cmTHE EXTERNAL DIAMETER OF SHAFT = 14.3  cmPOWER ABSORBED IN FRICTION = 1.8  kW


## Ex14-pg111¶

In [14]:
##CHAPTER 3 ILLUSRTATION 14 PAGE NO 111
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
PI=3.147
P=10000.##                               POWER TRRANSMITTED BY CLUTCH IN WATTS
N=3000.##                                SPEED IN rpm
p=.09##                                 AXIAL PRESSURE IN N/mm**2
##d1=1.4d2                              RELATION BETWEEN DIAMETERS
K=1.4##                                 D1/D2
n=2.
U=.3##                                  COEFFICIENT OF FRICTION
##==========================================================================================
T=P*60000./1000./(2.*PI*N)##                     ASSUMING UNIFORM WEAR            TORQUE IN N-m

##===========================================================================================
print'%s %.1f %s %.1f %s '%('THE INTERNAL RADIUS =',r2*100,' cm' 'THE EXTERNAL RADIUS =',K*r2*100,' cm')


THE INTERNAL RADIUS = 5.8  cmTHE EXTERNAL RADIUS = 8.1  cm


## Ex15-pg111¶

In [15]:
##CHAPTER 3 ILLUSRTATION 14 PAGE NO 111
##TITLE:FRICTION

##===========================================================================================
##INPUT DATA
PI=3.147
n1=3.##                          NO OF DICS ON DRIVING SHAFTS
n2=2.##                          NO OF DICS ON DRIVEN SHAFTS
d1=30.##                         DIAMETER OF DRIVING SHAFT IN cm
d2=15.##                         DIAMETER OF DRIVEN SHAFT IN cm
r1=d1/2.
r2=d2/2.
U=.3##                          COEFFICIENT FRICTION
P=30000.##                       TANSMITTING POWER IN WATTS
N=1800.##                        SPEED IN rpm
##===========================================================================================
##CALCULATION
n=n1+n2-1.##                     NO OF PAIRS OF CONTACT SURFACES
T=P*60000./(2.*PI*N)##            TORQUE IN N-m

MAX AXIAL INTENSITY OF PRESSURE = 0.033  N/mm**2