# Example 5.1¶

In [1]:
import math
Wa=1000.0 #weight of block a
Wb=2000.0  #weight of block b
uab=1.0/4.0   #coefficient of friction between A and B
ubg=1.0/3.0    #coefficient of friction between ground and B
#When P is horizontal
#considering  equilibrium of block A
N1=Wa         #Normal Reaction on block A from block B
F1=uab*N1     #limiting Friction between A and B
T=F1         #tension
#considering  equilibrium of block B
N2=N1+ Wb         #Normal Reaction on block B from Ground
F2=ubg*N2     #limiting Friction between A and ground
P=F1+F2
print "P=",P,"N"
#When P is inclined at angle o
o=30.0*3.14/180.0
#considering  equilibrium of block A
N1=Wa         #Normal Reaction on block A from block B
F1=uab*N1     #limiting Friction between A and B
T=F1         #tension
#considering  equilibrium of block B
#from
#N2+Psin30=N1+Wb
#Pcos30=F1+F2
#F1=ubg*N2
N2=(N1+Wb-F1*math.tan(o))/(1+ubg*math.tan(o))
P=(N1+Wb-N2)/math.sin(o)
print "P=",P,"N"

P= 1250.0 N
P= 1210.36288071 N


# Example 5.2¶

In [2]:
import math
Wa=300.0 #weight of upper block
Wb=900.0  #weight of lower block
u1=1.0/3.0   #coefficient of friction between upper block and lower block
u2=1.0/3.0    #coefficient of friction between ground and  lower block
#using
#N1=Wacoso          Normal Reaction
#F1=u1*N1           Friction
#N2=Wbcoso+N1
#F2=u2*N2
o=math.atan((u1*Wa+u2*Wb+u2*Wa)/Wb)*180/3.14
print o,"°"

29.0693410161 °


# Example 5.3¶

In [3]:
import math
W=500.0    #weight of block
F1=200.0     #force up the inclined plane when block is moving down
F2=300.0     #force up the inclined plane when block is at rest
#When block starts moving down the plane
#sum of all forces perpendicular to the plane = 0
#N =Wcoso
#sum of all forces parallel to the plane = 0
#Fr+F1=Wsino
#sino-ucoso=F1/w    1
#When block starts moving up the plane
#sum of all forces perpendicular to the plane = 0
#N =Wcoso
#sum of all forces parallel to the plane = 0
#Wsino+Wucoso=F2
#using these equations
o=math.asin((F1*0.5/W)+(F2*0.5/W))   #angle of inclination
print "Angle of inclination is ",(o*180/3.14)
#using 1
u=math.sin(o)-F1/W
print "coefficient of friction is",round(u,3)

Angle of inclination is  30.0152164356
coefficient of friction is 0.1


# Example 5.4¶

In [4]:
import math
uag=0.5   #coefficient of friction between block A and the plane
ubg=0.2   #coefficient of friction between block B and the plane
Wb=500.0  #weight of block B
Wa=1000.0  #weight of block A
#Considering equilibrium of block A,
#sum of all forces along the plane is 0
#N1=Wacoso ,Fr=uagN1
#sum of all forces perpendicaular to  the plane is 0
#T=uagWacoso-Wasino
#Considering equilibrium of block A,
#sum of all forces along the plane is 0
#N2=Wbcoso ,Fr=uagN2
#sum of all forces perpendicaular to  the plane is 0
#T=Wbsino-ubgwbsino
o=math.atan((uag*Wa+ubg*Wb)/(Wa+Wb))*180.0/3.14
print "Angle of Inclination",o;

Angle of Inclination 21.8124674778


# Example 5.5¶

In [5]:
import math
Wl=750.0  #weight of lower block
Wu=500.0  #weight of upper block
o1=60.0*3.14/180.0   #angle of inclined plane
o2=30.0 *3.14/180.0 # anlge at which pull is applied
u=0.2     #coefficient of friction
#for 750 N block
#Σ Forces normal to the plane = 0
N1=Wl*math.cos(o1)
F1=u*N1
#Σ Forces parallel to the plane = 0
T=F1+Wl*math.sin(o1)
#Σ Forces horizontal to the plane = 0
P=(T+u*Wu)/(math.cos(o2)+u*math.sin(o2))
print P,"N"

853.305553493 N


# Example 5.6¶

In [6]:
import math
o1=60.0*3.14/180.0    #angle of inclination of plane AC
o2=30.0*3.14/180.0    #angle of inclination of plane BC
Wbc=1000.0  #weight of block on plane BC
ubc=0.28    #coefficient of friction between the load and the plane BC
uac=0.20   #coefficient of friction between the load and the plane AC
#for least weight
N1=Wbc*math.cos(o2)                            #Normal Reaction
F1=ubc*N1                                       #frictional Force
T=Wbc*math.sin(o2)-F1                          #Tension
#for block on plane AC
#N2=Wcoso1
#F2=uac*N2
#T=F2+W sino2
W=T/(uac*math.cos(o1)+math.sin(o1))
print "Least Weight is",W,"N"
#for greatest weight
N1=Wbc*math.cos(o2)                            #Normal Reaction
F1=ubc*N1                                       #frictional Force
T=Wbc*math.sin(o2)+F1                          #Tension
#for block on plane AC
#N2=Wcoso1
#F2=uac*N2
#T=F2+W sino2
W=T/(-1*uac*math.cos(o1)+math.sin(o1))
print "Greatest Weight is",W,"N"

Least Weight is 266.34090474 N
Greatest Weight is 969.473014916 N


# Example 5.7¶

In [7]:
import math
u=0.4     #The coefficient of friction on the horizontal plane
oi=30     #angle of inclined plane
o=20.0  #The limiting angle of  friction for block B on the inclined plane
wb=5000.0   #weight of block b
ub=math.tan(o*3.14/180.0)            #coefficcient of friction on plane
#for block B
#N1 N2 N3 are normal reaction
#F1 F2  are frictional forces
#F1=ub*N1
#N1 sinoi + F1 cos oi=wb
N1=wb/(math.sin(oi*3.14/180.0)+ub*math.cos(oi*3.14/180.0))
F1=ub*N1
C=N1*math.cos(oi*3.14/180.0)-F1*math.sin(oi*3.14/180.0)

#force balance on A in horizontal balance
F2=C
N2=F2/u
#force balance on A in vertical balance
W=N2
print "Weight ",W,"N"

Weight  10498.172578 N


# Example 5.8¶

In [8]:
import math
w=20000.0    #weight of upper block
o=15.0      #The angle of friction for all surfaces of contact
u=math.tan(o)  #coefficient of friction
#R1 R2 are forces
Or1=15.0     #angle force R1 makes with x axis
Or2=35.0       #angle force R2 makes with Y axis
R2=w*math.sin((90-Or1)*3.14/180.0)/math.sin((90+Or1+Or2)*3.14/180.0)
#applyig lamis theorem on block B
Or1=15.0     #angle force R3 makes with Y axis
Or2=35.0       #angle force R2 makes with Y axis
P=R2*math.sin((180-Or1-Or2)*3.14/180.0)/math.sin((90+Or1)*3.14/180.0)
print "Force =",P,"N"

Force = 23812.7516422 N


# Example 5.9¶

In [9]:
import math
w=160.0    #weight of block,KN
u=0.25             #coefficient of friction
phi=math.atan(u)

#The free body diagrams of wedges A, B and block C .The problem being symmetric, the reactions R1 and R2 on wedges A and B are equal. The system of forces on block C andon wedge A are shown in the form convenient for applying Lami’s theorem
R1=w*math.sin(math.pi-(16*math.pi/180)-phi)/math.sin(2*(phi+math.pi*16/180))
#consider the equillibrium of the wedge A ,Ny lamis's theorem,we get
P=R1*math.sin(math.pi-phi-phi-(16*math.pi/180))/math.sin((math.pi/2)+phi)
print"P=",round(P,2),"KN"

P= 66.26 KN


# Example 5.10¶

In [10]:
import math
u1=0.2    #coefficient of friction between the wall and the ladder
u2=0.3    #coefficient of friction between floor and the ladder
wm=600.0     #weight of man
lm=3.0      #distance of man
#sum of all moment about A =0
Nb=(w*l/2*math.cos(o)+wm*lm*math.cos(o))/(l*(math.sin(o)+u1*math.cos(o)))     # normal reaction from wall
Fb=u1*Nb     #friction from wall
#force balance in vertical direction
Na=(w+wm-Fb)      # normal reaction from ground
Fa=u2*Na      #friction from ground
P=Nb-Fa
print "Force required is ",P,"N"

Force required is  62.0836173323 N


# Example 5.11¶

In [11]:
import math
u1=0.4    #coefficient of friction between the wall and the ladder
u2=0.25    #coefficient of friction between floor and the ladder
#force balancing
#Na Nb   normal reaction at A and B
#Fa Fb    friction  at A and B
#Fa=u2*Na
#Fb=u1*Nb
#Na+Fb=w+wl
#Fa=Nb
Nb=(wl+w)*u2/(1+u2*u1)
Na=Nb/u2
Fa=u2*Na
Fb=u1*Nb
#sum of all moments about a is =0
temp=((w*l*0.5)+(wl*ll)-(Fb*l))/(Nb*l)
o=math.atan(temp)*180/3.14
print "Angle of inclination is ",o,"degrees"

Angle of inclination is  71.6013500101 degrees


# Example 5.12¶

In [12]:
import math
o=45.0*3.14/180.0       #angle of inclination
u=0.5                        #coefficient of friction
r=1.5                    #ratio of mans weight to ladders weight
o1=45.0*math.pi/180.0           #angle of inclination
#from law of friction
#Fa = μNa
#Fb = μNb
#Fa – Nb = 0
#Na + Fb = W + r W
#ΣMA = 0
o=(((u*u+u)*(1+r)/((1+u)))-1.0/2.0)/r
print "length will",o,"times"

length will 0.5 times


# Example 5.13¶

In [13]:
import math
n=1.25                 #number of turns
o=2*3.14*n              #angle of contact
u=0.3                 #coefficient of friction
t=600.0                #force at the other end of the rope
#if the impending motion of the weight be downward.
W=T2=t*2.71**(u*o)
print "Maximum weight is ",W
#if the impending motion of weight be upwards
W=T1=t*2.71**(-1*u*o)
print "Minimum weight is ",W

Maximum weight is  6277.60420331
Minimum weight is  57.3467183245


# Example 5.14¶

In [14]:
import math
ur=0.20       #The coefficient of friction  between the rope and the fixed drum
uo=0.30        #The coefficient of friction  between other surfaces
cosa=4.0/5.0 #cos of angle of inclination
sina=3.0/5.0 #sin of angle of inclination
Ww=1000.0  #weight
o=3.14   #angle of contact of rope with pulley
#for unknown weight
#force balance perpendicular to the plane
#N1 = W cos α
#fr=uoN1
#force balance along the plane
#T1 = F1 + W sin α
#for 1000 N body
#force balance perpendicular to the plane
#N2=N1+Wwcosa
#fr2=uoN2
#force balance along the plane
#T2= Wwsina -F1 -F2
#T2=T1*e^(ur*o)
W=(Ww*sina-uo*Ww*cosa)/(((uo*cosa+sina)*(2.71**(uo*o)))+(uo*cosa+uo*cosa))
print "Weight is ",W,"N"

Weight is  136.9599857 N


# Example 5.15¶

In [15]:
import math
u=0.3           #coefficient of friction
l=300            #length of lever arm
M=300000.0      #torque
o=r*3.14/180.0
l2=50.0
#using
#T2 = T1e^(μθ)         T1 and T2 are tension
#(T2-T1)r=M
T1=M/(r*(2.71**(u*o)-1))
T2=(2.71**(u*o))*T1
#Consider the lever arm. Taking moment about the hinge
p=T2*l2/l                     #force P applied at the end of the brake lever
print "force P applied at the end of the brake lever",p

force P applied at the end of the brake lever 274.480678202


# Example 5.16¶

In [16]:
import math
d1=500.0         #diameter of a shaft
d2=100.0        #diameter of a shaft
D=3000.0                   #distance between shafts in mm
T=1000.0              #Maximum permissible tension in the belt
U=0.25           #coefficient of friction between the belt and the pulley
R=220.0              #revlution per minute of larger shaft
O1=O2=3.14+2*math.asin((d1+d2)/(2*D))
#Length of belt = Arc length DC + Arc length FE + 2BG
L=(d1/2+d2/2)*O1+2*D*math.cos(math.asin((d1+d2)/(2*D)))
print  "Length of belt is ",L,"mm"
T1=T/(2.71**(U*O1))
Velocity_of_the_belt =d1/2*(R*2*3.14/60.0)
Power_transmitted=(T-T1)*Velocity_of_the_belt
print "Power Transmitted",Power_transmitted,"Watt"

Length of belt is  6972.02507534 mm
Power Transmitted 3252832.96438 Watt


# Example 5.17¶

In [17]:
import math
d1=500.0         #diameter of a shaft
d2=100.0        #diameter of a shaft
D=3000.0                   #distance between shafts in mm
T=1000.0              #Maximum permissible tension in the belt
U=0.25           #coefficient of friction between the belt and the pulley
R=220.0              #revlution per minute of larger shaft
O1=3.14+2*math.asin((d1-d2)/(2*D))
O2=3.14-2*math.asin((d1-d2)/(2*D))
#Length of belt = Arc length DC + Arc length FE + 2BG
L=(d1/2*O1+d2/2*O2)+2*D*math.cos(math.asin((d1-d2)/(2*D)))
print  "Length of belt is ",L,"mm"
T1=T/(2.71**(U*O2))
Velocity_of_the_belt =d1/2*(R*2*3.14/60.0)
Power_transmitted=(T-T1)*Velocity_of_the_belt
print "Power Transmitted",Power_transmitted,"Watt"

Length of belt is  6955.3382782 mm
Power Transmitted 3035637.41075 Watt