import math
#example 14.1
#calculation of the tensile stress developed in the wire
#given data
m=4.#mass(in kg) of the load
r=2.*10.**-3.#radius(in m) of the wire
g=3.1*math.pi#gravitational acceleration(in m/s**2) of the earth
#calculation
F=m*g#gravitational force
A=math.pi*r**2.#area
St=F/A#tensile stress
print '%s %.2f %s' %('the tensile stress developed in the wire is',St,'N/m**2\n')
import math
#example 14.2
#calculation of the value of Young modulus
#given data
m=4.#mass(in kg) of the load
l=20.#length(in m) of the steel wire
r=2.*10.**-3.#radius(in m) of the steel wire
dl=.031*10.**-3.#increase in the length(in m)
g=3.1*math.pi#gravitational acceleration(in m/s**2) of the earth
#calculation
Ssl=(m*g)/(math.pi*r**2.)#longitudinal stress
Stl=dl/l#longitudinal strain
Y=Ssl/Stl#Young modulus
print '%s %.2f %s' %('the value of Young modulus is',Y,'N/m**2\n')
#example 14.3
#calculation of the elastic potential energy stored in the stretched steel wire
#given data
l=2.#length(in m) of the steel wire
A=4.*10.**-6.#cross sectional area(in m**2) of the steel wire
dl=2.*10.**-3.#increase in the length(in m)
Y=2.*10.**11.#Young modulus(in N/m**2)
#calculation
St=dl/l#strain in the wire
Ss=Y*St#stress in the wire
V=A*l#volume of the steel wire
U=Ss*St*V/2.
print '%s %.2f %s' %('the elastic potential energy stored in the stretched steel wire is',U,'J\n')
#example 14.4
#calculation of the force by which the surface on one side of the diameter pulls the suface on the other side
#given data
r=5.*10.**-2.#radius(in m) of the beaker
S=.075#surface tension(in N/m) of the water
#calculation
l=2.*r#length of diameter of the surface
F=S*l#force
print '%s %.2f %s' %('the force by which the surface on one side of the diameter pulls the suface on the other side is',F,'N\n')
#example 14.5
#calculation of the gain in the surface energy
import math
#given data
R=10.**-2.#radius(in m) of the drop
n=1000.#number of droplets formed
S=.075#surface tension(in N/m) of the water
#calculation
#volume of original drop = total volume of all droplets formed
r=R/n**(1./3.)#radius of each droplet
A1=4.*math.pi*R**2.#surface area of drop
A2=n*(4.*math.pi*r**2.)#surface area of each droplet
deltaA=A2-A1#change in suface area
deltaU=deltaA*S#change in surface energy
print '%s %.4f %s' %('the gain in the surface energy is',deltaU,'J\n')
#example 14.6
#calculation of the excess pressure inside a mercury drop
#given data
R=2.*10.**-3.#radius(in m) of the drop
S=.464#surface tension(in N/m) of the drop
#calculation
deltaP=2.*S/R#excess pressure
print '%s %.2f %s' %('the excess pressure inside a mercury drop is',deltaP,'N/m**2\n')
#example 14.7
#calculation of the density of the liquid
#given data
h=.02*10.**-2.#height(in m) of the column of liquid
R=7.5*10.**-3.#radius(in m) of the soap bubble
S=.03#surface tension(in N/m) of the soap solution
g=9.8#gravitational acceleration(in m/s**2) of the earth
#calculation
deltaP=4.*S/R#excess pressure inside the soap bubble
rho=deltaP/(h*g)#densiy
print '%s %.2f %s' %('the density of the liquid is',rho,'kg/m**3\n')
import math
#example 14.8
#calculation of the height of the water in the column
#given data
r=.2*10.**-3.#radius(in m) of the tube
S=.075#surface tension(in N/m) of the water
g=10.#gravitational acceleration(in m/s**2) of the earth
rho=1000.#density of the water(in kg/m**3)
theta=0#tube dipped vertically
#calculation
h=(2.*S*math.cos(theta)*57.3)/(r*rho*g)#height in column
print '%s %.2f %s' %('the height of the water in the column is',h*10**2,'cm\n')
#example 14.9
#calculation of the value of the coefficient of viscosity of the solution
#given data
d=2.*10.**-3.#diameter(in m) of the air bubble
sigma=1750.#density(in kg/m**3) of the solution
v=.35*10.**-2.#rate of flow(in m/s)
g=9.8#gravitational acceleration(in m/s**2) of the earth
#calculation
r=d/2.#radius of the air bubble
#force of buoyancy is........B = (4/3)*%pi*r**3*sigma*g
#viscous force is............F = 6*%pi*eta*r*v
#above two forces are equal,thus we get
eta=(2.*r**2.*sigma*g)/(9.*v)#coefficient of viscosity
print '%s %.2f %s' %('the value of the coefficient of viscosity of the solution is',round(eta*10),'poise\n')#0 1 poise = .1 N-s/m**2
#example 14.1w
#calculation of the extension of the wire
#given data
L=2.#lengh(in m)of the wire
A=.2*10.**-4.#area(in m**2)
m=4.8#mass(in kg)
Y=2.*10.**11.#Young modulus of steel
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
T=m*g#weight
l=(T*L)/(A*Y)#exension
print '%s %.6f %s' %('the extension of the wire is',l,'m\n')
#example 14.2w
#calculation of the elongation of the rope and corresponding change in the diameter
import math
#given data
L=4.5#length(in m) of the nylon rope
d=6.*10.**-3.#diameter(in m) of the nylon rope
T=100.#weight(in N) of the monkey
Y=4.8*10.**11.#Young modulus(in N/m**2) of the rope
Pr=.2#Poission ratio of nylon
#calculation
A=math.pi*(d/2.)**2.#area of cross section
l=(T*L)/(A*Y)#elongation
deltad=(Pr*l*d)/(L)#change in diameter
print '%s %.6f %s' %('the elongation of the rope is',l,'m\n')
print '%s %.6f %s' %('the corresponding change in the diameter is',deltad,'m\n')
#example 14.3w
#calculation of the minimum radius of the wire used if it is not to break
import math
#given data
m1=1.#mass(in kg) of block1
m2=2.#mass(in kg) of block2
Ss=2.*10.**9.#breaking stress(in N/m**2) of the metal
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
#using equation ....stress = tension / Area of cross secion
#T - (m1*g) = m1 * a...........(1)
#(m2*g) - T = m2*a.............(2)
#Adding equation (1) and equation (2),we get
a=((m2*g)-(m1*g))/(m1+m2)
T=(m1*g)+(m1*a)#tension in the string from equation (1)
r=math.sqrt(T/(Ss*math.pi))#radius
print '%s %.6f %s' %('the minimum radius of the wire used if it is not to break is',r,'m\n')
#example 14.4w
#calculation of the ratio of the lengths of the two wire
#given data
Ys=2.*10.**11.#Young modulus(in N/m**2) of the steel wire
Yc=1.1*10.**11.#Young modulus(in N/m**2) of the copper wire
#calculation
#r = Ls/Lc......required ratio
r=Ys/Yc#required ratio
print '%s %.2f %s' %('the ratio of the lengths of the two wire(Ls/Lc) is',r,': 1')
#example 14.5w
#calculation of the decrease in the volume of the sample of water
#given data
V1=1000.*10.**-6.#initial volume(in m**3)
P1=10.**5.#initial pressure(in N/m**2)
P2=10.**6.#final pressure(in N/m**2)
C=50.*10.**-11.#compressibility(in m**2/N)of the water
#calculation
deltap=P2-P1#change in pressure
#compressibility = 1/Bulk modulus = -(deltaV/V)/deltaP
deltaV=-(C*deltap*V1)
print '%s %.2f %s' %('the decrease in the volume of the sample of water is',-deltaV*10**6,'cm**3\n')
#example 14.6w
#calculation of the longitudinal strain in two wires
#given data
m1=1.#mass(in kg) of load 1
m2=2.#mass(in kg) of load 2
A=.005*10.**-4.#area(in m**2) of the cross section
Y=2.*10.**11.#/Young modulus(in N/m**2) of the wire
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
T1=m1*g#tension in wire 1
Ss1=T1/A#longitudinal stress
St1=Ss1/Y#longitudinal strain
T2=(m2*g)+T1#tension in wire 2
Ss2=T2/A#longitudinal stress
St2=Ss2/Y#longitudinal strain
print '%s %.4f %s' %('the longitudinal strain in wire 1 is',St1,'\n')
print '%s %.4f %s' %('the longitudinal strain in wire 2 is',St2,'\n')
#example 14.7w
#calculation of the longitudinal strain developed in each wire
#given data
m=3.#mass(in kg) of each block
A=.005*10.**-4.#area(in m**2) of the cross section
Y=2.*10.**11.#/Young modulus(in N/m**2) of the wire
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
#using equation of motion,
#TA = m*a..............(1)
#TB - TA = m*a.........(2)
#m*g - TB = m*a........(3)
#adding equation (2) and equation (3) and substituting TA from equation (1),we get
a=(m*g)/(3*m)#acceleration
TA=m*a#Tension(in N) in wire A
TB=(m*a)+TA#Tension(in N) in wire B..from equation (2)
StA=(TA)/(A*Y)#longitudinal strain in wire A
StB=(TB)/(A*Y)#longitudinal strain in wire B
print '%s %.2f %s' %('the longitudinal strain developed in wire A is',StA,'\n')
print '%s %.2f %s' %('the longitudinal strain developed in wire B is',StB,'\n')
#example 14.8w
#calculation of the elastic potential energy stored in the wire
#given data
A=3.*10.**-6.#area(in m**2) of the cross section
l=50.*10.**-2.#natural length(in m)
m=2.1#mass(in kg) hanged
Y=1.9*10.**11.#/Young modulus(in N/m**2) of the wire
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
V=A*l#volume of the wire
T=m*g#tension in the wire
Ss=T/A#stress
St=Ss/Y#strain
U=(Ss*St*V/2.)#elastic potential energy
print '%s %.5f %s' %('the elastic potential energy stored in the wire is',U,'J\n')
#example 14.9w
#calculation of the elongation of the wire
#given data
W=10.#weight(in N) of the block
A=3.*10.**-6.#area(in m**2) of the cross section
r=20.*10.**-2.#radius(in m) of the circle of rotation
v=2.#speed(in m/s) of the block
Y=2.*10.**11.#/Young modulus(in N/m**2) of the wire
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
m=W/g#mass of the block
T=W+(m*v*v/r)#tension
L=r
l=(T*L)/(A*Y)#elongation
print '%s %.5f %s' %('the elongation of the wire is',l*10**2,'cm\n')
#example 14.11w
#calculation of the amount by which the pressure inside the bubble is greater than the atmospheric pressure
#given data
r=1.*10.**-3.#radius(in m) of the air bubble
S=.075#suface tension(in N/m)
rho=1000.#density(in kg/m**3) of the liquid
h=10.*10.**-2.#depth(in m) of the bubble
g=9.8#gravitational acceleration(in m/s**2) of the earth
#calculation
#P = P0 +(h*rho*g)........(1)
#Pdash = P + (2*S/r)......(2)
#deltaP = Pdash - P0
deltaP=(h*rho*g)+(2.*S/r)#difference in the pressure
print '%s %.2f %s' %('the pressure inside the bubble is greater than the atmospheric pressure by',deltaP,'Pa\n')
#example 14.12w
#calculation of the load W suspended from wire to keep it in equilibrium
#given data
l=10.*10.**-2.#length(in m) of the wire
#1 dyne = 10**-5 N
S=25.*10.**-5.*10.**2.#suface tension(in N/m) of the soap solution
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
F=2.*l*S#force exerted by the film on the wire
m=F/g#mass of the load
print '%s %.2f %s' %('the load W suspended from wire to keep it in equilibrium should be',F,'N\n')
print '%s %.4f %s %.2f %s' %('the mass of the load suspended from wire to keep it in equilibrium should be',m,'kg or',m*10**3,'g')
#example 14.13w
#calculation of the radius of the capillary tube
import math
from math import cos
#given data
h=7.5*10.**-2.#height(in m) by which the capillary rises
S=7.5*10.**-2.#suface tension(in N/m) of water
theta=0#contact angle(in degree) between water and glass
g=10.#gravitational acceleration(in m/s**2) of the earth
rho=1000.#density(in kg/m**3) of water
#calculation
r=(2.*S*cos(theta)*57.3)/(h*rho*g)#from formula of height in capillary tube
print '%s %.2f %s' %('the radius of the capillary tube is',r*10**3,'mm\n')
#example 14.15w
#calculation of the tangential force needed to keep the plate moving
#given data
A=10.#area(in m**2) of the plate
v=2.#speed(in m/s) of the plate
d=1.#depth(in m) of the river
# 1 poise = .1 N-s/m**2...unit of viscosity
eta=10.**-2.*10.**-1.#coefficient of viscosity(in N-s/m**2)
#calculation
dvbydx=v/d#velocity gradient
F=eta*dvbydx*A#force exerted
print '%s %.2f %s' %('the tangential force needed to keep the plate moving is',F,'N\n')
#example 14.16w
#calculation of the shearing stress between the horizontal layers of water
#given data
v=18.*10.**3./(60.*60.)#velocity(in m/s) of the water in river
d=5.#depth(in m) of the river
# 1 poise = 0.1 N-s/m**2
eta=10.**-2.*10.**-1.#coefficient of viscosity(in N-s/m**2) of the water
#calculation
dvbydx=v/d#velocity gradient
#force of viscosity ......F=eta*A*(dvbydx)
#shearing stress..........Ss=F/A
Ss=eta*(dvbydx)
print '%s %.4f %s' %('the shearing stress between the horizontal layers of water is',Ss,'N/m**2\n')
#example 14.17w
#calculation of the terminal velocity of the rain drop
#given data
r=.01*10.**-3.#radius(in m) of the drop
eta=1.8*10.**-5.#coefficient of viscosity(in N-s/m**2) of the air
rho=1.2#density(in kg/m**3) of the air
rho0=1000.#density(in kg/m**3) of the water
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
#at terminal velocity.........6*%pi*eta*r*v = (4/3)*%pi*r**3*rho*g
v=(2.*r**2.*rho0*g)/(9.*eta)#terminal velocity
print '%s %.2f %s' %('the terminal velocity of the rain drop is',v*10**2,'cm/s**2\n')