In [1]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.1
# calculation of the angle made by the contact force with the vertical and the magnitude of contact force
# given data
import math
M=.4# mass(in kg) of the body
f=3.# frictional force(in N)
g=10.# gravitational acceleration(in m/s^2) of the earth
# calculation
N=M*g# formula of normal force
theta=36.9;#atand(f/N)# angle made by the contact force with the vertical
F=math.sqrt((N*N)+(f*f))
print'the angle made by the contact force with the vertical is',theta,'degree','\nthe magnitude of contact force is',F,'N'
```

In [2]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.2
# calculation of the force of friction exerted by the horizontal surface on the box
# given data
M=20.# mass(in kg) of the box
muk=.25# coefficient of kinetic friction
g=9.8# gravitational acceleration(in m/s^2) of the earth
# calculation
fk=muk*M*g# formula of kinetic friction
print'the force of friction exerted by the horizontal surface on the box,in opposite direction to the pull is',fk,'N'
```

In [3]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.3
# calculation of the force of friction exerted by the horse and condition of boy for sliding back
# given data
M=30.# mass(in kg) of the boy
a=2.# average acceleration(in m/s^2) of the horse
g=10.# gravitational acceleration(in m/s^2) of the earth
# calculation
fs=M*a# Newton's second law
musmax=fs/(M*g)# equation of static friction
print'the force of friction exerted by the horse on the boy is',fs,'N'
print'for the boy sliding back during acceleration, the value of coefficient of static friction is less than',musmax
```

In [4]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.4
# calculation of coefficient of static friction and kinetic friction between the block and the plank
# given data
theta1=18# angle of plank(in degree) with horizontal when block starts slipping
theta2=15# angle of plank(in degree) with horizontal when block slips with uniform speed
# calculation
mus=0.325;#tand(theta1)# formula of coefficient of static friction
muk=0.268;#tand(theta2)# formula of coefficient of kinetic friction
print'the coefficient of static friction between the block and the plank is tan(',theta1,')=',mus
print'the coefficient of kinetic friction between the block and the plank is tan(',theta2,')=',muk
```

In [5]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.1w
# calculation of the maximum angle to prevent slipping
# given data
mus=.3# coefficient of static friction
# calculation
thetamax=16.7;#atand(mus)
print'the maximum angle to prevent slipping is',thetamax
```

In [6]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.2w
# calculation of frictional force and minimum value of coefficient of static friction
# given data
m=4.# mass(in kg) of the block
f=20.# frictional force(in N)=horizontal force(in N)
g=10.# gravitational acceleration(in m/s^2) of the earth
# calculation
N=m*g# normal force
musmin=f/N
print'the frictional force on the block,in opposite direction to the applied force is',f,'N'
print'the coefficient of static friction between the block and the table is greater than or equal to',musmin
```

In [7]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.3w
# calculation of the maximum value of mass of the block
# given data
mus=.2# coefficient of static friction between the block and the table
M=2.# mass(in kg) of one block
g=10.# gravitational acceleration(in m/s^2) of the earth
# calculation
N=M*g# normal force
# T=m*g tension in the string (1)
# fs=mus*N frictional force (2)
# f=T from equlibrium equation of 2 kg block (3)
# from above equations,we get
m=(mus*N)/g
print'the maximum value of mass of the block is',m,'kg'
```

In [8]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.5w
# calculation of the coefficient of kinetic friction
# given data
theta=30.# angle(in degree)f the incline
g=10.#gravitational acceleration(in m/s^2) of the earth
# calculation
a=g/4.# acceleration(in m/s^2) of the block.....given
# f=m*g/4................taking parallel components to the incline
# N=m*g*cosd(theta)......taking vertical components to the incline
# from above equations,we get
muk=0.289;#1./(4.*cosd(theta))# muk=f/N equation of static friction
print'the coefficient of kinetic friction is',muk
```

In [9]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.6w
# calculation of the values of coefficient of static and kinetic friction
# given data
M=2.5# mass(in kg) of the block
F=15.# horizontal force(in N)
g=10.# gravitational acceleration(in m/s^2) of the earth
x=10.# displacement(in m) of the block
t=5.# time(in s) required by the block
# calculation
mus=F/(M*g)
a=(2.*x)/(t*t)# acceleration of the block from equation of uniform linear motion
# F-muk*M*g=M*a.....newton's second law
muk=(F-(M*a))/(M*g)
print'the coefficient of static friction between the block and the surface is',mus
print'the coefficient of kinetic friction between the block and the surface is',muk
```

In [10]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 6.10w
# calculation of mimimum and maximum values of m(mass) and the acceleration if given a gentle push
# given data
import math
from math import sqrt
mus=.28# the value of coefficient of static friction between the block and the surface
muk=.25# the value of coefficient of kinetic friction between the block and the surface
M=2.# mass(in kg) of one block
g=9.8# gravitational acceleration(in m/s^2) of the earth
# calculation
# T=(M*g*(1-mus))/sqrt(2)................taking components along incline for block1......(1)
# T=(M*g*(1+mus))/sqrt(2)................taking components along incline for block2......(2)
# from above equations,we get
m1=((1.-mus)*M)/(1.+mus)# minimum value of m...............................................(3)
m2=((1.+mus)*M)/(1.-mus)# maximum value of m obtained by taking reverse direction of friction in above equations
# (M*g/sqrt(2)) - T = M*a.........newton's second law for M block........................(4)
# T - (m*g/sqrt(2)) = m*a.........newton's second law for m block........................(5)
# adding equations (4) and (5)
# ((M*g*(1-muk))/sqrt(2)) - ((m*g*(1+muk))/sqrt(2)) = (M+m)*a
a=(((M*(1.-muk))-(m1*(1.+muk)))*g)/(sqrt(2.)*(M+m1))# calculating acceleration for minimum value of m if gently pushed......given
print'the minimum value of m for which the system remains at rest is',m1,'kg'
print'the maximum value of m for which the system remains at rest is',m2,'kg'
print'the acceleration of either block for minimum value of m and if gently pushed up the incline is',a,'m/s**2'
```