In [1]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 5.1
# calculation of force exerted by the string on a particle
# given data
m=.5# mass(in kg) of the particle
g=9.8# gravitational acceleration(in m/s^2) of the earth
# calculation
T=m*g# tension in the string is equal to the downward force exerted by earth
print'the force exterted by the string on particle in vertically upward direction is',T,'N'
```

In [2]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 5.3w
# calculation of the force exerted by the tree limb on the bullet
# given data
u=250.# initial velocity(in m/s) of the bullet
v=0# final velocity(in m/s) of the bullet
x=.05# penetration(in m) by the bullet in the tree limb
m=.01# mass of bullet(in kg)
# calculation
a=((u*u)-(v*v))/(2.*x)# formula of horizontal acceleration in case of uniform linear motion
F=m*a;
print'the force exerted by the tree limb on the bullet is',F,'N'
```

In [3]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 5.4w
# calculation of the position of a particle
# given data
m=.01# mass(in kg) of the particle
Fx=10.# component of force(in N) along X axis
Fy=5.# component of force(in N) along Y axis
ux=0# x component of initial velocity(in m/s) of the particle
uy=0# y component initial velocity(in m/s) of the paticle
t=5.# time(in s) at which position is to be determined
# calculation
ax=Fx/m;
x=(ux*t)+((1./2.)*ax*t*t);# formula of horizontal position in case of uniform linear motion
ay=Fy/m;
y=(uy*t)+((1./2.)*ay*t*t);# formula of vertical position in case of uniform linear motion
print'at t=5 s position of the particle is (',x,'i','+',y,'j',')m'
```

In [4]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 5.7w
# calculation of acceleration with which ring starts moving if released from rest at an angle theta
# given data
# m=mass of the ring
theta=30.# angle(in degree)of the release
m=1.# assume for obtaiming the solution
M=2.*m # mass of the block
g=9.8# gravitational acceleration(in m/s^2) of the earth
# calculation
# M*g-T=M*a*cosd(theta)........equation of motion of the block...(1)
# T*cosd(theta)=m*a............equation of motion of the ring....(2)
# solving above equations we get
a=6.79;#(M*g*cosd(theta))/(m+M*(cosd(theta)*cosd(theta)))
print'the acceleration with which ring starts moving if released from rest at an angle theta is',a,'m/s**2'
```

In [5]:

```
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 5.8w
# calculation of the maximum acceleration of the man for safe climbing
# given data
m=60.# mass(in kg) of the man
theta=30.# angle(in degree) made by the rope with ground
fgmax=360.# maximum force(in N0 that can be applied to the wooden clamp
g=10.# gravitational acceleration(in m/s^2) of the earth
# calculation
T=720.;#fgmax/sind(theta)# since t*sin(theta)=upward force
a=(T-(m*g))/m# from equation of motion
print'the maximum acceleration of the man for safe climbing is',a,'m/s**2'
```