CHAPTER05 : NEWTONS LAWS OF MOTION¶

Example E1 : Pg 65¶

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'

the force exterted by the string on particle in vertically upward direction is 4.9 N


WORKED EXAMPLES¶

Example E3W : Pg 72¶

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'

the force exerted by the tree limb on the bullet is 6250.0 N


Example E4w : Pg 73¶

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'

at t=5 s position of the particle is ( 12500.0 i + 6250.0 j )m


Example E7w : Pg 73¶

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'

the acceleration with which ring starts moving if released from rest at an angle theta is 6.79 m/s**2


Example E8w : Pg 74¶

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'

the maximum acceleration of the man for safe climbing is 2.0 m/s**2