Solutions Manual For Engineering Mechanics Manoj Kumar Harbola IIT Kanpur

Solutions Manual
For

Engineering
Mechanics

Manoj Kumar Harbola
IIT Kanpur




1

, Chapter 1


Rotational speed of the earth is very small (about 7 105 radians per second). Its
effect on particle motion over small distances is therefore negligible. This will not be
true for intercontinental missiles.
The net force on the (belt+person) system is zero. This can be seen as follows. To
pull the rope up, the person also pushes the ground and therefore the belt on which he
is standing. This gives zero net force on the belt. For the person, the ground pushes
him up on the feet but the belt pulls him down when he pulls it, giving a zero net
force on him.
A vector between coordinates (x1, y1, z1) and (x2,y2,z2) is given by

(x2  x1 )iˆ  ( y2  y1 ) ˆj  (z2  z1 )kˆ . Thus (i), (ii) and (iv) are equal.

The vectors are (i) 2iˆ  3 ˆj  5kˆ (ii) 4iˆ  3 ĵ  k̂ (iii) 2iˆ  9 ĵ  5k̂ (iv)
 3iˆ  3 ĵ  2k̂

(i) The resultant vectors are 6iˆ  6kˆ , 4iˆ  6 ĵ  10k̂ and  iˆ  3kˆ
(ii) The resultant vectors are 2iˆ  6 ĵ  4k̂ , 2iˆ  6 ĵ  4k̂ and 7iˆ  3kˆ
1.6

A
⃗ ⃗
⃗ AB
B ⃗
B A B
A


On each reflection, the sign of the vector component perpendicular to the reflecting
mirror changes.
z




y
1.8 v

2 x
O

, The fly is flying along the vector from (2.5, 2, 0) to (5, 4, 4). This vector is

2.5iˆ  2 ˆj  4kˆ 2.5iˆ  2 ˆj  4kˆ
2.5iˆ  2 ˆj  4kˆ . The unit vector in this direction is  .
6.25  4  16 26.25

The velocity of the fly is therefore

2.5iˆ  2 ĵ  4k̂
0.5   0.25iˆ  .20 ĵ  0.39k̂ .
26.25


⃗ ⃗
After time t, the position vectors rA and rB of particles A and B, respectively, are
⃗ ⃗
rA  l sin t iˆ  l cost ˆj rB  l sin t iˆ  l cost ˆj
⃗ ⃗
Their velocities v A and v B are given by differentiating these vectors with respect to
time to get
⃗ ⃗
v A  l cost iˆ  l sin t ĵ v B  l cost iˆ  l sin t ĵ
⃗ ⃗ ⃗
Velocity v AB of A with respect to B is obtained by subtracting v B from v A
⃗ ⃗ ⃗
v AB  v A  v B  2l cost iˆ

For rotation about the z-axis by an angle 
v x'  v x cos  v y sin v y '  v x sin  v y cos v z'  v z

It is given that   30 ◦ . Therefore
 
v   v 
1 1
v x'  x 3  vy v y'  x  vy 3  v z'  v z
2 2




3

, Component of a vector A along an axis is given by its projection on that axis. This is
obtained by taking the dot product of the vector with the unit vector along that axis.
Thus
⃗ ⃗ ⃗ ⃗ ⃗ ⃗
A  A  iˆ  A cos ˆ ˆ
x 1 Ay  A  j  A cos 2 Az  A  k  A cos3

Also 
⃗2
A  A2  A2  A2
x y z



Substituting the expression for Ax, Ay and Az completes the proof.
(i) Dot product of two vectors A and B is
⃗ ⃗
A  B  Ax Bx  Ay By  Az Bz

This gives the dot product of the first vector of problem 1.4 each of the other vectors
to be 4, 2 and 25.
⃗
(ii) Cross product between two vectors A and B is
B  (Ay Bz  Az By )iˆ  (Az Bx  Ax Bz )ˆj  (Ax By  Ay Bx )kˆ
A�
⃗
Taking A to be the fourth vector and B to be the first, second and the third vector
gives the cross products to be
 9iˆ  11 ĵ  3k̂

 9iˆ  5 ĵ  21k̂

 33iˆ  11 ĵ  24k̂

If the angle between two vectors is , the cosine of this angle is given by

A B
cos  ⃗ ⃗
AB. Thus

4
Between (i) and (ii) cos   0.127    82.7 ◦
38 26
 2
Between (i) and (iii) cos   0.037    88.2 ◦
38 110
 25
Between (i) and (iv) cos    0.865    149.8 ◦
38 22
40
Between (ii) and (iii) cos   0.748    41.6 ◦
26 110
5
Between (ii) and (iv) cos   0.209    102.1◦
26 22



4