Solutions Manual For Engineering Mechanics Manoj Kumar Harbola

Solutions Manual For Engineering Mechanics Manoj Kumar Harbola IIT Kanpur 1Chapter 1 1.1 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. 1.2 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. 1.3 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. 1.4 The vectors are (i) 2iˆ  3 ˆj  5kˆ (ii) 4iˆ  3 ˆj  kˆ (iii) 2iˆ  9 ˆj  5kˆ (iv)  3iˆ  3 ˆj  2kˆ 1.5 (i) The resultant vectors are 6iˆ  6kˆ , 4iˆ  6 ˆj 10kˆ and  iˆ  3kˆ (ii) The resultant vectors are 2iˆ  6 ˆj  4kˆ , 2iˆ  6 ˆj  4kˆ and 7iˆ  3kˆ 1.6 1.7 On each reflection, the sign of the vector component perpendicular to the reflecting mirror changes. 1.8 2 A B A B   B   A B A    z x O v yThe fly is flying along the vector from (2.5, 2, 0) to (5, 4, 4). This vector is 2.5iˆ  2 ˆj  4kˆ . The unit vector in this direction is 26.25 2.5ˆ 2 ˆ 4 ˆ 6.25 4 16 2.5iˆ 2 ˆj 4kˆ i  j  k      . The velocity of the fly is therefore i j k 0.25iˆ .20 ˆj 0.39kˆ 26.25 2.5ˆ 2 ˆ 4 ˆ 0.5      . 1.9 After time t, the position vectors rA and rB of particles A and B, respectively, are rA  l sint iˆ  l cost ˆj rB  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 ˆj vB  l cost iˆ  l sint ˆj Velocity vAB of A with respect to B is obtained by subtracting vB from vA l t i AB A B v  v  v  2  cos ˆ 1.10 For rotation about the z-axis by an angle  v v cos v sin x'  x  y v y'  vx sin  v y cos vz'  vz It is given that   30 . Therefore x v x 3 v y  12 v '    v v 3 12 v y'   x  y vz'  vz 31.11 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 Ax  Aiˆ  A cos1 Ay  A ˆj  A cos2 Az  A kˆ  A cos3 Also 2 2 2 2 A  Ax  Ay  Az  Substituting the expression for Ax, Ay and Az completes the proof. 1.12 (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 A B A B A B i A B A B j A B A B k r        r ( ) ( ) ( ) y z z y z x x z x y y x ˆ ˆ ˆ 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 11ˆj  3kˆ  9iˆ  5 ˆj  21kˆ  33iˆ 11ˆj  24kˆ 1.13 If the angle between two vectors is , the cosine of this angle is given by A B A B      cos  . Thus Between (i) and (ii) 0.127 82.7 38 26 4 cos      Between (i) and (iii) 0.037 88.2 38 110 2 cos      Between (i) and (iv) 0.865 149.8 38 22 25 cos         Between (ii) and (iii) 0.748 41.6 26 110 40 cos      Between (ii) and (iv) 0.209 102.1 26 22 5 cos         4Between (iii) and (iv) 0.380 67.6 38 22 11 cos      1.14 Vector A can be written as A  Acos1iˆ  cos2 ˆj  cos3kˆ Similarly B  Bcos 1iˆ  cos 2 ˆj  cos 3kˆ Taking the dot product between the two vectors and using the formula A B A B      cos  , where  is the angle between the two vectors, we get the answer. 1.15 Magnitude A  B  A 2  B 2  2A Bcos Similarly A  B  A 2  B 2  2 A  Bcos Equating the two gives A B  0 which implies that the two vectors are perpendicular to each other. 1.16 B C  ByCz  BzCy iˆ  BzCx  BxCz  ˆj  BxCy  ByCx kˆ A  B C  Ax ByCz  Bz C y   Ay  BzCx  BxCz   Az BxC y  ByCx  This comes out to be equal to C A B and B C A 1.17 From the expression for A B C it is clear that it is equal to the determinant x y z x y z x y z C C C B B B A A A Interchange of two rows in a determinant changes the sign of the determinant. This implies x y z x y z x y z x y z x y z x y z x y z x y z x y z C C C B B B A A A C C C A A A B B B A A A C C C B B B    thereby proving the equalities in problem 1.16.