Motion in a Plane Vector

Motion in a Plane VECTOR SCALAR AND VECTOR QUANTITIES The physical quantities are two types : scalar and vector. Scaler quantities : The quantities which have only magnitude, and no direction, are called ‘scalar quantities’, e.g. mass, distance, time, speed, volume, density, pressure, work, energy, power, charge, electric current, temperature, potential, specific heat, frequency etc. Vector quantities : Certain quantities have both magnitude and direction, e.g. position, displacement, velocity, acceleration, force, weight, momentum, impulse, electric field, magnetic field, current density, etc. Such quantities are called ‘vector quantities’ TYPES OF VECTOR (a) Polar - Vectors : have starting point (like displacement) or a point of application (like force) (b) Axial - Vectors : Rotational effects are represented by axial vectors. They are along axis of rotation, direction denoted by right hand thumb rule or right hand screw rule. Ex. Angular displacement, angular velocity, torque, angular momentum. SOME OTHER TYPES OF VECTOR [a] Negative of a vector : It has direction just opposite to given vector and have same magnitude fig.(a) Ex. 0 fig. (a) fig. (b) fig. (c) fig. (d) Motion in a Plane [b] Zero vector or null vector : A vector will zero magnitude having no specific direction is called zero vector fig.(b)   (i) Multiplying a vector by zero. i.e. 0( A )  0 (ii) By adding a negative vector to the given vector.    A  ( A )  0 VECTOR ADDITION (i) Law of Triangle : If two sides of triangle are   shown by two continuous vectors ( A and B ) then third side of triangle in opposite direction shown  resultant of two vectors ( C ) [c] Equal vector : Two vectors are called equal (or equivalent) vectors if they have equal magnitude,    and same direction fig.(c) A  B  C [d] Collinear vectors : Two vectors acting along same straight line or along parallel straight line in same direction or in opposite direction fig.(d) [e] Coplanar vectors : Three (or more) vectors are called coplanar vectors if they lie in the same plane or are parallel to the same plane. Two (free) vectors are always coplanar.       A VECTOR WITH SCALAR If a vector multiplied by a scalar quantity then we get a new vector in the same direction but having new magnitude = scalar quantity × magnitude of the   COMPONENT OF A VECTOR ALONG ANOTHER VECTOR   Two vector A and B there as shown below to get   component of B along A   MULTIPLICATION OF  OC  OA  AC  C  A  B C  A B shift B || to it self and made it coinitial with A , draw  a perpendicular from B to OA let is meets A at   M is angle in between A and B .   Component of B , along A  OM = OB cos  = OB cos vector quantity fig. 3   a  3a  Note : (i)  = 0, OM maximum = zero UNIT VECTOR A vector having unit magnitude. It is used to denote the direction of a given vector.  A  â.A  â is unit vector along the direction of A SUBTRACTION OF VECTORS Vector which is want to subtracted just change direction of that vector and then add.     A  B  A  ( B ) fig. (1) Ex. As shown in fig.(1) if OA is displacement of vector in any orbiter direction then â is unit vector in the same direction. (ii)  = 90º,OM Motion in a Plane Important results   Ex.1 Given a  3 î  2 ĵ  k̂ and b  î  ĵ  3k̂  Determine      (i) a  b  (b)   180º  anti || then Rmin  P  Q  (ii) a  b     (ii) a  b  (3 î  2 ĵ  k̂ )  ( î  ĵ  3k̂ )   (d) | P || Q | a | R | 2P cos  3 î  2 ĵ  k̂  î  ĵ  3k̂   = 2 î  ĵ  4k̂  4 î  3 ĵ  2k̂ Q (c)   90º  P  Q R  a 2  b 2 here tan     P Sol. (i) a  b  (3 î  2 ĵ  k̂ ) + ( î  ĵ  3k̂ )  3 î  2 ĵ  k̂  î  ĵ  3k̂  (a)   0  P || Q then R max  P  Q    (e) | P || Q | and (i)  = 60º R= 3P (ii)  = 90º R= 2P OF VECTOR   According to this law if two vectors P and Q are represented by two adjacent sides of a parallelogram both pointing outwards as shown in fig. The diagonal drawn through the intersection of  2 and tan  =  2 PARALLELOGRAM LAW To determine magnitude & direction of resultant vector, when two vectors act at an angle .  2  (iii)  = 120º R = P (f) If three vectors of equal magnitudes makes an angle of 120º with each other then the resultant vector will be zero. R = 0 (g) If n vectors of equal magnitudes makes the angle the of equal measure with each other then the resultant vector will be zero.  the two vectors represents the resultant R . SCALAR PRODUCT OR DOT PRODUCT OF TWO VECTORS    R  P Q From triangle OCM   OC2 = OM2 + CM2 If  is the angle between A and B . = (P + Q cos)2 + (Q sin )2 Then A (B cos) = A . B , A and B are the magnitude   of A and B = P2 + Q2 cos2 + 2PQ cos  + Q2 sin2  Since Q2 (cos2 + sin2) = Q2 2 2 2   The quantity AB cos is a scalar quantity.  R = P + Q + 2PQ cos  B cosis the component of vector B in the direction R  P 2  Q 2  2PQ cos  of A . (Magnitude of resultant vector) and tan = or CM OM = Q sin  P  Q cos   Q sin     tan 1    P  Q cos   (Angle of resultant vector with P)  Hence, the scalar product of two vectors is equal to the product of the magnitude of one vector and the component of the second vector in the direction of the first vector.     Ex. [1] W  F . S [2] P  F . V   [3]   B . A Motion in a Plane   Propoerties of scalar product   A . B  AB cos  (a) The scalar product is commutative.       A . B = A(B cos) = (A cos) i.e. A . B  B . A      3 46   B = B.A (b) The scalar product is distributive  19 cos   A .B cos   AB   19   cos 1   3 46     Ex.3 A particle, under constant force î  ĵ  2k̂ gets   displaced from point A(2, –1, 3) to B(4, 3, 2) . Find the work done by the force - A .( B  C)  A . B  A . C (c) The scalar product of two mutually Sol. Force = î  ĵ  2k̂ perpendiucular vectors is zero.      A . B  AB cos 90º  0 A . B  0 for AB  dispalcement  î . ĵ  0, ĵ k̂  0,  d  AB  (4 î  3 ĵ  2k̂ )  (2î  ĵ  3k̂ )  (2î  4 ĵ  k̂ ) k̂ , î  0 (d) The scalar product of two parallel vectors is equal to the product of their magnitudes.   cos 0º = 1 A . B  AB cos 0º  AB work done = F.d = ( î  ĵ  2k̂ ) . (2 î  4 ĵ  k̂ ) = 1 x 2 + 1 x 4 + (–2) x (–1) = 2 + 4 + 2 = 8 units (e) The scalar product of a vector with itseft is equal Ex.4 Show that the vectors a to the square of the magnitude of the vector. = 3 î  2 ĵ  k̂, b  î  3 ĵ  5k̂ , c  2 î  ĵ  4k̂   A . A  A A cos 0 º  A 2 î . î  1 , ĵ . ĵ  1 , form a right angled triangle. Sol. We have b + c k̂ . k̂  1 (f) The scalar product of two vectors is equal to the sum of the products of their corresponding x,   y, z components. Let A and B be two vectors.   A  A x î  A y ĵ  A z k̂ B  B x î  B y ĵ  B z k̂ Their scalar product is given by  a, b, c are coplanar Hence no two of these vectors are parallel, therfore, the given vectors form a triangle. a . c = (3 î  2 ĵ  k̂ ) . (2 î  ĵ  4k̂ )   A . B  ( A x î  A y ĵ  A zk̂ ).(B x î  B y ĵ  B z k̂ ) =3x2 –2x1 –4x1=0   Hence the given vectors form a right angled triangle. A . B  A xB x  A y B y  A zB z Ex.2 = ( î  3 ĵ  5k̂ )  (2 î  ĵ  4k̂ )  3 î  2 ĵ  k̂  a  Two vector A  î  2 ĵ  2k̂ Find (i) Dot product (ii) Angle between them A  (1)2  (2)2  (2)2  3 B  (1)2  (3)2  (6)2  46   A . B  (i  2 ĵ  2k̂ ).(i  3 ĵ  6k̂ ) = 1 × 1 + 2 × 3 + 2 × 6 = 19 (ii) Angle between them  B  î  3 ĵ  6k̂ Ex.5 Find the value of  so that the two vectors 2 î  3 ĵ  k̂ and  4 î  6 ĵ  k̂ are (i) parallel (ii) perpendicular to each other Sol. Let a  2 î  3 ĵ  k̂ and b  4 î  6 ĵ  k̂ (i) a and b are parallel to each other a1 a 2 a 3   b1 b 2 b 3 i.e. if 1 2 3    4 6 (ii) a and b are perpendicular to each other if a.b = 0 i.e. if 2(–4) + 3(–6) + (–1) () = 0  = –8 – 18 = –26 Motion in a Plane CROSS PRODUCT OR VECTOR PRODUCT OF TWO VECTORS     A B sin  n̂  A  B n̂  to plane of A and B Direction of n̂ is given by right hand thumb rule.   Curl the fingers of your right hand from A to B . Then the direction of the erect thumb will point in         (i)   r  F (ii) J  r  p       (iii) v    r (iv) a    r Properties of vector product (a) The vector product is ‘not’ commutative i.e   k̂  ĵ   î k̂  î  ĵ î  k̂   ĵ (ii) î  î  0 ĵ  ĵ  0 k̂  k̂  0 The magnitude of each of the vectors î, ĵ and k̂ is 1 and the angle between any two of them is 90º. Therefore, we write î  ĵ  (1) (1) sin 90º n̂  n̂ , where n̂ is a unit a vector perpendicular to the plane   A  A x î  A y ĵ  A z k̂ B  B x î  B y ĵ  B z k̂   ĵ  k̂  î of î and ĵ i.e. it is just the third unit vector k̂ . (g) The vector product of two vectors in terms of their x, y & z components can be expressed as a   determinant. Let A and B be two vectors. Let us write their rectangular components : the direction A B . Example : ˆj  î  k̂ (i)  Cross product of A and B inclined to each other at an angle is defined as :  î  ĵ  k̂      A  B  (A x î  A y ĵ  A zk̂ ) × ( B x î  B y ĵ  B zk̂ ) = (A yBz  By A z ) î  ( A xBz  Bx Az ) ĵ  (A xBy  Bx A z )k̂ î A  B  Ax Bx   ĵ Ay By k̂ Az Bz  A B   B  A A B  B A (b) The vector product is distributive i.e. Ex.6 The vector from origin to the points A and B are a = 3 î  6 ĵ  2k̂ and b = 2 î  ĵ  2k̂ respectively. Find (c) The magnitude of the vector product of two the area of vectors mutually at right angles is equal to the product (i) the triangle OAB of the magnitudes of the vectors. (ii) the parallelogram formed by OA and OB as     adjacent sides. A  B  AB sin 90º n̂  AB n̂, | A  B | AB Sol. Given OA = a = 3 î  6 ĵ  2k̂ and OB = b (if  = 90º)        A  ( B  C)  A  B  A  C (d) The vector product of two parallel vectors is a null vector (or zero vector)    A  B  AB (sin 0) n̂  0 or 0 (e) The vector product of a vector by itself is a null vector (zero vector)   = 2 î  ĵ  2k̂ î ĵ k̂  (a  b )  3  6 2 2 1 2  = (12 – 2) î – (–6 – 4) ĵ + (3 + 12) k̂  10 î  10 ĵ  15k̂  |a × b| = 10 2  10 2  15 2 = A  B  AA (sin 0) n̂  0 or 0 (f) The vector product of unit orthogonal vectors î, ĵ, k̂ have the following relations in the right handed coordinate system (i) Area of OAB = 1 2 |a × b| = 425 = 5 17 5 17 sq. units 2 (ii) Area of parallelogram formed by OA and OB as adjacent sides = |a × b| = 5 17 sq. units. Motion in a Plane Ex.7 Find a × b and b × a if (i) a  3k̂  4 ĵ, b  î  ĵ  k̂ (ii) a = (2, –1, 1) ; b = (3, 4, –1) î ĵ k̂ 0 4 3 Sol. (i) a × b = 1 1 3 b×a= Ex.8 A rope is stretched between two poles. A 50 N boy hangs from it, as shown in fig. Find the tensions in the two parts of the rope.  7î  3 ĵ  4k̂ î ĵ k̂ 1 1 1  7 î  3 ĵ  4k̂ 0 4 3 î ĵ k̂ 2 1 1 (ii) a × b =  3 î  5 ĵ  11k̂ 3 4 1 î ĵ k̂ 3 4 1 b×a= = 3 î  5 ĵ  11k̂ 2 1 1 Tensor Sol. In fig.  = 90º + 15º = 105º  and  = 180º – (30º + 15º) = 135º Using Lami’s Theorem, we have T1 T W  2  sin  sin  sin   If a physical quantity itself has no direction but has different values in different directions, then it is said to be TENSOR. = 90º + 30º = 120º sin  sin 105 º T1 = W × sin  = 50 × sin 135 º = 50 × sin 75 º 50  0.9659  sin 45 º 0.7071 W sin  = 68.3 N 50  sin120 º 50  sin 60 º T2 = sin  = = Ex. (a) Moment of inertia : It is not a vector as its sin 135 º sin 45 º direction is not be specified but has different values 50  0.8660 = 61.24 N = in different directions. Hence it is neither a scalar 0.7071 nor a vector but a Tensor. Ex.9 Three forces, 3N, 4N, 5N are acting on a body (b) Stress : It is defined as force per unit area (acting together and the body is in equilibrium . If the angle at a point). It has no direction so it is not a vector between 3N and 4N forces is 90º, then what is the but at a point it has different values for different areas angle between 3N and 5N forces? so it is not a scalar also. It is actually a tensor. Sol. The forces are shown in fig. Then according to Lami’s theorem LAMI’S THEOREM If three forces acting at a point are in equilibrium then each force is proportional to sine of the angle between the other two. or A sin  B C = sin  = sin  4 5 = sin(  90º ) sin(    ) Thus sin ( – ) =  4 5 4  1 –  = sin  5  = 53º   (remember the 3, 4, 5 triangle) Thus  = 180 – 53º = 127º Motion in a Plane Ex.10 The concurrent forces 8N, 6N and 15N are acting on an object. Is it possible to arrange the direction of forces in such a way that the object remain in equilibrium ? Sol. Consider 8N and 6N as, say F1 and F2 forces. Then Fmax = F1 + F2 = 14N Fmin = F1 – F2 = 2N For equilibrium, magnitude of the third force should satisfy, thus, 14  F3  2 which the F3 = 15 newton does not satisfy, Therefore, equilibrium under 8N, 6N, 15 N forces is not possible SPECIAL POINTS ABOUT VECTOR  * If a vector A is multiplied by zero, we get a vector whose magnitude is zero called null vector or Zero vector.   * The unit of n A is same as that of A , if n is a pure real number. * The unit of vector does not change on being multiplied by a dimensionless scalar.   * The unit of n A is different from that of A , if n is a dimensional scalar. * The multilication of velocity vector by time gives us displacement. * Sum of non-coplanar forces can never be zero. * Minimum number of equal forces required for a zero resultant is two. * Minimum number of unequal forces required for a zero resultant is three.