know parallelogram law of vector addition class 11 derivation magnitude method ?
Scalar and Vector Quantities
A scalar quantity has only magnitude but no direction, such as distance, speed, mass, area, volume, time, work, energy, power, temperature, specific heat, charge, potential, etc. A vector quantity has both magnitude and direction, such as displacement, velocity, acceleration, force, momentum, torque, electric field, magnetic field, etc.
Parallelogram Law of Vector Addition
The procedure of finding the resultant of two vectors is known as the parallelogram law of vector addition and may be stated as follows. If the two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point. Figure 2.1 show two vectors A and B of magnitudes A and B inclined at an angle α. The magnitude R of the resultant vector R is given by
The angle β which the resultant vector R subtends with vector A is given by
In vector notation, the resultant vector is written as R = A + B.
(i) When the two vectors are in the same direction, i.e. α = 0°, then
Therefore, the magnitude of the resultant is equal to the sum of the magnitudes of the two vectors. Also tan β = 0 or β = 0, i.e. the direction of the resultant is the same as that of either vector.
(ii) When the two vectors are in opposite directions, i.e., α = 180°, then
(iii) When the two vectors are at right angles to each other, i.e. α = 90°, then
Note: Rmax = A + B and Rmin = A – B
Triangle Law of Vector Addition
The parallelogram law of vector addition yields the triangle law of vector addition. In Fig. 2.1, vector QP = vector OS = B. In triangle OQP, vector OP = R. Hence, the triangle law of vector addition may be stated as follows: If the two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite order.
Subtraction of Vectors
Suppose we wish to subtract a vector B from a vector A. Since A – B = A + (– B) the subtraction of vector B from vector A is equivalent to the addition of vector – B to vector A. Hence the procedure to find (A – B) is as follows:
Choose a convenient scale and draw the vectors A and B as shown in Fig. 2.2 (a). If B is to be subtracted from A, draw the vector negative of B, i.e. draw the vector – B [see Fig. 2.2 (b)]. Now shift the vector – B parallel to itself so that the tail of – B is at the head of A. Vector C is the sum of vectors A and – B, i.e. [see Fig. 2.2 (c)] C = A + (– B) = A – B
The Unit Vector A vector whose magnitude is unity is called a unit vector. A unit vector is represented by A cap where
The unit vectors along x, y and z axes of a rectangular co-ordinate system are represented by i cap , j cap and k cap respectively
Resolution of a Vector into Rectangular Components
Consider a vector A in the x–y plane making an angle q with the x-axis. The x and y components of A are Ax and
Ay. The magnitudes of Ax and Ay are (Fig. 2.3)
Ax = A cos q, along x-direction
and Ay = A sin q, along y-direction
Thus Ax = Ax i = (A cos q) i
and Ay = Ay j = (A sin q) j
From parallelogram law A = Ax + Ay = (A cos q) i + (A sin q) j The magnitude of A in terms of the magnitudes of its rectangular components is
note : The magnitude of a component along + x direction or + y direction is taken to be positive while the magnitude of a component along – x direction or – y direction is taken to be negative.
Addition of Vectors Using Components
Two or more vectors are added by using components as follows:
(a) Resolve each vector into its rectangular components.
(b) Add the magnitudes of all the x-components taking into account their signs. Rx = sum of x-components of all the vectors
(c) Similarly Ry = sum of y-components of all the vectors
(d) Magnitude of resultant is
(e) The angle q which the resultant subtends with the x-axis is given by
Scalar or Dot Product
The scalar (or dot) product of two vectors A and B is defined as the product of the magnitudes of A and B and the cosine of the angle between them, i.e. A . B = AB cos θ The scalar product is represented by putting a dot between the two vectors. The scalar product of two vectors is a scalar quantity
The scalar product is represented by putting a dot between the two vectors. The scalar product of two vectors is a scalar quantity.
Properties of the Scalar Product
(i) The scalar product is commutative, i.e. A . B = B . A
(iii) If two vectors A and B are perpendicular to each other, then θ = 90° and A . B = AB cos 90° = 0.
Note that A . B can be zero when neither A nor B is zero.
(iv) For unit vectors i , j and k along the three field axes x, y and z, we have
Some Examples of Scalar Product
(i) Work done is W = F . s where F is the force vector and s is the displacement vector.
(ii) Power consumed is P = F . v where F is the force vector and v is the velocity vector.
(iii) Electric current is I = J . A where J is the current density vector and A is the area vector.
(iv) Magnetic flux is Θ = B. A where B is the magnetic field vector and A is the area vector.
Vector or Cross Product
If the smaller angle between two vectors A and B is θ, then the vector or cross product of vectors A and B is defined as A x B = (AB sin θ) n where A and B are the magnitudes of vectors A and B and n is a unit vector perpendicular to the plane containing A and B. The vector product of two vectors A and B is equal to a vector C, i.e.
A x B = C
The magnitude of vector C is given by C = AB sin θ The direction of C is perpendicular to the plane formed by A and B and is given by the right hand screw rule
Properties of a Vector Product
(i) Vector product is anticommutative, i.e. (A x B) = – (B x A)
(ii) A x A = 0, i.e. the vector product of a vector by itself is zero. This is because, in this case, θ = 0, and hence sin θ = 0. Therefore A x A = AA sin θ = 0 Hence, the condition for two vectors to be parallel (θ = 0°) or antiparallel (θ = 180°) is that their vector product should be zero.
If A x B = 0, it means either
(i) A is zero or,
(ii) B = 0 or
(iii) the angle θ between them is 0° or 180°.
(iii) The distributive law holds for both scalar and vector products, i.e.
A . (B + C) = A . B + A . C
A x (B + C) = A x B + A x C
(vi) i , j and k are the three mutually perpendicular unit vectors at the origin O and along OX, OY and OZ respectively; the right-hand rule gives:
Some Examples of Vector Product
(i) Torque = r x F, where r is the position vector and F is the force vector.
(ii) Linear velocity = w x r where w is the angular frequency vector and r is the position vector.
(iii) Angular momentum = r x p where r is the position vector and p is the linear momentum vector.
Position Vector and Displacement Vector
The position vector of a particle describes its instantaneous position with respect to the origin of the chosen frame of reference. It is a vector joining the origin to the particle and is denoted by vector r. For one-dimensional motion (say along x-axis), r = x i , where x is the distance of the particle from origin O. For two-dimensional motion (say in the x–y plane), r = xi + y j + , where (x, y) are the x and y coordinates of the particle. For three-dimensional motion, r = x i + y j + z k
If r1 is the position vector of a particle at time t1, and r2 at time t2, then the displacement vector is given by s = r2 – r1 Vector s is the resultant of vectors r2 and – r1
The displacement vector is a vector joining the initial and the final positions of the particle after a given interval of time and its direction is from the initial to the final position.
Instantaneous Velocity and Average Velocity
The rate of change of displacement with time at a given instant is called instantaneous velocity and is given by
The average velocity in a given time interval is defined as
The rate of change of velocity with time at a given instant is called instantaneous acceleration and is given by
Equations of One Dimensional Motion with Constant Acceleration
Let x0 be the position of a particle at time t = 0 and let u be its velocity at t = 0. It is given a constant acceleration a for time t. As a result it moves in a straight line to a position x and acquires a velocity v. The particle suffers a displacement s = x – x0 in time t. The equations of motion of the particle are
1. While solving numerical problems of bodies moving in a straight horizontal direction, we will consider only the magnitudes of u, v, a and s and take care of their direction by assigning positive or negative sign to the quantity. For example, + a will mean acceleration – a will mean retardation (or deceleration).
2. In the case of a body falling vertically under gravity or projected vertically upwards, we use the following sign conventions. Quantities directed vertically upwards are taken to be positive and those directed vertically downwards are taken to be negative. Since the acceleration due to gravity is directed downward for a body moving vertically up or falling vertically down, we take a = – g = – 9.8 ms–2 in Eqs.
3. Displacement in the nth second is given by sn = displacement in n seconds – displacement in (n – 1) seconds
(i) If a body moving with constant acceleration, starts from A with initial velocity u and reaches B with a velocity v, then the velocity midway between A and B is
(ii) A body starting from rest has an acceleration a for a time t1 and comes to rest under a retardation b for a time t2. If s1 and s2 are the distances travelled in t1 and t2,
(iii) At time t = 0 a body is thrown vertically upwards with a velocity u at time t = T, another body is thrown vertically upwards with the same velocity u. The two bodies will meet at time
(iv) A body is dropped from rest and at the same time another body is thrown downward with a velocity u from the same point, then (a) the acceleration of each body is g, (b) their relative velocity is always u, (c) their separation will be x after a time t = x/ u
(v) From the top of a building, body A is thrown upwards with a certain speed, body B is thrown downwards with the same speed and body C is dropped from rest from the same point. If t1, t2 and t3 are their respective times of reach the ground, then
(vi) A body of mass m is dropped from a height h on a heap of sand. If it penetrates a depth x in the sand, (a) the average retardation in sand is given by
because loss in PE (mgh) = work done against the resistive force of sand (max). (b) total average force exerted by sand is F = mg + ma = m(g + a)
(vii) A body is thrown vertically upward with a velocity u. If the resistive force due to air-friction produces a constant acceleration (or retardation) a (< g) (a) the net acceleration during upward motion = g + a, (b) the net acceleration during downward motion = g – a, (c) the maximum height attained is
Some useful tips
(i) If a body, starting from rest, moves with a constant acceleration, the distances covered by it in 1s, 2s, 3s …… are in the ratio 12 : 22 : 32 … = 1 : 4 : 9 ……
(ii) If a body starts from rest and moves with a contant acceleration, the distance covered by it in the lst, 2nd, 3rd….seconds are in the ratio 1 : 3 : 5…
(iii) A body is projected upward with a certain speed. If air resistance is nelected, the speed with which it hits the ground = the speed of projection.
(iv) If a body is projected upwards with a velocity u, the maximum height attained is proportional to u2 and the time of ascent is proportional to u.
(v) For a freely falling body, (a) velocity ∝ time (b) distance fallen ∝ (time)2
Relative Velocity in One Dimension
If two bodies A and B are moving in a straight line with velocities vA and vB respectively, the relative velocity of A with respect to B is defined as vAB = vA – vB The relative velocity of B with respect to A will be vBA = vB – vA
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