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# Parallelogram law of vector addition Questions and Answers

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## Rectangular components of a vector

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Note: vectors are shown in bold. scalars are shown in normal type

The diagram above shows two vectors **A** and **B** with angle p between them.

**R** is the resultant of **A** and **B**

**R** = **A** + **B**

This is the resultant in vector

R is the magnitude of vector **R**

Similarly A and B are the magnitudes of vectors **A** and **B**

R = √(A^{2} + B^{2} 2ABCos p) or [A^{2} + B^{2} 2ABCos p]^{1/2}

To give the direction of **R** we find the angle q that **R** makes with **B**

Tan q = (A Sin p)/(B + A Cos q)

A vector is completely defined only if both magnitude and direction are given.

Two forces of 3 N and 4 N are acting at a point such that the angle between them is 60 degrees. Find the resultant force

Magnitude R of the resultant force is R = √(3^{2} + 4^{2} + 2 x 3 x 4 Cos 60 deg)

= √(9 + 16 + 12) = √(37 = 6.08 N

Direction of R is given by finding the angle q

tan q = (3 Sin 60 deg)/(4 + 3 Cos 60 deg) = 0.472

q = tan^{-1} 0.472

= 25.3 deg

Thus **R** is 6.08 N in magnitude and is at an angle of 25.3 deg to the 4 N force.

A car goes 5 km east 3 km south, 2 km west and 1 km north. Find the resultant displacement.

First we will make the vector diagram

O to A 5 km east

A to B 3 km south

B to C 2 km west

c to D 1 km north

Net displacement is **OD**

Along the horizontal direction: 5 km east - 2 km west = 3 km east

Along the vertical direction: 3 km south - 1 km north = 2 km south

OD = √(3^{2} + 2^{2} + 2 x 3 x Cos 90 deg)

= √(3^{2} + 2^{2})

= 3.6 km

tan p = 2/3

or p = tan^{-1}2/3 = 34 deg

Thus resultant displacement is 3.6 km, 34 deg south of east.

To find the component of a vector along a given axis, we drop a perpendicular on the given axis from the vector

For example **OA** is the given vector. We have to find its component along the the horizontal axis. Let us
call it x-axis. We drop a perpendicular AB from A onto the x-axis. The length OB is the component of OA along
x-axis. If OA makes angle p with the horizontal axis, then in triangle OAB, OB/OA = Cos P or OB = OA Cos P.

Remember that component of a vector is a scalar quantity. If the component is along the negative direction, we put a (-) sign with it.)

Usually we resolve the vector into components along mutually perpendicular components.

OB is the x component OB = OA Cos p.

Similarly component along the vertical direction or the y axis is OC

OCAB is a rectangle.

So OC = AB

look at triangle OAB again,

AB/OA = Sin p

=> AB = OA Sin p = OC

Thus y component OC = OA Sin p.

Note that p is the angle with the horizontal axis.

Find the x and y components of a 25 m displacement at an angle of 210 deg.

OA is the displacement vector. The angle with the horizontal axis is 210 deg - 180 deg = 30 deg

x component = OB = -25 Cos 30 deg = -21.7

y component = AB = -25 Sin 30 deg = -12.5 m

Note that each component is pointing along the negative coordinate direction and thus we must take it as negative.

Now we will solve a problem using the component method

Find the resultant of the following two displacements: 2 m at 30 deg and 4 m at 120 deg. The angles are taken relative to the x axis.

Rx = 2 Cos 30 deg - 4 Cos 60 deg = - 0.268 m

Ry = 2 Sin 30 deg + 4 Sin 60 degg = 4.46 m

R = √(Rx^{2} + Ry^{2})

= √(-0.268^{2} + 4.46^{2}) = 4.47 m

tan q = Ry/Rx = 4.46/0.268

=> q = 86.6 deg

p = 180 deg - 86.6 deg = 93.4 deg