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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
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    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
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    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
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  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
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    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
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    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
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    1 Quiz



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This is used when two vectors are inclined at an angle to each other and it states that:

If two forces acting at a point are represented in magnitude and direction by the sides of the parallelogram drawn from that point, their resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from that point.

Derivation of the Law:

Let θ be the angle between B and D and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OE represents the resultant of B and D.

So, we have

Screenshot 2022 07 01 at 19.32.41

R = B + D

Now, expand O to F and draw EF perpendicular to OF.

Screenshot 2022 07 01 at 19.40.03

From ΔOFE,

OE2 = OF2 + EF2

OE2 = (OD + DF)2 + EF2 ……..(i)

In ΔDEF,

cos θ = \( \frac{DF}{DE}\)

∴ DF = DE cos θ

But DE = OB = B

∴ DF = OB cos θ = B cos θ

sin θ = \( \frac{EF}{DE}\)

∴ EF = DE sin θ

But DE = OB = B

∴ EF = OB sin θ = B sin θ

substituting the value of DF and EF in (i), we get:

OE2 = (D + DF)2 + EF2 ……..(i)

From the diagram OE = R

∴ R2 = (D + B cos θ)2 + (B sin θ)2

which can also be written as

⇒ R2 = (D + B cos θ)(D + B cos θ) + (B sin θ)(B sin θ)

expand the brackets

⇒ R2 = D2 + DB cos θ + DB cos θ + B2 cos2 θ + B2 sin2 θ

⇒ R2 = D2 + 2DB cos θ + B2 cos2 θ + B2 sin2 θ

collect like terms (i.e. B)

⇒ R2 = D2 + B2 cos2 θ + B2 sin2 θ + 2DB cos θ

⇒ R2 = D2 + B2 ( cos2 θ + sin2 θ ) + 2DB cos θ

but cos2 θ + sin2 θ = 1

∴ R2 = D2 + B2(1) + 2DB cos θ

⇒ R2 = D2 + B2 + 2DB cos θ

⇒ R = \( \scriptsize \sqrt{D^2 \: + \: B^2 \: + \: 2DB \:cos \: \theta}\)

Direction of the Resultant Vector:

Let \( \scriptsize \alpha \) be the angle made by resultant R with D.

Screenshot 2022 07 01 at 19.42.51

Then, \( \scriptsize tan \alpha = \normalsize \frac{EF}{OF} = \frac{EF}{OD \: + \: DF} \)

or, \( \scriptsize tan \alpha = \normalsize \frac{Bsin \theta}{D \: + \: Bcos \theta} \)

∴ \( \scriptsize \alpha = tan^{-1} \left(\normalsize \frac{Bsin \theta}{D \: + \: Bcos \theta} \right) \)

Example 1:

Two forces of magnitude 4N and 5N are inclined at an angle of 60° with each other. Calculate the magnitude of resultant and the angle made by resultant with 6N force.

Solution:

If B = 4N and D = 5N and θ = 60°,

We can use the formula;

⇒ R2 = D2 + B2 + 2DBcos θ

Substitute the values into the equation,

R2 = 42 + 52 + 2 x 4 x 5 x cos60°

R2 = 16 + 25 + 2 x 20 x cos 60°

R2 = 41 + 40 x 0.5

R2 = 41 + 20

R2 = 61

R = \( \scriptsize \sqrt {61} \)

R   = 7.81N

To find the angle made by the resultant we use the formula;

⇒ \( \scriptsize \alpha = tan^{-1} \left(\normalsize \frac{4 sin 60^o}{6 \: + \: 4cos 60} \right) \)

∴ \( \scriptsize \alpha = tan^{-1} \left(\normalsize \frac{4 \: \times \: 0.866}{6 \: + \: 4 \: \times \: 0.5} \right) \)

⇒ \( \scriptsize \alpha = tan^{-1} \left(\normalsize \frac{3.464}{8} \right) \)

⇒ \( \scriptsize \alpha = tan^{-1} \left(\scriptsize 0.433 \right) \)

∴ \( \scriptsize \alpha = 23.41^o \)

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