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SS2: PHYSICS - 1ST TERM

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



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A vector, V, generally has two components, namely the vertical component of a vector Fy, and the horizontal component of a vector, Fx.

The component of a vector is the effective value of the vector in a particular direction.

The horizontal component of a vector is the effective value of the vector in the horizontal direction (Fx)

trian e1603727128951

Use  SOH CAH TOA to calculate the horizontal component Fx

Fx = Adjacent, F = Hypotenuse, Therefore we use CAH

Cos θ = \( \frac{Adj}{Hyp} \)

Cos θ = \( \frac{F_X}{F} \)

Fx = F Cos θ

The effective value of any force or vector in the horizontal direction (Fx) or component is FCos(Fx)

The vertical component of a force is the effective value of a force/vector in the vertical direction (Fy)

trian e1603727128951

Use  SOH CAH TOA to calculate the vertical component Fy

Fy = Opposite, F = Hypotenuse, Therefore we use SOH

Sin θ = \( \frac{opp}{hyp} \)

Sin θ = \( \frac{F_y}{F} \)

Fy = F Sin θ

The effective value of any vector or force in vertical direction Fy is F Sin θ

VECTOR2 e1603728032238

The magnitude of the resultant vector;

R2 = \( \scriptsize F_y^2 + F_x^2 \)

R = \( \scriptsize \sqrt{F_y^2 + F_x^2} \)

The direction of the force

tan θ = \( \frac{F_y}{F_x} \)

Example 1:

A force of 30 N acts at an angle of 60° to the horizontal. Calculate the components of the vector.

exmple e1606759700617

Solution:

Values given in question: F = 30N, θ = 60°

Vertical component Fy = F x Sin θ 

Substitution:    Fy = 30 x Sin 60°
.                          Fy = 30 x 0.8660
                           Fy = 25.98 N

Horizontal component Fx = F x Cos θ 

Substitution:    Fx = 30 x Cos 60°
.                          Fx = 30 x 0.5
                           Fx = 15 N

Example 2:

A wire is tied to a nail on a vertical wall such that the wire makes an angle of 60° to the vertical wall. If a 20 N force is applied at the end of the rope, calculate:
(a) the force that will make the nail bend.
(b) the force that will pull the nail out from the wall.

Solution:

nail e1606762684276

Force F on the rope = 20 N, angle of rope to the vertical wall = 60°

a. The force that will bend the nail is the Vertical component Fy

Vertical component Fy = F x Sin θ 

Substitution:    Fy = 20 x Sin 60°
.                          Fy = 20 x 0.8660
                           Fy = 17.32 N

b. The force that will pull the nail out from the wall is the Horizontal component Fx

Horizontal component Fx = F x Cos θ 

Substitution:    Fx = 20 x Cos 60°
.                          Fx = 20 x 0.5
                           Fx = 10 N

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