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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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Lesson 1, Topic 5
In Progress

Two or More Vectors Acting at a Point

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When there are two or more vectors acting at a point, each of the vectors is resolved into the vertical and horizontal components.

After you have calculated the vertical component and the horizontal component of each vector, you then proceed to add all the vertical components to get a single vertical component.

In a similar fashion, you are also to add all the horizontal components to get a single horizontal component.

The directions of the vectors must not be ignored. If the calculated vertical component is on the negative side of the y-axis, your calculated vertical component must be negative ( -ve / minus ). Also, if your calculated horizontal component is on the negative side of the x-axis, then your calculated horizontal component must be negative ( – ve / minus ).

This is shown below. Consider forces F1, F2, F3 and F4 acting at a point.

Vect0 e1603731310453
FθVertical
Component (Fy)
Horizontal
Component(Fx)
F1θ1F1 sin θ1F1 cos θ1
F2θ2F2 sin θ2-F2 cos θ2
F3θ3-F3 sin θ3-F3 cos θ3
F3θ4-F4 sin θ4F4 cos θ4
FyFx

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

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

tan θ = \( \frac{F_y}{F_x} \)


Example:

VECTOR6 e1606758374735

Find the Magnitude and the Direction of the Resultant Forces.

FθVertical Component (Fy)Horizontal Component (Fx)
10N30°F1 sin θ1
= 10 sin 30°
= 10 x 0.5
= 5.0N
F1cos 30
= 10 x 0.8660
= 8.67N
15N60°F2 sin θ2
= 15sin 60°
= 15 x 0.8660
= 12.99N
-F2 cos θ2
= -15cos 60
= -15 x 0.5
= -7.5N
12N45°-F3 sin θ3
= -12sin 45°
= -12 x 7.7071
= -8.49N
-F3 cos θ3
= -12cos 45°
= -12 x 0.7071 
= -8.49N
14N50°-F4 sin θ4
= -14sin 50°
= -14 x 7.7660
= -10.73N
F4 cos θ4
= 14cos 50
= 14 x 0.6428
= 9.00N
Fy = -1.23NFx = 1.68N
xvc e1603734454619

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

R = \( \scriptsize 1.68^2 + (-1.23)^2 \)

R2 = 1.5129 + 2.8224

R = \( \scriptsize \sqrt {4.3353} \)

R = 2.082N

The direction of the force

tan θ = \( \frac{F_y}{F_x} \)

tan θ = \( \frac{-1.23}{1.68} \)

θ = \( \scriptsize tan^{-1} (-0.732) \\ = \scriptsize\: – 36.2^o \)

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