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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
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    1 Quiz



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Lesson 4, Topic 4
In Progress

Conditions of Equilibrium under the Action of Parallel Coplanar Forces

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Coplanar forces are forces that lie in the same plane and parallel forces are forces whose lines of action are parallel to each other.

The conditions for equilibrium of coplanar parallel forces are;

1. The algebraic sum of the forces acting on a body in any given direction must be zero. In other words, the sum of the upward forces must be equal to the sum of the downward forces.

2. The algebraic sum of the moments of all the forces about any point on a body must be zero. i.e. the total clockwise moments about a point must be equal to the total anticlockwise moments about the same point.

Principle of Moment:

The principle of moments states that if a body is in equilibrium, the sum of clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

Example 1:

A 100m uniform rod has a load of 40N suspended at 20cm from one end. If the fulcrum is at the centre of gravity. Calculate the force that must be applied at its other end to keep it in horizontal equilibrium.

Coplanar forces eg 1

Solution:

Taking moments about O,

Clockwise moments about O = Anticlockwise moments about the same point.

\( \scriptsize F \: \times \: 50 = 40 \: \times \: 30 \\ \scriptsize F = \normalsize \frac{40 \: \times \: 30}{50} \\ \scriptsize F = 24 N \)

Example 2:

A light rod CD, sits on a pivot AB. A load of 30N hangs at 4m from the pivot support at A. Find the value of the reaction forces at A and B as shown

Coplanar 3

Solution:

Note: Because it’s a light rod we can ignore its weight which is meant to act at its centre.

From conditions of equilibrium of parallel forces, total upward forces = total downward forces.

E + F = 30N.

Using the principle of moments,

The total clockwise moments = total anticlockwise moments

Taking moments about A,

30 x 4 = F x (8 + 4)

30 x 4 = F x 12

F = \( \frac{30\: \times\: 4}{12} \)

F = 10N

∴ E = 30 – 10 = 20N

Alternative method:

We can take moments about B

E x 12 = 30 x 8

E = \( \frac{30 \: \times \: 8}{12} \)

E = 20N

F = 30 – 20 = 10N

This proves that we can take moments about A or B and get the same results.

Example 3:

The diagram below shows a light rod of negligible weight which is balanced under the action of different forces acting on it.

Determine the
(i) Value of x.
(ii) Length of the rod.

eg6

Solution:

(i) Total Clockwise moments = Total Anticlockwise moments

∴ 15 × x + [5 × (20 + x)] = 10 × 20 + [10 × (40 + 20)]

Open the brackets

15× + 100 + 5x = 200 + 600

Collect like terms

20x + 100 = 800

20x = 800 – 100

20x = 700

x = \( \frac{700}{20}\)

x = 35cm

(ii) Length of rod = 40 + 20 + 35 + 20

= 115cm

Evaluation Questions:

1. Distinguish between equilibrant and resultant forces
2. What is dynamic equilibrium?
3. State the conditions of equilibrium under the action of coplanar forces.
4. A metre rule is balanced at the 45cm mark. When a body of mass 50g is suspended at the 8cm mark, the balance point is found to be the 25cm mark. Find the mass of the metre rule.

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