# Turning effect of a force | High School Physics Form 2

#### PHYSICS REVISION QUESTIONS

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## Turning Effects: Introduction

- The turning effect of a body is called the moment of that force.
- The turning effect produced depends on both the size of the force and the distance from the pivot.
- The moment of a force about a point is the product of the force applied and the perpendicular distance from the pivot (or turning point) to the line of action of the force.
Hence, Moments of a force = Force x perpendicular distance from pivot.

## The Law of Moments

- It states that when a body is in balance or in equilibrium, the sum of the clockwise moments equals the sum of anti-clockwise moments.
- The SI units of the moments of a force is Newton metre (Nm).

#### Practice Example 1

1. A uniform rod of negligible mass balances when a weight of 3 N is at A, weight of 3 N is at B and a weight of W is at C. What is the value of weight W?

##### Solution
Taking moments about the fulcrum, O then
Anti-clockwise moments = (3 x 1) + (3 x 3)
= 3 + 9 = 12 Nm
Anti-clockwise moments = clockwise moments
3 W = 12 Nm
W = 4 N

#### Practice Example 2

2. The following bar is of negligible weight. Determine the value of ‘x’ if the bar is balanced.

##### Solution
The distance from the turning point to the line of action can be determined as;

Clockwise moments = 10 × 30 = 300 N cm,
Anticlockwise moments = 10 × ‘x’ = 10 x. N cm.
Using the principle of moments
Anti-clockwise moments = clockwise moments
10 x = 300, hence x = 30 cm.

#### Practice Example 3

3. Study the diagram below and determine the value of X and hence the length of the bar.

##### Solution
Clockwise moments = 15x N + 5(X x 20) N
Anticlockwise moments = (20 x 10) + (60 × 10) N cm = 800 N cm.
Anti-clockwise moments = clockwise moments
800 N cm = 15X + 5X + 100
800 n cm = 20X + 100
20X = 700
X = 35 cm.
Therefore, the length of the bar = 40 + 20 + 35 + 20 = 115 cm.

## The Lever

-A lever is any device which can turn about a pivot or fulcrum.
-The applied force is called the effort and is used to overcome the resisting force called the load.

#### Practice Example 1

Consider the following diagram. (The bar is of negligible mass). Determine the effort applied.

##### Solution
Taking moments about O. Then, clockwise moments = effort x 200 cm.
Anticlockwise moments = 200 x 30 cm.
Effort = (200 x 30)/ 200 = 30 N.

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