Chapter 5: Turning effect of forces

Week 4

Introduction

Everyday examples

joint

Definition The turning effect of a force is called the

moment of a force (or torque). The moment of a force about a pivot is the

product of force and the perpendicular distance from the line of action of the force to the pivot.

SI unit is newton metre (N m).

moment of a force = F x d

Perpendicular distance?

Pivot Pivot

Perpendicular distance?

Pivot

2 Directions

Clockwise moment Anticlockwise moment

Anticlockwise or Clockwise moment?

Example 1i) Calculate the moment of the 10 N force about the pivot.

Clockwise moment = Fd = 10(0.20 ) = 2.00 N m

Example 1ii) Calculate the moment of the 10 N force about the pivot.

Clockwise moment = Fd = 10(0.15)

= 1.50 N m

Example 2The moment required to open a door is 40 N m. What is the minimum force required to open the door if the distance between the handle and the hinges is 80 cm?

See-saw

The see-saw is 4 m long.Suppose Ali of weight 800 N sits on the left end of the see-saw. What is the moment of his weight about the pivot?

See-sawSuppose Sam of weight 800 N sits on the right end of the see-saw. What is the moment of his weight about the pivot?

The anticlockwise moment is equal to the clockwise moment and the see-saw would be balanced. This is called the Principle of Moments.

The Principle of Moments states that when a body is in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot.

Conditions for equilibrium The resultant force on the object is zero. The sum of clockwise moments is equal to

the sum of anticlockwise moments.

Principle of Moments

Example 3

A daughter of mass 30.0 kg is sitting at one end of the see-saw of length 4.0 m. Where should the mother of mass 60.0 kg sit so that the see-saw would be in equilibrium?

By the Principle of Moments, taking moments about the pivot,Sum of anticlockwise moments = Sum of clockwise moments

30(10)(2) = 60(10)dd = 1.00 m

The mother should sit 1.00 m away from the pivot on the other side of the see-saw.

The diagram shows a decoration, which is made by suspending objects P, Q and R from light rods M and N. The weights of P, Q and R are such that the rods are horizontal. Calculate the weights of P and R if the weight of Q is 1.0 N.

Test yourself

What have you learnt today? moment of a force = F x d

The Principle of Moments states that when a body is in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot.

Q1A uniform metre rule balances horizontally about a pivot at its midpoint. A mass of weight 0.750 N is suspended at the 20.0 cm mark and another object of unknown mass is suspended at the 65.0 cm mark where the metre rule is horizontal. Determine the mass of the object.

Q2A metre rule is hung freely at the 50.0 cm mark. A 1.00 kg mass and a 400 g mass are hung at the 66.0 cm and the 90.0 cm mark respectively. Where should a 1.60 kg mass be hung so that the metre rule is at equilibrium?

Ruler demo The weight of the ruler caused a moment

about the pivot and hence it falls off.

So far, we have always take the point of the weight to be from the centre of the ruler.

But it may not be at the centre always!

Try finding the CG of the broom and pen!

Test Yourself

Test Yourself

Centre of Gravity (CG) The centre of gravity of any object is defined

as the point through which its whole weight appears to act.

How to find CG of irregular object?

Finding CG of irregular-shaped lamina

1. Make 3 small holes near the edge of the lamina.

2. Suspend the lamina through each hole using a pin.

3. Hang a plumbline on the pin in front of the lamina.

4. Slightly displace the lamina and release it.

5. When it comes to a stop, make 3 dots on the lamina along the plumbline.

6. Repeat for the other 2 holes.7. The point of intersection of the 3

lines is the position of the centre of gravity.

Example 4A non-uniform metre rule is supported on a pivot at the 40 cm mark. When a mass of 48 g is suspended at the 15 cm mark, the metre balances horizontally. If the weight of the metre rule is 0.80 N, calculate the position of centre of gravity of the metre rule. By Principle of moments, taking moments about the pivot,Sum of anticlockwise moments = sum of clockwise moments(0.048)(10)(25) = (0.80)dd = 15.0 cmPosition of centre of gravity = 40 +15.0

= 55.0 cm mark

Example 5A uniform plank of weight 80 N rests centrally on a pair of trestles placed 2.0 m apart as shown in the diagram. A pile of bricks weighing 60 N is placed on the plank so that its weight acts at a point of 0.5 m from trestle B. (a) Find the upward forces exerted on the plank by each trestle.(b) The bricks are moved towards trestle B. What is the distance

from the centre of the plank such that the plank just lifts from trestle A?

A B

Example 5(a) By Principle of moments, taking moments about A,Sum of anticlockwise moments = sum of clockwise moments(2.0) = (80)(1.0) + (60)(1.5)

Since resultant force on the plank = 0,Sum of upward forces = sum of downward forces+85.0 = 80 + 60 = 55.0 N

(b) By Principle of moments, taking moments about B,Sum of anticlockwise moments = sum of clockwise moments80(1.0) = 60dd = 1.33 mDistance from centre of plank = 1 + 1.33

= 2.33 m

Example 6A non-uniform plank of mass 20 kg is set up as shown. When a block is placed at A, the spring balance reads 180 N in order to keep the plank horizontal. When the same block is placed at B, the spring balance reads 720 N. a) Find the mass of the block.b) How far is the centre of gravity of the plank from the pivot?

A B1 m 6 m

8 m

Block Spring balance

Example 6Let x be the distance of the centre of gravity of the plank from the pivot.By Principle of moments, taking moments about the pivot,Sum of anticlockwise moments = sum of clockwise momentsW(1) + 180(6) = 20(10)xW + 1080 = 200x --------- (1)720(6) = 8W + 200x4320 = 8W + 200x ---------- (2)Solve both equations simultaneously… (you should know how to do)

x = 7.20 mW = 360 Nm = = = 36.0 kg

Stability The stability of an object is a measure of its

ability to return to its original position after it is slightly displaced.

Stable Equilibrium If the cone is slightly tilted,

Its centre of gravity rises before returning to its original height.

The line of action through its weight still lies within its base.

The moment of its weight about the contact point C causes the cone to return to its original position.

C

Unstable equilibrium If the cone is slightly tilted,

Its centre of gravity drops. The line of action through its weight lies outside its

base. The moment of its weight about the contact point

C causes the cone to topple.

C

Neutral equilibrium If the cone is slightly displaced,

Its centre of gravity remains at the same height. The moment of its weight about the point C is

zero. It stays in the position to which it is displaced.

C

Stable, Neutral or Unstable?

Factors affecting stability

Factors affecting stability

How much can they tilt?

Factors affecting stability Position of centre of gravity Base area

To increase stability of an object Centre of gravity should be as low as possible Base area should be as wide as possible

Balancing toys Where is the centre of gravity of these

objects?

Watch this: Balancing 2 forks on a toothpick http://www.youtube.com/watch?v=czsoNaf4MeY

What you have learnt (a) describe the moment of a force in terms of its turning

effect and relate this to everyday examples (b) recall and apply the relationship moment of a force (or

torque) = force × perpendicular distance from the pivot to new situations or to solve related problems (c) state the principle of moments for a body in equilibrium (d) apply the principle of moments to new situations or to

solve related problems (e) show understanding that the weight of a body may be

taken as acting at a single point known as its centre of gravity (f) describe qualitatively the effect of the position of the

centre of gravity on the stability of objects