FORCES 5. Turning forces and moments

 from spanners to wheelbarrows and equilibrium situations

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

 This page will help you answer questions such as:

 What is a moment? What is a mechanical advantage?

 How do you calculate the turning effect of a force?

 Why are the turning effects of a force so important?

 Where do we apply the advantages of the turning effect of a force?



An introduction to moments and mechanical forces of rotation

Forces can cause an object to rotate and the turning effect of the force is called a moment.

The magnitude of a moment can be easily calculated from the formula:

M = F x d, where M = the moment of a force (Nm), F = force applied (N)

and d (m) is the perpendicular distance from the pivot point to the line of action of the force.

This is illustrated by the simple diagram of the spanner above. The turning force is F x d.

The pivot point is the central axis of the bolt on which the nut is being turned by the spanner.

To tighten or loosen the nut you apply a force, to best advantage, at 90o to the spanner itself.

Applying the force at any other angle less than 90o reduces d and so reduces the effective moment of the force.

You determine the force F by how hard you push/pull the end of the spanner, but d is a fixed distance for give spanner.

This is one of many situations where you are applying a force to increase the effect of your hand muscles.

The size of the moment increases with increase in distance d or applied force F.

The longer the spanner, the greater the turning force generated, the greater the mechanical advantage.


Moment calculations and balancing situation (equilibrium)

The left diagram illustrates a balanced situation (equilibrium) where a ruler is pivoted in the middle and two weights w1 and w2 are placed at distances d1 and d2 from the pivot point. Remember weight = force in newtons.

The weights hang vertically so the force due to gravity is acting perpendicularly (at 90o) to the ruler

For the ruler to be balanced in a perfect horizontal position the two turning forces must be equal.

Here we use the terms clockwise moment and anticlockwise moment for the two turning effects of the forces involved.

anticlockwise moment = w1 x d1 (left-hand side of pivot), clockwise moment = w2 x d2 (right-hand side of pivot)

so when w1d1 = w2d2

the ruler is balanced horizontally, at equilibrium when the turning effects of the forces are equal.

 

This situation conforms with the principle of moments which states that when the total sum of the anti clockwise moments is equal to the total sum of the clockwise moments the system is in equilibrium and the object (system) will NOT turn. When a system is stable (no movement) or balanced it is said to be in equilibrium as all the forces acting on the system cancel each other out.

You see this when you do a simple experiment balancing a rule on a pencil and putting small weights on either side until balanced.

Similarly, when a the nut of a bolt is tightened, there comes a point when the moment you are applying is balanced by the opposing moment of the bolt and nut and you cannot tighten the nut anymore.

 


Examples of simple calculations using the above situation.

Predict what happens in the following situations (a) to (c)

1 kg = 1000 g and 100 cm = 1 m and for simplicity assume g = 10 N/kg (weight = mass x force of gravity)

Q1(a) Suppose d1 = 20 cm, w1 = mass of 25 g, d2 = 10 cm, w2 = mass of 50 g

but is it balanced?

anticlockwise moment = d1w1 = (20/100) x (10 x 25/1000) = 0.05 Nm

clockwise moment = d2w2 = (10/100) x (10 x 50/1000) = 0.05 Nm

In this case the anticlockwise moment = clockwise moment, so the ruler is balanced horizontally.

identical twins on a seesaw will be balanced !!

  BUT, by applying an alternating extra muscle turning force, you can have great fun !!!

Q1(b) Suppose d1 is 14 cm, w1 = mass of 52 g, d2 = 12 cm, w2 = mass of 60 g

but is it balanced?

anticlockwise moment = d1w1 = (14/100) x (10 x 52/1000) = 0.073 Nm

clockwise moment = d2w2 = (12/100) x (10 x 60/1000) = 0.072 Nm

In this case the anticlockwise moment > clockwise moment, so the ruler will rotate anticlockwise.

Q1(c) Suppose d1 is 2.5 m, w1 = mass of 55 kg, d2 = 3.0 m, w2 = mass of 50 kg

but is it balanced?

anticlockwise moment = d1w1 = 2.5 x (10 x 55) = 1375 Nm

clockwise moment = d2w2 = 3.0 x (10 x 50) = 1500 Nm

In this case the clockwise moment > clockwise moment, so the ruler will turn clockwise.

 

An example of using the principle of moments - old fashioned kitchen scales

The beam of the scales should be horizontal when the bowl and weights plate are empty (d1 = d2, w1 = w2).

When the object to be weighed is placed in the dish, the scales tip anticlockwise down on the left.

You then add weights until the beam is horizontally balanced again, thus giving the weight of the material e.g. flour in the bowl.


More complex calculations


Q2

If w1 is 12.5 N and 3.5 m from the pivot point, what weight w2 is required if placed at 2.5 m from the pivot to balance the beam?

anticlockwise moment = 12.5 x 3.5 = 43.75 Nm

clockwise moment = w2 x 2.5

To balance the moments must be equal so:

w2 x 2.5 = 43.75, therefore w2 = 43.75/2.5 = 17.5 m


Q3

A beam is placed evenly on a pivot point (fulcrum).

On one side a 10 N weight is placed 2 m from the pivot point and a 40 N weight a further 4 m from the pivot point.

How far from the pivot point must the centre of gravity of an 80 N weight be placed to perfectly balance the beam horizontally?

The principle of moments states that the sum of the clockwise moments must equal the sum of the anticlockwise moments to attain equilibrium. A moment (Nm) = F (N) x d (m)

The sum of the clockwise moments = (10 x 2) + (40 x {2 + 4}) = 20 + 240 = 260 Nm

To balance this the anticlockwise moment must = 240 Nm, 240 = 80 x d, d = 260/80 = 3.25

Therefore the 80 N weight must be placed on the left 3.25 m from the pivot point.

 


Q4 This calculation is the sort of thing civil engineers and architects have to consider in the construction of 'modern' buildings.

A 5.0 m aluminium beam is suspended by a steel cable from a concrete beam and 3.0 m along rests on a steel pole.

Assuming gravitational field force is 9.8 N/kg, calculate T, the tension in newtons in the supporting steel cable.

The weight of the aluminium beam = 9.8 x 200 = 1960 N.

moment = force x perpendicular distance from turning point.

You consider the weight of the aluminium beam to act through its centre of mass 2.5 m from either end, but it is 0.5 m from the steel rod which is effectively the pivot point on which you base your moment calculations.

Its actually the same situation as a wheelbarrow described further down the page!

anticlockwise moment = beam weight x distance from steel pole to centre of mass of aluminium beam = 1960 x 0.5 = 980 Nm

clockwise moment = tension in the steel cable x distance from steel cable to steel pole = T x 3.0

at equilibrium, i.e. balanced, the clockwise moment = anticlockwise moment

therefore: T x 3.0 = 980, so T = 980/3 = 327 N (3 sf)


Some simple applications of turning effects of forces

(a) A hole punch of some description

This machine can punch holes in a material. The pivot point (turning point) is on the left.

We can analyse this situation in terms of turning forces.

Applying the principle of moments: F1 x d1 = F2 x d2

Rearranging the equations gives: F1 = F2 x d2 / d1

Therefore by making d2 'long' and d1 'short' you considerably multiply the force F1 compared to F2.

So you are able to easily punch holes in a strong material e.g. sheet of metal.

For example, suppose d2 is 0.5 m (50 cm) and d1 0.05 m (5 cm)

F1 = F2 x 0.5/0.05, so F1 = 10 x F2

So the force you manually apply is multiplied 10 ten times, not bad for a little effort!

In other words you need less force to get the same moment.

 

(b) Scissors

When you press the scissor hands together you create a powerful turning force effect close to the pivot point.

That's why you apply the blades to whatever you are cutting as near as possible to the pivot point.

You don't cut using the ends of the scissor blades where you gain little mechanical advantage i.e no multiply of the force you apply. Its the same principle as described in the whole punch machine described in (a) above.

 

(c) Levers

You can use a broad bladed screwdriver to get the lid off a can of paint. The pivot point is the rim of the can.

The length of the screwdriver to the pivot point (d2) is much greater than tip of the screwdriver beyond the rim (d1).

F1 x d1 = F2 x d2, F1 = F2 x d2/d1, so if d2 is much bigger than d1, you get a great magnification of the force you apply (F2) to give a much greater up force (F1) to force the lid off.

Another example of needing less force to get the same moment to do the job of opening the can.

 

(d)

Spanners have long handles to give a strong turning force effect.

 

(e) Cork screw

The radius of the handle is much greater than the boring rod. The great difference in radius gives you a much greater torque (turning force effect) to bore into the cork stopper of a wine bottle.

 

(f) Screw driver

 The argument for a screwdriver is the same as for the corkscrew above. The greater the diameter of the screwdriver 'handle' compared to the diameter of the screw head, the greater the force (torque) you can apply to drive a screw into wood.

 

(g) Wheelbarrow

The handles of the wheelbarrow are much further away from the wheel axis  than the centre of gravity of the full wheelbarrow is the yellow blob!). The wheel axis is the pivot point about which you calculate the two moments involved.

F1 is the weight of the loaded wheelbarrow acting from its centre of mass (centre of gravity).

F2 is the force you exert to lift the loaded wheelbarrow.

 The two moments are as follows:

 The 'weight' moment F1 x d1 is a small moment to manage the weight of the wheelbarrow.

  (F1 acts down from the centre of mass/gravity)

 However the 'lift' moment is F2 x d2 and so a smaller force F2 is needed operating at the longer perpendicular distance d2 to lift up the wheelbarrow and its load.

  F1 x d1 = F2 x d2, so F2 = F1 x d1/d2

  So F2 is < F1

 The magnitude of the lifting force F2 is much less than the weight of the load, so you can lift the barrow and move it along. Another example of needing less force to get the same moment to do the job of lifting the wheel barrow to move it along.

 Six year old granddaughter Niamh can barely lift the wheelbarrow off the ground (just a few cm) but Granny Molly has no trouble lifting the barrow to move it along.

At an earlier age Niamh wasn't quite as interested in science!


Gears and cog wheels - a means of transmitting rotational effects

Cog wheels have teeth so that when several of them are combined together a rotational force can be transmitted when fitted in contact with each other. Through the interlocking, one cog wheel can turn another in the opposite direction. By using different size cog wheels differing in the number of teeth, you can increase or decrease the turning effect of the force applied.

If you transfer the force from a larger cog wheel (gear wheel of more teeth) to a smaller cog wheel (gear wheel of less teeth) you decrease the moment as you have decreased the distance from the pivot point.

The smaller cog wheel will be made to turn faster than the larger cog wheel.

This is a way to increase rotational speeds in machines.

If the first cog wheel has 20 teeth and the second cog wheel 5 teeth, one rotation of the first wheel causes the smaller wheel to rotate 20/5 times = 4.

If you transfer the force from a smaller cog wheel (gear wheel of less teeth) to a larger cog wheel (gear wheel of more teeth) you increase the moment as you have increased the distance from the pivot point.

The larger cog wheel will turn more slowly than the smaller one.

This is the way a relatively low powered machine can be made to lift heavy weights.

If the first cog wheel has 8 teeth and the second wheel 56 teeth, you need to turn the first wheel 56/8 = 7 times to completely rotate the second wheel.

 

(a) An old fashioned manual drill

The large cog wheel turns a smaller cog wheel at much greater speed.

The force is transmitted from one cog wheel to another.

Since a larger cog wheel of more teeth drives a smaller cog wheel of less teeth, the output is high turning speed of the drill.

 

(b) Gearing in mill wheel systems

Complex machines such as you find in older flour mills and textile mills, use gears to utilise the power of the e.g. water wheel, to transmit a force to drive the machinery with the required speed and power.

A slow rotating water wheel driving a cog wheel system ( a gear system) can produce high speeds of rotation to drive a spinning machine.

 

(d) ?

?

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