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Forces 5 Turning forces: 5.5 Applications of turning effects of forces including principle of the lever

Doc Brown's Physics exam study revision notes: Describing the applications of turning effects of forces, principle of the lever opening cans, punch press machines, moment calculations, how scissors, spanners, screwdrivers and wheelbarrows work

Index of physics notes: 5. Turning forces, calculating moments, problem solving

This page contains online questions only. Jot down your answers and check them against the worked out answers at the end of the page


5.5 Some applications of turning effects of forces including levers

Many of the examples described below involve a lever, which is a means of increasing the rotational effect of a force.

You push down on one end of a lever, and the rotation about the pivot point can cause the other end to rise with a greater force.

The diagram below illustrates the principle of lever to gain a mechanical advantage - its all about the d1/d2 ratio.

F is the force involved (N) and d (m) is the (shortest) perpendicular distance from the pivot point to the point where the force is applied OR generated.

F1 = the applied input force, d1 = the distance from the pivot to where the force F1 is perpendicularly applied

F2 = the output force, d2 = the distance from the pivot to where the force F2 is perpendicularly generated

When balanced, i.e. just before something is 'levered' and caused to move

input moment = F1 x d1 = output moment = F2 x d2

From F1 x d1 = F2 x d2, rearranging gives F2 = F1 x d1/d2

Therefore by making d1 much bigger than d2 you can produce a much bigger output force compared to the original input force.

Generally speaking you make the distance d1 much bigger than distance d2 - you can see this using scissors, levering a lid off a can  and its a very similar situation when using a fork to lift deep tough roots out of the soil or moving a heavy stone with a pole.

So, levers are very useful because they make lifting and moving things much easier by reducing the input force necessary to perform the task.

 

Q1 Example of a lever question

Suppose a heavy manhole cover requires a force of 160N to open it. Imagine you have a steel bar 1 metre long pivoted at a point 0.1 m (10 cm) from the end to lever open the heavy manhole cover.

If you press down with a force of 20 N, will the upward force you generate be enough to lift the manhole cover?

referring to the diagram again

input moment = F1 x d1 = 20 x 0.9 = 18 Nm

output moment = F2 x d2 = F2 x 0.1 Nm

Since F1 x d1 = F2 x d2

F2 = (F1 x d1) / d2 = 18 / 0.1 = 180 N

The output force 180 N exceeds the 160 N force required, so the manhole cover can be lifted.

 

Q2 A hole punch of some description

Punch press machine

The diagram shows a simple machine to punch holes in a sheet of material.

The lever is pulled down to produce the hole.

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.

Calculations based on the punch machine illustrated above

You can treat the moments as Ncm, but you may be asked to give the answer in Nm.

(a) If d1 is 10 cm, d2 120 cm and the pull down force F2 required is 2 N, what punch force F1 is produced?

moment = force x perpendicular distance to the turning point:

moment 1 = moment 2 = F1 x d1 = F2 x d2

so F1 x 10 = 2 x 120 = 240 Ncm (2.4 Nm)

therefore F1 = 240/10 = F1 = 24 N

(b) If d2 is 80 cm and the pull force F2 used is 4 N, what is the maximum length d1 can be to produce a minimum required punch force F1 of 40 N?

moment = force x perpendicular distance to the turning point:

moment 1 = moment 2 = F1 x d1 = F2 x d2,

so 40 x d1 = 4 x 80 = 320 Ncm, so d1 = 320/40 = d1 = 8 cm

(c) If d1 is 9 cm, d2 108 cm and the punch force F1 required is 60 N, what is the minimum pull down force F2 required?

moment = force x perpendicular distance to the turning point:

moment 1 = moment 2 = F1 x d1 = F2 x d2

9 x 60 = F2 x 108 = 540 Ncm

so F2 = 530/108 = F2 = 5 N

(d) When d1 is 12 cm, the punch force required F1 is 36 N if a pull-down force F2 of 3N is applied. What is the minimum length of the lever d2 required to punch the hole?

moment = force x perpendicular distance to the turning point:

moment 1 = moment 2 = F1 x d1 = F2 x d2,

so 36 x 12 = 3 x d2 = 432 Ncm, so d2 = 36 x 12 / 3 = d2 = 144 cm

 

Scissors

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

F1 x d1 = F2 x d2, rearranging gives F2 = F1 x d1/d2

So, by making d1 >> d2, you create a much bigger out force F2, sufficient to crisply cut through paper or card.

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 significant multiplication of the force you apply. Its the same principle as described in the whole punch machine described in (a) above.

 

Levering the lid of a can

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.

 

The relatively long handle of a spanner

 

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

Generally speaking, the larger the nut to be tightened, the longer the spanner.

Spanners were discussed in detail at the start of the page.

 

A 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.

 

The screwdriver

 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.

 

A wheelbarrow

The handles of the wheelbarrow are much further away from the wheel axis than the centre of gravity of the full wheelbarrow (shown by 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.

Note that since Work = force x distance, you can think of this lift as doing work against the Earth's gravitational field.

In the case of moments involving lifting objects vertically, each moment = gain in GPE = mgh,

GPE = gravitational potential energy (J), m = mass (kg), g = 9.8 N/kg, h = height lifted in m, and  for more on GPE

see also

Mass and the effect of gravity force on it - weight, (mention of work done and GPE )

 The crucial experimental result is 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!

Index of Forces 5. Turning forces, calculating moments, problem solving


Keywords, phrases and learning objectives for turning forces

Be able to describe, explain and apply the principle of moments to applications of the turning effects of forces including the principle of lever opening cans, operation of a  punch press machine, moment calculations of scissors, spanners and screwdriver and a wheelbarrow.


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Index Forces 5. Turning forces, calculating moments, problem solving

Worked out ANSWERS to the moment calculations

Q1 Example of a lever question

Suppose a heavy manhole cover requires a force of 160N to open it. Imagine you have a steel bar 1 metre long pivoted at a point 0.1 m (10 cm) from the end to lever open the heavy manhole cover.

If you press down with a force of 20 N, will the upward force you generate be enough to lift the manhole cover?

Worked out ANSWERS

 

Q2 A hole punch of some description

A punch press machine

The diagram shows a simple machine to punch holes in a sheet of material.

The lever is pulled down to produce the hole.

The pivot point (turning point) is on the left.

(a) If d1 is 10 cm, d2 120 cm and the pull down force F2 required is 2 N, what punch force F1 is produced?

Worked out ANSWERS

(b) If d2 is 80 cm and the pull force F2 used is 4 N, what is the maximum length d1 can be to produce a minimum required punch force F1 of 40 N?

Worked out ANSWERS

(c) If d1 is 9 cm, d2 108 cm and the punch force F1 required is 60 N, what is the minimum pull down force F2 required?

Worked out ANSWERS

(d) When d1 is 12 cm, the punch force required F1 is 36 N if a pull-down force F2 of 3N is applied.

 What is the minimum length of the lever d2 required to punch the hole?

Worked out ANSWERS

 

Index Forces 5. Turning forces, calculating moments, problem solving

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