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. If a resultant force acts on an object about a fixed turning point (the pivot) it will cause the object to rotate e.g. turning a nut with a spanner, applying a screwdriver, opening a door fixed on hinges. The pivot might also be called the fulcrum. The rotational or turning effect, the moment, has a magnitude easily calculated from the formula:
The size of the moment increases with increase in distance d or applied force F.
More on the physics of unscrewing a tight nut!
Relative comments on the three 'moment' situations A to C
Some simple moment calculations Q1
Q2 A force of 20 N is applied to a door causing a moment of 5 Nm.
Q3 What force must applied to a 30 cm long spanner to generate a moment of 6.0 Nm?
Q4 ? 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 90^{o}) 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 (lefthand side of pivot), clockwise moment = w2 x d2 (righthand side of pivot)
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.
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
Q1(b) Suppose d1 is 14 cm, w1 = mass of 52 g, d2 = 12 cm, w2 = mass of 60 g
Q1(c) Suppose d1 is 2.5 m, w1 = mass of 55 kg, d2 = 3.0 m, w2 = mass of 50 kg
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?
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.
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
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.
When balanced, i.e. just before something is 'levered' and caused to move
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.
(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.
(b) 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.
(c) 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.
(d) Spanners have long handles to give a strong turning force effect. Spanners were discussed in detail at the start of the page.
(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.
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 are circular discs with teeth and components of many machines in transport and industry. When several of them are combined together (linked in contact) a rotational force can be transmitted when fitted in contact with each other.
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 of the 2nd as you have decreased the distance from the applied force to axle pivot point.
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 applied force to the axle pivot point.
Examples of gear wheel (cogwheel) applications (a) An old fashioned manual drill
(b) Gearing in mill wheel systems
(c) Clocks
(d) ?
Gear wheel questions
Forces revision notes index FORCES 2. Mass and the effect of gravity force on it  weight, (mention of work done and GPE) FORCES 3. Calculating resultant forces using vector diagrams and work done FORCES 4. Elasticity and energy stored in a spring FORCES 5. Turning forces and moments  from spanners to wheelbarrows and equilibrium situations FORCES 6. Pressure in liquid fluids and hydraulic systems IGCSE revision notes moment calculations uses of turning forces KS4 physics Science notes on moment calculations uses of turning forces GCSE physics guide notes on moment calculations uses of turning forces for schools colleges academies science course tutors images pictures diagrams for moment calculations uses of turning forces science revision notes on moment calculations uses of turning forces for revising physics modules physics topics notes to help on understanding of moment calculations uses of turning forces university courses in physics careers in science physics jobs in the engineering industry technical laboratory assistant apprenticeships engineer internships in physics USA US grade 8 grade 9 grade10 AQA GCSE 91 physics science revision notes on moment calculations uses of turning forces GCSE notes on moment calculations uses of turning forces Edexcel GCSE 91 physics science revision notes on moment calculations uses of turning forces for OCR GCSE 91 21st century physics science notes on moment calculations uses of turning forces OCR GCSE 91 Gateway physics science revision notes on moment calculations uses of turning forces WJEC gcse science CCEA/CEA gcse science

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