Topic 7.4
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Putting one’s foot down
Aims
In this activity you will read an article called ‘Forces can be larger than expected’ and use
your knowledge of vectors to understand and explain the meaning of various terms. You will
show your understanding by drawing a force diagram and calculating forces using moments
and vector addition. You will gain experience in naming forces and drawing them in the
correct position and direction.
Forces can be larger than expected
Read the text that follows and then answer the questions.
Free body diagrams
Drawing a free body diagram is a technique for investigating parts of a system rather than the
whole of the system. The part in question is treated as the ‘free body’ and all the forces acting
on it are drawn across the boundary surface that separates it from the rest of the body.
Consider the free body analysis of the foot of someone standing on tiptoe on one leg. The
‘free body’ is the foot and the boundary of the free body is through the junction of the foot
bones with the leg bones and the Achilles tendon. This tendon is pulled by the calf muscle.
When the muscles are tightened, they produce a pulling force or tension T. In this example, it
is an upward force. The ground also pushes upwards on the toe, with a force usually known
as the reaction force R. If the foot is in equilibrium then the leg bones must push downwards
with a force F. This is shown above in the ‘free body force analysis’ diagram.
If the person is standing still, their weight W must be equal to R. This can be shown with a
free body analysis of the whole body.
At equilibrium, F = T + R. The pull of the calf muscle adds to the reaction force so that the
force in the leg bones is larger than the weight. If the person is jumping upwards then the
reaction R must be greater than the person’s weight and the force in the leg bones is even
larger.
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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Frictional forces
Frictional forces are necessary during walking and running to prevent
the foot slipping.
Heel-strike occurs as the heel first touches the ground.
The force acting up on the leg is the resultant of the frictional force F
and the reaction force R of the ground.
Because of the geometry of the vector diagram, when the two forces F
and R are added together the resultant force is larger than either.
This means that the force acting up on the leg is larger than the simple reaction force acting
upwards from the ground. On an average walking surface the largest that F can be before the
foot starts to slip is 15% greater than R.
Questions
1 The article mentions three reasons why the total force in the leg bones can be larger than
the weight of a person. What are they?
2 What is meant by ‘equilibrium’ (beginning of paragraph 4 of ‘Free body diagrams’) and
what are the conditions for equilibrium?
3 The diagram below shows the situation when a person carries a weight in one hand. C is
the centre of gravity of the hand and forearm.
(a) On the diagram, draw a free body force analysis for the hand and forearm. Your
diagram should show four labelled forces. Show clearly the direction of the force on
the forearm caused by the bone in the upper arm, the humerus.
(b) Considering the elbow as a pivot, suggest why the force in the biceps muscle is larger
than the total weight of the arm and the weight being carried.
4 When a ballerina stands on tiptoe the reaction force R is 400 N. The dimensions of her
foot are shown in the ‘free body force analysis’ diagram in the article.
Calculate:
(a) the force T in the Achilles tendon
(b) the force F in the leg bones.
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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5 (a) Draw a vector diagram showing how the force acting on the leg during heel-strike is
related to the frictional force and the reaction force on the ground.
(b) Explain why a vector diagram shows that ‘when the two forces F and R are added
together the resultant force is larger than either’. (Hint: see second-to-last paragraph.)
6 The leg of a man first makes an angle of 30° with the vertical during heel-strike. The
reaction force upwards due to the ground is 1000 N.
You may assume that, at heel-strike, the resultant force acting up on the leg acts along the
direction of the leg.
(a) Calculate the frictional force between the heel and the ground.
(b) Calculate the force acting up on the leg.
Is the heel likely to slip when it hits at this angle? (Hint: use the last sentence in the
article.)
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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Topic 7.4
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Aims

To check comprehension of concepts of forces and simple ideas of equilibrium.

To draw a free body force diagram and use moments to calculate forces.

To draw a vector diagram and use this to calculate forces.
Teaching notes
Students should have covered moments and addition of vectors before tackling this exercise.
To draw a free body force diagram, they may need to be shown how to isolate part of the
body and consider the forces acting on the various parts in isolation. When taking moments,
students should be encouraged to state clearly the point about which they take moments.
Some students will need help in deciding the point about which to take moments where the
‘pivot’ is not clear.
Answers
7 The force in the muscle adds to the force due to the ground; when jumping upwards, extra
force is needed for the acceleration; when landing, frictional force adds to the reaction
force of the ground.
8 Equilibrium is a stable condition, where the total force on the body and the total moment
on the body are both zero.
9 (a) The diagram should show: three forces on the forearm acting downwards (the weight
of the arm acting at C, the weight of the object in the hand and the reaction force on
the humerus, which acts downwards at the elbow); and one force acting upwards,
along the biceps muscle.
(c) The biceps muscle is closer to the elbow so, to create the same moment, the force in
the biceps muscle must be larger than the total weight of the arm and the weight
being carried. Considered another way, there is a reaction force downwards at the
elbow which adds to the weight and must be balanced by the force in the biceps
muscle.
10 (a) The principle of moments states that if a body is in equilibrium, the sum of the
clockwise moments equals the sum of the anticlockwise moments. Stating this
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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another way, the sum of the moments about any point should be zero. Taking
moments about the junction of the leg and foot bones (that is, where F acts):
400  15 = T  6
So, T = 1000 N
(d) Moments can be taken about where the tendon meets the foot bone or about the toe
but it is easier just to recognise that the upward forces must balance the downward
force.
F = T + R = 1000 + 400 = 1400 N
11 (a) The vector diagram should comprise a simple right-angled triangle with F horizontal,
R vertical and the force acting on the leg as the hypotenuse. The arrows on the two
forces added together must be ‘head to tail’ as shown in the following diagram.
Alternatively, a parallelogram of forces can be drawn.
(e) The two forces are at right angles and so the resultant is the hypotenuse in a rightangled triangle. The hypotenuse is always larger than either of the sides of a triangle.
12 (a) In the vector triangle, F = 1000 tan 30° = 577 N
(f) Resultant force acting up on the leg =
1000
= 1155 N
cos 30
or Pythagoras’ theorem may be used.
(g) 15% larger means 1.15 times larger so R + 15% = 1000  1.15 = 1150 N.
Since F = 580 N, it is less than R + 15% so the heel is not likely to slip at this angle.
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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