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Forces 5 Turning forces: 5.7
Turning forces and
the physics of the human skeleton and muscle system
Doc Brown's Physics exam study revision notes
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
5.7
Turning forces and
the physics of the human skeleton and muscle system
1. The knee joint
Cartilage of joints
in your skeletal-muscle system, has many functions, including
the ability to resist compressive forces, enhance bone
resilience, and provide support on bony areas where there is a
need for flexibility e.g.
The cartilage is a softer material between
the bones of a joint and acts as a shock absorber to
minimise the chance of bone fracture damage.
Cartilage being a softer material between
the bones of a joint, reduces friction that would cause pain
and wear of the bones.
Cartilage connective tissue also acts as a
binding agent to hold the bones in place, but not too
rigidly, the bone-cartilage system must be flexible.
2. The arm joint
The comments on cartilage in 1. apply here too.
The arm joint is a lever system operated by the
expansion and contraction of muscles e.g.
to bend and raise your forearm the biceps
contract and the triceps relax, and,
to lower or straighten your forearm the
biceps relax and the triceps contract
3. The pelvic hip joint
The comments on cartilage in 1. apply here too.
Weight lifted and force
calculations based on the action of muscles
Some
example questions (answers)
ALL based on the diagram
above for the antagonistic biceps and triceps muscles of the elbow
joint
Assume the gravitational field constant is
10 N/kg Question 1
If d1 = 0.40 m, d2 = 5 cm, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a weight of 20 N.
Question 2
If d1 = 0.36 m, d2 = 0.06 m, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a
mass of 3 kg.
Question 3
If d1 = 45 cm, d2 = 5 cm, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a weight of 15 N.
Question 4
If d1 = 48 cm, d2 = 6 cm, using the above
diagram for help, and applying the principle of moments,
calculate the minimum force needed by the biceps contraction to
lift a weight of 2 kg.
Question 5
If d1 = 0.50 m, d2 = 0.05 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 150 N, calculate the maximum weight the person's arm can raise
and hold.
Question 6
If d1 = 0.60 m, d2 = 0.06 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 180 N, calculate the maximum
mass the person's arm can raise
and hold.
Question 7
If d1 = 0.50 m, d2 = 0.05 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 200 N, calculate the maximum weight the person's arm can raise
and hold.
Question 8
If d1 = 0.40 m, d2 = 0.06 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 240 N, calculate the maximum
mass the person's arm can raise
and hold.
WORKED out answers
Keywords, phrases and learning objectives for turning forces
Be able to describe and explain the application of turning forces and the physics of the human skeleton
and muscle system e.g. knee joints, arm joints and pelvic hip joints,
and where necessary be able to do calculations on the forces
involved.
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Index Forces 5.
Turning forces, calculating moments, problem solving
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Worked out
ANSWERS to the moment calculations Remember 1 kg mass is equivalent to 10 N on the
Earth's surface.
Q1 F1 x d1 = F2 x d2, convert cm to m, 20
x 0.40 = F2 x 0.05, F2 = 20 x 0.40 / 0.05 =
160 N
Q2 3 kg = 3 x 10 = 30 N, F1 x d1 = F2 x d2, convert cm to m, 30
x 0.36 = F2 x 0.06 x F2, F2 = 30 x 0.36 / 0.06 =
180 N
Q3 F1 x d1 = F2 x d2, convert cm to m, 15
x 0.45 = F2 x 0.05, F2 = 15 x 0.45 / 0.05 =
135 N
Q4 2 kg = 2 x 10 = 20 N, F1 x d1 = F2 x d2, convert cm to m, 20
x 0.48 = F2 x 0.06, F2 = 20 x 0.48 / 0.06 =
160 N
Q5 F1 x d1 = F2 x d2, F1 x 0.50 = 150 x
0.05, F1 = 150 x 0.05 / 0.50 =
15 N
Q6 F1 x d1 = F2 x d2, F1 x 0.60 = 180 x
0.06, F1 = 180 x 0.06 / 0.60 = 18 N, mass = 18 / 10 =
1.8 kg
Q7 F1 x d1 = F2 x d2, F1 x 0.50 = 200 x
0.05, F1 = 200 x 0.05 / 0.50 =
20 N
Q8 F1 x d1 = F2 x d2, F1 x 0.40 = 240 x
0.06, F1 = 240 x 0.06 / 0.40 = 36 N, mass = 36 /10 =
3.6 kg
Index Forces 5.
Turning forces, calculating moments, problem solving
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