SITEMAP   School Physics Notes: Forces 5.1 Turning forces and skeletal muscles

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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

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

Repeat of calculations in biology notes from Part 3 of

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.

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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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

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