As part of an exercise program, a 75 kg person does toe raises in which he raises his entire body weight on the ball of one foot. The Achilles tendon pulls straight upward on the heel bone of his foot. This tendon is 25 cm long and has a cross sectional area of 78 mm^2 and a Young's Modulus of 1470 MPa. (a) Make a free body diagram of the person's foot (everything below the ankle joint). You can neglect the weight of the foot. (b) what force does the Achilles tendon exert on the heel during the exercise? (c) By how many millimeters does the exercise stretch his Achilles tendon?
(a)
Free Body Diagram:
Person's Foot
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Achilles Tendon (pulling upward)
(b) The force exerted by the Achilles tendon on the heel during the exercise is equal to the person's body weight, or 75 kg x 9.8 m/s^2 = 735 N.
(c) The amount of stretch in the Achilles tendon can be calculated using Hooke's Law, which states that the amount of stretch is equal to the force divided by the Young's Modulus. Therefore, the amount of stretch in the Achilles tendon is 735 N / 1470 MPa = 0.5 mm.
(a) To make a free body diagram of the person's foot, we need to consider the forces acting on it. Since we are neglecting the weight of the foot, the forces acting on the foot are the force exerted by the Achilles tendon and the reaction force exerted by the ground. Let's label these forces:
- F_Achilles: Force exerted by the Achilles tendon on the heel bone.
- F_ground: Reaction force exerted by the ground on the foot.
Since the exercise involves raising the body weight on the ball of one foot, we can assume that the ground reaction force, F_ground, acts vertically upwards. The force exerted by the Achilles tendon, F_Achilles, acts in the opposite direction.
(b) To determine the force exerted by the Achilles tendon on the heel during the exercise, we need to balance the forces acting on the foot. Since the person is lifting their entire body weight on the ball of one foot, the resultant force should be zero (in equilibrium). Thus, we have:
F_ground + F_Achilles = 0
Since we want to find the force exerted by the Achilles tendon (F_Achilles), we rearrange the equation:
F_Achilles = -F_ground
The magnitude of the force exerted by the Achilles tendon is equal to the magnitude of the ground reaction force. Therefore, the force exerted by the Achilles tendon on the heel during the exercise is the same as the person's body weight.
(c) To calculate how much the exercise stretches the Achilles tendon, we can use Hooke's Law, which states that the force exerted by a spring (or tendon) is directly proportional to its extension. Hooke's Law can be expressed as:
F = k * Δl
Where:
F is the force applied to the tendon,
k is the spring constant (a measure of the tendon's stiffness),
Δl is the change in length of the tendon.
In this case, the force exerted by the Achilles tendon (F) is equal to the person's body weight, which we can find using their mass (m) and the acceleration due to gravity (g):
F = m * g
To find the change in length of the tendon (Δl), we can use the equation:
Δl = (F * L) / (A * E)
Where:
L is the original length of the tendon,
A is the cross-sectional area of the tendon,
E is the Young's Modulus of the tendon.
Given that the original length of the Achilles tendon (L) is 25 cm (or 0.25 m), the cross-sectional area (A) is 78 mm^2 (or 78 * 10^-6 m^2), and the Young's Modulus (E) is 1470 MPa (or 1470 * 10^6 Pa), we can substitute these values into the equation to find Δl.
(a) A free body diagram of the person's foot (everything below the ankle joint) can be represented as follows:
UPWARD FORCE
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+-----+-----+
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| Foot |
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+-----+-----+
\
\
\
GRAVITY
(75 kg)
The upward force represents the force exerted by the Achilles tendon on the heel, while the downward force represents the gravitational force acting on the person's foot.
(b) To calculate the force exerted by the Achilles tendon on the heel during the exercise, we can use the equation:
Stress = Force / Area
Given that the cross-sectional area of the Achilles tendon is 78 mm^2 = 78 × 10^-6 m^2, and the Young's Modulus is 1470 MPa = 1470 × 10^6 Pa, we can rearrange the equation to solve for force:
Force = Stress × Area
Since the stress is unknown, we need to calculate it using the applied force (resultant force from exercise) and the initial length (no stretching).
(c) The amount of stretching δl can be calculated using Hooke's Law:
Stress = Young's Modulus × Strain
By rearranging the equation, we can solve for strain:
Strain = Stress / Young's Modulus
Then,
δl = Strain × Initial Length
Let's calculate the force exerted by the Achilles tendon and the amount of stretching it causes.
Please provide the applied force or the height the person raises his body weight.