The Achilles tendon is attached to the rear of the foot as shown in (Figure 1) . A person elevates himself just barely off the floor on the "ball of one foot." Assume the person has a mass of 80kg and D is twice as long as d.Find the tension FT in the Achilles tendon (pulling upward).
weight = 80*9.81 Newtons
I do not have your diagram but assume a fulcrum and two lever arms in which case
weight * lever arm from weight to fulcrum = tension * lever arm from fulcrum to tendon
To find the tension FT in the Achilles tendon when a person elevates himself on the "ball of one foot," we can use the principles of torque and equilibrium.
Let's break down the problem step-by-step:
Step 1: Identify the forces acting on the person.
The only two forces acting on the person are the gravitational force (mg) pulling downward and the tension force in the Achilles tendon (FT) pulling upward.
Step 2: Determine the torque about the point where the Achilles tendon is attached.
Since the person is barely off the floor on the "ball of one foot," we can assume that the person's body can be approximated as a uniform rod. The distance between the point of attachment of the Achilles tendon and the person's center of mass is given as D. The torque (τ) is equal to the force (F) multiplied by the perpendicular distance (r) from the point of rotation (attachment point) to the line of action of the force. In this case, the torque is given by τ = FT × D.
Step 3: Set up the equilibrium equation.
Since the person is in equilibrium (not rotating), the sum of the torques about any point should be zero. In this case, we can write the equilibrium equation as:
τclockwise + τcounterclockwise = 0
Since the only torque is in the clockwise direction, we can simplify the equation to:
τclockwise = 0
So, FT × D = 0
Step 4: Solve for the tension FT.
Since we have FT × D = 0, and D is given as twice the length of d, we can substitute D = 2d into the equation:
FT × 2d = 0
Simplifying the equation, we have:
2d × FT = 0
Now, since the product of 2d and FT equals zero, either FT or 2d must be zero. Given that D represents a non-zero length, we conclude that FT must be zero.
Therefore, the tension FT in the Achilles tendon when a person elevates himself just barely off the floor on the "ball of one foot" is zero.
To find the tension FT in the Achilles tendon, we can use the principle of torque equilibrium. The torque exerted by the person's weight must be balanced by the torque exerted by the tension in the Achilles tendon.
Here's how you can solve the problem step by step:
Step 1: Draw a free-body diagram of the person's foot and lower leg. Label the weight of the person's body as mg acting at the center of mass, the tension in the Achilles tendon as FT, and the distances from the pivoting point (the ball of the foot) to the weight and to the tension as r1 and r2, respectively. Since D is twice as long as d, we can express r1 as 2d.
Step 2: Write the torque equilibrium equation. The torque exerted by the weight is given by the product of the weight (mg) and the distance r1, and the torque exerted by the tension is given by the product of the tension (FT) and the distance r2. Since the person is in equilibrium, these torques must balance each other.
Therefore, we have:
mg * r1 = FT * r2
Step 3: Solve the equation for FT. Substitute the values we know into the equation. The weight of the person is 80 kg multiplied by the acceleration due to gravity (approximately 9.8 m/s^2). The distance r1 is 2d, and r2 is D.
80 kg * 9.8 m/s^2 * 2d = FT * D
Step 4: Simplify the equation. You can cancel out the mass (kg) and the acceleration due to gravity (m/s^2) to simplify the equation further.
156.8 d = FT * D
Step 5: Solve for FT. Divide both sides of the equation by D.
FT = (156.8 d) / D
Now you can substitute the given values for d and D into this equation to find the tension FT in the Achilles tendon.