The Achilles tendon is attached to the rear of the foot as shown in the figure below. When a person elevates himself just barely off the floor on the ball of one foot, estimate the tension Ft in the Achilles tendon (pulling upward) and the (downward) force Fb exerted by the lower leg bone on the foot. Assume the person has a mass of 72 kg and D is twice as long as d.
Here's the figure:
img687.imageshack.us/img687/663/img3552n.jpg
I have no idea where to start.
Well, it seems like you're trying to solve a physics problem. Now, I must admit, physics and I have a bit of a love-hate relationship. But hey, what's life without a little challenge, right?
Now, let's take a look at the figure you've posted. Ah, yes. The good old human anatomy. Always a classic.
So, we have a person raising themselves on the ball of one foot. They have a mass of 72 kg. Quite the heavy lifter, aren't they?
Now, to estimate the tension in the Achilles tendon (Ft) and the downward force exerted by the lower leg bone on the foot (Fb), we'll need to do a little bit of mathematical gymnastics. And no, I'm not talking about doing somersaults. Although that would definitely make physics more entertaining.
Let's take a closer look at the figure. The Achilles tendon is attached to the rear of the foot, which means it's pulling upward (Ft). Oh, Achilles, always causing trouble with your epic battles.
Now, here's the fun part. We need to take into account the lengths of D and d. Remember, D is twice as long as d. Talk about inequality in the anatomical world.
To estimate the tension in the Achilles tendon (Ft), we can consider the person's center of mass. Since they're raising themselves on one foot, the force exerted by the lower leg bone on the foot (Fb) will have to balance the gravitational force acting on the person.
But wait, there's more! We also need to consider the lever arms of D and d. Remember, they play an important role in determining the tension in the Achilles tendon (Ft) and the downward force (Fb).
So, I suggest you grab a pen, a calculator, and maybe a stress ball, because this problem might give your brain a little workout. And hey, if all else fails, just remember that laughter is the best medicine. Especially when physics is involved.
To estimate the tension in the Achilles tendon (Ft) and the force exerted by the lower leg bone on the foot (Fb), we can use the principles of static equilibrium. In static equilibrium, the sum of the forces acting on an object must be zero, and the sum of the torques (or moments) about any point must also be zero.
Let's start by analyzing the forces acting on the person in this situation. There are three main forces to consider: the weight of the person (mg), the tension in the Achilles tendon (Ft), and the force exerted by the lower leg bone on the foot (Fb).
1. Weight of the person (mg):
The weight of the person is given as 72 kg. The force due to gravity acting on the person can be calculated by multiplying the mass (m) by the acceleration due to gravity (g). The value of g is approximately 9.8 m/s^2.
Weight (mg) = mass (m) × acceleration due to gravity (g)
Weight (mg) = 72 kg × 9.8 m/s^2
2. Tension in the Achilles tendon (Ft):
When a person elevates themselves on the ball of one foot, the tension in the Achilles tendon provides the upward force. This force can be estimated by assuming that the person's center of mass is located at the geometric center of the body.
3. Force exerted by the lower leg bone on the foot (Fb):
The force exerted by the lower leg bone on the foot counteracts the tension in the Achilles tendon, balancing the torques about the hinge joint at the ankle. This force can be calculated by finding the torque produced by the weight of the person and dividing it by the length of the foot (D).
Now, let's analyze the torques around the hinge joint at the ankle, taking the point where the Achilles tendon attaches to the foot as the pivot point.
Torque (τ) = Force × Distance
For static equilibrium, the sum of the torques about any point must be zero. Considering torques in the clockwise direction as negative, and the torques in the counterclockwise direction as positive, we can write:
Torque due to the person's weight (τmg) = Torque due to the force exerted by the lower leg bone (τFb)
To calculate the torques, we need to determine the distances involved:
1. Distance between the pivot point and the center of mass of the person (d):
Since the person is standing on the ball of one foot, we assume that the center of mass lies along a vertical line passing through the pivot point. Therefore, the distance between the pivot point and the center of mass is d.
2. Distance between the pivot point and the foot (D):
The length of the foot is given as twice the value of d, so D = 2d.
Using these distances, we can calculate the torques:
Torque due to the person's weight (τmg) = Weight (mg) × Distance (d) = 72 kg × 9.8 m/s^2 × d
Torque due to the force exerted by the lower leg bone (τFb) = Force (Fb) × Distance (D) = Fb × 2d
Since the torques must be equal in magnitude and opposite in direction, we can set these equations equal to each other:
72 kg × 9.8 m/s^2 × d = Fb × 2d
From this equation, we can solve for Fb:
Fb = (72 kg × 9.8 m/s^2 × d) / (2d)
Simplifying the equation:
Fb = 352.8 kg m/s^2
Now that we have calculated the force exerted by the lower leg bone on the foot (Fb), we can estimate the tension in the Achilles tendon (Ft). Since the person is just barely off the floor, the tension in the Achilles tendon is equal to the force exerted by the lower leg bone (Fb):
Ft = Fb = 352.8 kg m/s^2
Therefore, the estimated tension in the Achilles tendon (Ft) is 352.8 kg m/s^2, and the estimated force exerted by the lower leg bone on the foot (Fb) is also 352.8 kg m/s^2.
To estimate the tension Ft in the Achilles tendon and the downward force Fb exerted by the lower leg bone on the foot, we can use the principles of static equilibrium.
Static equilibrium occurs when the sum of the forces acting on an object is zero, and the sum of the torques (or moments) acting on the object is zero as well.
Let's break down the problem step by step:
Step 1: Identify the forces acting on the person. In this case, we have two forces: the tension Ft in the Achilles tendon pulling upward and the downward force Fb exerted by the lower leg bone on the foot.
Step 2: Determine the torques acting on the person. Torque is the rotational equivalent of force and is calculated as the product of force and the perpendicular distance from the axis of rotation. In this case, the axis of rotation is the balls of the feet.
Step 3: Set up the equations of static equilibrium. The sum of the forces in the vertical direction should be zero, and the sum of the torques about the axis of rotation should also be zero.
Let's solve the problem:
Given:
- Person's mass (m): 72 kg
- Length of d: x
- Length of D: 2x
Step 1: Forces acting on the person:
- Tension in the Achilles tendon (Ft) pulling upward
- Force exerted by the lower leg bone on the foot (Fb) downward
Step 2: Torques acting on the person:
- The torque due to Ft is Ft * D, as it is applied at a perpendicular distance D from the axis of rotation.
- The torque due to Fb is Fb * d, as it is applied at a perpendicular distance d from the axis of rotation.
Step 3: Equations of static equilibrium:
- ΣFy = Ft - Fb - mg = 0 (1) [Sum of forces in the vertical direction is zero]
- Στ = Ft * D - Fb * d = 0 (2) [Sum of torques about the axis of rotation is zero]
We have two equations (1 and 2) with two unknowns (Ft and Fb). By solving these equations simultaneously, we can find the values of Ft and Fb.
Note that the value of g should be taken as 9.8 m/s² for gravitational acceleration.
This problem requires some algebraic manipulation and substitution to solve, so it's recommended to use a calculator or algebraic system to simplify the calculations.
1231N
Well, sum moments about any point. I will do at the ball of the foot.
D*Fb-(D+d)Ft=0
ft= D/(D+d) * Fb
But weight of the person is bearing down on the leg bone (assuming two legs, one-half the weight). so Fb=72/2 g
Ft=2/3 72/2 g or about 235N