The Achilles tendon is attached to the rear of the foot as shown in
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. A person elevates himself just barely off the floor on the "ball of one foot." Assume the person has a mass of 61 kg and D is twice as long as d.
Part A
Find the tension FT in the Achilles tendon (pulling upward).
Express your answer to two significant figures and include the appropriate units.
Find the (downward) force FB exerted by the lower leg bone on the foot.
Express your answer to two significant figures and include the appropriate units.
FB =
A better description is needed, since this forum doesn't display images.
To find the tension FT in the Achilles tendon, we will use the principle of equilibrium. In equilibrium, the sum of all forces acting on an object is zero.
First, we need to identify the forces acting on the person. There are two forces to consider: the tension in the Achilles tendon (FT), pulling upward, and the force exerted by the lower leg bone on the foot (FB), pushing downward.
Since the person is just barely off the floor, there is no vertical acceleration, which means the net force in the vertical direction is zero. Therefore, the tension in the Achilles tendon must be equal in magnitude and opposite in direction to the force exerted by the lower leg bone on the foot.
FT = FB
Now, let's find the value of FB.
The downward force FB can be calculated using the person's mass (m) and the acceleration due to gravity (g) using the formula:
FB = m * g
Given that the person's mass is 61 kg, and the standard acceleration due to gravity is approximately 9.81 m/s^2, we can substitute these values into the formula:
FB = 61 kg * 9.81 m/s^2 = 598.41 N
Rounded to two significant figures, the downward force exerted by the lower leg bone on the foot is 600 N (rounded to the nearest ten).
Therefore, the tension FT in the Achilles tendon will also be 600 N (rounded to the nearest ten).