Question

Question: A \({\rm{75}}\;{\rm{kg}}\) man stands on his toes by exerting an upward force through the Achilles tendon, as inFigure. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint.

Short answer

(a) The total force in the Achilles tendon is 2205 N.

(b) The force at the pivot is 2940 N.

Step-by-step solution

Definition of the force

A force is an external factor that can change the rest or motion of a body. It has a size and a general direction.

Free-body diagram

The diagram for the leg muscles is shown below:

The mass of the man is $m=75\text{kg}$.

Calculation of the force on Achilles’ tendon

(a)

The net torque about the pivot point is zero in equilibrium. We can write

$F_A\times 4=N\times 12$

In this, the normal force is equal to the weight. So,

$\begin{array}{rcl}N& =& mg\\ & =& 75\text{kg}\times 9.8{\text{m/s}}^2\\ & =& 735\text{N}\end{array}$

Now, we can write,

$$\begin{gathered} F_A\times 4=735\times 12 \\ {F}_A=\frac{735\times 12}{4} \\ {F}_A=2205\text{N} \end{gathered}$$

Hence, the force 2205N should be upward.

Calculation of the force on the pivot

(b)

For the equilibrium under the forces,

$\begin{array}{rcl}{F}_P& =& F_A+N\\ & =& 2205+735\\ & =& 2940\text{N}\end{array}$

Hence, the force on the pivot is 2940N.