Question

Question: Fig. 12-31 shows the anatomical structures in the lower leg and foot that are involved in standing on tiptoe, with the heel raised slightly off the floor so that the foot effectively contacts the floor only at point P. Assume distance a = 0 .5 cm , distanceb = 15 cm, and the person’s weight W = 900 N. Of the forces acting on the foot, what are the (a) magnitude and (b) direction (up or down) of the force at point Afrom the calf muscle and the (c) magnitude and (d) direction (up or down) of the force at point Bfrom the lower leg bones?

Short answer

Answer:

1. The magnitude of force at point A, $F_A=2.7\times {10}^3\text{N}$.
2. The direction of force at point A is upward.
3. The magnitude of force at point B, $F_B=3.6\times {10}^3\text{N}$.

The direction of force at point B is downward.

Step-by-step solution

Understanding the given information  

$$\begin{gathered} a=5.0\text{cm} \\ b=15.0\text{cm} \\ W=900\text{N} \end{gathered}$$

Concept and formula used in the given question

By applying equilibrium conditions, you can write the equations for force and torque in terms of unknown forces and distances. Then by solving these two equations, you can get the value of unknown forces. The direction of force can be determined by checking the sign of value of forces. The equations used are given below.

Static Equilibrium conditions:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

(a) Calculation for the magnitude of the force at point A from the calf muscle

Applying static equilibrium conditions to FBD:

$$\begin{gathered} \sum {\mathrm{F}}_{\mathrm{y}}=0 \\ {\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}+\mathrm{W}=0······\left(1\right) \\ \sum \mathrm{\tau }=0 \end{gathered}$$

Using point B as a pivot point,

$\left(W\times b\right)-(F_A\times a)=0······\left(2\right)$

From equation (2):

$\left(900\times 15\right)--\left(F_A\times 5\right)=0$

Hence,

$$\begin{gathered} F_A=2700\text{N} \\ =2.7\times {10}^3\text{N} \end{gathered}$$

The magnitude of force at point A is$F_A=2.7\times {10}^3\text{N}$ .

(b) Calculation for the direction (up or down) of the force at point A from the calf muscle 

The positive sign indicates that the force at point A is directed upward

(c) Calculation for the magnitude of the force at point B from the lower leg bones

Plugging in the value of in equation (1):

$2700+F_B+900=0$

Hence,

$F_B=-3600\text{N}$

The magnitude of force at point B is $F_B=3.6\times {10}^3\text{N}$.

(d) Calculation for the direction (up or down) of the force at point B from the lower leg bones 

The negative sign indicates that the force at point B is directed downwards.